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Critical Relativity Theory
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patdolan
2021-11-16 03:43:30 UTC
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Critical Theory comes in several flavors: Critical Law Theory, Critical Literature Theory, Critical History Theory, Critical Race Theory. To these we now add Critical Relativity Theory!

In the spirit of Derrida we shall deconstruct the clumsy reasoning of special relativity and separate said reasoning from the algebraic symbols and equations that express it, keeping in mind that mathematics is just another form of rhetorical expression wherein falsity can be expressed every bit as plausibly as the truth.

“There is only the text.”— J. Derrida

According to special relativity two observers in motion with respect to each other will disagree on each other’s length. They will also disagree on the proper flow of time. But they will always agree on the velocity they have with respect to one another. This is exceedingly strange. How can it be that two relative quantities, space and time, combine to produce an absolute quantity called relative velocity? It is true that SR does have a formula for calculating coordinate velocity; just like it has formulas for calculating coordinate space and coordinate time. But the Einstein velocity addition formula ONLY applies to a third object in motion wrt a pair of FoRs. If that third object happens to be at rest wrt one of the FoRs then Einsteinian velocity reduces to Galilean velocity, albeit subject to the speed limit c. Relativists simply assumed without further justification that if FoR-1 measures a velocity v between itself and FoR-2 then FoR-2 must measure the same numerical value v for the velocity between itself and FoR-1. Trivial? Nope. Seemingly trivial assumptions can be monumental when constructing a theory of motion. But a 26 year old would probably not yet have the requisite philosophical sophistication needed to recognize this.

Einstein’s choice to make velocity strictly Galilean when calculating the velocity between pairs of FoRs ( yes, it is a choice because it does not follow from either the first or the second postulates ) can be expressed mathematically as

∆x’/∆t’ = ∆x/∆t = v (1)

I now raise equation (1) to the level of a postulate and declare it to be the third, and heretofore hidden, postulate of special relativity. In recognizing its own structural Galileanism through this new postulate, special relativity can finally claim to be woke.

The problem with the third postulate is that even though it is already assumed in every equation of special relativity, it turns out to be true only when v = c ( the second postulate ) or when v = 0. The third postulate can be demonstrated invalid for all values of v in between. The invalidity of the third postulate causes special relativity fall on it’s algebraic face. Big Bang move over…Not recognizing and acknowledging the third postulate was Einstein’s biggest blunder.

Time for some examples.


DIRK & DONO

Consider two FoRs whose x-axes are parallel and lie very close to one another. The relative velocity between these two FoRs is .866c (γ = 2). Dirk assumes the Lotus position at the origin of one FoR whilst Dono assumes the fetal position at the origin of the other. Both Dirk and Dono and their clocks are glued to the origins of their respective FoRs.

Dirk opens one eye and takes note of the meter marks on Dono’s contracted x-axis as they whiz by. Dirk apprehends that Dono’s meter marks are contracted to only half as long as his own meter marks. Dirk opens the other eye and observes that Dono’s clock is ticking at only half the rate of his own clock. Dirk begins to count Dono’s meter marks as they race past Dirk’s position. After one year (by Dirk’s clock) of counting Dono’s meter marks, Dirk has tallied 1.64e+16 Dono meters ( 9.5e+15 meter marks/Ly x .866Ly x γ ). Dirk has also observed that only 0.5 years have elapsed on Dono’s clock. Dirk now calculates what his coordinate velocity should be according to Dono

( 1.64e+16 Dono meters ) / ( 1.58e+7 Dono seconds ) = 1.04e+9 m/s = 3.5c (2)

[ shortcut: ∆x’/∆t’ = (∆x/∆t)(γ^2) = v(γ^2) ] (3)

“Stop!” You cry, “Dirk’s and Dono’s relative velocity was already stipulated to be an absolute .866c with respect to one another, when measured in either FoR.”

That is true, according to (1). But remember that (1) is an arbitrary choice made by Einstein when he built his theory. It is no more legitimate a choice than the Dolan FoR coordinate velocity transform (3) for determining one’s FoR coordinate velocity. It is also no less inconsistent. For we immediately see by inspection that the Dolan FoR coordinate velocity is always greater than the relative velocity by a factor of γ^2. But Einstein’s choice for requiring Galilean FoR pair velocities clangs with just as much antinomy as Dolan’s transform, as we shall see. Special relativity’s dirty little secret is that it’s hidden third postulate (1) destroys the theory from within.

I can see by Dono’s tightening fetal position that he still doesn’t believe me. Very well. We shall prove SR’s mathematical inconsistency in the next example.


SPECIAL RELATIVITY COLLAPSES UNDER THE WEIGHT OF ITS ALGEBRAIC ORIGINAL SIN

Consider a pair of FoRs whose relative velocity v is some value other than c and other than zero. This is expressed mathematically as

v = ∆x’/∆t’ = ∆x/∆t != c (4)

and

v = ∆x’/∆t’ = ∆x/∆t != 0 (5)

The Lorentz transforms allow us to construct the FoR coordinate velocities for pairs of FoRs

∆x’ = γ( ∆x - v∆t )

∆t’ = γ( ∆t - ∆xv/c^2 )

∆x’/∆t’ = [ γ( ∆x - v∆t ) ] / [ γ( ∆t - ∆xv/c^2 ) ] (6)

We now endeavor to solve (6) for v in hopes of demonstrating the internal consistency of special relativity, i.e, that v = v for all pair of FoRs. We saw how this was not the case in DIRK & DONO.

The reader will find that trying to solve for v is hopeless unless we make the substitution

v = ∆x/∆t or v = ∆x’/∆t’

It does not matter which we use—either substitution is permitted by the third postulate (1).

With the substitution made, we eventually arrive at the preposterous results

∆x’/∆t’ = c (7)

or

∆x’/∆t’ = 0 (8)

The derivation is left as an exercise for the reader. For those needing help with the derivation ( I'm looking at you, Dono ) I will be happy to provide it in another post.

The laughable results (7) and (8) directly contradict our assumptions (4) and (5). Furthermore, the results impose the requirement that v is not even a variable—v turns out to be a constant always equal to c or zero. Absolutely absurd.

QED.

Algebraic relativity is thus reduced to ridiculous rubble by means mathematical reductio ad absurdum. The root cause of special relativity’s spectacular algebraic failure lies in the propositional calculus. I am happy to expatiate on that subject in another post if there is interest.

Let’s do one more example—this one ripped from the headlines of experimental physics.


MUONS, SCHMUONS!

Imagine that you are a hadron in deep space minding your own business when all of a sudden you turn to see a Lorentz-flattened earth coming at you at a velocity of .866c. In the impending collision the first thing to strike you is an air molecule high in the earth’s atmosphere. That molecules knocks the stuffing out of you. What is left of you is now a muon which means that you only have 2.2 microseconds more to live.

It turns out that the surface of the flattened earth is exactly 571.56 meters away from you, according to your own muon FoR. By a remarkable coincidence there is a flattened scintillator and a flattened clock in a flattened laboratory directly below you on the surface of the flattened earth. You note the time on the flattened lab clock. Because you are a muon, you are your own 2.2 microsecond alarm clock.

The surface of the flattened earth continues to speed towards you at .866c. 2.2 microseconds later the flattened earth, lab and scintillator smash into you. Just as you expire in the scintillator you note that only 1.1 microseconds has elapsed on the lab clock. Nothing strange here; the labs clock is traveling at γ=2 with respect to you so it only logs half as much elapsed time as you.

One of the lab scientists now cries out “Just a minute! That’s not how this story goes. First of all, it is the hadron that is Lorentz-flattened, not us. And second of all, that hadron used it’s flattened FoR to determine that is was only 571.56 meters away from our scintillator at the point it became a muon. Our unflattened earth FoR clearly indicated that the hadron was actually 1143.12 meters away when it became a muon. That’s why it took 4.4 microseconds, traveling at .866 c, to reach our scintillator. And that’s what our lab clock shows. Not 1.1 microseconds, like that dumb muon is claiming.”

Special relativity leaves us in a quandary. How much time actually did elapse on the lab clock from the moment of the muon’s inception to the moment of its scintillating demise? Was it 1.1 microseconds? Or was it 4.4 microseconds? It has to be one or the other, it can’t be both. However, different elapsed times on the same lab clock can be calculated by the muon and by the earthbound scientists with equal justification. Special relativity is incapable of providing a unique elapsed time on the lab clock. I challenge anyone to show that special relativity can provide a unique elapsed time in the preceding case.

Special relativity’s greatest experimental confirmation turns out to be it’s second greatest falsification. Gravity has been special relativity’s greatest falsification since 1907 ( check out the author’s brilliant post on special relativity vs. Kepler’s 3rd law ).
Richard Hertz
2021-11-16 04:38:32 UTC
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Critical Theory comes in several flavors: Critical Law Theory, Critical Literature Theory, Critical History Theory, Critical Race Theory. To these we now add Critical Relativity Theory!
In the spirit of Derrida we shall deconstruct the clumsy reasoning of special relativity and separate said reasoning from the algebraic symbols and equations that express it, keeping in mind that mathematics is just another form of rhetorical expression wherein falsity can be expressed every bit as plausibly as the truth.
“There is only the text.”— J. Derrida
According to special relativity two observers in motion with respect to each other will disagree on each other’s length. They will also disagree on the proper flow of time. But they will always agree on the velocity they have with respect to one another. This is exceedingly strange. How can it be that two relative quantities, space and time, combine to produce an absolute quantity called relative velocity? It is true that SR does have a formula for calculating coordinate velocity; just like it has formulas for calculating coordinate space and coordinate time. But the Einstein velocity addition formula ONLY applies to a third object in motion wrt a pair of FoRs. If that third object happens to be at rest wrt one of the FoRs then Einsteinian velocity reduces to Galilean velocity, albeit subject to the speed limit c. Relativists simply assumed without further justification that if FoR-1 measures a velocity v between itself and FoR-2 then FoR-2 must measure the same numerical value v for the velocity between itself and FoR-1. Trivial? Nope. Seemingly trivial assumptions can be monumental when constructing a theory of motion. But a 26 year old would probably not yet have the requisite philosophical sophistication needed to recognize this.
Einstein’s choice to make velocity strictly Galilean when calculating the velocity between pairs of FoRs ( yes, it is a choice because it does not follow from either the first or the second postulates ) can be expressed mathematically as
∆x’/∆t’ = ∆x/∆t = v (1)
I now raise equation (1) to the level of a postulate and declare it to be the third, and heretofore hidden, postulate of special relativity. In recognizing its own structural Galileanism through this new postulate, special relativity can finally claim to be woke.
The problem with the third postulate is that even though it is already assumed in every equation of special relativity, it turns out to be true only when v = c ( the second postulate ) or when v = 0. The third postulate can be demonstrated invalid for all values of v in between. The invalidity of the third postulate causes special relativity fall on it’s algebraic face. Big Bang move over…Not recognizing and acknowledging the third postulate was Einstein’s biggest blunder.
Time for some examples.
DIRK & DONO
Consider two FoRs whose x-axes are parallel and lie very close to one another. The relative velocity between these two FoRs is .866c (γ = 2). Dirk assumes the Lotus position at the origin of one FoR whilst Dono assumes the fetal position at the origin of the other. Both Dirk and Dono and their clocks are glued to the origins of their respective FoRs.
Dirk opens one eye and takes note of the meter marks on Dono’s contracted x-axis as they whiz by. Dirk apprehends that Dono’s meter marks are contracted to only half as long as his own meter marks. Dirk opens the other eye and observes that Dono’s clock is ticking at only half the rate of his own clock. Dirk begins to count Dono’s meter marks as they race past Dirk’s position. After one year (by Dirk’s clock) of counting Dono’s meter marks, Dirk has tallied 1.64e+16 Dono meters ( 9.5e+15 meter marks/Ly x .866Ly x γ ). Dirk has also observed that only 0.5 years have elapsed on Dono’s clock. Dirk now calculates what his coordinate velocity should be according to Dono
( 1.64e+16 Dono meters ) / ( 1.58e+7 Dono seconds ) = 1.04e+9 m/s = 3.5c (2)
[ shortcut: ∆x’/∆t’ = (∆x/∆t)(γ^2) = v(γ^2) ] (3)
“Stop!” You cry, “Dirk’s and Dono’s relative velocity was already stipulated to be an absolute .866c with respect to one another, when measured in either FoR.”
That is true, according to (1). But remember that (1) is an arbitrary choice made by Einstein when he built his theory. It is no more legitimate a choice than the Dolan FoR coordinate velocity transform (3) for determining one’s FoR coordinate velocity. It is also no less inconsistent. For we immediately see by inspection that the Dolan FoR coordinate velocity is always greater than the relative velocity by a factor of γ^2. But Einstein’s choice for requiring Galilean FoR pair velocities clangs with just as much antinomy as Dolan’s transform, as we shall see. Special relativity’s dirty little secret is that it’s hidden third postulate (1) destroys the theory from within.
I can see by Dono’s tightening fetal position that he still doesn’t believe me. Very well. We shall prove SR’s mathematical inconsistency in the next example.
SPECIAL RELATIVITY COLLAPSES UNDER THE WEIGHT OF ITS ALGEBRAIC ORIGINAL SIN
Consider a pair of FoRs whose relative velocity v is some value other than c and other than zero. This is expressed mathematically as
v = ∆x’/∆t’ = ∆x/∆t != c (4)
and
v = ∆x’/∆t’ = ∆x/∆t != 0 (5)
The Lorentz transforms allow us to construct the FoR coordinate velocities for pairs of FoRs
∆x’ = γ( ∆x - v∆t )
∆t’ = γ( ∆t - ∆xv/c^2 )
∆x’/∆t’ = [ γ( ∆x - v∆t ) ] / [ γ( ∆t - ∆xv/c^2 ) ] (6)
We now endeavor to solve (6) for v in hopes of demonstrating the internal consistency of special relativity, i.e, that v = v for all pair of FoRs. We saw how this was not the case in DIRK & DONO.
The reader will find that trying to solve for v is hopeless unless we make the substitution
v = ∆x/∆t or v = ∆x’/∆t’
It does not matter which we use—either substitution is permitted by the third postulate (1).
With the substitution made, we eventually arrive at the preposterous results
∆x’/∆t’ = c (7)
or
∆x’/∆t’ = 0 (8)
The derivation is left as an exercise for the reader. For those needing help with the derivation ( I'm looking at you, Dono ) I will be happy to provide it in another post.
The laughable results (7) and (8) directly contradict our assumptions (4) and (5). Furthermore, the results impose the requirement that v is not even a variable—v turns out to be a constant always equal to c or zero. Absolutely absurd.
QED.
Algebraic relativity is thus reduced to ridiculous rubble by means mathematical reductio ad absurdum. The root cause of special relativity’s spectacular algebraic failure lies in the propositional calculus. I am happy to expatiate on that subject in another post if there is interest.
Let’s do one more example—this one ripped from the headlines of experimental physics.
MUONS, SCHMUONS!
Imagine that you are a hadron in deep space minding your own business when all of a sudden you turn to see a Lorentz-flattened earth coming at you at a velocity of .866c. In the impending collision the first thing to strike you is an air molecule high in the earth’s atmosphere. That molecules knocks the stuffing out of you. What is left of you is now a muon which means that you only have 2.2 microseconds more to live.
It turns out that the surface of the flattened earth is exactly 571.56 meters away from you, according to your own muon FoR. By a remarkable coincidence there is a flattened scintillator and a flattened clock in a flattened laboratory directly below you on the surface of the flattened earth. You note the time on the flattened lab clock. Because you are a muon, you are your own 2.2 microsecond alarm clock.
The surface of the flattened earth continues to speed towards you at .866c. 2.2 microseconds later the flattened earth, lab and scintillator smash into you. Just as you expire in the scintillator you note that only 1.1 microseconds has elapsed on the lab clock. Nothing strange here; the labs clock is traveling at γ=2 with respect to you so it only logs half as much elapsed time as you.
One of the lab scientists now cries out “Just a minute! That’s not how this story goes. First of all, it is the hadron that is Lorentz-flattened, not us. And second of all, that hadron used it’s flattened FoR to determine that is was only 571.56 meters away from our scintillator at the point it became a muon. Our unflattened earth FoR clearly indicated that the hadron was actually 1143.12 meters away when it became a muon. That’s why it took 4.4 microseconds, traveling at .866 c, to reach our scintillator. And that’s what our lab clock shows. Not 1.1 microseconds, like that dumb muon is claiming.”
Special relativity leaves us in a quandary. How much time actually did elapse on the lab clock from the moment of the muon’s inception to the moment of its scintillating demise? Was it 1.1 microseconds? Or was it 4.4 microseconds? It has to be one or the other, it can’t be both. However, different elapsed times on the same lab clock can be calculated by the muon and by the earthbound scientists with equal justification. Special relativity is incapable of providing a unique elapsed time on the lab clock. I challenge anyone to show that special relativity can provide a unique elapsed time in the preceding case.
Special relativity’s greatest experimental confirmation turns out to be it’s second greatest falsification. Gravity has been special relativity’s greatest falsification since 1907 ( check out the author’s brilliant post on special relativity vs. Kepler’s 3rd law ).
Dono is a fucking slug, which moves at 0.000000000000000000000000000000000000000000001c, and this when is in a hurry.

You have to select another imbecile here. I propose Moroney, the FAKE EE.
Odd Bodkin
2021-11-16 14:54:09 UTC
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Post by patdolan
Critical Theory comes in several flavors: Critical Law Theory, Critical
Literature Theory, Critical History Theory, Critical Race Theory. To
these we now add Critical Relativity Theory!
In the spirit of Derrida we shall deconstruct the clumsy reasoning of
special relativity and separate said reasoning from the algebraic symbols
and equations that express it, keeping in mind that mathematics is just
another form of rhetorical expression wherein falsity can be expressed
every bit as plausibly as the truth.
“There is only the text.”— J. Derrida
According to special relativity two observers in motion with respect to
each other will disagree on each other’s length. They will also disagree
on the proper flow of time. But they will always agree on the velocity
they have with respect to one another. This is exceedingly strange. How
can it be that two relative quantities, space and time, combine to
produce an absolute quantity called relative velocity? It is true that SR
does have a formula for calculating coordinate velocity; just like it has
formulas for calculating coordinate space and coordinate time. But the
Einstein velocity addition formula ONLY applies to a third object in
motion wrt a pair of FoRs. If that third object happens to be at rest
wrt one of the FoRs then Einsteinian velocity reduces to Galilean
velocity, albeit subject to the speed limit c. Relativists simply
assumed without further justification that if FoR-1 measures a velocity v
between itself and FoR-2 then FoR-2 must measure the same numerical value
v for the velocity between itself and FoR-1. Trivial? Nope. Seemingly
trivial assumptions can be monumental when constructing a theory of
motion. But a 26 year old would probably not yet have the requisite
philosophical sophistication needed to recognize this.
This was discussed recently here about a month or so ago, which you can
find by searching for postulate in thread titles. You have missed all the
clear explanation why this is a non-issue.

Rather than assuming you’re the first to arrive at this thought, why don’t
you take the time to research if anyone else has already thought about it?

Oh I know why. Because then you wouldn’t get an attention fix.
Post by patdolan
Einstein’s choice to make velocity strictly Galilean when calculating the
velocity between pairs of FoRs ( yes, it is a choice because it does not
follow from either the first or the second postulates ) can be expressed mathematically as
∆x’/∆t’ = ∆x/∆t = v (1)
I now raise equation (1) to the level of a postulate and declare it to be
the third, and heretofore hidden, postulate of special relativity. In
recognizing its own structural Galileanism through this new postulate,
special relativity can finally claim to be woke.
The problem with the third postulate is that even though it is already
assumed in every equation of special relativity, it turns out to be true
only when v = c ( the second postulate ) or when v = 0. The third
postulate can be demonstrated invalid for all values of v in between.
The invalidity of the third postulate causes special relativity fall on
it’s algebraic face. Big Bang move over…Not recognizing and
acknowledging the third postulate was Einstein’s biggest blunder.
Time for some examples.
DIRK & DONO
Consider two FoRs whose x-axes are parallel and lie very close to one
another. The relative velocity between these two FoRs is .866c (γ = 2).
Dirk assumes the Lotus position at the origin of one FoR whilst Dono
assumes the fetal position at the origin of the other. Both Dirk and
Dono and their clocks are glued to the origins of their respective FoRs.
Dirk opens one eye and takes note of the meter marks on Dono’s contracted
x-axis as they whiz by. Dirk apprehends that Dono’s meter marks are
contracted to only half as long as his own meter marks. Dirk opens the
other eye and observes that Dono’s clock is ticking at only half the rate
of his own clock. Dirk begins to count Dono’s meter marks as they race
past Dirk’s position. After one year (by Dirk’s clock) of counting
Dono’s meter marks, Dirk has tallied 1.64e+16 Dono meters ( 9.5e+15 meter
marks/Ly x .866Ly x γ ). Dirk has also observed that only 0.5 years have
elapsed on Dono’s clock. Dirk now calculates what his coordinate
velocity should be according to Dono
( 1.64e+16 Dono meters ) / ( 1.58e+7 Dono seconds ) = 1.04e+9 m/s = 3.5c (2)
[ shortcut: ∆x’/∆t’ = (∆x/∆t)(γ^2) = v(γ^2) ] (3)
“Stop!” You cry, “Dirk’s and Dono’s relative velocity was already
stipulated to be an absolute .866c with respect to one another, when
measured in either FoR.”
That is true, according to (1). But remember that (1) is an arbitrary
choice made by Einstein when he built his theory. It is no more
legitimate a choice than the Dolan FoR coordinate velocity transform (3)
for determining one’s FoR coordinate velocity. It is also no less
inconsistent. For we immediately see by inspection that the Dolan FoR
coordinate velocity is always greater than the relative velocity by a
factor of γ^2. But Einstein’s choice for requiring Galilean FoR pair
velocities clangs with just as much antinomy as Dolan’s transform, as we
shall see. Special relativity’s dirty little secret is that it’s hidden
third postulate (1) destroys the theory from within.
I can see by Dono’s tightening fetal position that he still doesn’t
believe me. Very well. We shall prove SR’s mathematical inconsistency
in the next example.
SPECIAL RELATIVITY COLLAPSES UNDER THE WEIGHT OF ITS ALGEBRAIC ORIGINAL SIN
Consider a pair of FoRs whose relative velocity v is some value other
than c and other than zero. This is expressed mathematically as
v = ∆x’/∆t’ = ∆x/∆t != c (4)
and
v = ∆x’/∆t’ = ∆x/∆t != 0 (5)
The Lorentz transforms allow us to construct the FoR coordinate
velocities for pairs of FoRs
∆x’ = γ( ∆x - v∆t )
∆t’ = γ( ∆t - ∆xv/c^2 )
∆x’/∆t’ = [ γ( ∆x - v∆t ) ] / [ γ( ∆t - ∆xv/c^2 ) ] (6)
We now endeavor to solve (6) for v in hopes of demonstrating the internal
consistency of special relativity, i.e, that v = v for all pair of FoRs.
We saw how this was not the case in DIRK & DONO.
The reader will find that trying to solve for v is hopeless unless we make the substitution
v = ∆x/∆t or v = ∆x’/∆t’
It does not matter which we use—either substitution is permitted by the
third postulate (1).
With the substitution made, we eventually arrive at the preposterous results
∆x’/∆t’ = c (7)
or
∆x’/∆t’ = 0 (8)
The derivation is left as an exercise for the reader. For those needing
help with the derivation ( I'm looking at you, Dono ) I will be happy to
provide it in another post.
The laughable results (7) and (8) directly contradict our assumptions (4)
and (5). Furthermore, the results impose the requirement that v is not
even a variable—v turns out to be a constant always equal to c or zero. Absolutely absurd.
QED.
Algebraic relativity is thus reduced to ridiculous rubble by means
mathematical reductio ad absurdum. The root cause of special relativity’s
spectacular algebraic failure lies in the propositional calculus. I am
happy to expatiate on that subject in another post if there is interest.
Let’s do one more example—this one ripped from the headlines of experimental physics.
MUONS, SCHMUONS!
Imagine that you are a hadron in deep space minding your own business
when all of a sudden you turn to see a Lorentz-flattened earth coming at
you at a velocity of .866c. In the impending collision the first thing
to strike you is an air molecule high in the earth’s atmosphere. That
molecules knocks the stuffing out of you. What is left of you is now a
muon which means that you only have 2.2 microseconds more to live.
It turns out that the surface of the flattened earth is exactly 571.56
meters away from you, according to your own muon FoR. By a remarkable
coincidence there is a flattened scintillator and a flattened clock in a
flattened laboratory directly below you on the surface of the flattened
earth. You note the time on the flattened lab clock. Because you are a
muon, you are your own 2.2 microsecond alarm clock.
The surface of the flattened earth continues to speed towards you at
.866c. 2.2 microseconds later the flattened earth, lab and scintillator
smash into you. Just as you expire in the scintillator you note that
only 1.1 microseconds has elapsed on the lab clock. Nothing strange
here; the labs clock is traveling at γ=2 with respect to you so it only
logs half as much elapsed time as you.
One of the lab scientists now cries out “Just a minute! That’s not how
this story goes. First of all, it is the hadron that is
Lorentz-flattened, not us. And second of all, that hadron used it’s
flattened FoR to determine that is was only 571.56 meters away from our
scintillator at the point it became a muon. Our unflattened earth FoR
clearly indicated that the hadron was actually 1143.12 meters away when
it became a muon. That’s why it took 4.4 microseconds, traveling at .866
c, to reach our scintillator. And that’s what our lab clock shows. Not
1.1 microseconds, like that dumb muon is claiming.”
Special relativity leaves us in a quandary. How much time actually did
elapse on the lab clock from the moment of the muon’s inception to the
moment of its scintillating demise? Was it 1.1 microseconds? Or was it
4.4 microseconds? It has to be one or the other, it can’t be both.
However, different elapsed times on the same lab clock can be calculated
by the muon and by the earthbound scientists with equal justification.
Special relativity is incapable of providing a unique elapsed time on the
lab clock. I challenge anyone to show that special relativity can
provide a unique elapsed time in the preceding case.
Special relativity’s greatest experimental confirmation turns out to be
it’s second greatest falsification. Gravity has been special
relativity’s greatest falsification since 1907 ( check out the author’s
brilliant post on special relativity vs. Kepler’s 3rd law ).
--
Odd Bodkin — Maker of fine toys, tools, tables
patdolan
2021-11-16 16:03:30 UTC
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On Tuesday, November 16, 2021 at 6:54:15 AM UTC-8, ***@gmail.com wrote:

Bodkin, I searched on the word "postulate" as you suggested. From Aug 1, 2021 until today. I am the only poster that the search returned.
Odd Bodkin
2021-11-16 16:43:18 UTC
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Post by patdolan
Bodkin, I searched on the word "postulate" as you suggested. From Aug 1,
2021 until today. I am the only poster that the search returned.
Search for “unstated special relativity axiom”.

Would the next spoon feeding require a rubber-coated spoon?
--
Odd Bodkin -- maker of fine toys, tools, tables
patdolan
2021-11-16 17:07:25 UTC
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Post by Odd Bodkin
Bodkin, I searched on the word "postulate" as you suggested. From Aug 1,
2021 until today. I am the only poster that the search returned.
Search for “unstated special relativity axiom”.
Would the next spoon feeding require a rubber-coated spoon?
--
Odd Bodkin -- maker of fine toys, tools, tables
Bodkin,

I give Ricardo Jimenez priority as being first to publish. But I reserver for myself the claim of being first to rigorously prove what the brilliant and far-sighted Ricardo only suspected.

PS--I certainly enjoyed Ricardo exposing that blowhard Robert's circular reasoning on this issue. And rotchm! rotchm had an absolute panic attack and started delivering a course on vector algebra in order to avoid the subject entirely.

I should like to meet the Jimenez fellow online sometime.

PPS--how's the re-think going Dirk? Let me know if you get stuck again.
Odd Bodkin
2021-11-16 17:10:37 UTC
Reply
Permalink
Post by patdolan
Post by Odd Bodkin
Bodkin, I searched on the word "postulate" as you suggested. From Aug 1,
2021 until today. I am the only poster that the search returned.
Search for “unstated special relativity axiom”.
Would the next spoon feeding require a rubber-coated spoon?
--
Odd Bodkin -- maker of fine toys, tools, tables
Bodkin,
I give Ricardo Jimenez priority as being first to publish.
And you seem not to have read anything in the thread showing Ricardo where
he was astray.
Post by patdolan
But I reserver for myself the claim of being first to rigorously prove
what the brilliant and far-sighted Ricardo only suspected.
PS--I certainly enjoyed Ricardo exposing that blowhard Robert's circular
reasoning on this issue. And rotchm! rotchm had an absolute panic
attack and started delivering a course on vector algebra in order to
avoid the subject entirely.
I should like to meet the Jimenez fellow online sometime.
PPS--how's the re-think going Dirk? Let me know if you get stuck again.
--
Odd Bodkin -- maker of fine toys, tools, tables
patdolan
2021-11-16 17:21:13 UTC
Reply
Permalink
Post by Odd Bodkin
Post by patdolan
Post by Odd Bodkin
Bodkin, I searched on the word "postulate" as you suggested. From Aug 1,
2021 until today. I am the only poster that the search returned.
Search for “unstated special relativity axiom”.
Would the next spoon feeding require a rubber-coated spoon?
--
Odd Bodkin -- maker of fine toys, tools, tables
Bodkin,
I give Ricardo Jimenez priority as being first to publish.
And you seem not to have read anything in the thread showing Ricardo where
he was astray.
That's is correct Bodkin. Because RJ appears to have answered every objection and parried every passe'. Do you disagree?

Bodkin: Yes.
Dolan: Where?
Bodkin: I won't spoon feed you.
Post by Odd Bodkin
Post by patdolan
But I reserver for myself the claim of being first to rigorously prove
what the brilliant and far-sighted Ricardo only suspected.
PS--I certainly enjoyed Ricardo exposing that blowhard Robert's circular
reasoning on this issue. And rotchm! rotchm had an absolute panic
attack and started delivering a course on vector algebra in order to
avoid the subject entirely.
I should like to meet the Jimenez fellow online sometime.
PPS--how's the re-think going Dirk? Let me know if you get stuck again.
--
Odd Bodkin -- maker of fine toys, tools, tables
Odd Bodkin
2021-11-16 17:22:37 UTC
Reply
Permalink
Post by patdolan
Post by Odd Bodkin
Post by patdolan
Post by Odd Bodkin
Bodkin, I searched on the word "postulate" as you suggested. From Aug 1,
2021 until today. I am the only poster that the search returned.
Search for “unstated special relativity axiom”.
Would the next spoon feeding require a rubber-coated spoon?
--
Odd Bodkin -- maker of fine toys, tools, tables
Bodkin,
I give Ricardo Jimenez priority as being first to publish.
And you seem not to have read anything in the thread showing Ricardo where
he was astray.
That's is correct Bodkin. Because RJ appears to have answered every
objection and parried every passe'. Do you disagree?
Yes. Read the conversation.
Post by patdolan
Bodkin: Yes.
Dolan: Where?
Bodkin: I won't spoon feed you.
Well, I’m not going to read for you.
Post by patdolan
Post by Odd Bodkin
Post by patdolan
But I reserver for myself the claim of being first to rigorously prove
what the brilliant and far-sighted Ricardo only suspected.
PS--I certainly enjoyed Ricardo exposing that blowhard Robert's circular
reasoning on this issue. And rotchm! rotchm had an absolute panic
attack and started delivering a course on vector algebra in order to
avoid the subject entirely.
I should like to meet the Jimenez fellow online sometime.
PPS--how's the re-think going Dirk? Let me know if you get stuck again.
--
Odd Bodkin -- maker of fine toys, tools, tables
--
Odd Bodkin -- maker of fine toys, tools, tables
Dono.
2021-11-16 15:44:03 UTC
Reply
Permalink
snip imbecilities<
Dirk Van de moortel
2021-11-16 15:47:35 UTC
Reply
Permalink
Post by patdolan
Critical Theory comes in several flavors: Critical Law Theory,
Critical Literature Theory, Critical History Theory, Critical Race
Theory. To these we now add Critical Relativity Theory!
In the spirit of Derrida we shall deconstruct the clumsy reasoning of
special relativity and separate said reasoning from the algebraic
symbols and equations that express it, keeping in mind that
mathematics is just another form of rhetorical expression wherein
falsity can be expressed every bit as plausibly as the truth.
“There is only the text.”— J. Derrida
According to special relativity two observers in motion with respect
to each other will disagree on each other’s length. They will also
disagree on the proper flow of time. But they will always agree on
the velocity they have with respect to one another. This is
exceedingly strange. How can it be that two relative quantities,
space and time, combine to produce an absolute quantity called
relative velocity? It is true that SR does have a formula for
calculating coordinate velocity; just like it has formulas for
calculating coordinate space and coordinate time. But the Einstein
velocity addition formula ONLY applies to a third object in motion
wrt a pair of FoRs. If that third object happens to be at rest wrt
one of the FoRs then Einsteinian velocity reduces to Galilean
velocity, albeit subject to the speed limit c. Relativists simply
assumed without further justification that if FoR-1 measures a
velocity v between itself and FoR-2 then FoR-2 must measure the same
numerical value v for the velocity between itself and FoR-1. Trivial?
Nope. Seemingly trivial assumptions can be monumental when
constructing a theory of motion. But a 26 year old would probably
not yet have the requisite philosophical sophistication needed to
recognize this.
Einstein’s choice to make velocity strictly Galilean when calculating
the velocity between pairs of FoRs ( yes, it is a choice because it
does not follow from either the first or the second postulates ) can
be expressed mathematically as
∆x’/∆t’ = ∆x/∆t = v (1)
You can only write
∆x’/∆t’ = v
for events taking place on an object at rest in FoR-1, i.o.w.
for events taking place a the same location in For-1, i.o.w.
for events that satisfy
∆x = 0.

You can only write
∆x/∆t = v
for events taking place on an object at rest in FoR-2, i.o.w.
for events taking place a the same location in For-2, i.o.w.
for events that satisfy
∆x' = 0.

Unless v = 0 (and thus FoR-1 = FoR-2), there are no objects
that are at rest in both FoR-1 and FoR-2, i.o.w. there are
no distinct events that happen at the same place in both
FoR-1 and FoR-2.

So, yes
0/∆t’ = 0/∆t = 0
for all values of ∆t' and ∆t.
A profound discovery, congratulations!

So when you write
∆x’/∆t’ = ∆x/∆t = v ,
you clearly have no idea what you are talking about.

But by the way, perhaps you just want to say that the origins
of FoR-1 and FoR-2 have mutual velocity v w.r.t. each other.
So you want to say that
∆x’/∆t’ = v
for events taking place on an object at rest in FoR-1 (∆x=0)
and that
∆x/∆t = v
for events taking place on an object at rest in FoR-2 (∆x'=0)
*without* combining the equations?

In that case you have the x and x' axes pointing in different
directions, and the standard Lorentz transformation then takes
the form
∆x’ = - g ( ∆x - v ∆t )
∆t’ = g ( ∆t - v/c^2 ∆x )
and the inverse
∆x = - g ( ∆x' - v ∆t' )
∆t = g ( ∆t' - v/c^2 ∆x' )
where
g = 1 / sqrt( 1 - v^2/c^2 )

See if this helps.

Dirk Vdm
Wade Earl
2021-11-16 15:57:27 UTC
Reply
Permalink
n that case you have the x and x' axes pointing in different directions,
and the standard Lorentz transformation then takes the form
∆x’ = - g ( ∆x - v ∆t ) ∆t’ = g ( ∆t - v/c^2 ∆x )
and the inverse
∆x = - g ( ∆x' - v ∆t' ) ∆t = g ( ∆t' - v/c^2 ∆x' )
where
g = 1 / sqrt( 1 - v^2/c^2 )
See if this helps.
not sure, you are working in 3D, meanwhile the g is 1D.
patdolan
2021-11-16 16:27:41 UTC
Reply
Permalink
Post by Dirk Van de moortel
So when you write
∆x’/∆t’ = ∆x/∆t = v ,
you clearly have no idea what you are talking about.
c is a velocity too.

∆x'/∆t' = ∆x/∆t = c, aka the second postulate. Now do I know what I am talking about?

Prove to this forum that a pair of observers glued to the origins of their respective FoRs measure the same relative velocity between their FoRs. I have disproved this in three different ways.

You need to think about this a little more, Dirk. Why don't you try assuming the Lotus position. See if that helps.
Post by Dirk Van de moortel
See if this helps.
Dirk Vdm
patdolan
2021-11-16 16:33:04 UTC
Reply
Permalink
Post by patdolan
Post by Dirk Van de moortel
So when you write
∆x’/∆t’ = ∆x/∆t = v ,
you clearly have no idea what you are talking about.
c is a velocity too.
∆x'/∆t' = ∆x/∆t = c, aka the second postulate. Now do I know what I am talking about?
Prove to this forum that a pair of observers glued to the origins of their respective FoRs measure the same relative velocity between their FoRs. I have disproved this in three different ways.
You need to think about this a little more, Dirk. Why don't you try assuming the Lotus position. See if that helps.
Post by Dirk Van de moortel
See if this helps.
Dirk Vdm
Oh, and nice pick up on my removing the "-" from my derivation. It is the one and only liberty I took for the sake of clarity. So, yes, you did at least prove that the x-axes do point in opposite directions. At least you can feel good about that.
Dono.
2021-11-16 16:34:40 UTC
Reply
Permalink
Post by patdolan
I have disproved this in three different ways.
Three different way, one common imbecility
patdolan
2021-11-16 17:50:13 UTC
Reply
Permalink
Post by patdolan
So when you write ∆x’/∆t’ = ∆x/∆t = v , you clearly have no idea
what you are talking about.
c is a velocity too.
∆x'/∆t' = ∆x/∆t = c, aka the second postulate. Now do I know what I
am talking about?
Prove to this forum that a pair of observers glued to the origins of
their respective FoRs measure the same relative velocity between
their FoRs. I have disproved this in three different ways.
After all these years, you still haven't got the faintest clue.
In three different ways you've shown again what a MEGA-duncehead
you are.
Dirk Vdm
That's all ya got Dirk? Okaaaaaaaay.....my paper has just passed it's first (sub-)peer review. On to bigger venues!
Dirk Van de moortel
2021-11-16 18:24:07 UTC
Reply
Permalink
On Tuesday, November 16, 2021 at 9:33:01 AM UTC-8, Dirk Van de
Post by patdolan
So when you write ∆x’/∆t’ = ∆x/∆t = v , you clearly have no
idea what you are talking about.
c is a velocity too.
∆x'/∆t' = ∆x/∆t = c, aka the second postulate. Now do I know what
I am talking about?
Prove to this forum that a pair of observers glued to the origins
of their respective FoRs measure the same relative velocity
between their FoRs. I have disproved this in three different
ways.
After all these years, you still haven't got the faintest clue. In
three different ways you've shown again what a MEGA-duncehead you
are.
Make that four.

Dirk Vdm
Dirk Vdm
That's all ya got Dirk? Okaaaaaaaay.....my paper has just passed
it's first (sub-)peer review. On to bigger venues!
Thomas 'PointedEars' Lahn
2021-11-17 23:09:36 UTC
Reply
Permalink
Post by patdolan
Critical Theory comes in several flavors: Critical Law Theory, Critical
Literature Theory, Critical History Theory, Critical Race Theory. To
these we now add Critical Relativity Theory!
In the spirit of Derrida we shall deconstruct the clumsy reasoning of
special relativity
LOL.
Post by patdolan
and separate said reasoning from the algebraic symbols and equations
that express it,
Since it is fundamentally based in geometry, namely frames of reference
which can understood as (moving) coordinate systems, that approach is
hopeless.
Post by patdolan
keeping in mind that mathematics is just another form of rhetorical
expression wherein falsity can be expressed every bit as plausibly
as the truth.
Not true. Natural language is ambiguous and rather loose; it is why
(despite knowing logic) it is so easy to commit fallacies using natural
language. Mathematics, in its symbols, terms, and reasoning is (given a
context) unambiguous and unforgivingly strict.

In the words of Richard Feynman:

‘You might say, “All right, then, there’s no explanation of the law;
at least tell me what the law is — why not tell me in words instead of
in the symbols? Mathematics is just a language, and I want to be able
to translate the language." And, in fact, I can — and with patience,
I think I partly did. I could go a little further and explain more
detail — that this means if it’s twice as far away the force is
one-fourth as much, and so on — and can convert all these into words.
I would be, in other words, kind to the layman, as they all sit, hopeful
that you will explain something. Various different people get different
reputations for their skill at explaining to the layman in layman’s
language these difficult and abstruse subjects.

The layman then searches for book after book with the hope that he will
avoid the complexity which ultimately sets in, even by the best expositor
of this type. He reads the things, hoping — he finds, as he reads, a
generally increased confusion, one complicated statement after the other,
one difficult-to-understand thing after the other, all apparently
disconnected from one another — and it becomes a little obscure, and
he hopes that maybe in some other book there’s some explanation which
avoids — I mean, the man almost made it, you see — maybe another fellow
makes it right.

I don’t think it’s possible, because there’s another feature: mathematics
is not just a language; mathematics is a language plus reasoning; it’s
like a language plus logic. Mathematics is a tool for reasoning. It’s in
fact a big collection of the results of some person’s careful thought and
reasoning. […]’

<https://www.feynmanlectures.caltech.edu/fml.html#2>
Post by patdolan
“There is only the text.”— J. Derrida
LOL.
Post by patdolan
According to special relativity two observers in motion with respect to
each other will disagree on each other’s length. They will also disagree
on the proper flow of time. But they will always agree on the velocity
they have with respect to one another.
No, the velocity that observer A ascribes to observer B is necessarily the
arithmetic inverse of the velocity that observer ascribes to observer B,
provided that they are both looking in the same direction and choose their
coordinate systems such that the positive part in on the same (usually
right-hand) side.
Post by patdolan
This is exceedingly strange.
There is nothing strange about it.
Post by patdolan
How can it be that two relative quantities, space and time, combine to
produce an absolute quantity called relative velocity?
It is not absolute at all. If it were absolute, then every observer would
assign the same velocity to the same object, regardless of their relative
state of motion. The very point of the theories of relativity is that this
is not so.

However, if you consider the relevant equations of the (inverse) Lorentz
transformation for relative (inertial) motion *only* along the x-axis,

t = γ (t' + v/c² x')
x = γ (x' + v t'),

where for simplicity we say that v would be the velocity (usually just a
speed, but this particular motion is one-dimensional) that an observer that
we define as stationary would assign to a clock that, at rest in its own
frame shows time t, then you can see that for the temporal interval

Δt = γ (Δt' + v/c² Δx') = γ Δt'

and for the spatial interval

Δx = γ (Δx' + v Δt') = γ v Δt',

then the velocity of that moving clock for the stationary observer is still

Δx/Δt = v.

Similarly for an observer at rest in the "moving" frame, when they are
talking about a clock that is at rest in the stationary frame, they
transform the coordinates by the (original) Lorentz transformation

t' = γ (t − v/c² x)
x' = γ (x − v t),

– you can obtain that by solving the equations for the other quantity, not
just by arbitrary substitution – which for (their) time intervals becomes

Δt' = γ (Δt − v/c² Δx) = γ Δt
Δx' = γ (Δx − v Δt) = −γ v Δt.

Then the velocity that they assign to the stationary clock is

Δx'/Δt' = −v,

as we would expect.
Post by patdolan
It is true that SR does have a formula for calculating coordinate
velocity; just like it has formulas for calculating coordinate space
and coordinate time. But the Einstein velocity addition formula ONLY
applies to a third object in motion wrt a pair of FoRs.
But the second FoR is implied when we are talking about a velocity. There
has to be some frame of reference relative to which we are defining the
value.
Post by patdolan
If that third object happens to be at rest wrt one of the FoRs then
Einsteinian velocity reduces to Galilean velocity,
Yes, but that is a rather pointless statement because “at rest” means that
the relative velocity is zero.

It is more useful to realize that as one of the relative speeds *approaches*
zero the composed velocity *approaches* the Galilean term:

u(v, u') = (v + u')/(1 + v u'/c²)

lim_{v u' → 0} u(v, u') = v + u',

as v u'/c² → 0 then due to the large value of c.

[This equation did not simply fell out of the sky. It is, again, a
(mathematical) *consequence* of the Lorentz transformation, and Einstein
derives it in “On the Electrodynamics of Moving Bodies” (1905). I have
also derived it here many times.]
Post by patdolan
albeit subject to the speed limit c.
That is a contradiction in terms.
Post by patdolan
Relativists simply assumed without further justification that if FoR-1
measures a velocity v between itself and FoR-2 then FoR-2 must measure
the same numerical value v for the velocity between itself and FoR-1.
No.
Post by patdolan
Einstein’s choice to make velocity strictly Galilean when calculating the
velocity between pairs of FoRs ( yes, it is a choice because it does not
follow from either the first or the second postulates ) can be expressed
mathematically as
∆x’/∆t’ = ∆x/∆t = v (1)
Wrong, see above.
Post by patdolan
I now raise equation (1) to the level of a postulate and declare it to be
the third, and heretofore hidden, postulate of special relativity.
But you do not understand special relativity correctly (yet).
Your/an argument/conclusions from your ignorance is a useless fallacy.
Post by patdolan
[tl;dr]
PointedEars
--
“Nature uses only the longest threads to weave her patterns
so that each small piece of her fabric reveals the organization
of the entire tapestry.”
—Richard Feynman, theoretical physicist, “Messenger Lecture” 1 (1964)
Thomas 'PointedEars' Lahn
2021-11-17 23:34:59 UTC
Reply
Permalink
Post by Thomas 'PointedEars' Lahn
Post by patdolan
According to special relativity two observers in motion with respect to
each other will disagree on each other’s length. They will also disagree
on the proper flow of time. But they will always agree on the velocity
they have with respect to one another.
No, the velocity that observer A ascribes to observer B is necessarily the
arithmetic inverse of the velocity that observer ascribes to observer B,
I meant to write

„No, the velocity that observer A ascribes to observer B is necessarily the
arithmetic inverse of the velocity that observer B ascribes to observer A,“
Post by Thomas 'PointedEars' Lahn
provided that they are both looking in the same direction and choose their
coordinate systems such that the positive part in on the same (usually
right-hand) side.
PointedEars
--
Heisenberg is out for a drive when he's stopped by a traffic cop.
The officer asks him "Do you know how fast you were going?"
Heisenberg replies "No, but I know where I am."
(from: WolframAlpha)
Maciej Wozniak
2021-11-18 06:58:30 UTC
Reply
Permalink
Post by Thomas 'PointedEars' Lahn
Post by Thomas 'PointedEars' Lahn
Post by patdolan
According to special relativity two observers in motion with respect to
each other will disagree on each other’s length. They will also disagree
on the proper flow of time. But they will always agree on the velocity
they have with respect to one another.
No, the velocity that observer A ascribes to observer B is necessarily the
arithmetic inverse of the velocity that observer ascribes to observer B,
I meant to write
„No, the velocity that observer A ascribes to observer B is necessarily the
arithmetic inverse of the velocity that observer B ascribes to observer A,“
In the meantime in the real world, however, forbidden by
your moronic religion GPS clocks keep measuring
t'=t, just like all serious clocks always did.
patdolan
2021-11-18 02:07:07 UTC
Reply
Permalink
Post by Thomas 'PointedEars' Lahn
Post by patdolan
Critical Theory comes in several flavors: Critical Law Theory, Critical
Literature Theory, Critical History Theory, Critical Race Theory. To
these we now add Critical Relativity Theory!
In the spirit of Derrida we shall deconstruct the clumsy reasoning of
special relativity
LOL.
Post by patdolan
and separate said reasoning from the algebraic symbols and equations
that express it,
Since it is fundamentally based in geometry, namely frames of reference
which can understood as (moving) coordinate systems, that approach is
hopeless.
Post by patdolan
keeping in mind that mathematics is just another form of rhetorical
expression wherein falsity can be expressed every bit as plausibly
as the truth.
Not true. Natural language is ambiguous and rather loose; it is why
(despite knowing logic) it is so easy to commit fallacies using natural
language. Mathematics, in its symbols, terms, and reasoning is (given a
context) unambiguous and unforgivingly strict.
1^4 = i^4

sqrt[ 1^4 ] = sqrt[ i^4 ]

+/-( 1^2 ) = +/-( i^2 )

+/-( 1 ) = +/-( -1 )

+/- 1 = -/+ 1
Post by Thomas 'PointedEars' Lahn
‘You might say, “All right, then, there’s no explanation of the law;
at least tell me what the law is — why not tell me in words instead of
in the symbols?
I did this. See MUONS, SCHMUONS! at the bottom of my post. Read through it slow, Long Ears. Then take up my challenge if you dare: Prove that relativity provides only one and unique answer for the elapsed time on the lab clock. I can assure the world and posterity that you can't and won't do prove one, unique time.

Mathematics is just a language, and I want to be able
Post by Thomas 'PointedEars' Lahn
to translate the language." And, in fact, I can — and with patience,
I think I partly did. I could go a little further and explain more
detail — that this means if it’s twice as far away the force is
one-fourth as much, and so on — and can convert all these into words.
I would be, in other words, kind to the layman, as they all sit, hopeful
that you will explain something. Various different people get different
reputations for their skill at explaining to the layman in layman’s
language these difficult and abstruse subjects.
The layman then searches for book after book with the hope that he will
avoid the complexity which ultimately sets in, even by the best expositor
of this type. He reads the things, hoping — he finds, as he reads, a
generally increased confusion, one complicated statement after the other,
one difficult-to-understand thing after the other, all apparently
disconnected from one another — and it becomes a little obscure, and
he hopes that maybe in some other book there’s some explanation which
avoids — I mean, the man almost made it, you see — maybe another fellow
makes it right.
I don’t think it’s possible, because there’s another feature: mathematics
is not just a language; mathematics is a language plus reasoning; it’s
like a language plus logic. Mathematics is a tool for reasoning. It’s in
fact a big collection of the results of some person’s careful thought and
reasoning. […]’
<https://www.feynmanlectures.caltech.edu/fml.html#2>
Post by patdolan
“There is only the text.”— J. Derrida
LOL.
Post by patdolan
According to special relativity two observers in motion with respect to
each other will disagree on each other’s length. They will also disagree
on the proper flow of time. But they will always agree on the velocity
they have with respect to one another.
No, the velocity that observer A ascribes to observer B is necessarily the
arithmetic inverse of the velocity that observer ascribes to observer B,
provided that they are both looking in the same direction and choose their
coordinate systems such that the positive part in on the same (usually
right-hand) side.
Post by patdolan
This is exceedingly strange.
There is nothing strange about it.
Post by patdolan
How can it be that two relative quantities, space and time, combine to
produce an absolute quantity called relative velocity?
It is not absolute at all. If it were absolute, then every observer would
assign the same velocity to the same object, regardless of their relative
state of motion. The very point of the theories of relativity is that this
is not so.
However, if you consider the relevant equations of the (inverse) Lorentz
transformation for relative (inertial) motion *only* along the x-axis,
t = γ (t' + v/c² x')
x = γ (x' + v t'),
where for simplicity we say that v would be the velocity (usually just a
speed, but this particular motion is one-dimensional) that an observer that
we define as stationary would assign to a clock that, at rest in its own
frame shows time t, then you can see that for the temporal interval
Δt = γ (Δt' + v/c² Δx') = γ Δt'
and for the spatial interval
Δx = γ (Δx' + v Δt') = γ v Δt',
then the velocity of that moving clock for the stationary observer is still
Δx/Δt = v.
Similarly for an observer at rest in the "moving" frame, when they are
talking about a clock that is at rest in the stationary frame, they
transform the coordinates by the (original) Lorentz transformation
t' = γ (t − v/c² x)
x' = γ (x − v t),
– you can obtain that by solving the equations for the other quantity, not
just by arbitrary substitution – which for (their) time intervals becomes
Δt' = γ (Δt − v/c² Δx) = γ Δt
Δx' = γ (Δx − v Δt) = −γ v Δt.
Then the velocity that they assign to the stationary clock is
Δx'/Δt' = −v,
as we would expect.
Post by patdolan
It is true that SR does have a formula for calculating coordinate
velocity; just like it has formulas for calculating coordinate space
and coordinate time. But the Einstein velocity addition formula ONLY
applies to a third object in motion wrt a pair of FoRs.
But the second FoR is implied when we are talking about a velocity. There
has to be some frame of reference relative to which we are defining the
value.
Post by patdolan
If that third object happens to be at rest wrt one of the FoRs then
Einsteinian velocity reduces to Galilean velocity,
Yes, but that is a rather pointless statement because “at rest” means that
the relative velocity is zero.
It is more useful to realize that as one of the relative speeds *approaches*
u(v, u') = (v + u')/(1 + v u'/c²)
lim_{v u' → 0} u(v, u') = v + u',
as v u'/c² → 0 then due to the large value of c.
[This equation did not simply fell out of the sky. It is, again, a
(mathematical) *consequence* of the Lorentz transformation, and Einstein
derives it in “On the Electrodynamics of Moving Bodies” (1905). I have
also derived it here many times.]
Post by patdolan
albeit subject to the speed limit c.
That is a contradiction in terms.
Post by patdolan
Relativists simply assumed without further justification that if FoR-1
measures a velocity v between itself and FoR-2 then FoR-2 must measure
the same numerical value v for the velocity between itself and FoR-1.
No.
Post by patdolan
Einstein’s choice to make velocity strictly Galilean when calculating the
velocity between pairs of FoRs ( yes, it is a choice because it does not
follow from either the first or the second postulates ) can be expressed
mathematically as
∆x’/∆t’ = ∆x/∆t = v (1)
Wrong, see above.
Post by patdolan
I now raise equation (1) to the level of a postulate and declare it to be
the third, and heretofore hidden, postulate of special relativity.
But you do not understand special relativity correctly (yet).
Your/an argument/conclusions from your ignorance is a useless fallacy.
Post by patdolan
[tl;dr]
PointedEars
--
“Nature uses only the longest threads to weave her patterns
so that each small piece of her fabric reveals the organization
of the entire tapestry.”
—Richard Feynman, theoretical physicist, “Messenger Lecture” 1 (1964)
Thomas, I have no interest in debating the algebra with you. You are not appreciating the meta-algebra involved. Nor will you anytime soon. Just address the MUONS, SCHMUONS! challenge. Or go silent. You have always been a second stringer in this forum.
Thomas 'PointedEars' Lahn
2021-11-19 23:00:28 UTC
Reply
Permalink
Post by patdolan
Post by Thomas 'PointedEars' Lahn
[…] mathematics is just another form of rhetorical expression wherein
falsity can be expressed every bit as plausibly as the truth.
Not true. Natural language is ambiguous and rather loose; it is why
(despite knowing logic) it is so easy to commit fallacies using natural
language. Mathematics, in its symbols, terms, and reasoning is (given a
context) unambiguous and unforgivingly strict.
1^4 = i^4
That much is true.
Post by patdolan
sqrt[ 1^4 ] = sqrt[ i^4 ]
+/-( 1^2 ) = +/-( i^2 )
+/-( 1 ) = +/-( -1 )
+/- 1 = -/+ 1
Your logic is flawed.
Post by patdolan
Post by Thomas 'PointedEars' Lahn
‘You might say, “All right, then, there’s no explanation of the law;
at least tell me what the law is — why not tell me in words instead of
in the symbols?
I did this.
I know that you did. Feynman explains there why your request cannot be
fulfilled.
Post by patdolan
See MUONS, SCHMUONS! at the bottom of my post. Read through it slow,
Long Ears.
I have better things to do than discussing with you at this low a level.
Post by patdolan
[tl;dr]
PointedEars
--
«Nec fasces, nec opes, sola artis sceptra perennant.»
(“Neither high office nor power, only the scepters of science survive.”)

—Tycho Brahe, astronomer (1546-1601): inscription at Hven
patdolan
2021-11-20 03:04:53 UTC
Reply
Permalink
Post by Thomas 'PointedEars' Lahn
Post by patdolan
Post by Thomas 'PointedEars' Lahn
[…] mathematics is just another form of rhetorical expression wherein
falsity can be expressed every bit as plausibly as the truth.
Not true. Natural language is ambiguous and rather loose; it is why
(despite knowing logic) it is so easy to commit fallacies using natural
language. Mathematics, in its symbols, terms, and reasoning is (given a
context) unambiguous and unforgivingly strict.
1^4 = i^4
That much is true.
Post by patdolan
sqrt[ 1^4 ] = sqrt[ i^4 ]
+/-( 1^2 ) = +/-( i^2 )
+/-( 1 ) = +/-( -1 )
+/- 1 = -/+ 1
Your logic is flawed.
How, long ears. How?
Post by Thomas 'PointedEars' Lahn
Post by patdolan
Post by Thomas 'PointedEars' Lahn
‘You might say, “All right, then, there’s no explanation of the law;
at least tell me what the law is — why not tell me in words instead of
in the symbols?
I did this.
I know that you did. Feynman explains there why your request cannot be
fulfilled.
Post by patdolan
See MUONS, SCHMUONS! at the bottom of my post. Read through it slow,
Long Ears.
I have better things to do than discussing with you at this low a level.
Post by patdolan
[tl;dr]
PointedEars
--
«Nec fasces, nec opes, sola artis sceptra perennant.»
(“Neither high office nor power, only the scepters of science survive.”)
—Tycho Brahe, astronomer (1546-1601): inscription at Hven
Paul Alsing
2021-11-20 06:24:31 UTC
Reply
Permalink
Post by Thomas 'PointedEars' Lahn
Post by patdolan
1^4 = i^4
That much is true.
Post by patdolan
sqrt[ 1^4 ] = sqrt[ i^4 ]
Your logic is flawed.
How, long ears. How?
Because i is not a number, it is a definition. The term *i* is "defined" to be 'the square root of -1', and therefore cannot be manipulated the way you are claiming. It is not a real number, it is an imaginary number... just like infinity, which is a definition and absolutely NOT a number. You are pretty much dead in the water... as everyone already knows...
Maciej Wozniak
2021-11-20 07:06:29 UTC
Reply
Permalink
Post by Paul Alsing
Post by Thomas 'PointedEars' Lahn
Post by patdolan
1^4 = i^4
That much is true.
Post by patdolan
sqrt[ 1^4 ] = sqrt[ i^4 ]
Your logic is flawed.
How, long ears. How?
Because i is not a number, it is a definition. The term *i* is "defined" to be 'the square root of -1',
:) And how is "the square root" defined, Al,
poor halfbrain?
patdolan
2021-11-20 13:15:13 UTC
Reply
Permalink
Post by Paul Alsing
Post by Thomas 'PointedEars' Lahn
Post by patdolan
1^4 = i^4
That much is true.
Post by patdolan
sqrt[ 1^4 ] = sqrt[ i^4 ]
Your logic is flawed.
How, long ears. How?
Because i is not a number, it is a definition. The term *i* is "defined" to be 'the square root of -1', and therefore cannot be manipulated the way you are claiming. It is not a real number, it is an imaginary number... just like infinity, which is a definition and absolutely NOT a number. You are pretty much dead in the water... as everyone already knows...
[ This is going to be fun, tee-hee-hee...] So Paul A. #2, when you multiply a number by a definition is the product a number or another definition?
Python
2021-11-20 13:18:35 UTC
Reply
Permalink
Post by Paul Alsing
Post by Thomas 'PointedEars' Lahn
Post by patdolan
1^4 = i^4
That much is true.
Post by patdolan
sqrt[ 1^4 ] = sqrt[ i^4 ]
Your logic is flawed.
How, long ears. How?
Because i is not a number, it is a definition. The term *i* is "defined" to be 'the square root of -1', and therefore cannot be manipulated the way you are claiming. It is not a real number, it is an imaginary number... just like infinity, which is a definition and absolutely NOT a number. You are pretty much dead in the water... as everyone already knows...
There is a more rigorous definition for i than "the square root of -1"
in term of equivalence classes of polynomials. Of course patdolan
claims are dead in the water since in C sqrt is a multi-valued
function (ask Demented Woz for details, he discovered this concept
recently).
Maciej Wozniak
2021-11-20 13:22:25 UTC
Reply
Permalink
Post by Python
Post by Paul Alsing
Post by Thomas 'PointedEars' Lahn
Post by patdolan
1^4 = i^4
That much is true.
Post by patdolan
sqrt[ 1^4 ] = sqrt[ i^4 ]
Your logic is flawed.
How, long ears. How?
Because i is not a number, it is a definition. The term *i* is "defined" to be 'the square root of -1', and therefore cannot be manipulated the way you are claiming. It is not a real number, it is an imaginary number... just like infinity, which is a definition and absolutely NOT a number. You are pretty much dead in the water... as everyone already knows...
There is a more rigorous definition for i than "the square root of -1"
in term of equivalence classes of polynomials. Of course patdolan
claims are dead in the water since in C sqrt is a multi-valued
function (ask Demented Woz for details, he discovered this concept
recently).
It's not true it was recently, but anyway poor halfbrain Al
may always start asking a wiser one about some explainations.
patdolan
2021-11-20 15:01:55 UTC
Reply
Permalink
Post by Python
Post by Paul Alsing
Post by Thomas 'PointedEars' Lahn
Post by patdolan
1^4 = i^4
That much is true.
Post by patdolan
sqrt[ 1^4 ] = sqrt[ i^4 ]
Your logic is flawed.
How, long ears. How?
Because i is not a number, it is a definition. The term *i* is "defined" to be 'the square root of -1', and therefore cannot be manipulated the way you are claiming. It is not a real number, it is an imaginary number... just like infinity, which is a definition and absolutely NOT a number. You are pretty much dead in the water... as everyone already knows...
There is a more rigorous definition for i than "the square root of -1"
in term of equivalence classes of polynomials. Of course patdolan
claims are dead in the water since in C sqrt is a multi-valued
function (ask Demented Woz for details, he discovered this concept
recently).
I have accounted for both branches of the double branch sqrt function in my derivation and conclusion and made the correct branch cuts:

+/-1 = -/+1

Just as I would have accounted for all three branches of the cube root function, four branches for the...

I am an algebraic King Cobra. You are just a python. Never mess with the king.
Paul Alsing
2021-11-20 18:06:56 UTC
Reply
Permalink
Post by patdolan
Post by Python
Post by Paul Alsing
Post by Thomas 'PointedEars' Lahn
Post by patdolan
1^4 = i^4
That much is true.
Post by patdolan
sqrt[ 1^4 ] = sqrt[ i^4 ]
Your logic is flawed.
How, long ears. How?
Because i is not a number, it is a definition. The term *i* is "defined" to be 'the square root of -1', and therefore cannot be manipulated the way you are claiming. It is not a real number, it is an imaginary number... just like infinity, which is a definition and absolutely NOT a number. You are pretty much dead in the water... as everyone already knows...
There is a more rigorous definition for i than "the square root of -1"
in term of equivalence classes of polynomials. Of course patdolan
claims are dead in the water since in C sqrt is a multi-valued
function (ask Demented Woz for details, he discovered this concept
recently).
+/-1 = -/+1
Just as I would have accounted for all three branches of the cube root function, four branches for the...
I am an algebraic King Cobra. You are just a python. Never mess with the king.
Well, the *king* is dead... dead wrong!
patdolan
2021-11-20 18:20:14 UTC
Reply
Permalink
Post by Paul Alsing
Post by patdolan
Post by Python
Post by Paul Alsing
Post by Thomas 'PointedEars' Lahn
Post by patdolan
1^4 = i^4
That much is true.
Post by patdolan
sqrt[ 1^4 ] = sqrt[ i^4 ]
Your logic is flawed.
How, long ears. How?
Because i is not a number, it is a definition. The term *i* is "defined" to be 'the square root of -1', and therefore cannot be manipulated the way you are claiming. It is not a real number, it is an imaginary number... just like infinity, which is a definition and absolutely NOT a number. You are pretty much dead in the water... as everyone already knows...
There is a more rigorous definition for i than "the square root of -1"
in term of equivalence classes of polynomials. Of course patdolan
claims are dead in the water since in C sqrt is a multi-valued
function (ask Demented Woz for details, he discovered this concept
recently).
+/-1 = -/+1
Just as I would have accounted for all three branches of the cube root function, four branches for the...
I am an algebraic King Cobra. You are just a python. Never mess with the king.
Well, the *king* is dead... dead wrong!
Very cheeky, Paul A#2. I recall that Paul A#1 used to be quite cheeky also until he was put in his place. Paul A#1 lives in a country that still respects Kings and Queens.
Thomas 'PointedEars' Lahn
2021-11-20 22:54:55 UTC
Reply
Permalink
Post by Paul Alsing
Post by Thomas 'PointedEars' Lahn
Post by patdolan
1^4 = i^4
That much is true.
Post by patdolan
sqrt[ 1^4 ] = sqrt[ i^4 ]
Your logic is flawed.
How, long ears. How?
Because i is not a number, it is a definition.
No; that statement is

1. utterly wrong;
2. not the reason.
Post by Paul Alsing
The term *i* is "defined" to be 'the square root of -1',
No, it is NOT; that is a common misconception (unfortunately, it can also be
found in otherwise great textbooks such as Griffiths’ “Introduction to
Quantum Mechanics”, at least in the second edition).

Instead, 𝕚 is defined to be the solution (*historically*: “root”) of the
equation

x² + 1 = 0.

[This equation does not have a solution in the real numbers, because the
square of every real number is non-negative. Thus, the set of complex
numbers were invented as a solution to the problem, as it had been done
with ℝ for ℚ, ℚ for ℤ, and ℤ for ℕ, before.]

This means that 𝕚 is defined as a number whose square is −1: 𝕚² ≔ −1. This
is not the same thing because there are *two* complex numbers which satisfy
this condition: 𝕚 and −𝕚.

By definition, 𝕚² = −1. Then (−𝕚)² = (−1 · 𝕚)² = (−1)² i² = 1 · (−1) = −1.


Concisely, we can write this as √(−1) = ±𝕚. However, care must be taken how
the “±” notation is used; and *that* is why Pat Dolan’s logic is flawed.
Post by Paul Alsing
It is not a real number,
True.
Post by Paul Alsing
it is an imaginary number...
Yes, but: “imaginary” does NOT mean “not real” in the sense of “does not
exist” in mathematics; it means that it is not an element of the set of real
numbers, ℝ: 𝕚 ∉ ℝ. It is an element of a superset of the real numbers, the
set of complex numbers ℂ ⊃ ℝ, where ℂ ≔ {a + b 𝕚 | a, b ∈ ℝ; 𝕚² = −1}:
𝕚 ∈ ℂ∖ℝ.


PointedEars
--
Q: How many theoretical physicists specializing in general relativity
does it take to change a light bulb?
A: Two: one to hold the bulb and one to rotate the universe.
(from: WolframAlpha)
Thomas 'PointedEars' Lahn
2021-11-20 23:03:28 UTC
Reply
Permalink
Post by Thomas 'PointedEars' Lahn
Instead, 𝕚 is defined to be the solution (*historically*: “root”) of the
equation
x² + 1 = 0.
[This equation does not have a solution in the real numbers, because
[the square of every real number is non-negative. Thus, the set of
complex numbers were invented as a solution to the problem, as it
had been done with ℝ for ℚ, ℚ for ℤ, and ℤ for ℕ, before.]
This means that 𝕚 is defined as a number whose square is −1: 𝕚² ≔ −1.
This is not the same thing because there are *two* complex numbers which
satisfy this condition: 𝕚 and −𝕚.
By definition, 𝕚² = −1. Then (−𝕚)² = (−1 · 𝕚)² = (−1)² i² = 1 · (−1) =
−1. ∎
Therefore, I should have been more precise and should have written
“_a_ solution” instead.


PointedEars
--
A neutron walks into a bar and inquires how much a drink costs.
The bartender replies, "For you? No charge."

(from: WolframAlpha)
patdolan
2021-11-20 23:29:11 UTC
Reply
Permalink
Post by Thomas 'PointedEars' Lahn
Post by Paul Alsing
Post by Thomas 'PointedEars' Lahn
Post by patdolan
1^4 = i^4
That much is true.
Post by patdolan
sqrt[ 1^4 ] = sqrt[ i^4 ]
Your logic is flawed.
How, long ears. How?
Because i is not a number, it is a definition.
No; that statement is
1. utterly wrong;
2. not the reason.
Post by Paul Alsing
The term *i* is "defined" to be 'the square root of -1',
No, it is NOT; that is a common misconception (unfortunately, it can also be
found in otherwise great textbooks such as Griffiths’ “Introduction to
Quantum Mechanics”, at least in the second edition).
Instead, 𝕚 is defined to be the solution (*historically*: “root”) of the
equation
x² + 1 = 0.
[This equation does not have a solution in the real numbers, because the
square of every real number is non-negative. Thus, the set of complex
numbers were invented as a solution to the problem, as it had been done
with ℝ for ℚ, ℚ for ℤ, and ℤ for ℕ, before.]
This means that 𝕚 is defined as a number whose square is −1: 𝕚² ≔ −1. This
is not the same thing because there are *two* complex numbers which satisfy
this condition: 𝕚 and −𝕚.
By definition, 𝕚² = −1. Then (−𝕚)² = (−1 · 𝕚)² = (−1)² i² = 1 · (−1) = −1.

Concisely, we can write this as √(−1) = ±𝕚. However, care must be taken how
the “±” notation is used; and *that* is why Pat Dolan’s logic is flawed.
Okay Long Ears, do you your hero Feynman say. Don't just say it. Show it! Show how you rescue +/-1=-/+1 by the using the notation "with care". Go.
Post by Thomas 'PointedEars' Lahn
Post by Paul Alsing
It is not a real number,
True.
Post by Paul Alsing
it is an imaginary number...
Yes, but: “imaginary” does NOT mean “not real” in the sense of “does not
exist” in mathematics; it means that it is not an element of the set of real
numbers, ℝ: 𝕚 ∉ ℝ. It is an element of a superset of the real numbers, the
𝕚 ∈ ℂ∖ℝ.
PointedEars
--
Q: How many theoretical physicists specializing in general relativity
does it take to change a light bulb?
A: Two: one to hold the bulb and one to rotate the universe.
(from: WolframAlpha)
Paul Alsing
2021-11-21 03:05:51 UTC
Reply
Permalink
Post by Thomas 'PointedEars' Lahn
Post by Paul Alsing
Post by Thomas 'PointedEars' Lahn
Post by patdolan
1^4 = i^4
That much is true.
Post by patdolan
sqrt[ 1^4 ] = sqrt[ i^4 ]
Your logic is flawed.
How, long ears. How?
Because i is not a number, it is a definition.
No; that statement is
1. utterly wrong;
2. not the reason.
Post by Paul Alsing
The term *i* is "defined" to be 'the square root of -1'
No, it is NOT; that is a common misconception (unfortunately, it can also be
found in otherwise great textbooks such as Griffiths’ “Introduction to
Quantum Mechanics”, at least in the second edition).
Well, you have a LOT of web pages that you need to contact because I can show you a whole lot of them that disagree with you...

https://en.wikipedia.org/wiki/Imaginary_number

"i, which is defined by its property i2 = −1"

https://en.wikipedia.org/wiki/Imaginary_unit

"The imaginary unit or unit imaginary number (i) is a solution to the quadratic equation x2 + 1 = 0"

https://byjus.com/maths/value-of-i/

"The value of i is √-1"

https://mathworld.wolfram.com/i.html

"The imaginary number i (also called the imaginary unit) is defined as the square root of -1"

https://math.hmc.edu/funfacts/i-to-the-i-is-a-real-number/

"the “imaginary” number i has the property that the square of i is -1"

https://www.mathopenref.com/i.html

"It (i) stands for the square root of negative one"

... and there are thousands more, but at this point in time, I'm bored...
Thomas 'PointedEars' Lahn
2021-11-21 03:38:23 UTC
Reply
Permalink
Post by Paul Alsing
Post by Thomas 'PointedEars' Lahn
Post by Paul Alsing
The term *i* is "defined" to be 'the square root of -1'
No, it is NOT; that is a common misconception (unfortunately, it can also
be found in otherwise great textbooks such as Griffiths’ “Introduction to
Quantum Mechanics”, at least in the second edition).
Well, you have a LOT of web pages that you need to contact because I can
show you a whole lot of them that disagree with you...
As you can see above, either you misunderstand your own sources or they are
wrong.
Post by Paul Alsing
https://en.wikipedia.org/wiki/Imaginary_number
"i, which is defined by its property i2 = −1"
This is the same as I said.
Post by Paul Alsing
https://en.wikipedia.org/wiki/Imaginary_unit
"The imaginary unit or unit imaginary number (i) is a solution to the
quadratic equation x2 + 1 = 0"
This is the same as I said.
Post by Paul Alsing
https://byjus.com/maths/value-of-i/
"The value of i is √-1"
This is wrong.
Post by Paul Alsing
https://mathworld.wolfram.com/i.html
"The imaginary number i (also called the imaginary unit) is defined as the
square root of -1"
This is wrong, unfortunately. It should be noted, though, that Wolfram
MathWorld is but compilation of statements from textbooks, and some
textbooks oversimplify there, as I just said.
Post by Paul Alsing
https://math.hmc.edu/funfacts/i-to-the-i-is-a-real-number/
"the “imaginary” number i has the property that the square of i is -1"
This is the same as I said.
Post by Paul Alsing
https://www.mathopenref.com/i.html
"It (i) stands for the square root of negative one"
This is wrong.
Post by Paul Alsing
... and there are thousands more, but at this point in time, I'm bored...
No, you are incompetent and can’t read.


PointedEars
--
Q: What happens when electrons lose their energy?
A: They get Bohr'ed.

(from: WolframAlpha)
Thomas 'PointedEars' Lahn
2021-11-21 03:42:20 UTC
Reply
Permalink
Post by Paul Alsing
Post by Thomas 'PointedEars' Lahn
Post by Paul Alsing
The term *i* is "defined" to be 'the square root of -1'
No, it is NOT; that is a common misconception (unfortunately, it can also
be found in otherwise great textbooks such as Griffiths’ “Introduction to
Quantum Mechanics”, at least in the second edition).
Well, you have a LOT of web pages that you need to contact because I can
show you a whole lot of them that disagree with you...
As you can see above, either you misunderstand your own sources or they are
wrong.
Post by Paul Alsing
https://en.wikipedia.org/wiki/Imaginary_number
"i, which is defined by its property i2 = −1"
Copy & pray aside, this is the same as I said.
Post by Paul Alsing
https://en.wikipedia.org/wiki/Imaginary_unit
"The imaginary unit or unit imaginary number (i) is a solution to the
quadratic equation x2 + 1 = 0"
Copy & pray aside, this is the same as I said.
Post by Paul Alsing
https://byjus.com/maths/value-of-i/
"The value of i is √-1"
This is wrong.
Post by Paul Alsing
https://mathworld.wolfram.com/i.html
"The imaginary number i (also called the imaginary unit) is defined as the
square root of -1"
This is wrong, unfortunately. The argument made there smells of theory
finding and is questionable.

It should be noted, though, that Wolfram MathWorld is mainly a compilation
of statements from textbooks, and some textbooks oversimplify there, as I
just said.
Post by Paul Alsing
https://math.hmc.edu/funfacts/i-to-the-i-is-a-real-number/
"the “imaginary” number i has the property that the square of i is -1"
This is the same as I said.
Post by Paul Alsing
https://www.mathopenref.com/i.html
"It (i) stands for the square root of negative one"
This is wrong.
Post by Paul Alsing
... and there are thousands more, but at this point in time, I'm bored...
No, you are incompetent and can’t read.


PointedEars
--
Q: What happens when electrons lose their energy?
A: They get Bohr'ed.

(from: WolframAlpha)
Thomas 'PointedEars' Lahn
2021-11-21 03:47:03 UTC
Reply
Permalink
Post by Paul Alsing
Post by Thomas 'PointedEars' Lahn
Post by Paul Alsing
The term *i* is "defined" to be 'the square root of -1'
No, it is NOT; that is a common misconception (unfortunately, it can also
be found in otherwise great textbooks such as Griffiths’ “Introduction to
Quantum Mechanics”, at least in the second edition).
Well, you have a LOT of web pages that you need to contact because I can
show you a whole lot of them that disagree with you...
As you can see below, either you misunderstand your own sources or they are
wrong.
Post by Paul Alsing
https://en.wikipedia.org/wiki/Imaginary_number
"i, which is defined by its property i2 = −1"
Copy & pray aside, this is the same as I said.
Post by Paul Alsing
https://en.wikipedia.org/wiki/Imaginary_unit
"The imaginary unit or unit imaginary number (i) is a solution to the
quadratic equation x2 + 1 = 0"
Copy & pray aside, this is the same as I said.
Post by Paul Alsing
https://byjus.com/maths/value-of-i/
"The value of i is √-1"
This is wrong.
Post by Paul Alsing
https://mathworld.wolfram.com/i.html
"The imaginary number i (also called the imaginary unit) is defined as the
square root of -1"
This is wrong, unfortunately. The argument made there smells of theory
finding and is questionable.

It should be noted, though, that Wolfram MathWorld is mainly a compilation
of statements from textbooks, and some textbooks oversimplify there, as I
just said.
Post by Paul Alsing
https://math.hmc.edu/funfacts/i-to-the-i-is-a-real-number/
"the “imaginary” number i has the property that the square of i is -1"
This is the same as I said.
Post by Paul Alsing
https://www.mathopenref.com/i.html
"It (i) stands for the square root of negative one"
This is wrong.
Post by Paul Alsing
... and there are thousands more, but at this point in time, I'm bored...
No, you are incompetent and can’t read.


PointedEars
--
Q: What happens when electrons lose their energy?
A: They get Bohr'ed.

(from: WolframAlpha)
patdolan
2021-11-21 03:54:55 UTC
Reply
Permalink
Post by Thomas 'PointedEars' Lahn
Post by Paul Alsing
Post by Thomas 'PointedEars' Lahn
Post by Paul Alsing
The term *i* is "defined" to be 'the square root of -1'
No, it is NOT; that is a common misconception (unfortunately, it can also
be found in otherwise great textbooks such as Griffiths’ “Introduction to
Quantum Mechanics”, at least in the second edition).
Well, you have a LOT of web pages that you need to contact because I can
show you a whole lot of them that disagree with you...
As you can see below, either you misunderstand your own sources or they are
wrong.
Post by Paul Alsing
https://en.wikipedia.org/wiki/Imaginary_number
"i, which is defined by its property i2 = −1"
Copy & pray aside, this is the same as I said.
Post by Paul Alsing
https://en.wikipedia.org/wiki/Imaginary_unit
"The imaginary unit or unit imaginary number (i) is a solution to the
quadratic equation x2 + 1 = 0"
Copy & pray aside, this is the same as I said.
Post by Paul Alsing
https://byjus.com/maths/value-of-i/
"The value of i is √-1"
This is wrong.
Post by Paul Alsing
https://mathworld.wolfram.com/i.html
"The imaginary number i (also called the imaginary unit) is defined as the
square root of -1"
This is wrong, unfortunately. The argument made there smells of theory
finding and is questionable.
It should be noted, though, that Wolfram MathWorld is mainly a compilation
of statements from textbooks, and some textbooks oversimplify there, as I
just said.
Post by Paul Alsing
https://math.hmc.edu/funfacts/i-to-the-i-is-a-real-number/
"the “imaginary” number i has the property that the square of i is -1"
This is the same as I said.
Post by Paul Alsing
https://www.mathopenref.com/i.html
"It (i) stands for the square root of negative one"
This is wrong.
Ya know what is REALLY WRONG LongEars is what you typed earlier:

"Not true. Natural language is ambiguous and rather loose; it is why
(despite knowing logic) it is so easy to commit fallacies using natural
language. Mathematics, in its symbols, terms, and reasoning is (given a
context) unambiguous and unforgivingly strict. "

Man! were you wrong about mathematics being unambiguous and unforgivingly strict. This just might make you the chump #2 of the day, behind chump #1 Bodkin.
Post by Thomas 'PointedEars' Lahn
Post by Paul Alsing
... and there are thousands more, but at this point in time, I'm bored...
No, you are incompetent and can’t read.
PointedEars
--
Q: What happens when electrons lose their energy?
A: They get Bohr'ed.
(from: WolframAlpha)
Paul Alsing
2021-11-21 16:54:35 UTC
Reply
Permalink
Post by Thomas 'PointedEars' Lahn
Post by Paul Alsing
Post by Thomas 'PointedEars' Lahn
Post by Paul Alsing
The term *i* is "defined" to be 'the square root of -1'
No, it is NOT; that is a common misconception (unfortunately, it can also
be found in otherwise great textbooks such as Griffiths’ “Introduction to
Quantum Mechanics”, at least in the second edition).
Well, you have a LOT of web pages that you need to contact because I can
show you a whole lot of them that disagree with you...
As you can see below, either you misunderstand your own sources or they are
wrong.
Post by Paul Alsing
https://en.wikipedia.org/wiki/Imaginary_number
"i, which is defined by its property i2 = −1"
Copy & pray aside, this is the same as I said.
Post by Paul Alsing
https://en.wikipedia.org/wiki/Imaginary_unit
"The imaginary unit or unit imaginary number (i) is a solution to the
quadratic equation x2 + 1 = 0"
Copy & pray aside, this is the same as I said.
Post by Paul Alsing
https://byjus.com/maths/value-of-i/
"The value of i is √-1"
This is wrong.
Post by Paul Alsing
https://mathworld.wolfram.com/i.html
"The imaginary number i (also called the imaginary unit) is defined as the
square root of -1"
This is wrong, unfortunately. The argument made there smells of theory
finding and is questionable.
It should be noted, though, that Wolfram MathWorld is mainly a compilation
of statements from textbooks, and some textbooks oversimplify there, as I
just said.
Post by Paul Alsing
https://math.hmc.edu/funfacts/i-to-the-i-is-a-real-number/
"the “imaginary” number i has the property that the square of i is -1"
This is the same as I said.
Post by Paul Alsing
https://www.mathopenref.com/i.html
"It (i) stands for the square root of negative one"
This is wrong.
Post by Paul Alsing
... and there are thousands more, but at this point in time, I'm bored...
No, you are incompetent and can’t read.
PointedEars
--
Q: What happens when electrons lose their energy?
A: They get Bohr'ed.
(from: WolframAlpha)
Like I said... you have a LOT of web pages and textbooks to set straight... so you better get busy!
Thomas 'PointedEars' Lahn
2021-11-21 22:40:25 UTC
Reply
Permalink
Post by Paul Alsing
Post by Thomas 'PointedEars' Lahn
Post by Paul Alsing
... and there are thousands more, but at this point in time, I'm bored...
No, you are incompetent and can’t read.
Like I said... you have a LOT of web pages and textbooks to set
straight... so you better get busy!
Fallacy: Shifting the burden of proof.

I do not have to find sources to support my claim; *you* made the claim.
I showed you the problems with your claims, and their solution.

And apparently it has escaped your attention that most of the sources that
you quoted so far say *exactly* what I said, and are explicitly or
implicitly contradicting what you claimed, which is contradictory *in
itself* [that on the one hand 𝕚 would NOT be a number, but merely
“imaginary”; and that on the other hand 𝕚 *would* be a number defined as
√(−1).]

Apparently you are that incompetent that not only you do not realize when
you are contradicting yourself, but also that you do not realize that yourr
own sources are contradicting yourself

In that case this discussion would be over as you would have have
demonstrated that you had disconnected from reality, and there is no
argument by which you can be convinced.


PointedEars
--
Heisenberg is out for a drive when he's stopped by a traffic cop.
The officer asks him "Do you know how fast you were going?"
Heisenberg replies "No, but I know where I am."
(from: WolframAlpha)
Paul Alsing
2021-11-22 02:48:33 UTC
Reply
Permalink
Post by Thomas 'PointedEars' Lahn
Post by Paul Alsing
Post by Thomas 'PointedEars' Lahn
Post by Paul Alsing
... and there are thousands more, but at this point in time, I'm bored...
No, you are incompetent and can’t read.
Like I said... you have a LOT of web pages and textbooks to set
straight... so you better get busy!
Fallacy: Shifting the burden of proof.
I do not have to find sources to support my claim; *you* made the claim.
I showed you the problems with your claims, and their solution.
And I have shown you hundreds of sources that agree with me... and *you* have provided exactly *zero* sources that agree with you!

I stand by my claim that a vast majority of textbooks in the world state that the definition of "i" is "the square root of -1"... if you think that I am incorrect I expect you to provide evidence otherwise... which you have failed to do. Your babble is not evidence.

As always, either put up or shut up.

Without evidence, you are done here. But then again, you have been done here more often than not.
Thomas 'PointedEars' Lahn
2021-11-22 04:30:43 UTC
Reply
Permalink
Post by Paul Alsing
Post by Thomas 'PointedEars' Lahn
On Saturday, November 20, 2021 at 7:47:06 PM UTC-8, Thomas
Post by Thomas 'PointedEars' Lahn
Post by Paul Alsing
... and there are thousands more, but at this point in time, I'm bored...
No, you are incompetent and can’t read.
Like I said... you have a LOT of web pages and textbooks to set
straight... so you better get busy!
Fallacy: Shifting the burden of proof.
I do not have to find sources to support my claim; *you* made the claim.
I showed you the problems with your claims, and their solution.
And I have shown you hundreds of sources that agree with me...
No, you have NOT. You are delusional.

You have shown exactly 6 (in words: SIX) sources, and 3 OF THEM SAY EXACTLY
WHAT I HAVE SAID/EXPLAINED (in a nutshell: “𝕚 *is* a number defined by
𝕚² ≔ −1, which is NOT EQUIVALENT TO 𝕚 ≔ √(−1)”), which I have pointed out
to you.

The other 3 say something WRONG (see above), and one of the three (Wolfram
MathWorld) uses a questionable argument; both of which I have also pointed
out to you.


PointedEars
--
«Nec fasces, nec opes, sola artis sceptra perennant.»
(“Neither high office nor power, only the scepters of science survive.”)

—Tycho Brahe, astronomer (1546-1601): inscription at Hven
Paul Alsing
2021-11-22 04:43:41 UTC
Reply
Permalink
Post by Thomas 'PointedEars' Lahn
WHAT I HAVE SAID/EXPLAINED (in a nutshell: “𝕚 *is* a number defined by
𝕚² ≔ −1, which is NOT EQUIVALENT TO 𝕚 ≔ √(−1)”), which I have pointed out
to you.
No, those are exactly the same thing... and i is not a number, it is a definition!

Here you go, 8,030,000,000 results

https://www.google.com/search?q=definition+of+i+in+math&sxsrf=AOaemvKtYDv1GdhjVTo2hiJJ9ZYTp7sP1w%3A1637555975409&ei=Bx-bYe6sGN3AytMPgLCh0Aw&ved=0ahUKEwiuzNqxk6v0AhVdoHIEHQBYCMoQ4dUDCA4&uact=5&oq=definition+of+i+in+math&gs_lcp=Cgdnd3Mtd2l6EAMyBQgAEIAEMgQIABAeOgcIABBHELADOgYIABAHEB46BAgAEA1KBQg8EgEySgQIQRgASgQIRhgAUKAIWKcWYPIXaAJwAngAgAGcAogBiQeSAQUwLjQuMZgBAKABAcgBCMABAQ&sclient=gws-wiz

"The imaginary unit or unit imaginary number (i) is a solution to the quadratic equation x2 + 1 = 0. ... Here, the term "imaginary" is used because there is no real number having a negative square."

Again... " there is no real number having a negative square"

Get a life...
Thomas 'PointedEars' Lahn
2021-11-22 05:46:58 UTC
Reply
Permalink
Post by Paul Alsing
Post by Thomas 'PointedEars' Lahn
WHAT I HAVE SAID/EXPLAINED (in a nutshell: “𝕚 *is* a number defined by
𝕚² ≔ −1, which is NOT EQUIVALENT TO 𝕚 ≔ √(−1)”), which I have pointed
out to you.
No, those are exactly the same thing...
No, they are not.
Post by Paul Alsing
and i is not a number, it is a definition!
Oh for crying out loud. Where have you studied mathematics – Instagram
University?

ALL numbers are defined.

For example, 1 (or 0) is DEFINED to be a natural NUMBER by the first Peano
axiom, and 2 (or 1) is defined as the number that is the successor of 1 (or
0), which by the sixth Peano axiom makes it a natural NUMBER, which is an
element of the set of natural NUMBERS: 1, 2 ∈ ℕ.

The set of integers (“whole NUMBERS”) is DEFINED
as ℤ ≔ {−n, …, −1, 0, 1, …, n | n ∈ ℕ ∪ {0}}.

The set of rational numbers is DEFINED as ℚ ≔ {p/q | p, q ∈ ℤ, q ≠ 0},
as I already told you before.

The set of real numbers ℝ is DEFINED in various ways, for example via
Dedekind cuts based on ℚ.

And finally, the set of complex NUMBERS is DEFINED as

ℂ ≔ {a + b 𝕚 | a, b ∈ ℝ; 𝕚² = −1},

as I already told you before.

This CLEARLY makes 𝕚 a complex NUMBER. It makes it also an imaginary
NUMBER, because its real part is 0 (and “imaginary” means NOTHING ELSE in
mathematics):

z ∈ ℂ, a = 0, b = 1 ⇒ z = 0 + 1 · 𝕚 = 𝕚 ∈ (𝕀 ≔ ℂ∖ℝ).

SHUT UP and STUDY.
Post by Paul Alsing
Here you go, 8,030,000,000 results
The number of Google search hits for a set of keywords means NOTHING as
MOST people HAVE NO CLUE, and are publishing MOSTLY BULLSHIT on the Web.

That would be so even if your search keywords would be correct, which they
are not.
https://www.google.com/search?q=definition+of+i+in+math&sxsrf=AOaemvKtYDv1GdhjVTo2hiJJ9ZYTp7sP1w%3A1637555975409&ei=Bx-bYe6sGN3AytMPgLCh0Aw&ved=0ahUKEwiuzNqxk6v0AhVdoHIEHQBYCMoQ4dUDCA4&uact=5&oq=definition+of+i+in+math&gs_lcp=Cgdnd3Mtd2l6EAMyBQgAEIAEMgQIABAeOgcIABBHELADOgYIABAHEB46BAgAEA1KBQg8EgEySgQIQRgASgQIRhgAUKAIWKcWYPIXaAJwAngAgAGcAogBiQeSAQUwLjQuMZgBAKABAcgBCMABAQ&sclient=gws-wiz
Post by Paul Alsing
"The imaginary unit or unit imaginary number (i) is a solution to the
quadratic equation x2 + 1 = 0. ...
That (if “x2” is to be “x²”) means that 𝕚² = −1, WHICH IS EXACTLY WHAT I
SAID. And that does NOT mean that 𝕚 = √(−1); THAT IS SIMPLY NOT THE SAME
THING.
Post by Paul Alsing
Here, the term "imaginary" is used because there is no real number having
a negative square."
That’s BULLSHIT (see above for the correct explanation).
Post by Paul Alsing
Again... " there is no real number having a negative square"
Hopeless (Dunning–Kruger) case.


PointedEars
--
Q: What did the nuclear physicist post on the laboratory door
when he went camping?
A: 'Gone fission'.
(from: WolframAlpha)
Paul Alsing
2021-11-22 06:04:05 UTC
Reply
Permalink
Post by Thomas 'PointedEars' Lahn
That (if “x2” is to be “x²”) means that 𝕚² = −1, WHICH IS EXACTLY WHAT I
SAID. And that does NOT mean that 𝕚 = √(−1); THAT IS SIMPLY NOT THE SAME
THING.
Well, yeah, it is...
patdolan
2021-11-22 11:35:58 UTC
Reply
Permalink
Post by Thomas 'PointedEars' Lahn
That (if “x2” is to be “x²”) means that 𝕚² = −1, WHICH IS EXACTLY WHAT I
SAID. And that does NOT mean that 𝕚 = √(−1); THAT IS SIMPLY NOT THE SAME
THING.
Well, yeah, it is...
How is it possible that Paul A#2 and Long Ears could have a disagreement over i when:

"Mathematics, in its symbols, terms, and reasoning is (given a
context) unambiguous and unforgivingly strict. "
patdolan
2021-11-22 04:46:32 UTC
Reply
Permalink
Post by Paul Alsing
Post by Thomas 'PointedEars' Lahn
On Saturday, November 20, 2021 at 7:47:06 PM UTC-8, Thomas
Post by Thomas 'PointedEars' Lahn
Post by Paul Alsing
... and there are thousands more, but at this point in time, I'm bored...
No, you are incompetent and can’t read.
Like I said... you have a LOT of web pages and textbooks to set
straight... so you better get busy!
Fallacy: Shifting the burden of proof.
I do not have to find sources to support my claim; *you* made the claim.
I showed you the problems with your claims, and their solution.
And I have shown you hundreds of sources that agree with me...
No, you have NOT. You are delusional.
You have shown exactly 6 (in words: SIX) sources, and 3 OF THEM SAY EXACTLY
WHAT I HAVE SAID/EXPLAINED (in a nutshell: “𝕚 *is* a number defined by
𝕚² ≔ −1, which is NOT EQUIVALENT TO 𝕚 ≔ √(−1)”), which I have pointed out
to you.
The other 3 say something WRONG (see above), and one of the three (Wolfram
MathWorld) uses a questionable argument; both of which I have also pointed
out to you.
How is it possible that Wolfram could use a questionable argument when

"Mathematics, in its symbols, terms, and reasoning is (given a
context) unambiguous and unforgivingly strict. "
PointedEars
--
«Nec fasces, nec opes, sola artis sceptra perennant.»
(“Neither high office nor power, only the scepters of science survive.”)
—Tycho Brahe, astronomer (1546-1601): inscription at Hven
Odd Bodkin
2021-11-20 15:56:28 UTC
Reply
Permalink
Post by patdolan
Post by Thomas 'PointedEars' Lahn
Post by patdolan
Post by Thomas 'PointedEars' Lahn
[…] mathematics is just another form of rhetorical expression wherein
falsity can be expressed every bit as plausibly as the truth.
Not true. Natural language is ambiguous and rather loose; it is why
(despite knowing logic) it is so easy to commit fallacies using natural
language. Mathematics, in its symbols, terms, and reasoning is (given a
context) unambiguous and unforgivingly strict.
1^4 = i^4
That much is true.
Post by patdolan
sqrt[ 1^4 ] = sqrt[ i^4 ]
+/-( 1^2 ) = +/-( i^2 )
+/-( 1 ) = +/-( -1 )
+/- 1 = -/+ 1
Your logic is flawed.
How, long ears. How?
If you’re invoking the square root on complex numbers then you have to
follow the rules of that function on the foliated complex plane, not the
rules you would apply on the real number line.

See a book on complex variables and the square root foliation.

Idiot.

Half-ignorant idiot.

Blissfully ignorant of being ignorant, half-ignorant idiot.
Post by patdolan
Post by Thomas 'PointedEars' Lahn
Post by patdolan
Post by Thomas 'PointedEars' Lahn
‘You might say, “All right, then, there’s no explanation of the law;
at least tell me what the law is — why not tell me in words instead of
in the symbols?
I did this.
I know that you did. Feynman explains there why your request cannot be
fulfilled.
Post by patdolan
See MUONS, SCHMUONS! at the bottom of my post. Read through it slow,
Long Ears.
I have better things to do than discussing with you at this low a level.
Post by patdolan
[tl;dr]
PointedEars
--
«Nec fasces, nec opes, sola artis sceptra perennant.»
(“Neither high office nor power, only the scepters of science survive.”)
—Tycho Brahe, astronomer (1546-1601): inscription at Hven
--
Odd Bodkin — Maker of fine toys, tools, tables
Thomas 'PointedEars' Lahn
2021-11-20 23:25:11 UTC
Reply
Permalink
Post by Odd Bodkin
If you’re invoking the square root on complex numbers then you have to
follow the rules of that function on the foliated complex plane, not the
rules you would apply on the real number line.
See a book on complex variables and the square root foliation.
I did follow the foliation rules to the letter. I made two branch cuts on
the complex plane and solved. You didn't recognize this. […]
Blind leading the blind :-D


PointedEars
--
Q: Who's on the case when the electricity goes out?
A: Sherlock Ohms.

(from: WolframAlpha)
Maciej Wozniak
2021-11-18 06:57:16 UTC
Reply
Permalink
Post by Thomas 'PointedEars' Lahn
Post by patdolan
Critical Theory comes in several flavors: Critical Law Theory, Critical
Literature Theory, Critical History Theory, Critical Race Theory. To
these we now add Critical Relativity Theory!
In the spirit of Derrida we shall deconstruct the clumsy reasoning of
special relativity
LOL.
Post by patdolan
and separate said reasoning from the algebraic symbols and equations
that express it,
Since it is fundamentally based in geometry, namely frames of reference
which can understood as (moving) coordinate systems, that approach is
hopeless.
Post by patdolan
keeping in mind that mathematics is just another form of rhetorical
expression wherein falsity can be expressed every bit as plausibly
as the truth.
Not true. Natural language is ambiguous and rather loose; it is why
(despite knowing logic) it is so easy to commit fallacies using natural
language. Mathematics, in its symbols, terms, and reasoning is (given a
context) unambiguous and unforgivingly strict.
Wishes.
Post by Thomas 'PointedEars' Lahn
I don’t think it’s possible, because there’s another feature: mathematics
is not just a language; mathematics is a language plus reasoning
So is natural language. It's just that it can use complicated terms
and complicated rules.
Post by Thomas 'PointedEars' Lahn
No, the velocity that observer A ascribes to observer B is necessarily the
arithmetic inverse of the velocity that observer ascribes to observer B,
In the meantime in the reral world, both observers observe
GPS clocks measuring t'=t, just like all serious clocks
always did.
Post by Thomas 'PointedEars' Lahn
It is not absolute at all. If it were absolute, then every observer would
assign the same velocity to the same object, regardless of their relative
state of motion. The very point of the theories of relativity is that this
is not so.
Your tales of observers assigning this and that didn't match the real
observers even in Galileo's time. Your little theories are - simply - too
primitive; or, as you wish, the real observers are too complcated.
Post by Thomas 'PointedEars' Lahn
“Nature uses only the longest threads to weave her patterns
so that each small piece of her fabric reveals the organization
of the entire tapestry.”
—Richard Feynman, theoretical physicist, “Messenger Lecture” 1 (1964)
Mystical bullshit.
Richard Hachel
2021-11-18 22:46:10 UTC
Reply
Permalink
Post by Thomas 'PointedEars' Lahn
u(v, u') = (v + u')/(1 + v u'/c²)
Yes, linear addition (because µ=0° and cosµ=1).

and perpendicular addition µ=90° cosµ=0 --->
u=sqrt|v²+u'²-v²u'²/c²]

-----

otherwise : speeds general addition :

u(v, u') = sqrt[(v+cosµ.u')²+(sinµ.u'sqrt(1-v²/c²)²)] /
(1+cosµ.u'.v/c²)

R.H.
Dono.
2021-11-18 22:52:09 UTC
Reply
Permalink
Post by Richard Hachel
Post by Thomas 'PointedEars' Lahn
u(v, u') = (v + u')/(1 + v u'/c²)
Yes, linear addition (because µ=0° and cosµ=1).
and perpendicular addition µ=90° cosµ=0 --->
u=sqrt|v²+u'²-v²u'²/c²]
-----
u(v, u') = sqrt[(v+cosµ.u')²+(sinµ.u'sqrt(1-v²/c²)²)] /
(1+cosµ.u'.v/c²)
R.H.
No cretinoid, this not not the way velocity composition works. You need to stop making up shit (and eating it)
Richard Hachel
2021-11-18 23:21:00 UTC
Reply
Permalink
Post by Dono.
Post by Richard Hachel
Post by Thomas 'PointedEars' Lahn
u(v, u') = (v + u')/(1 + v u'/c²)
Yes, linear addition (because µ=0° and cosµ=1).
and perpendicular addition µ=90° cosµ=0 --->
u=sqrt|v²+u'²-v²u'²/c²]
-----
u(v, u') = sqrt[(v+cosµ.u')²+(sinµ.u'sqrt(1-v²/c²)²)] /
(1+cosµ.u'.v/c²)
R.H.
No cretinoid, this not not the way velocity composition works. You need to stop
making up shit (and eating it)
LOL.

R.H.
Python
2021-11-19 01:57:11 UTC
Reply
Permalink
Post by Dono.
Post by Richard Hachel
Post by Thomas 'PointedEars' Lahn
u(v, u') = (v + u')/(1 + v u'/c²)
Yes, linear addition (because µ=0° and cosµ=1).
and perpendicular addition µ=90° cosµ=0 ---> u=sqrt|v²+u'²-v²u'²/c²]
-----
u(v, u') = sqrt[(v+cosµ.u')²+(sinµ.u'sqrt(1-v²/c²)²)] / (1+cosµ.u'.v/c²)
R.H.
No cretinoid, this not not the way velocity composition works. You
need to stop making up shit (and eating it)
LOL.
Ouais, elle est peu chelou ta formule avec une racine dont tu prends
le carré "sqrt(1-v²/c²)²" imbriquée dans une autre racine carrée...

ça ressemble de loin à la bonne, tu t'es pas vautré en recopiant par
hasard?
Maciej Wozniak
2021-11-19 06:13:01 UTC
Reply
Permalink
Post by Python
Post by Dono.
Post by Richard Hachel
Post by Thomas 'PointedEars' Lahn
u(v, u') = (v + u')/(1 + v u'/c²)
Yes, linear addition (because µ=0° and cosµ=1).
and perpendicular addition µ=90° cosµ=0 ---> u=sqrt|v²+u'²-v²u'²/c²]
-----
u(v, u') = sqrt[(v+cosµ.u')²+(sinµ.u'sqrt(1-v²/c²)²)] / (1+cosµ.u'.v/c²)
R.H.
No cretinoid, this not not the way velocity composition works. You
need to stop making up shit (and eating it)
LOL.
Ouais, elle est peu chelou ta formule avec une racine dont tu prends
le carré "sqrt(1-v²/c²)²" imbriquée dans une autre racine carrée...
sqrt, of course, is a multivalue function, returning a set of
numbers; right, poor halfbrain?
Python
2021-11-19 11:20:58 UTC
Reply
Permalink
Post by Maciej Wozniak
Post by Python
Post by Dono.
Post by Richard Hachel
Post by Thomas 'PointedEars' Lahn
u(v, u') = (v + u')/(1 + v u'/c²)
Yes, linear addition (because µ=0° and cosµ=1).
and perpendicular addition µ=90° cosµ=0 ---> u=sqrt|v²+u'²-v²u'²/c²]
-----
u(v, u') = sqrt[(v+cosµ.u')²+(sinµ.u'sqrt(1-v²/c²)²)] / (1+cosµ.u'.v/c²)
R.H.
No cretinoid, this not not the way velocity composition works. You
need to stop making up shit (and eating it)
LOL.
Ouais, elle est peu chelou ta formule avec une racine dont tu prends
le carré "sqrt(1-v²/c²)²" imbriquée dans une autre racine carrée...
sqrt, of course, is a multivalue function, returning a set of
numbers; right,
A multivalued function is NOT a function returning a set of numbers.
BTW, this is off-topic.
Post by Maciej Wozniak
poor halfbrain?
Nice sig, Woz.
patdolan
2021-11-19 14:56:39 UTC
Reply
Permalink
Post by Python
Post by Dono.
Post by Richard Hachel
Post by Thomas 'PointedEars' Lahn
u(v, u') = (v + u')/(1 + v u'/c²)
Yes, linear addition (because µ=0° and cosµ=1).
and perpendicular addition µ=90° cosµ=0 ---> u=sqrt|v²+u'²-v²u'²/c²]
-----
u(v, u') = sqrt[(v+cosµ.u')²+(sinµ.u'sqrt(1-v²/c²)²)] / (1+cosµ.u'.v/c²)
R.H.
No cretinoid, this not not the way velocity composition works. You
need to stop making up shit (and eating it)
LOL.
Ouais, elle est peu chelou ta formule avec une racine dont tu prends
le carré "sqrt(1-v²/c²)²" imbriquée dans une autre racine carrée...
ça ressemble de loin à la bonne, tu t'es pas vautré en recopiant par
hasard?
Ben non, c'est la même que je donne depuis des décennies déjà.
J'avais recommandé à tous les physiciens relativistes de l'apprendre par
coeur.
Comme je leur avais recommandé, de bien examiner ma description du
Langevin, mes transformations de Lorentz "revisitées", et ma façon
littéraire de définir l'anisochronie spatiale, et la notion de
simultanéité.
Le temps n'érode pas les choses justes.
Il en va de même pour la contraction des distances et la dilatation des
durées.
J'ai toujours demandé aux physicien d'abandonner leurs deux équations
simplificatrices (et qui ne marchent que pour les mouvements transversaux)
t'=t/sqrt(1-v²/c²) et x'=x.sqrt(1-v²/c²), comme je leur ai demandé de
porter aux distances ce qu'ils donnaient pour les longueurs (d'où
l'explication de l'effet-zoom relativiste que personne n'a jamais décrit
correctement sinon moi).
x'=x.sqrt(1-v²/c²)/(1+cosµ.v/c)
t'=t(1+cosµ.v/c)/sqrt(1-v²/c²)
Tu peux dire que ça rejoint de loin "les vraies" ou que c'est du
recopiage.
Ca n'a aucune espèce d'intérêt.
L'intérêt n'est pas là. Il est qu'il faut les apprendre par coeur, et
qu'il est inadmissible aujourd'hui,
cent-vingt ans après les premières idées de Poincaré sur cette
magnifique théorie, les étudiants ne les connaissent pas encore, voire
les prennent à la plaisanterie.
R.H.
This just goes to prove that whether it's French, English, German, Swahili, or any other language, relativity just doesn't make sense. C'est pas?
Odd Bodkin
2021-11-19 15:15:06 UTC
Reply
Permalink
Post by patdolan
Post by Python
Post by Dono.
Post by Richard Hachel
Post by Thomas 'PointedEars' Lahn
u(v, u') = (v + u')/(1 + v u'/c²)
Yes, linear addition (because µ=0° and cosµ=1).
and perpendicular addition µ=90° cosµ=0 ---> u=sqrt|v²+u'²-v²u'²/c²]
-----
u(v, u') = sqrt[(v+cosµ.u')²+(sinµ.u'sqrt(1-v²/c²)²)] / (1+cosµ.u'.v/c²)
R.H.
No cretinoid, this not not the way velocity composition works. You
need to stop making up shit (and eating it)
LOL.
Ouais, elle est peu chelou ta formule avec une racine dont tu prends
le carré "sqrt(1-v²/c²)²" imbriquée dans une autre racine carrée...
ça ressemble de loin à la bonne, tu t'es pas vautré en recopiant par
hasard?
Ben non, c'est la même que je donne depuis des décennies déjà.
J'avais recommandé à tous les physiciens relativistes de l'apprendre par
coeur.
Comme je leur avais recommandé, de bien examiner ma description du
Langevin, mes transformations de Lorentz "revisitées", et ma façon
littéraire de définir l'anisochronie spatiale, et la notion de
simultanéité.
Le temps n'érode pas les choses justes.
Il en va de même pour la contraction des distances et la dilatation des
durées.
J'ai toujours demandé aux physicien d'abandonner leurs deux équations
simplificatrices (et qui ne marchent que pour les mouvements transversaux)
t'=t/sqrt(1-v²/c²) et x'=x.sqrt(1-v²/c²), comme je leur ai demandé de
porter aux distances ce qu'ils donnaient pour les longueurs (d'où
l'explication de l'effet-zoom relativiste que personne n'a jamais décrit
correctement sinon moi).
x'=x.sqrt(1-v²/c²)/(1+cosµ.v/c)
t'=t(1+cosµ.v/c)/sqrt(1-v²/c²)
Tu peux dire que ça rejoint de loin "les vraies" ou que c'est du
recopiage.
Ca n'a aucune espèce d'intérêt.
L'intérêt n'est pas là. Il est qu'il faut les apprendre par coeur, et
qu'il est inadmissible aujourd'hui,
cent-vingt ans après les premières idées de Poincaré sur cette
magnifique théorie, les étudiants ne les connaissent pas encore, voire
les prennent à la plaisanterie.
R.H.
This just goes to prove that whether it's French, English, German,
Swahili, or any other language, relativity just doesn't make sense. C'est pas?
Poor Pat. Still find it confusing?
Have you considered reading a good book about it?
Or do you think that a little chat by the fireplace should do it?
--
Odd Bodkin -- maker of fine toys, tools, tables
patdolan
2021-11-19 16:07:08 UTC
Reply
Permalink
Post by Odd Bodkin
Post by patdolan
Post by Python
Post by Dono.
Post by Richard Hachel
Post by Thomas 'PointedEars' Lahn
u(v, u') = (v + u')/(1 + v u'/c²)
Yes, linear addition (because µ=0° and cosµ=1).
and perpendicular addition µ=90° cosµ=0 ---> u=sqrt|v²+u'²-v²u'²/c²]
-----
u(v, u') = sqrt[(v+cosµ.u')²+(sinµ.u'sqrt(1-v²/c²)²)] / (1+cosµ.u'.v/c²)
R.H.
No cretinoid, this not not the way velocity composition works. You
need to stop making up shit (and eating it)
LOL.
Ouais, elle est peu chelou ta formule avec une racine dont tu prends
le carré "sqrt(1-v²/c²)²" imbriquée dans une autre racine carrée...
ça ressemble de loin à la bonne, tu t'es pas vautré en recopiant par
hasard?
Ben non, c'est la même que je donne depuis des décennies déjà.
J'avais recommandé à tous les physiciens relativistes de l'apprendre par
coeur.
Comme je leur avais recommandé, de bien examiner ma description du
Langevin, mes transformations de Lorentz "revisitées", et ma façon
littéraire de définir l'anisochronie spatiale, et la notion de
simultanéité.
Le temps n'érode pas les choses justes.
Il en va de même pour la contraction des distances et la dilatation des
durées.
J'ai toujours demandé aux physicien d'abandonner leurs deux équations
simplificatrices (et qui ne marchent que pour les mouvements transversaux)
t'=t/sqrt(1-v²/c²) et x'=x.sqrt(1-v²/c²), comme je leur ai demandé de
porter aux distances ce qu'ils donnaient pour les longueurs (d'où
l'explication de l'effet-zoom relativiste que personne n'a jamais décrit
correctement sinon moi).
x'=x.sqrt(1-v²/c²)/(1+cosµ.v/c)
t'=t(1+cosµ.v/c)/sqrt(1-v²/c²)
Tu peux dire que ça rejoint de loin "les vraies" ou que c'est du
recopiage.
Ca n'a aucune espèce d'intérêt.
L'intérêt n'est pas là. Il est qu'il faut les apprendre par coeur, et
qu'il est inadmissible aujourd'hui,
cent-vingt ans après les premières idées de Poincaré sur cette
magnifique théorie, les étudiants ne les connaissent pas encore, voire
les prennent à la plaisanterie.
R.H.
This just goes to prove that whether it's French, English, German,
Swahili, or any other language, relativity just doesn't make sense. C'est pas?
Poor Pat. Still find it confusing?
Have you considered reading a good book about it?
Or do you think that a little chat by the fireplace should do it?
--
Odd Bodkin -- maker of fine toys, tools, tables
So Bodkin...why don't you have a try at solving the MUON, SCHMUON challenge. Not worth your time? Or perhaps the challenge is just too much for you; that's what I and everyone else suspect. To remind you of the challenge: Prove that special relativity provides only one, unique solution for how much time elapses on the laboratory clock. Choose which one.
Odd Bodkin
2021-11-19 17:19:42 UTC
Reply
Permalink
Post by patdolan
Post by Odd Bodkin
Post by patdolan
Post by Python
Post by Dono.
Post by Richard Hachel
Post by Thomas 'PointedEars' Lahn
u(v, u') = (v + u')/(1 + v u'/c²)
Yes, linear addition (because µ=0° and cosµ=1).
and perpendicular addition µ=90° cosµ=0 ---> u=sqrt|v²+u'²-v²u'²/c²]
-----
u(v, u') = sqrt[(v+cosµ.u')²+(sinµ.u'sqrt(1-v²/c²)²)] / (1+cosµ.u'.v/c²)
R.H.
No cretinoid, this not not the way velocity composition works. You
need to stop making up shit (and eating it)
LOL.
Ouais, elle est peu chelou ta formule avec une racine dont tu prends
le carré "sqrt(1-v²/c²)²" imbriquée dans une autre racine carrée...
ça ressemble de loin à la bonne, tu t'es pas vautré en recopiant par
hasard?
Ben non, c'est la même que je donne depuis des décennies déjà.
J'avais recommandé à tous les physiciens relativistes de l'apprendre par
coeur.
Comme je leur avais recommandé, de bien examiner ma description du
Langevin, mes transformations de Lorentz "revisitées", et ma façon
littéraire de définir l'anisochronie spatiale, et la notion de
simultanéité.
Le temps n'érode pas les choses justes.
Il en va de même pour la contraction des distances et la dilatation des
durées.
J'ai toujours demandé aux physicien d'abandonner leurs deux équations
simplificatrices (et qui ne marchent que pour les mouvements transversaux)
t'=t/sqrt(1-v²/c²) et x'=x.sqrt(1-v²/c²), comme je leur ai demandé de
porter aux distances ce qu'ils donnaient pour les longueurs (d'où
l'explication de l'effet-zoom relativiste que personne n'a jamais décrit
correctement sinon moi).
x'=x.sqrt(1-v²/c²)/(1+cosµ.v/c)
t'=t(1+cosµ.v/c)/sqrt(1-v²/c²)
Tu peux dire que ça rejoint de loin "les vraies" ou que c'est du
recopiage.
Ca n'a aucune espèce d'intérêt.
L'intérêt n'est pas là. Il est qu'il faut les apprendre par coeur, et
qu'il est inadmissible aujourd'hui,
cent-vingt ans après les premières idées de Poincaré sur cette
magnifique théorie, les étudiants ne les connaissent pas encore, voire
les prennent à la plaisanterie.
R.H.
This just goes to prove that whether it's French, English, German,
Swahili, or any other language, relativity just doesn't make sense. C'est pas?
Poor Pat. Still find it confusing?
Have you considered reading a good book about it?
Or do you think that a little chat by the fireplace should do it?
--
Odd Bodkin -- maker of fine toys, tools, tables
So Bodkin...why don't you have a try at solving the MUON, SCHMUON challenge.
There’s no challenge to solve here. What you have done is simply to lay out
your poorly grasped understanding of what relativity says (which is wrong),
and inviting people to try to teach you the right way to understand how
muons behave. So the only CHALLENGE here is to erase your confusion. This
is why I suggested that maybe a less silly approach of actually reading a
good book about it.

Really the puzzle for me is, why do you think it’s more fun to come here,
demonstrate how poorly you understand relativity, and then slap on a bunch
of bluster and bravado while you dare people to make you understand it — as
opposed to just sitting down with a decent book, which is done without
exposing yourself as a blustering idiot.
Post by patdolan
Not worth your time? Or perhaps the challenge is just too much for you;
Prove that special relativity provides only one, unique solution for how
much time elapses on the laboratory clock. Choose which one.
--
Odd Bodkin -- maker of fine toys, tools, tables
Odd Bodkin
2021-11-19 18:02:53 UTC
Reply
Permalink
Post by Odd Bodkin
Post by patdolan
Post by Odd Bodkin
Post by patdolan
Post by Python
Post by Dono.
Post by Richard Hachel
Post by Thomas 'PointedEars' Lahn
u(v, u') = (v + u')/(1 + v u'/c²)
Yes, linear addition (because µ=0° and cosµ=1).
and perpendicular addition µ=90° cosµ=0 ---> u=sqrt|v²+u'²-v²u'²/c²]
-----
u(v, u') = sqrt[(v+cosµ.u')²+(sinµ.u'sqrt(1-v²/c²)²)] / (1+cosµ.u'.v/c²)
R.H.
No cretinoid, this not not the way velocity composition works. You
need to stop making up shit (and eating it)
LOL.
Ouais, elle est peu chelou ta formule avec une racine dont tu prends
le carré "sqrt(1-v²/c²)²" imbriquée dans une autre racine carrée...
ça ressemble de loin à la bonne, tu t'es pas vautré en recopiant par
hasard?
Ben non, c'est la même que je donne depuis des décennies déjà.
J'avais recommandé à tous les physiciens relativistes de l'apprendre par
coeur.
Comme je leur avais recommandé, de bien examiner ma description du
Langevin, mes transformations de Lorentz "revisitées", et ma façon
littéraire de définir l'anisochronie spatiale, et la notion de
simultanéité.
Le temps n'érode pas les choses justes.
Il en va de même pour la contraction des distances et la dilatation des
durées.
J'ai toujours demandé aux physicien d'abandonner leurs deux équations
simplificatrices (et qui ne marchent que pour les mouvements transversaux)
t'=t/sqrt(1-v²/c²) et x'=x.sqrt(1-v²/c²), comme je leur ai demandé de
porter aux distances ce qu'ils donnaient pour les longueurs (d'où
l'explication de l'effet-zoom relativiste que personne n'a jamais décrit
correctement sinon moi).
x'=x.sqrt(1-v²/c²)/(1+cosµ.v/c)
t'=t(1+cosµ.v/c)/sqrt(1-v²/c²)
Tu peux dire que ça rejoint de loin "les vraies" ou que c'est du
recopiage.
Ca n'a aucune espèce d'intérêt.
L'intérêt n'est pas là. Il est qu'il faut les apprendre par coeur, et
qu'il est inadmissible aujourd'hui,
cent-vingt ans après les premières idées de Poincaré sur cette
magnifique théorie, les étudiants ne les connaissent pas encore, voire
les prennent à la plaisanterie.
R.H.
This just goes to prove that whether it's French, English, German,
Swahili, or any other language, relativity just doesn't make sense. C'est pas?
Poor Pat. Still find it confusing?
Have you considered reading a good book about it?
Or do you think that a little chat by the fireplace should do it?
--
Odd Bodkin -- maker of fine toys, tools, tables
So Bodkin...why don't you have a try at solving the MUON, SCHMUON challenge.
There’s no challenge to solve here. What you have done is simply to lay out
your poorly grasped understanding of what relativity says (which is wrong),
and inviting people to try to teach you the right way to understand how
muons behave. So the only CHALLENGE here is to erase your confusion. This
is why I suggested that maybe a less silly approach of actually reading a
good book about it.
Really the puzzle for me is, why do you think it’s more fun to come here,
demonstrate how poorly you understand relativity, and then slap on a bunch
of bluster and bravado while you dare people to make you understand it — as
opposed to just sitting down with a decent book, which is done without
exposing yourself as a blustering idiot.
Post by patdolan
Not worth your time? Or perhaps the challenge is just too much for you;
Prove that special relativity provides only one, unique solution for how
much time elapses on the laboratory clock. Choose which one.
--
Odd Bodkin -- maker of fine toys, tools, tables
I understand relativity far better than you Bodkin. MUONS, SCHMUONS
demonstrates exactly how much better I understand it than either you or Einstein.
No, Pat, you don’t. And your burlap gauntlet, which you have thrown
vigorously to the ground with a lot of chest-thumping, is a rather silly
token of hubris.

So I’ll ask you one more time. Why don’t you spend more time reading a good
book about this subject and a little less time making a fool of yourself on
an internet backwater? What’s really your objective?Entertaining yourself
with a little rain dance? Or actually coming to a better understanding of
relativity? (As you probably know, rain dances are not effective.)
I request a voice vote of this forum. Who knows relativity better? Me
or Einstein & Bodkin?
--
Odd Bodkin -- maker of fine toys, tools, tables
patdolan
2021-11-19 18:22:56 UTC
Reply
Permalink
Post by Odd Bodkin
Post by Odd Bodkin
Post by patdolan
Post by Odd Bodkin
Post by patdolan
Post by Python
Post by Dono.
Post by Richard Hachel
Post by Thomas 'PointedEars' Lahn
u(v, u') = (v + u')/(1 + v u'/c²)
Yes, linear addition (because µ=0° and cosµ=1).
and perpendicular addition µ=90° cosµ=0 ---> u=sqrt|v²+u'²-v²u'²/c²]
-----
u(v, u') = sqrt[(v+cosµ.u')²+(sinµ.u'sqrt(1-v²/c²)²)] / (1+cosµ.u'.v/c²)
R.H.
No cretinoid, this not not the way velocity composition works. You
need to stop making up shit (and eating it)
LOL.
Ouais, elle est peu chelou ta formule avec une racine dont tu prends
le carré "sqrt(1-v²/c²)²" imbriquée dans une autre racine carrée...
ça ressemble de loin à la bonne, tu t'es pas vautré en recopiant par
hasard?
Ben non, c'est la même que je donne depuis des décennies déjà.
J'avais recommandé à tous les physiciens relativistes de l'apprendre par
coeur.
Comme je leur avais recommandé, de bien examiner ma description du
Langevin, mes transformations de Lorentz "revisitées", et ma façon
littéraire de définir l'anisochronie spatiale, et la notion de
simultanéité.
Le temps n'érode pas les choses justes.
Il en va de même pour la contraction des distances et la dilatation des
durées.
J'ai toujours demandé aux physicien d'abandonner leurs deux équations
simplificatrices (et qui ne marchent que pour les mouvements transversaux)
t'=t/sqrt(1-v²/c²) et x'=x.sqrt(1-v²/c²), comme je leur ai demandé de
porter aux distances ce qu'ils donnaient pour les longueurs (d'où
l'explication de l'effet-zoom relativiste que personne n'a jamais décrit
correctement sinon moi).
x'=x.sqrt(1-v²/c²)/(1+cosµ.v/c)
t'=t(1+cosµ.v/c)/sqrt(1-v²/c²)
Tu peux dire que ça rejoint de loin "les vraies" ou que c'est du
recopiage.
Ca n'a aucune espèce d'intérêt.
L'intérêt n'est pas là. Il est qu'il faut les apprendre par coeur, et
qu'il est inadmissible aujourd'hui,
cent-vingt ans après les premières idées de Poincaré sur cette
magnifique théorie, les étudiants ne les connaissent pas encore, voire
les prennent à la plaisanterie.
R.H.
This just goes to prove that whether it's French, English, German,
Swahili, or any other language, relativity just doesn't make sense. C'est pas?
Poor Pat. Still find it confusing?
Have you considered reading a good book about it?
Or do you think that a little chat by the fireplace should do it?
--
Odd Bodkin -- maker of fine toys, tools, tables
So Bodkin...why don't you have a try at solving the MUON, SCHMUON challenge.
There’s no challenge to solve here. What you have done is simply to lay out
your poorly grasped understanding of what relativity says (which is wrong),
and inviting people to try to teach you the right way to understand how
muons behave. So the only CHALLENGE here is to erase your confusion. This
is why I suggested that maybe a less silly approach of actually reading a
good book about it.
Really the puzzle for me is, why do you think it’s more fun to come here,
demonstrate how poorly you understand relativity, and then slap on a bunch
of bluster and bravado while you dare people to make you understand it — as
opposed to just sitting down with a decent book, which is done without
exposing yourself as a blustering idiot.
Post by patdolan
Not worth your time? Or perhaps the challenge is just too much for you;
Prove that special relativity provides only one, unique solution for how
much time elapses on the laboratory clock. Choose which one.
--
Odd Bodkin -- maker of fine toys, tools, tables
I understand relativity far better than you Bodkin. MUONS, SCHMUONS
demonstrates exactly how much better I understand it than either you or Einstein.
No, Pat, you don’t. And your burlap gauntlet, which you have thrown
vigorously to the ground with a lot of chest-thumping, is a rather silly
token of hubris.
So I’ll ask you one more time. Why don’t you spend more time reading a good
book about this subject and a little less time making a fool of yourself on
an internet backwater? What’s really your objective?Entertaining yourself
with a little rain dance? Or actually coming to a better understanding of
relativity? (As you probably know, rain dances are not effective.)
You are quite a dance expert yourself, Bodkin. You specialty is the fan dance but you never get quite naked when it comes to showing what you can and cannot do with relativity.
Post by Odd Bodkin
I request a voice vote of this forum. Who knows relativity better? Me
or Einstein & Bodkin?
--
Odd Bodkin -- maker of fine toys, tools, tables
Odd Bodkin
2021-11-19 18:45:20 UTC
Reply
Permalink
Post by patdolan
Post by Odd Bodkin
Post by Odd Bodkin
Post by patdolan
Post by Odd Bodkin
Post by patdolan
Post by Python
Post by Dono.
Post by Richard Hachel
Post by Thomas 'PointedEars' Lahn
u(v, u') = (v + u')/(1 + v u'/c²)
Yes, linear addition (because µ=0° and cosµ=1).
and perpendicular addition µ=90° cosµ=0 ---> u=sqrt|v²+u'²-v²u'²/c²]
-----
u(v, u') = sqrt[(v+cosµ.u')²+(sinµ.u'sqrt(1-v²/c²)²)] / (1+cosµ.u'.v/c²)
R.H.
No cretinoid, this not not the way velocity composition works. You
need to stop making up shit (and eating it)
LOL.
Ouais, elle est peu chelou ta formule avec une racine dont tu prends
le carré "sqrt(1-v²/c²)²" imbriquée dans une autre racine carrée...
ça ressemble de loin à la bonne, tu t'es pas vautré en recopiant par
hasard?
Ben non, c'est la même que je donne depuis des décennies déjà.
J'avais recommandé à tous les physiciens relativistes de l'apprendre par
coeur.
Comme je leur avais recommandé, de bien examiner ma description du
Langevin, mes transformations de Lorentz "revisitées", et ma façon
littéraire de définir l'anisochronie spatiale, et la notion de
simultanéité.
Le temps n'érode pas les choses justes.
Il en va de même pour la contraction des distances et la dilatation des
durées.
J'ai toujours demandé aux physicien d'abandonner leurs deux équations
simplificatrices (et qui ne marchent que pour les mouvements transversaux)
t'=t/sqrt(1-v²/c²) et x'=x.sqrt(1-v²/c²), comme je leur ai demandé de
porter aux distances ce qu'ils donnaient pour les longueurs (d'où
l'explication de l'effet-zoom relativiste que personne n'a jamais décrit
correctement sinon moi).
x'=x.sqrt(1-v²/c²)/(1+cosµ.v/c)
t'=t(1+cosµ.v/c)/sqrt(1-v²/c²)
Tu peux dire que ça rejoint de loin "les vraies" ou que c'est du
recopiage.
Ca n'a aucune espèce d'intérêt.
L'intérêt n'est pas là. Il est qu'il faut les apprendre par coeur, et
qu'il est inadmissible aujourd'hui,
cent-vingt ans après les premières idées de Poincaré sur cette
magnifique théorie, les étudiants ne les connaissent pas encore, voire
les prennent à la plaisanterie.
R.H.
This just goes to prove that whether it's French, English, German,
Swahili, or any other language, relativity just doesn't make sense. C'est pas?
Poor Pat. Still find it confusing?
Have you considered reading a good book about it?
Or do you think that a little chat by the fireplace should do it?
--
Odd Bodkin -- maker of fine toys, tools, tables
So Bodkin...why don't you have a try at solving the MUON, SCHMUON challenge.
There’s no challenge to solve here. What you have done is simply to lay out
your poorly grasped understanding of what relativity says (which is wrong),
and inviting people to try to teach you the right way to understand how
muons behave. So the only CHALLENGE here is to erase your confusion. This
is why I suggested that maybe a less silly approach of actually reading a
good book about it.
Really the puzzle for me is, why do you think it’s more fun to come here,
demonstrate how poorly you understand relativity, and then slap on a bunch
of bluster and bravado while you dare people to make you understand it — as
opposed to just sitting down with a decent book, which is done without
exposing yourself as a blustering idiot.
Post by patdolan
Not worth your time? Or perhaps the challenge is just too much for you;
Prove that special relativity provides only one, unique solution for how
much time elapses on the laboratory clock. Choose which one.
--
Odd Bodkin -- maker of fine toys, tools, tables
I understand relativity far better than you Bodkin. MUONS, SCHMUONS
demonstrates exactly how much better I understand it than either you or Einstein.
No, Pat, you don’t. And your burlap gauntlet, which you have thrown
vigorously to the ground with a lot of chest-thumping, is a rather silly
token of hubris.
So I’ll ask you one more time. Why don’t you spend more time reading a good
book about this subject and a little less time making a fool of yourself on
an internet backwater? What’s really your objective?Entertaining yourself
with a little rain dance? Or actually coming to a better understanding of
relativity? (As you probably know, rain dances are not effective.)
You are quite a dance expert yourself, Bodkin. You specialty is the fan
dance but you never get quite naked when it comes to showing what you can
and cannot do with relativity.
And this is where your game is silly. This isn’t a game of poker, with
hidden cards and a lot of bluffing. Relativity is fully documented and
thoroughly described in a century’s worth of textbooks. So when you say
something and declare “This is relativity”, it’s OBVIOUS when you say
something stupid. It’s OBVIOUS when you haven’t read textbooks.

So here you are, pretending to play poker, and you’ve laid out your cards,
a 2, a 5, a 7, a Jack, and a Queen, and you announce “Straight flush”. And
you do this in a room full of people who instantly know you are an idiot
and don’t know what you’re talking about. And then you double down and say,
“So what cards have YOU got, chump?”
Post by patdolan
Post by Odd Bodkin
I request a voice vote of this forum. Who knows relativity better? Me
or Einstein & Bodkin?
--
Odd Bodkin -- maker of fine toys, tools, tables
--
Odd Bodkin -- maker of fine toys, tools, tables
patdolan
2021-11-19 19:05:29 UTC
Reply
Permalink
Post by Odd Bodkin
Post by Odd Bodkin
Post by Odd Bodkin
Post by patdolan
Post by Odd Bodkin
Post by patdolan
Post by Python
Post by Dono.
Post by Richard Hachel
Post by Thomas 'PointedEars' Lahn
u(v, u') = (v + u')/(1 + v u'/c²)
Yes, linear addition (because µ=0° and cosµ=1).
and perpendicular addition µ=90° cosµ=0 ---> u=sqrt|v²+u'²-v²u'²/c²]
-----
u(v, u') = sqrt[(v+cosµ.u')²+(sinµ.u'sqrt(1-v²/c²)²)] / (1+cosµ.u'.v/c²)
R.H.
No cretinoid, this not not the way velocity composition works. You
need to stop making up shit (and eating it)
LOL.
Ouais, elle est peu chelou ta formule avec une racine dont tu prends
le carré "sqrt(1-v²/c²)²" imbriquée dans une autre racine carrée...
ça ressemble de loin à la bonne, tu t'es pas vautré en recopiant par
hasard?
Ben non, c'est la même que je donne depuis des décennies déjà.
J'avais recommandé à tous les physiciens relativistes de l'apprendre par
coeur.
Comme je leur avais recommandé, de bien examiner ma description du
Langevin, mes transformations de Lorentz "revisitées", et ma façon
littéraire de définir l'anisochronie spatiale, et la notion de
simultanéité.
Le temps n'érode pas les choses justes.
Il en va de même pour la contraction des distances et la dilatation des
durées.
J'ai toujours demandé aux physicien d'abandonner leurs deux équations
simplificatrices (et qui ne marchent que pour les mouvements transversaux)
t'=t/sqrt(1-v²/c²) et x'=x.sqrt(1-v²/c²), comme je leur ai demandé de
porter aux distances ce qu'ils donnaient pour les longueurs (d'où
l'explication de l'effet-zoom relativiste que personne n'a jamais décrit
correctement sinon moi).
x'=x.sqrt(1-v²/c²)/(1+cosµ.v/c)
t'=t(1+cosµ.v/c)/sqrt(1-v²/c²)
Tu peux dire que ça rejoint de loin "les vraies" ou que c'est du
recopiage.
Ca n'a aucune espèce d'intérêt.
L'intérêt n'est pas là. Il est qu'il faut les apprendre par coeur, et
qu'il est inadmissible aujourd'hui,
cent-vingt ans après les premières idées de Poincaré sur cette
magnifique théorie, les étudiants ne les connaissent pas encore, voire
les prennent à la plaisanterie.
R.H.
This just goes to prove that whether it's French, English, German,
Swahili, or any other language, relativity just doesn't make sense. C'est pas?
Poor Pat. Still find it confusing?
Have you considered reading a good book about it?
Or do you think that a little chat by the fireplace should do it?
--
Odd Bodkin -- maker of fine toys, tools, tables
So Bodkin...why don't you have a try at solving the MUON, SCHMUON challenge.
There’s no challenge to solve here. What you have done is simply to lay out
your poorly grasped understanding of what relativity says (which is wrong),
and inviting people to try to teach you the right way to understand how
muons behave. So the only CHALLENGE here is to erase your confusion. This
is why I suggested that maybe a less silly approach of actually reading a
good book about it.
Really the puzzle for me is, why do you think it’s more fun to come here,
demonstrate how poorly you understand relativity, and then slap on a bunch
of bluster and bravado while you dare people to make you understand it — as
opposed to just sitting down with a decent book, which is done without
exposing yourself as a blustering idiot.
Post by patdolan
Not worth your time? Or perhaps the challenge is just too much for you;
Prove that special relativity provides only one, unique solution for how
much time elapses on the laboratory clock. Choose which one.
--
Odd Bodkin -- maker of fine toys, tools, tables
I understand relativity far better than you Bodkin. MUONS, SCHMUONS
demonstrates exactly how much better I understand it than either you or Einstein.
No, Pat, you don’t. And your burlap gauntlet, which you have thrown
vigorously to the ground with a lot of chest-thumping, is a rather silly
token of hubris.
So I’ll ask you one more time. Why don’t you spend more time reading a good
book about this subject and a little less time making a fool of yourself on
an internet backwater? What’s really your objective?Entertaining yourself
with a little rain dance? Or actually coming to a better understanding of
relativity? (As you probably know, rain dances are not effective.)
You are quite a dance expert yourself, Bodkin. You specialty is the fan
dance but you never get quite naked when it comes to showing what you can
and cannot do with relativity.
And this is where your game is silly. This isn’t a game of poker, with
hidden cards and a lot of bluffing. Relativity is fully documented and
thoroughly described in a century’s worth of textbooks. So when you say
something and declare “This is relativity”, it’s OBVIOUS when you say
something stupid. It’s OBVIOUS when you haven’t read textbooks.
So here you are, pretending to play poker, and you’ve laid out your cards,
a 2, a 5, a 7, a Jack, and a Queen, and you announce “Straight flush”. And
you do this in a room full of people who instantly know you are an idiot
and don’t know what you’re talking about. And then you double down and say,
“So what cards have YOU got, chump?”
Bodkin, this is one of your most penetrating insights. It accurately portrays this, and every other difference of opinion in Science. In fact, it is Science itself! You are right, I do have an excellent hand. And I've laid it down on the table. I call.

But you have phrased it better. In your own words "So what cards do you have, chump?"
Post by Odd Bodkin
Post by Odd Bodkin
I request a voice vote of this forum. Who knows relativity better? Me
or Einstein & Bodkin?
--
Odd Bodkin -- maker of fine toys, tools, tables
--
Odd Bodkin -- maker of fine toys, tools, tables
Luigi Cotta
2021-11-19 21:28:19 UTC
Reply
Permalink
Post by patdolan
Bodkin, this is one of your most penetrating insights. It accurately
portrays this, and every other difference of opinion in Science. In
fact,
it is Science itself! You are right, I do have an excellent hand. And
I've laid it down on the table. I call.
But you have phrased it better. In your own words "So what cards do you have, chump?"
tensors, 4D.
Odd Bodkin
2021-11-19 21:54:41 UTC
Reply
Permalink
Post by patdolan
Post by Odd Bodkin
Post by Odd Bodkin
Post by Odd Bodkin
Post by patdolan
Post by Odd Bodkin
Post by patdolan
Post by Python
Post by Dono.
Post by Richard Hachel
Post by Thomas 'PointedEars' Lahn
u(v, u') = (v + u')/(1 + v u'/c²)
Yes, linear addition (because µ=0° and cosµ=1).
and perpendicular addition µ=90° cosµ=0 ---> u=sqrt|v²+u'²-v²u'²/c²]
-----
u(v, u') = sqrt[(v+cosµ.u')²+(sinµ.u'sqrt(1-v²/c²)²)] / (1+cosµ.u'.v/c²)
R.H.
No cretinoid, this not not the way velocity composition works. You
need to stop making up shit (and eating it)
LOL.
Ouais, elle est peu chelou ta formule avec une racine dont tu prends
le carré "sqrt(1-v²/c²)²" imbriquée dans une autre racine carrée...
ça ressemble de loin à la bonne, tu t'es pas vautré en recopiant par
hasard?
Ben non, c'est la même que je donne depuis des décennies déjà.
J'avais recommandé à tous les physiciens relativistes de l'apprendre par
coeur.
Comme je leur avais recommandé, de bien examiner ma description du
Langevin, mes transformations de Lorentz "revisitées", et ma façon
littéraire de définir l'anisochronie spatiale, et la notion de
simultanéité.
Le temps n'érode pas les choses justes.
Il en va de même pour la contraction des distances et la dilatation des
durées.
J'ai toujours demandé aux physicien d'abandonner leurs deux équations
simplificatrices (et qui ne marchent que pour les mouvements transversaux)
t'=t/sqrt(1-v²/c²) et x'=x.sqrt(1-v²/c²), comme je leur ai demandé de
porter aux distances ce qu'ils donnaient pour les longueurs (d'où
l'explication de l'effet-zoom relativiste que personne n'a jamais décrit
correctement sinon moi).
x'=x.sqrt(1-v²/c²)/(1+cosµ.v/c)
t'=t(1+cosµ.v/c)/sqrt(1-v²/c²)
Tu peux dire que ça rejoint de loin "les vraies" ou que c'est du
recopiage.
Ca n'a aucune espèce d'intérêt.
L'intérêt n'est pas là. Il est qu'il faut les apprendre par coeur, et
qu'il est inadmissible aujourd'hui,
cent-vingt ans après les premières idées de Poincaré sur cette
magnifique théorie, les étudiants ne les connaissent pas encore, voire
les prennent à la plaisanterie.
R.H.
This just goes to prove that whether it's French, English, German,
Swahili, or any other language, relativity just doesn't make sense. C'est pas?
Poor Pat. Still find it confusing?
Have you considered reading a good book about it?
Or do you think that a little chat by the fireplace should do it?
--
Odd Bodkin -- maker of fine toys, tools, tables
So Bodkin...why don't you have a try at solving the MUON, SCHMUON challenge.
There’s no challenge to solve here. What you have done is simply to lay out
your poorly grasped understanding of what relativity says (which is wrong),
and inviting people to try to teach you the right way to understand how
muons behave. So the only CHALLENGE here is to erase your confusion. This
is why I suggested that maybe a less silly approach of actually reading a
good book about it.
Really the puzzle for me is, why do you think it’s more fun to come here,
demonstrate how poorly you understand relativity, and then slap on a bunch
of bluster and bravado while you dare people to make you understand it — as
opposed to just sitting down with a decent book, which is done without
exposing yourself as a blustering idiot.
Post by patdolan
Not worth your time? Or perhaps the challenge is just too much for you;
Prove that special relativity provides only one, unique solution for how
much time elapses on the laboratory clock. Choose which one.
--
Odd Bodkin -- maker of fine toys, tools, tables
I understand relativity far better than you Bodkin. MUONS, SCHMUONS
demonstrates exactly how much better I understand it than either you or Einstein.
No, Pat, you don’t. And your burlap gauntlet, which you have thrown
vigorously to the ground with a lot of chest-thumping, is a rather silly
token of hubris.
So I’ll ask you one more time. Why don’t you spend more time reading a good
book about this subject and a little less time making a fool of yourself on
an internet backwater? What’s really your objective?Entertaining yourself
with a little rain dance? Or actually coming to a better understanding of
relativity? (As you probably know, rain dances are not effective.)
You are quite a dance expert yourself, Bodkin. You specialty is the fan
dance but you never get quite naked when it comes to showing what you can
and cannot do with relativity.
And this is where your game is silly. This isn’t a game of poker, with
hidden cards and a lot of bluffing. Relativity is fully documented and
thoroughly described in a century’s worth of textbooks. So when you say
something and declare “This is relativity”, it’s OBVIOUS when you say
something stupid. It’s OBVIOUS when you haven’t read textbooks.
So here you are, pretending to play poker, and you’ve laid out your cards,
a 2, a 5, a 7, a Jack, and a Queen, and you announce “Straight flush”. And
you do this in a room full of people who instantly know you are an idiot
and don’t know what you’re talking about. And then you double down and say,
“So what cards have YOU got, chump?”
Bodkin, this is one of your most penetrating insights. It accurately
portrays this, and every other difference of opinion in Science. In
fact, it is Science itself! You are right, I do have an excellent hand.
And I've laid it down on the table. I call.
You have a 2, 5, 7, J, Q hand which you are calling an “excellent hand”
Post by patdolan
But you have phrased it better. In your own words "So what cards do you have, chump?"
It’s pretty simple, Pat. You asked the question, what is the “real”
distance the muon travels through, and what is the “real” time interval of
time that it traveled through decay? This question in and of itself proves
to all at the table that you have no idea what you’re talking about.

One of the foundational observations in relativity is that, given two
events (in your case, the creation of the muon for one event, and the
collision of the muon with the earth for the other event), the distance
between those events is frame dependent and the time between those events
is frame dependent. There IS NO “real” distance between the events and
there IS NO “real” time interval between the events. The distances and time
intervals have values that are accidents of the choice of reference frame.

If you had read a book on special relativity — ANY book on special
relativity — this would have been pounded on as a central theme.

For you to pose a “challenge” about what the “real” values are just shows
(once again) that you opened your mouth when your foot was too close to it.


If you’d like a simple and accessible book to understanding special
relativity, I might suggest General Relativity from A to B. Start there and
maybe you will say fewer stupid things.
Post by patdolan
Post by Odd Bodkin
Post by Odd Bodkin
I request a voice vote of this forum. Who knows relativity better? Me
or Einstein & Bodkin?
--
Odd Bodkin -- maker of fine toys, tools, tables
--
Odd Bodkin -- maker of fine toys, tools, tables
--
Odd Bodkin -- maker of fine toys, tools, tables
Richard Hachel
2021-11-20 22:29:50 UTC
Reply
Permalink
LOL.
R.H.
Tu devrais inviter Pencho Valev à boire le thé chez toi. Ca ferait des
vacances à usenet.
Richard Hachel
2021-11-20 22:48:42 UTC
Reply
Permalink
LOL.
R.H.
Tu devrais inviter Pencho Valev à boire le thé chez toi. Ca ferait des
vacances à usenet.
Nan écoute pehache, soit un peu beau joueur.

Et même si je racontais des conneries comme Pencho Valev, comme tu dis,
pffff... Ca nuirait en quoi à usenet, et à un forum qui ne fait parfois
même plus un post par jour.

Je dis ça, je dis rien.

R.H.
robby
2021-11-21 09:15:00 UTC
Reply
Permalink
Ca nuirait en quoi à usenet, et à un forum qui ne fait parfois même
plus un post par jour.
justement, la proportion de pénibles n'en serait que plus grande, et de
ce fait d'autant plus repoussante.
--
Fabrice
patdolan
2021-11-21 12:50:40 UTC
Reply
Permalink
Post by robby
Ca nuirait en quoi à usenet, et à un forum qui ne fait parfois même
plus un post par jour.
justement, la proportion de pénibles n'en serait que plus grande, et de
ce fait d'autant plus repoussante.
Si vous avez des idées différentes des miennes, n'hésitez pas à les exprimer, robby.
Post by robby
--
Fabrice
Richard Hachel
2021-11-19 15:43:40 UTC
Reply
Permalink
Post by Dono.
Post by Richard Hachel
Post by Thomas 'PointedEars' Lahn
u(v, u') = (v + u')/(1 + v u'/c²)
Yes, linear addition (because µ=0° and cosµ=1).
and perpendicular addition µ=90° cosµ=0 --->
u=sqrt|v²+u'²-v²u'²/c²]
-----
u(v, u') = sqrt[(v+cosµ.u')²+(sinµ.u'sqrt(1-v²/c²))²] / (1+cosµ.u'.v/c²)
R.H.
No cretinoid, this not not the way velocity composition works. You need to stop
making up shit (and eating it)
Formule d'addition générale des vitesses relativistes.

<http://news2.nemoweb.net/jntp?***@jntp/Data.Media:1>

R.H.
Dono.
2021-11-19 16:07:50 UTC
Reply
Permalink
Post by Richard Hachel
Post by Dono.
Post by Richard Hachel
Post by Thomas 'PointedEars' Lahn
u(v, u') = (v + u')/(1 + v u'/c²)
Yes, linear addition (because µ=0° and cosµ=1).
and perpendicular addition µ=90° cosµ=0 --->
u=sqrt|v²+u'²-v²u'²/c²]
-----
u(v, u') = sqrt[(v+cosµ.u')²+(sinµ.u'sqrt(1-v²/c²))²] / (1+cosµ.u'.v/c²)
R.H.
No cretinoid, this not not the way velocity composition works. You need to stop
making up shit (and eating it)
Formule d'addition générale des vitesses relativistes.
Rubbish
patdolan
2021-11-20 03:03:39 UTC
Reply
Permalink
Post by patdolan
Critical Theory comes in several flavors: Critical Law Theory, Critical
Literature Theory, Critical History Theory, Critical Race Theory. To
these we now add Critical Relativity Theory!
In the spirit of Derrida we shall deconstruct the clumsy reasoning of
special relativity and separate said reasoning from the algebraic symbols
and equations that express it, keeping in mind that mathematics is just
another form of rhetorical expression wherein falsity can be expressed
every bit as plausibly as the truth.
“There is only the text.”— J. Derrida
According to special relativity two observers in motion with respect to
each other will disagree on each other’s length. They will also disagree
on the proper flow of time. But they will always agree on the velocity
they have with respect to one another. This is exceedingly strange. How
can it be that two relative quantities, space and time, combine to
produce an absolute quantity called relative velocity? It is true that SR
does have a formula for calculating coordinate velocity; just like it has
formulas for calculating coordinate space and coordinate time. But the
Einstein velocity addition formula ONLY applies to a third object in
motion wrt a pair of FoRs. If that third object happens to be at rest
wrt one of the FoRs then Einsteinian velocity reduces to Galilean
velocity, albeit subject to the speed limit c. Relativists simply
assumed without further justification that if FoR-1 measures a velocity v
between itself and FoR-2 then FoR-2 must measure the same numerical value
v for the velocity between itself and FoR-1. Trivial? Nope. Seemingly
trivial assumptions can be monumental when constructing a theory of
motion. But a 26 year old would probably not yet have the requisite
philosophical sophistication needed to recognize this.
Einstein’s choice to make velocity strictly Galilean when calculating the
velocity between pairs of FoRs ( yes, it is a choice because it does not
follow from either the first or the second postulates ) can be expressed mathematically as
∆x’/∆t’ = ∆x/∆t = v (1)
I now raise equation (1) to the level of a postulate and declare it to be
the third, and heretofore hidden, postulate of special relativity. In
recognizing its own structural Galileanism through this new postulate,
special relativity can finally claim to be woke.
The problem with the third postulate is that even though it is already
assumed in every equation of special relativity, it turns out to be true
only when v = c ( the second postulate ) or when v = 0. The third
postulate can be demonstrated invalid for all values of v in between.
The invalidity of the third postulate causes special relativity fall on
it’s algebraic face. Big Bang move over…Not recognizing and
acknowledging the third postulate was Einstein’s biggest blunder.
Time for some examples.
DIRK & DONO
Consider two FoRs whose x-axes are parallel and lie very close to one
another. The relative velocity between these two FoRs is .866c (γ = 2).
Dirk assumes the Lotus position at the origin of one FoR whilst Dono
assumes the fetal position at the origin of the other. Both Dirk and
Dono and their clocks are glued to the origins of their respective FoRs.
Dirk opens one eye and takes note of the meter marks on Dono’s contracted
x-axis as they whiz by. Dirk apprehends that Dono’s meter marks are
contracted to only half as long as his own meter marks. Dirk opens the
other eye and observes that Dono’s clock is ticking at only half the rate
of his own clock. Dirk begins to count Dono’s meter marks as they race
past Dirk’s position. After one year (by Dirk’s clock) of counting
Dono’s meter marks, Dirk has tallied 1.64e+16 Dono meters ( 9.5e+15 meter
marks/Ly x .866Ly x γ ). Dirk has also observed that only 0.5 years have
elapsed on Dono’s clock. Dirk now calculates what his coordinate
velocity should be according to Dono
( 1.64e+16 Dono meters ) / ( 1.58e+7 Dono seconds ) = 1.04e+9 m/s = 3.5c (2)
[ shortcut: ∆x’/∆t’ = (∆x/∆t)(γ^2) = v(γ^2) ] (3)
“Stop!” You cry, “Dirk’s and Dono’s relative velocity was already
stipulated to be an absolute .866c with respect to one another, when
measured in either FoR.”
That is true, according to (1). But remember that (1) is an arbitrary
choice made by Einstein when he built his theory. It is no more
legitimate a choice than the Dolan FoR coordinate velocity transform (3)
for determining one’s FoR coordinate velocity. It is also no less
inconsistent. For we immediately see by inspection that the Dolan FoR
coordinate velocity is always greater than the relative velocity by a
factor of γ^2. But Einstein’s choice for requiring Galilean FoR pair
velocities clangs with just as much antinomy as Dolan’s transform, as we
shall see. Special relativity’s dirty little secret is that it’s hidden
third postulate (1) destroys the theory from within.
I can see by Dono’s tightening fetal position that he still doesn’t
believe me. Very well. We shall prove SR’s mathematical inconsistency
in the next example.
SPECIAL RELATIVITY COLLAPSES UNDER THE WEIGHT OF ITS ALGEBRAIC ORIGINAL SIN
Consider a pair of FoRs whose relative velocity v is some value other
than c and other than zero. This is expressed mathematically as
v = ∆x’/∆t’ = ∆x/∆t != c (4)
and
v = ∆x’/∆t’ = ∆x/∆t != 0 (5)
The Lorentz transforms allow us to construct the FoR coordinate
velocities for pairs of FoRs
∆x’ = γ( ∆x - v∆t )
∆t’ = γ( ∆t - ∆xv/c^2 )
∆x’/∆t’ = [ γ( ∆x - v∆t ) ] / [ γ( ∆t - ∆xv/c^2 ) ] (6)
We now endeavor to solve (6) for v in hopes of demonstrating the internal
consistency of special relativity, i.e, that v = v for all pair of FoRs.
We saw how this was not the case in DIRK & DONO.
The reader will find that trying to solve for v is hopeless unless we make the substitution
v = ∆x/∆t or v = ∆x’/∆t’
It does not matter which we use—either substitution is permitted by the
third postulate (1).
With the substitution made, we eventually arrive at the preposterous results
∆x’/∆t’ = c (7)
or
∆x’/∆t’ = 0 (8)
The derivation is left as an exercise for the reader. For those needing
help with the derivation ( I'm looking at you, Dono ) I will be happy to
provide it in another post.
The laughable results (7) and (8) directly contradict our assumptions (4)
and (5). Furthermore, the results impose the requirement that v is not
even a variable—v turns out to be a constant always equal to c or zero. Absolutely absurd.
QED.
Algebraic relativity is thus reduced to ridiculous rubble by means
mathematical reductio ad absurdum. The root cause of special relativity’s
spectacular algebraic failure lies in the propositional calculus. I am
happy to expatiate on that subject in another post if there is interest.
Let’s do one more example—this one ripped from the headlines of experimental physics.
MUONS, SCHMUONS!
Imagine that you are a hadron in deep space minding your own business
when all of a sudden you turn to see a Lorentz-flattened earth coming at
you at a velocity of .866c. In the impending collision the first thing
to strike you is an air molecule high in the earth’s atmosphere. That
molecules knocks the stuffing out of you. What is left of you is now a
muon which means that you only have 2.2 microseconds more to live.
It turns out that the surface of the flattened earth is exactly 571.56
meters away from you, according to your own muon FoR. By a remarkable
coincidence there is a flattened scintillator and a flattened clock in a
flattened laboratory directly below you on the surface of the flattened
earth. You note the time on the flattened lab clock. Because you are a
muon, you are your own 2.2 microsecond alarm clock.
The surface of the flattened earth continues to speed towards you at
.866c. 2.2 microseconds later the flattened earth, lab and scintillator
smash into you. Just as you expire in the scintillator you note that
only 1.1 microseconds has elapsed on the lab clock.
You are an idiot. “You” the hadron cannot see the “lab clock” at t=0.
No synchronization required. The muon's clock doesn't even enter into the argument at all. Who said that the lab clock was ever at t=0 or any other value. E. L. A. P. S. E. D. T. I. M. E., Bodkin. Quantities of elapsed time are easily determined in special relativity. Synchronization of clocks doesn't even enter into the argument It's all about one clock, the lab clock. One clock, two observers (the muon and the lab scientists), TWO elapsed time. Quite brilliant on my part, n'es pas?

Tell us how synchronization has any bearing on the argument. The muon sees the lab clock tick off 1.1 microseconds on it's trip down. The lab scientists, using relativity and the standard muon life-span, calculate the same lab clock ticks off 4.4 seconds for the same trip.

Now take a deep breath and form an argument that addresses the facts of the case, not imaginary t=0 stuff.
can see the clock adjacent to you, which is 572 m above the lab clock. And,
by the way, the one adjacent to you and the lab clock are not synchronized,
according to “you”, and so “you” wouldn’t dream of subtracting the readings
on the two clocks at rest in the earth frame to come up with an elapsed
time according to the lab clocks.
“What?” Pat Dolan protests, “the lab clock and the clock up by the
atmosphere are perfectly synchronized beforehand. They are still
synchronized now.” But the question then is, “In what frame were they
synchronized?” Pat Dolan splutters, “What difference does THAT make?” And
the answer is, “It matters a lot. The same two clocks may be synchronized
in one frame, but they’re not synchronized in the muon’s frame, or any
other frame in fact. This is one of the foundational observations of
special relativity.”
“But, but, but….,” fumes Pat Dolan, “I had an EXCELLENT hand!”
No, Pat, you don’t have an excellent hand. You have crap cards, all because
you have never bothered to read a book that talks about muons IN DETAIL.
Instead, you came here with a poor and slippery grasp on a simple and
common example, and you thought somehow that something SO SIMPLE had been
missed in ALL THOSE presentations that you never read.
You, Pat, are a persistent idiot, thinking you have good cards, when you’ve
never even bothered to learn how to play poker.
Post by patdolan
Nothing strange here; the labs clock is traveling at γ=2 with respect to
you so it only logs half as much elapsed time as you.
One of the lab scientists now cries out “Just a minute! That’s not how
this story goes. First of all, it is the hadron that is
Lorentz-flattened, not us. And second of all, that hadron used it’s
flattened FoR to determine that is was only 571.56 meters away from our
scintillator at the point it became a muon. Our unflattened earth FoR
clearly indicated that the hadron was actually 1143.12 meters away when
it became a muon. That’s why it took 4.4 microseconds, traveling at .866
c, to reach our scintillator. And that’s what our lab clock shows. Not
1.1 microseconds, like that dumb muon is claiming.”
Special relativity leaves us in a quandary. How much time actually did
elapse on the lab clock from the moment of the muon’s inception to the
moment of its scintillating demise? Was it 1.1 microseconds? Or was it
4.4 microseconds? It has to be one or the other, it can’t be both.
However, different elapsed times on the same lab clock can be calculated
by the muon and by the earthbound scientists with equal justification.
Special relativity is incapable of providing a unique elapsed time on the
lab clock. I challenge anyone to show that special relativity can
provide a unique elapsed time in the preceding case.
Special relativity’s greatest experimental confirmation turns out to be
it’s second greatest falsification. Gravity has been special
relativity’s greatest falsification since 1907 ( check out the author’s
brilliant post on special relativity vs. Kepler’s 3rd law ).
--
Odd Bodkin -- maker of fine toys, tools, tables
Odd Bodkin
2021-11-20 15:43:44 UTC
Reply
Permalink
Post by patdolan
Post by patdolan
Critical Theory comes in several flavors: Critical Law Theory, Critical
Literature Theory, Critical History Theory, Critical Race Theory. To
these we now add Critical Relativity Theory!
In the spirit of Derrida we shall deconstruct the clumsy reasoning of
special relativity and separate said reasoning from the algebraic symbols
and equations that express it, keeping in mind that mathematics is just
another form of rhetorical expression wherein falsity can be expressed
every bit as plausibly as the truth.
“There is only the text.”— J. Derrida
According to special relativity two observers in motion with respect to
each other will disagree on each other’s length. They will also disagree
on the proper flow of time. But they will always agree on the velocity
they have with respect to one another. This is exceedingly strange. How
can it be that two relative quantities, space and time, combine to
produce an absolute quantity called relative velocity? It is true that SR
does have a formula for calculating coordinate velocity; just like it has
formulas for calculating coordinate space and coordinate time. But the
Einstein velocity addition formula ONLY applies to a third object in
motion wrt a pair of FoRs. If that third object happens to be at rest
wrt one of the FoRs then Einsteinian velocity reduces to Galilean
velocity, albeit subject to the speed limit c. Relativists simply
assumed without further justification that if FoR-1 measures a velocity v
between itself and FoR-2 then FoR-2 must measure the same numerical value
v for the velocity between itself and FoR-1. Trivial? Nope. Seemingly
trivial assumptions can be monumental when constructing a theory of
motion. But a 26 year old would probably not yet have the requisite
philosophical sophistication needed to recognize this.
Einstein’s choice to make velocity strictly Galilean when calculating the
velocity between pairs of FoRs ( yes, it is a choice because it does not
follow from either the first or the second postulates ) can be
expressed mathematically as
∆x’/∆t’ = ∆x/∆t = v (1)
I now raise equation (1) to the level of a postulate and declare it to be
the third, and heretofore hidden, postulate of special relativity. In
recognizing its own structural Galileanism through this new postulate,
special relativity can finally claim to be woke.
The problem with the third postulate is that even though it is already
assumed in every equation of special relativity, it turns out to be true
only when v = c ( the second postulate ) or when v = 0. The third
postulate can be demonstrated invalid for all values of v in between.
The invalidity of the third postulate causes special relativity fall on
it’s algebraic face. Big Bang move over…Not recognizing and
acknowledging the third postulate was Einstein’s biggest blunder.
Time for some examples.
DIRK & DONO
Consider two FoRs whose x-axes are parallel and lie very close to one
another. The relative velocity between these two FoRs is .866c (γ = 2).
Dirk assumes the Lotus position at the origin of one FoR whilst Dono
assumes the fetal position at the origin of the other. Both Dirk and
Dono and their clocks are glued to the origins of their respective FoRs.
Dirk opens one eye and takes note of the meter marks on Dono’s contracted
x-axis as they whiz by. Dirk apprehends that Dono’s meter marks are
contracted to only half as long as his own meter marks. Dirk opens the
other eye and observes that Dono’s clock is ticking at only half the rate
of his own clock. Dirk begins to count Dono’s meter marks as they race
past Dirk’s position. After one year (by Dirk’s clock) of counting
Dono’s meter marks, Dirk has tallied 1.64e+16 Dono meters ( 9.5e+15 meter
marks/Ly x .866Ly x γ ). Dirk has also observed that only 0.5 years have
elapsed on Dono’s clock. Dirk now calculates what his coordinate
velocity should be according to Dono
( 1.64e+16 Dono meters ) / ( 1.58e+7 Dono seconds ) = 1.04e+9 m/s = 3.5c (2)
[ shortcut: ∆x’/∆t’ = (∆x/∆t)(γ^2) = v(γ^2) ] (3)
“Stop!” You cry, “Dirk’s and Dono’s relative velocity was already
stipulated to be an absolute .866c with respect to one another, when
measured in either FoR.”
That is true, according to (1). But remember that (1) is an arbitrary
choice made by Einstein when he built his theory. It is no more
legitimate a choice than the Dolan FoR coordinate velocity transform (3)
for determining one’s FoR coordinate velocity. It is also no less
inconsistent. For we immediately see by inspection that the Dolan FoR
coordinate velocity is always greater than the relative velocity by a
factor of γ^2. But Einstein’s choice for requiring Galilean FoR pair
velocities clangs with just as much antinomy as Dolan’s transform, as we
shall see. Special relativity’s dirty little secret is that it’s hidden
third postulate (1) destroys the theory from within.
I can see by Dono’s tightening fetal position that he still doesn’t
believe me. Very well. We shall prove SR’s mathematical inconsistency
in the next example.
SPECIAL RELATIVITY COLLAPSES UNDER THE WEIGHT OF ITS ALGEBRAIC ORIGINAL SIN
Consider a pair of FoRs whose relative velocity v is some value other
than c and other than zero. This is expressed mathematically as
v = ∆x’/∆t’ = ∆x/∆t != c (4)
and
v = ∆x’/∆t’ = ∆x/∆t != 0 (5)
The Lorentz transforms allow us to construct the FoR coordinate
velocities for pairs of FoRs
∆x’ = γ( ∆x - v∆t )
∆t’ = γ( ∆t - ∆xv/c^2 )
∆x’/∆t’ = [ γ( ∆x - v∆t ) ] / [ γ( ∆t - ∆xv/c^2 ) ] (6)
We now endeavor to solve (6) for v in hopes of demonstrating the internal
consistency of special relativity, i.e, that v = v for all pair of FoRs.
We saw how this was not the case in DIRK & DONO.
The reader will find that trying to solve for v is hopeless unless we
make the substitution
v = ∆x/∆t or v = ∆x’/∆t’
It does not matter which we use—either substitution is permitted by the
third postulate (1).
With the substitution made, we eventually arrive at the preposterous results
∆x’/∆t’ = c (7)
or
∆x’/∆t’ = 0 (8)
The derivation is left as an exercise for the reader. For those needing
help with the derivation ( I'm looking at you, Dono ) I will be happy to
provide it in another post.
The laughable results (7) and (8) directly contradict our assumptions (4)
and (5). Furthermore, the results impose the requirement that v is not
even a variable—v turns out to be a constant always equal to c or zero.
Absolutely absurd.
QED.
Algebraic relativity is thus reduced to ridiculous rubble by means
mathematical reductio ad absurdum. The root cause of special relativity’s
spectacular algebraic failure lies in the propositional calculus. I am
happy to expatiate on that subject in another post if there is interest.
Let’s do one more example—this one ripped from the headlines of experimental physics.
MUONS, SCHMUONS!
Imagine that you are a hadron in deep space minding your own business
when all of a sudden you turn to see a Lorentz-flattened earth coming at
you at a velocity of .866c. In the impending collision the first thing
to strike you is an air molecule high in the earth’s atmosphere. That
molecules knocks the stuffing out of you. What is left of you is now a
muon which means that you only have 2.2 microseconds more to live.
It turns out that the surface of the flattened earth is exactly 571.56
meters away from you, according to your own muon FoR. By a remarkable
coincidence there is a flattened scintillator and a flattened clock in a
flattened laboratory directly below you on the surface of the flattened
earth. You note the time on the flattened lab clock. Because you are a
muon, you are your own 2.2 microsecond alarm clock.
The surface of the flattened earth continues to speed towards you at
.866c. 2.2 microseconds later the flattened earth, lab and scintillator
smash into you. Just as you expire in the scintillator you note that
only 1.1 microseconds has elapsed on the lab clock.
You are an idiot. “You” the hadron cannot see the “lab clock” at t=0.
No synchronization required. The muon's clock doesn't even enter into
the argument at all. Who said that the lab clock was ever at t=0 or any
other value. E. L. A. P. S. E. D. T. I. M. E., Bodkin.
Elapsed from what moment? How does the lab clock know when to start
ticking. It isn’t anywhere near the collision of the hadron and the
atmospheric molecule. Is it supposed to get an instantaneous signal from
that event?

This is what I mean when I say you haven’t bothered to read any book about
relativity. Your “argument” is based on two and only two clocks passing
each other, both deeming the other slowed, because that’s the comic book
version of relativity. But that isn’t what relativity says. Ask yourself
why relativity talks about a *lattice* of synchronized clocks in each
frame. Why would you need the lattice?

It’s so that there is a clock NEAR the event that you’re tracking the time
of.
Post by patdolan
Quantities of elapsed time are easily determined in special relativity.
Synchronization of clocks doesn't even enter into the argument It's all
about one clock, the lab clock. One clock, two observers (the muon and
the lab scientists), TWO elapsed time. Quite brilliant on my part, n'es pas?
Tell us how synchronization has any bearing on the argument. The muon
sees the lab clock tick off 1.1 microseconds on it's trip down.
No, it doesn’t. How is it supposed to “see” that? It isn’t ANYWHERE NEAR
the lab clock. The best the muon can do is either a) monitor its own clock,
because its own clock is at ONE location in its own frame, or b) have a
LATTICE of synchronized clocks traveling along with the muon, where one of
them is far away from the muon, near the lab clock at the time the muon is
created.

If you have two events that occur at DIFFERENT PLACES in some frame of
reference, you need two DIFFERENT clocks near those events. If those clocks
are synchronized in this frame, THEN you can measure the elapsed time
between the events.

You can’t measure the elapsed time between two events that occur in
different places with ONE clock.

You, Pat Dolan, don’t understand the first thing about relativity. And it
shows.

And yet you bluster simultaneously that a) you understand relativity very
well, and b) that it is incomprehensible.
Post by patdolan
The lab scientists, using relativity and the standard muon life-span,
calculate the same lab clock ticks off 4.4 seconds for the same trip.
Now take a deep breath and form an argument that addresses the facts of
the case, not imaginary t=0 stuff.
can see the clock adjacent to you, which is 572 m above the lab clock. And,
by the way, the one adjacent to you and the lab clock are not synchronized,
according to “you”, and so “you” wouldn’t dream of subtracting the readings
on the two clocks at rest in the earth frame to come up with an elapsed
time according to the lab clocks.
“What?” Pat Dolan protests, “the lab clock and the clock up by the
atmosphere are perfectly synchronized beforehand. They are still
synchronized now.” But the question then is, “In what frame were they
synchronized?” Pat Dolan splutters, “What difference does THAT make?” And
the answer is, “It matters a lot. The same two clocks may be synchronized
in one frame, but they’re not synchronized in the muon’s frame, or any
other frame in fact. This is one of the foundational observations of
special relativity.”
“But, but, but….,” fumes Pat Dolan, “I had an EXCELLENT hand!”
No, Pat, you don’t have an excellent hand. You have crap cards, all because
you have never bothered to read a book that talks about muons IN DETAIL.
Instead, you came here with a poor and slippery grasp on a simple and
common example, and you thought somehow that something SO SIMPLE had been
missed in ALL THOSE presentations that you never read.
You, Pat, are a persistent idiot, thinking you have good cards, when you’ve
never even bothered to learn how to play poker.
Post by patdolan
Nothing strange here; the labs clock is traveling at γ=2 with respect to
you so it only logs half as much elapsed time as you.
One of the lab scientists now cries out “Just a minute! That’s not how
this story goes. First of all, it is the hadron that is
Lorentz-flattened, not us. And second of all, that hadron used it’s
flattened FoR to determine that is was only 571.56 meters away from our
scintillator at the point it became a muon. Our unflattened earth FoR
clearly indicated that the hadron was actually 1143.12 meters away when
it became a muon. That’s why it took 4.4 microseconds, traveling at .866
c, to reach our scintillator. And that’s what our lab clock shows. Not
1.1 microseconds, like that dumb muon is claiming.”
Special relativity leaves us in a quandary. How much time actually did
elapse on the lab clock from the moment of the muon’s inception to the
moment of its scintillating demise? Was it 1.1 microseconds? Or was it
4.4 microseconds? It has to be one or the other, it can’t be both.
However, different elapsed times on the same lab clock can be calculated
by the muon and by the earthbound scientists with equal justification.
Special relativity is incapable of providing a unique elapsed time on the
lab clock. I challenge anyone to show that special relativity can
provide a unique elapsed time in the preceding case.
Special relativity’s greatest experimental confirmation turns out to be
it’s second greatest falsification. Gravity has been special
relativity’s greatest falsification since 1907 ( check out the author’s
brilliant post on special relativity vs. Kepler’s 3rd law ).
--
Odd Bodkin -- maker of fine toys, tools, tables
--
Odd Bodkin -- maker of fine toys, tools, tables
patdolan
2021-11-20 17:43:55 UTC
Reply
Permalink
Post by Odd Bodkin
Post by patdolan
Critical Theory comes in several flavors: Critical Law Theory, Critical
Literature Theory, Critical History Theory, Critical Race Theory. To
these we now add Critical Relativity Theory!
In the spirit of Derrida we shall deconstruct the clumsy reasoning of
special relativity and separate said reasoning from the algebraic symbols
and equations that express it, keeping in mind that mathematics is just
another form of rhetorical expression wherein falsity can be expressed
every bit as plausibly as the truth.
“There is only the text.”— J. Derrida
According to special relativity two observers in motion with respect to
each other will disagree on each other’s length. They will also disagree
on the proper flow of time. But they will always agree on the velocity
they have with respect to one another. This is exceedingly strange. How
can it be that two relative quantities, space and time, combine to
produce an absolute quantity called relative velocity? It is true that SR
does have a formula for calculating coordinate velocity; just like it has
formulas for calculating coordinate space and coordinate time. But the
Einstein velocity addition formula ONLY applies to a third object in
motion wrt a pair of FoRs. If that third object happens to be at rest
wrt one of the FoRs then Einsteinian velocity reduces to Galilean
velocity, albeit subject to the speed limit c. Relativists simply
assumed without further justification that if FoR-1 measures a velocity v
between itself and FoR-2 then FoR-2 must measure the same numerical value
v for the velocity between itself and FoR-1. Trivial? Nope. Seemingly
trivial assumptions can be monumental when constructing a theory of
motion. But a 26 year old would probably not yet have the requisite
philosophical sophistication needed to recognize this.
Einstein’s choice to make velocity strictly Galilean when calculating the
velocity between pairs of FoRs ( yes, it is a choice because it does not
follow from either the first or the second postulates ) can be
expressed mathematically as
∆x’/∆t’ = ∆x/∆t = v (1)
I now raise equation (1) to the level of a postulate and declare it to be
the third, and heretofore hidden, postulate of special relativity. In
recognizing its own structural Galileanism through this new postulate,
special relativity can finally claim to be woke.
The problem with the third postulate is that even though it is already
assumed in every equation of special relativity, it turns out to be true
only when v = c ( the second postulate ) or when v = 0. The third
postulate can be demonstrated invalid for all values of v in between.
The invalidity of the third postulate causes special relativity fall on
it’s algebraic face. Big Bang move over…Not recognizing and
acknowledging the third postulate was Einstein’s biggest blunder.
Time for some examples.
DIRK & DONO
Consider two FoRs whose x-axes are parallel and lie very close to one
another. The relative velocity between these two FoRs is .866c (γ = 2).
Dirk assumes the Lotus position at the origin of one FoR whilst Dono
assumes the fetal position at the origin of the other. Both Dirk and
Dono and their clocks are glued to the origins of their respective FoRs.
Dirk opens one eye and takes note of the meter marks on Dono’s contracted
x-axis as they whiz by. Dirk apprehends that Dono’s meter marks are
contracted to only half as long as his own meter marks. Dirk opens the
other eye and observes that Dono’s clock is ticking at only half the rate
of his own clock. Dirk begins to count Dono’s meter marks as they race
past Dirk’s position. After one year (by Dirk’s clock) of counting
Dono’s meter marks, Dirk has tallied 1.64e+16 Dono meters ( 9.5e+15 meter
marks/Ly x .866Ly x γ ). Dirk has also observed that only 0.5 years have
elapsed on Dono’s clock. Dirk now calculates what his coordinate
velocity should be according to Dono
( 1.64e+16 Dono meters ) / ( 1.58e+7 Dono seconds ) = 1.04e+9 m/s = 3.5c (2)
[ shortcut: ∆x’/∆t’ = (∆x/∆t)(γ^2) = v(γ^2) ] (3)
“Stop!” You cry, “Dirk’s and Dono’s relative velocity was already
stipulated to be an absolute .866c with respect to one another, when
measured in either FoR.”
That is true, according to (1). But remember that (1) is an arbitrary
choice made by Einstein when he built his theory. It is no more
legitimate a choice than the Dolan FoR coordinate velocity transform (3)
for determining one’s FoR coordinate velocity. It is also no less
inconsistent. For we immediately see by inspection that the Dolan FoR
coordinate velocity is always greater than the relative velocity by a
factor of γ^2. But Einstein’s choice for requiring Galilean FoR pair
velocities clangs with just as much antinomy as Dolan’s transform, as we
shall see. Special relativity’s dirty little secret is that it’s hidden
third postulate (1) destroys the theory from within.
I can see by Dono’s tightening fetal position that he still doesn’t
believe me. Very well. We shall prove SR’s mathematical inconsistency
in the next example.
SPECIAL RELATIVITY COLLAPSES UNDER THE WEIGHT OF ITS ALGEBRAIC ORIGINAL SIN
Consider a pair of FoRs whose relative velocity v is some value other
than c and other than zero. This is expressed mathematically as
v = ∆x’/∆t’ = ∆x/∆t != c (4)
and
v = ∆x’/∆t’ = ∆x/∆t != 0 (5)
The Lorentz transforms allow us to construct the FoR coordinate
velocities for pairs of FoRs
∆x’ = γ( ∆x - v∆t )
∆t’ = γ( ∆t - ∆xv/c^2 )
∆x’/∆t’ = [ γ( ∆x - v∆t ) ] / [ γ( ∆t - ∆xv/c^2 ) ] (6)
We now endeavor to solve (6) for v in hopes of demonstrating the internal
consistency of special relativity, i.e, that v = v for all pair of FoRs.
We saw how this was not the case in DIRK & DONO.
The reader will find that trying to solve for v is hopeless unless we
make the substitution
v = ∆x/∆t or v = ∆x’/∆t’
It does not matter which we use—either substitution is permitted by the
third postulate (1).
With the substitution made, we eventually arrive at the preposterous results
∆x’/∆t’ = c (7)
or
∆x’/∆t’ = 0 (8)
The derivation is left as an exercise for the reader. For those needing
help with the derivation ( I'm looking at you, Dono ) I will be happy to
provide it in another post.
The laughable results (7) and (8) directly contradict our assumptions (4)
and (5). Furthermore, the results impose the requirement that v is not
even a variable—v turns out to be a constant always equal to c or zero.
Absolutely absurd.
QED.
Algebraic relativity is thus reduced to ridiculous rubble by means
mathematical reductio ad absurdum. The root cause of special relativity’s
spectacular algebraic failure lies in the propositional calculus. I am
happy to expatiate on that subject in another post if there is interest.
Let’s do one more example—this one ripped from the headlines of experimental physics.
MUONS, SCHMUONS!
Imagine that you are a hadron in deep space minding your own business
when all of a sudden you turn to see a Lorentz-flattened earth coming at
you at a velocity of .866c. In the impending collision the first thing
to strike you is an air molecule high in the earth’s atmosphere. That
molecules knocks the stuffing out of you. What is left of you is now a
muon which means that you only have 2.2 microseconds more to live.
It turns out that the surface of the flattened earth is exactly 571.56
meters away from you, according to your own muon FoR. By a remarkable
coincidence there is a flattened scintillator and a flattened clock in a
flattened laboratory directly below you on the surface of the flattened
earth. You note the time on the flattened lab clock. Because you are a
muon, you are your own 2.2 microsecond alarm clock.
The surface of the flattened earth continues to speed towards you at
.866c. 2.2 microseconds later the flattened earth, lab and scintillator
smash into you. Just as you expire in the scintillator you note that
only 1.1 microseconds has elapsed on the lab clock.
You are an idiot. “You” the hadron cannot see the “lab clock” at t=0.
No synchronization required. The muon's clock doesn't even enter into
the argument at all. Who said that the lab clock was ever at t=0 or any
other value. E. L. A. P. S. E. D. T. I. M. E., Bodkin.
Elapsed from what moment? How does the lab clock know when to start
ticking. It isn’t anywhere near the collision of the hadron and the
atmospheric molecule. Is it supposed to get an instantaneous signal from
that event?
This is what I mean when I say you haven’t bothered to read any book about
relativity. Your “argument” is based on two and only two clocks passing
each other, both deeming the other slowed, because that’s the comic book
version of relativity. But that isn’t what relativity says. Ask yourself
why relativity talks about a *lattice* of synchronized clocks in each
frame. Why would you need the lattice?
It’s so that there is a clock NEAR the event that you’re tracking the time
of.
Quantities of elapsed time are easily determined in special relativity.
Synchronization of clocks doesn't even enter into the argument It's all
about one clock, the lab clock. One clock, two observers (the muon and
the lab scientists), TWO elapsed time. Quite brilliant on my part, n'es pas?
Tell us how synchronization has any bearing on the argument. The muon
sees the lab clock tick off 1.1 microseconds on it's trip down.
No, it doesn’t. How is it supposed to “see” that? It isn’t ANYWHERE NEAR
the lab clock. The best the muon can do is either a) monitor its own clock,
because its own clock is at ONE location in its own frame, or b) have a
LATTICE of synchronized clocks traveling along with the muon, where one of
them is far away from the muon, near the lab clock at the time the muon is
created.
If you have two events that occur at DIFFERENT PLACES in some frame of
reference, you need two DIFFERENT clocks near those events. If those clocks
are synchronized in this frame, THEN you can measure the elapsed time
between the events.
You can’t measure the elapsed time between two events that occur in
different places with ONE clock.
You, Pat Dolan, don’t understand the first thing about relativity. And it
shows.
And yet you bluster simultaneously that a) you understand relativity very
well, and b) that it is incomprehensible.
The lab scientists, using relativity and the standard muon life-span,
calculate the same lab clock ticks off 4.4 seconds for the same trip.
Now take a deep breath and form an argument that addresses the facts of
the case, not imaginary t=0 stuff.
can see the clock adjacent to you, which is 572 m above the lab clock. And,
by the way, the one adjacent to you and the lab clock are not synchronized,
according to “you”, and so “you” wouldn’t dream of subtracting the readings
on the two clocks at rest in the earth frame to come up with an elapsed
time according to the lab clocks.
“What?” Pat Dolan protests, “the lab clock and the clock up by the
atmosphere are perfectly synchronized beforehand. They are still
synchronized now.” But the question then is, “In what frame were they
synchronized?” Pat Dolan splutters, “What difference does THAT make?” And
the answer is, “It matters a lot. The same two clocks may be synchronized
in one frame, but they’re not synchronized in the muon’s frame, or any
other frame in fact. This is one of the foundational observations of
special relativity.”
“But, but, but….,” fumes Pat Dolan, “I had an EXCELLENT hand!”
No, Pat, you don’t have an excellent hand. You have crap cards, all because
you have never bothered to read a book that talks about muons IN DETAIL.
Instead, you came here with a poor and slippery grasp on a simple and
common example, and you thought somehow that something SO SIMPLE had been
missed in ALL THOSE presentations that you never read.
You, Pat, are a persistent idiot, thinking you have good cards, when you’ve
never even bothered to learn how to play poker.
Post by patdolan
Nothing strange here; the labs clock is traveling at γ=2 with respect to
you so it only logs half as much elapsed time as you.
One of the lab scientists now cries out “Just a minute! That’s not how
this story goes. First of all, it is the hadron that is
Lorentz-flattened, not us. And second of all, that hadron used it’s
flattened FoR to determine that is was only 571.56 meters away from our
scintillator at the point it became a muon. Our unflattened earth FoR
clearly indicated that the hadron was actually 1143.12 meters away when
it became a muon. That’s why it took 4.4 microseconds, traveling at .866
c, to reach our scintillator. And that’s what our lab clock shows. Not
1.1 microseconds, like that dumb muon is claiming.”
Special relativity leaves us in a quandary. How much time actually did
elapse on the lab clock from the moment of the muon’s inception to the
moment of its scintillating demise? Was it 1.1 microseconds? Or was it
4.4 microseconds? It has to be one or the other, it can’t be both.
However, different elapsed times on the same lab clock can be calculated
by the muon and by the earthbound scientists with equal justification.
Special relativity is incapable of providing a unique elapsed time on the
lab clock. I challenge anyone to show that special relativity can
provide a unique elapsed time in the preceding case.
Special relativity’s greatest experimental confirmation turns out to be
it’s second greatest falsification. Gravity has been special
relativity’s greatest falsification since 1907 ( check out the author’s
brilliant post on special relativity vs. Kepler’s 3rd law ).
--
Odd Bodkin -- maker of fine toys, tools, tables
--
Odd Bodkin -- maker of fine toys, tools, tables
The muon is bathed in the light cone from the lab clock a the moment of it's inception. The muon notes what the light cone is indicating at that moment while it still 572m/1143m away from the clock. The muon notes the light from the lab clock again when it is in the scintillator. This is T1. From T1 the muon subtracts 2.2 microseconds to get the elapsed time on the lab clock for the muon's trip. This elapsed time will be 1.1 microseconds according to relativity and the muon. No lattice of clocks, no synchronization no nothing. Easy-peasy Bodkin.
*the difference of the times is T1: light cone time at inception - light cone in scintillator = T1
patdolan
2021-11-20 17:55:31 UTC
Reply
Permalink
Post by patdolan
Post by Odd Bodkin
Post by patdolan
Critical Theory comes in several flavors: Critical Law Theory, Critical
Literature Theory, Critical History Theory, Critical Race Theory. To
these we now add Critical Relativity Theory!
In the spirit of Derrida we shall deconstruct the clumsy reasoning of
special relativity and separate said reasoning from the algebraic symbols
and equations that express it, keeping in mind that mathematics is just
another form of rhetorical expression wherein falsity can be expressed
every bit as plausibly as the truth.
“There is only the text.”— J. Derrida
According to special relativity two observers in motion with respect to
each other will disagree on each other’s length. They will also disagree
on the proper flow of time. But they will always agree on the velocity
they have with respect to one another. This is exceedingly strange. How
can it be that two relative quantities, space and time, combine to
produce an absolute quantity called relative velocity? It is true that SR
does have a formula for calculating coordinate velocity; just like it has
formulas for calculating coordinate space and coordinate time. But the
Einstein velocity addition formula ONLY applies to a third object in
motion wrt a pair of FoRs. If that third object happens to be at rest
wrt one of the FoRs then Einsteinian velocity reduces to Galilean
velocity, albeit subject to the speed limit c. Relativists simply
assumed without further justification that if FoR-1 measures a velocity v
between itself and FoR-2 then FoR-2 must measure the same numerical value
v for the velocity between itself and FoR-1. Trivial? Nope. Seemingly
trivial assumptions can be monumental when constructing a theory of
motion. But a 26 year old would probably not yet have the requisite
philosophical sophistication needed to recognize this.
Einstein’s choice to make velocity strictly Galilean when calculating the
velocity between pairs of FoRs ( yes, it is a choice because it does not
follow from either the first or the second postulates ) can be
expressed mathematically as
∆x’/∆t’ = ∆x/∆t = v (1)
I now raise equation (1) to the level of a postulate and declare it to be
the third, and heretofore hidden, postulate of special relativity. In
recognizing its own structural Galileanism through this new postulate,
special relativity can finally claim to be woke.
The problem with the third postulate is that even though it is already
assumed in every equation of special relativity, it turns out to be true
only when v = c ( the second postulate ) or when v = 0. The third
postulate can be demonstrated invalid for all values of v in between.
The invalidity of the third postulate causes special relativity fall on
it’s algebraic face. Big Bang move over…Not recognizing and
acknowledging the third postulate was Einstein’s biggest blunder.
Time for some examples.
DIRK & DONO
Consider two FoRs whose x-axes are parallel and lie very close to one
another. The relative velocity between these two FoRs is .866c (γ = 2).
Dirk assumes the Lotus position at the origin of one FoR whilst Dono
assumes the fetal position at the origin of the other. Both Dirk and
Dono and their clocks are glued to the origins of their respective FoRs.
Dirk opens one eye and takes note of the meter marks on Dono’s contracted
x-axis as they whiz by. Dirk apprehends that Dono’s meter marks are
contracted to only half as long as his own meter marks. Dirk opens the
other eye and observes that Dono’s clock is ticking at only half the rate
of his own clock. Dirk begins to count Dono’s meter marks as they race
past Dirk’s position. After one year (by Dirk’s clock) of counting
Dono’s meter marks, Dirk has tallied 1.64e+16 Dono meters ( 9.5e+15 meter
marks/Ly x .866Ly x γ ). Dirk has also observed that only 0.5 years have
elapsed on Dono’s clock. Dirk now calculates what his coordinate
velocity should be according to Dono
( 1.64e+16 Dono meters ) / ( 1.58e+7 Dono seconds ) = 1.04e+9 m/s = 3.5c (2)
[ shortcut: ∆x’/∆t’ = (∆x/∆t)(γ^2) = v(γ^2) ] (3)
“Stop!” You cry, “Dirk’s and Dono’s relative velocity was already
stipulated to be an absolute .866c with respect to one another, when
measured in either FoR.”
That is true, according to (1). But remember that (1) is an arbitrary
choice made by Einstein when he built his theory. It is no more
legitimate a choice than the Dolan FoR coordinate velocity transform (3)
for determining one’s FoR coordinate velocity. It is also no less
inconsistent. For we immediately see by inspection that the Dolan FoR
coordinate velocity is always greater than the relative velocity by a
factor of γ^2. But Einstein’s choice for requiring Galilean FoR pair
velocities clangs with just as much antinomy as Dolan’s transform, as we
shall see. Special relativity’s dirty little secret is that it’s hidden
third postulate (1) destroys the theory from within.
I can see by Dono’s tightening fetal position that he still doesn’t
believe me. Very well. We shall prove SR’s mathematical inconsistency
in the next example.
SPECIAL RELATIVITY COLLAPSES UNDER THE WEIGHT OF ITS ALGEBRAIC ORIGINAL SIN
Consider a pair of FoRs whose relative velocity v is some value other
than c and other than zero. This is expressed mathematically as
v = ∆x’/∆t’ = ∆x/∆t != c (4)
and
v = ∆x’/∆t’ = ∆x/∆t != 0 (5)
The Lorentz transforms allow us to construct the FoR coordinate
velocities for pairs of FoRs
∆x’ = γ( ∆x - v∆t )
∆t’ = γ( ∆t - ∆xv/c^2 )
∆x’/∆t’ = [ γ( ∆x - v∆t ) ] / [ γ( ∆t - ∆xv/c^2 ) ] (6)
We now endeavor to solve (6) for v in hopes of demonstrating the internal
consistency of special relativity, i.e, that v = v for all pair of FoRs.
We saw how this was not the case in DIRK & DONO.
The reader will find that trying to solve for v is hopeless unless we
make the substitution
v = ∆x/∆t or v = ∆x’/∆t’
It does not matter which we use—either substitution is permitted by the
third postulate (1).
With the substitution made, we eventually arrive at the preposterous results
∆x’/∆t’ = c (7)
or
∆x’/∆t’ = 0 (8)
The derivation is left as an exercise for the reader. For those needing
help with the derivation ( I'm looking at you, Dono ) I will be happy to
provide it in another post.
The laughable results (7) and (8) directly contradict our assumptions (4)
and (5). Furthermore, the results impose the requirement that v is not
even a variable—v turns out to be a constant always equal to c or zero.
Absolutely absurd.
QED.
Algebraic relativity is thus reduced to ridiculous rubble by means
mathematical reductio ad absurdum. The root cause of special relativity’s
spectacular algebraic failure lies in the propositional calculus. I am
happy to expatiate on that subject in another post if there is interest.
Let’s do one more example—this one ripped from the headlines of experimental physics.
MUONS, SCHMUONS!
Imagine that you are a hadron in deep space minding your own business
when all of a sudden you turn to see a Lorentz-flattened earth coming at
you at a velocity of .866c. In the impending collision the first thing
to strike you is an air molecule high in the earth’s atmosphere. That
molecules knocks the stuffing out of you. What is left of you is now a
muon which means that you only have 2.2 microseconds more to live.
It turns out that the surface of the flattened earth is exactly 571.56
meters away from you, according to your own muon FoR. By a remarkable
coincidence there is a flattened scintillator and a flattened clock in a
flattened laboratory directly below you on the surface of the flattened
earth. You note the time on the flattened lab clock. Because you are a
muon, you are your own 2.2 microsecond alarm clock.
The surface of the flattened earth continues to speed towards you at
.866c. 2.2 microseconds later the flattened earth, lab and scintillator
smash into you. Just as you expire in the scintillator you note that
only 1.1 microseconds has elapsed on the lab clock.
You are an idiot. “You” the hadron cannot see the “lab clock” at t=0.
No synchronization required. The muon's clock doesn't even enter into
the argument at all. Who said that the lab clock was ever at t=0 or any
other value. E. L. A. P. S. E. D. T. I. M. E., Bodkin.
Elapsed from what moment? How does the lab clock know when to start
ticking. It isn’t anywhere near the collision of the hadron and the
atmospheric molecule. Is it supposed to get an instantaneous signal from
that event?
This is what I mean when I say you haven’t bothered to read any book about
relativity. Your “argument” is based on two and only two clocks passing
each other, both deeming the other slowed, because that’s the comic book
version of relativity. But that isn’t what relativity says. Ask yourself
why relativity talks about a *lattice* of synchronized clocks in each
frame. Why would you need the lattice?
It’s so that there is a clock NEAR the event that you’re tracking the time
of.
Quantities of elapsed time are easily determined in special relativity.
Synchronization of clocks doesn't even enter into the argument It's all
about one clock, the lab clock. One clock, two observers (the muon and
the lab scientists), TWO elapsed time. Quite brilliant on my part, n'es pas?
Tell us how synchronization has any bearing on the argument. The muon
sees the lab clock tick off 1.1 microseconds on it's trip down.
No, it doesn’t. How is it supposed to “see” that? It isn’t ANYWHERE NEAR
the lab clock. The best the muon can do is either a) monitor its own clock,
because its own clock is at ONE location in its own frame, or b) have a
LATTICE of synchronized clocks traveling along with the muon, where one of
them is far away from the muon, near the lab clock at the time the muon is
created.
If you have two events that occur at DIFFERENT PLACES in some frame of
reference, you need two DIFFERENT clocks near those events. If those clocks
are synchronized in this frame, THEN you can measure the elapsed time
between the events.
You can’t measure the elapsed time between two events that occur in
different places with ONE clock.
You, Pat Dolan, don’t understand the first thing about relativity. And it
shows.
And yet you bluster simultaneously that a) you understand relativity very
well, and b) that it is incomprehensible.
The lab scientists, using relativity and the standard muon life-span,
calculate the same lab clock ticks off 4.4 seconds for the same trip.
Now take a deep breath and form an argument that addresses the facts of
the case, not imaginary t=0 stuff.
can see the clock adjacent to you, which is 572 m above the lab clock. And,
by the way, the one adjacent to you and the lab clock are not synchronized,
according to “you”, and so “you” wouldn’t dream of subtracting the readings
on the two clocks at rest in the earth frame to come up with an elapsed
time according to the lab clocks.
“What?” Pat Dolan protests, “the lab clock and the clock up by the
atmosphere are perfectly synchronized beforehand. They are still
synchronized now.” But the question then is, “In what frame were they
synchronized?” Pat Dolan splutters, “What difference does THAT make?” And
the answer is, “It matters a lot. The same two clocks may be synchronized
in one frame, but they’re not synchronized in the muon’s frame, or any
other frame in fact. This is one of the foundational observations of
special relativity.”
“But, but, but….,” fumes Pat Dolan, “I had an EXCELLENT hand!”
No, Pat, you don’t have an excellent hand. You have crap cards, all because
you have never bothered to read a book that talks about muons IN DETAIL.
Instead, you came here with a poor and slippery grasp on a simple and
common example, and you thought somehow that something SO SIMPLE had been
missed in ALL THOSE presentations that you never read.
You, Pat, are a persistent idiot, thinking you have good cards, when you’ve
never even bothered to learn how to play poker.
Post by patdolan
Nothing strange here; the labs clock is traveling at γ=2 with respect to
you so it only logs half as much elapsed time as you.
One of the lab scientists now cries out “Just a minute! That’s not how
this story goes. First of all, it is the hadron that is
Lorentz-flattened, not us. And second of all, that hadron used it’s
flattened FoR to determine that is was only 571.56 meters away from our
scintillator at the point it became a muon. Our unflattened earth FoR
clearly indicated that the hadron was actually 1143.12 meters away when
it became a muon. That’s why it took 4.4 microseconds, traveling at .866
c, to reach our scintillator. And that’s what our lab clock shows. Not
1.1 microseconds, like that dumb muon is claiming.”
Special relativity leaves us in a quandary. How much time actually did
elapse on the lab clock from the moment of the muon’s inception to the
moment of its scintillating demise? Was it 1.1 microseconds? Or was it
4.4 microseconds? It has to be one or the other, it can’t be both.
However, different elapsed times on the same lab clock can be calculated
by the muon and by the earthbound scientists with equal justification.
Special relativity is incapable of providing a unique elapsed time on the
lab clock. I challenge anyone to show that special relativity can
provide a unique elapsed time in the preceding case.
Special relativity’s greatest experimental confirmation turns out to be
it’s second greatest falsification. Gravity has been special
relativity’s greatest falsification since 1907 ( check out the author’s
brilliant post on special relativity vs. Kepler’s 3rd law ).
--
Odd Bodkin -- maker of fine toys, tools, tables
--
Odd Bodkin -- maker of fine toys, tools, tables
The muon is bathed in the light cone from the lab clock a the moment of it's inception. The muon notes what the light cone is indicating at that moment while it still 572m/1143m away from the clock. The muon notes the light from the lab clock again when it is in the scintillator. This is T1. From T1 the muon subtracts 2.2 microseconds to get the elapsed time on the lab clock for the muon's trip. This elapsed time will be 1.1 microseconds according to relativity and the muon. No lattice of clocks, no synchronization no nothing. Easy-peasy Bodkin.
*the difference of the times is T1: light cone time at inception - light cone in scintillator = T1
So on more time just to make it absolutely clear for LongEars:

The muon is bathed in the light cone from the lab clock a the moment of it's inception. The muon notes what the light cone is indicating at that moment while it still 572m/1143m away from the clock. The muon notes the light from the lab clock again when it is in the scintillator. This is T-inception. The muon notes the light from the lab clock again when it is in the scintillator. This is T-demise.

The muon subtracts another 2.2 microseconds to give the total elapsed time on the lab clock:

Total elapsed time on lab clock for relativistic muon trip from muon's pov = (T-demise - T-inception) - 2.2 microseconds = 1.1 microseconds
patdolan
2021-11-20 20:26:26 UTC
Reply
Permalink
Post by Odd Bodkin
Post by patdolan
Critical Theory comes in several flavors: Critical Law Theory, Critical
Literature Theory, Critical History Theory, Critical Race Theory. To
these we now add Critical Relativity Theory!
In the spirit of Derrida we shall deconstruct the clumsy reasoning of
special relativity and separate said reasoning from the algebraic symbols
and equations that express it, keeping in mind that mathematics is just
another form of rhetorical expression wherein falsity can be expressed
every bit as plausibly as the truth.
“There is only the text.”— J. Derrida
According to special relativity two observers in motion with respect to
each other will disagree on each other’s length. They will also disagree
on the proper flow of time. But they will always agree on the velocity
they have with respect to one another. This is exceedingly strange. How
can it be that two relative quantities, space and time, combine to
produce an absolute quantity called relative velocity? It is true that SR
does have a formula for calculating coordinate velocity; just like it has
formulas for calculating coordinate space and coordinate time. But the
Einstein velocity addition formula ONLY applies to a third object in
motion wrt a pair of FoRs. If that third object happens to be at rest
wrt one of the FoRs then Einsteinian velocity reduces to Galilean
velocity, albeit subject to the speed limit c. Relativists simply
assumed without further justification that if FoR-1 measures a velocity v
between itself and FoR-2 then FoR-2 must measure the same numerical value
v for the velocity between itself and FoR-1. Trivial? Nope. Seemingly
trivial assumptions can be monumental when constructing a theory of
motion. But a 26 year old would probably not yet have the requisite
philosophical sophistication needed to recognize this.
Einstein’s choice to make velocity strictly Galilean when calculating the
velocity between pairs of FoRs ( yes, it is a choice because it does not
follow from either the first or the second postulates ) can be
expressed mathematically as
∆x’/∆t’ = ∆x/∆t = v (1)
I now raise equation (1) to the level of a postulate and declare it to be
the third, and heretofore hidden, postulate of special relativity. In
recognizing its own structural Galileanism through this new postulate,
special relativity can finally claim to be woke.
The problem with the third postulate is that even though it is already
assumed in every equation of special relativity, it turns out to be true
only when v = c ( the second postulate ) or when v = 0. The third
postulate can be demonstrated invalid for all values of v in between.
The invalidity of the third postulate causes special relativity fall on
it’s algebraic face. Big Bang move over…Not recognizing and
acknowledging the third postulate was Einstein’s biggest blunder.
Time for some examples.
DIRK & DONO
Consider two FoRs whose x-axes are parallel and lie very close to one
another. The relative velocity between these two FoRs is .866c (γ = 2).
Dirk assumes the Lotus position at the origin of one FoR whilst Dono
assumes the fetal position at the origin of the other. Both Dirk and
Dono and their clocks are glued to the origins of their respective FoRs.
Dirk opens one eye and takes note of the meter marks on Dono’s contracted
x-axis as they whiz by. Dirk apprehends that Dono’s meter marks are
contracted to only half as long as his own meter marks. Dirk opens the
other eye and observes that Dono’s clock is ticking at only half the rate
of his own clock. Dirk begins to count Dono’s meter marks as they race
past Dirk’s position. After one year (by Dirk’s clock) of counting
Dono’s meter marks, Dirk has tallied 1.64e+16 Dono meters ( 9.5e+15 meter
marks/Ly x .866Ly x γ ). Dirk has also observed that only 0.5 years have
elapsed on Dono’s clock. Dirk now calculates what his coordinate
velocity should be according to Dono
( 1.64e+16 Dono meters ) / ( 1.58e+7 Dono seconds ) = 1.04e+9 m/s = 3.5c (2)
[ shortcut: ∆x’/∆t’ = (∆x/∆t)(γ^2) = v(γ^2) ] (3)
“Stop!” You cry, “Dirk’s and Dono’s relative velocity was already
stipulated to be an absolute .866c with respect to one another, when
measured in either FoR.”
That is true, according to (1). But remember that (1) is an arbitrary
choice made by Einstein when he built his theory. It is no more
legitimate a choice than the Dolan FoR coordinate velocity transform (3)
for determining one’s FoR coordinate velocity. It is also no less
inconsistent. For we immediately see by inspection that the Dolan FoR
coordinate velocity is always greater than the relative velocity by a
factor of γ^2. But Einstein’s choice for requiring Galilean FoR pair
velocities clangs with just as much antinomy as Dolan’s transform, as we
shall see. Special relativity’s dirty little secret is that it’s hidden
third postulate (1) destroys the theory from within.
I can see by Dono’s tightening fetal position that he still doesn’t
believe me. Very well. We shall prove SR’s mathematical inconsistency
in the next example.
SPECIAL RELATIVITY COLLAPSES UNDER THE WEIGHT OF ITS ALGEBRAIC ORIGINAL SIN
Consider a pair of FoRs whose relative velocity v is some value other
than c and other than zero. This is expressed mathematically as
v = ∆x’/∆t’ = ∆x/∆t != c (4)
and
v = ∆x’/∆t’ = ∆x/∆t != 0 (5)
The Lorentz transforms allow us to construct the FoR coordinate
velocities for pairs of FoRs
∆x’ = γ( ∆x - v∆t )
∆t’ = γ( ∆t - ∆xv/c^2 )
∆x’/∆t’ = [ γ( ∆x - v∆t ) ] / [ γ( ∆t - ∆xv/c^2 ) ] (6)
We now endeavor to solve (6) for v in hopes of demonstrating the internal
consistency of special relativity, i.e, that v = v for all pair of FoRs.
We saw how this was not the case in DIRK & DONO.
The reader will find that trying to solve for v is hopeless unless we
make the substitution
v = ∆x/∆t or v = ∆x’/∆t’
It does not matter which we use—either substitution is permitted by the
third postulate (1).
With the substitution made, we eventually arrive at the preposterous results
∆x’/∆t’ = c (7)
or
∆x’/∆t’ = 0 (8)
The derivation is left as an exercise for the reader. For those needing
help with the derivation ( I'm looking at you, Dono ) I will be happy to
provide it in another post.
The laughable results (7) and (8) directly contradict our assumptions (4)
and (5). Furthermore, the results impose the requirement that v is not
even a variable—v turns out to be a constant always equal to c or zero.
Absolutely absurd.
QED.
Algebraic relativity is thus reduced to ridiculous rubble by means
mathematical reductio ad absurdum. The root cause of special relativity’s
spectacular algebraic failure lies in the propositional calculus. I am
happy to expatiate on that subject in another post if there is interest.
Let’s do one more example—this one ripped from the headlines of experimental physics.
MUONS, SCHMUONS!
Imagine that you are a hadron in deep space minding your own business
when all of a sudden you turn to see a Lorentz-flattened earth coming at
you at a velocity of .866c. In the impending collision the first thing
to strike you is an air molecule high in the earth’s atmosphere. That
molecules knocks the stuffing out of you. What is left of you is now a
muon which means that you only have 2.2 microseconds more to live.
It turns out that the surface of the flattened earth is exactly 571.56
meters away from you, according to your own muon FoR. By a remarkable
coincidence there is a flattened scintillator and a flattened clock in a
flattened laboratory directly below you on the surface of the flattened
earth. You note the time on the flattened lab clock. Because you are a
muon, you are your own 2.2 microsecond alarm clock.
The surface of the flattened earth continues to speed towards you at
.866c. 2.2 microseconds later the flattened earth, lab and scintillator
smash into you. Just as you expire in the scintillator you note that
only 1.1 microseconds has elapsed on the lab clock.
You are an idiot. “You” the hadron cannot see the “lab clock” at t=0.
No synchronization required. The muon's clock doesn't even enter into
the argument at all. Who said that the lab clock was ever at t=0 or any
other value. E. L. A. P. S. E. D. T. I. M. E., Bodkin.
Elapsed from what moment? How does the lab clock know when to start
ticking. It isn’t anywhere near the collision of the hadron and the
atmospheric molecule. Is it supposed to get an instantaneous signal from
that event?
This is what I mean when I say you haven’t bothered to read any book about
relativity. Your “argument” is based on two and only two clocks passing
each other, both deeming the other slowed, because that’s the comic book
version of relativity. But that isn’t what relativity says. Ask yourself
why relativity talks about a *lattice* of synchronized clocks in each
frame. Why would you need the lattice?
It’s so that there is a clock NEAR the event that you’re tracking the time
of.
Quantities of elapsed time are easily determined in special relativity.
Synchronization of clocks doesn't even enter into the argument It's all
about one clock, the lab clock. One clock, two observers (the muon and
the lab scientists), TWO elapsed time. Quite brilliant on my part, n'es pas?
Tell us how synchronization has any bearing on the argument. The muon
sees the lab clock tick off 1.1 microseconds on it's trip down.
No, it doesn’t. How is it supposed to “see” that? It isn’t ANYWHERE NEAR
the lab clock. The best the muon can do is either a) monitor its own clock,
because its own clock is at ONE location in its own frame, or b) have a
LATTICE of synchronized clocks traveling along with the muon, where one of
them is far away from the muon, near the lab clock at the time the muon is
created.
If you have two events that occur at DIFFERENT PLACES in some frame of
reference, you need two DIFFERENT clocks near those events. If those clocks
are synchronized in this frame, THEN you can measure the elapsed time
between the events.
You can’t measure the elapsed time between two events that occur in
different places with ONE clock.
You, Pat Dolan, don’t understand the first thing about relativity. And it
shows.
And yet you bluster simultaneously that a) you understand relativity very
well, and b) that it is incomprehensible.
Bodkin, please declare to this forum that your argument also applies to the scientists in the lab--especially to the scientist in the lab of Dr's Frisch and Smith.



Frisch and Smith didn't have clocks up there alongside all the muons they measured either. So there is NO WAY according to Bodkin that Frisch and Smith could have possibly determined whether or not muons experienced relativistic time dilation.

Declare this, you checkmated chump.
Post by Odd Bodkin
The lab scientists, using relativity and the standard muon life-span,
calculate the same lab clock ticks off 4.4 seconds for the same trip.
Now take a deep breath and form an argument that addresses the facts of
the case, not imaginary t=0 stuff.
can see the clock adjacent to you, which is 572 m above the lab clock. And,
by the way, the one adjacent to you and the lab clock are not synchronized,
according to “you”, and so “you” wouldn’t dream of subtracting the readings
on the two clocks at rest in the earth frame to come up with an elapsed
time according to the lab clocks.
“What?” Pat Dolan protests, “the lab clock and the clock up by the
atmosphere are perfectly synchronized beforehand. They are still
synchronized now.” But the question then is, “In what frame were they
synchronized?” Pat Dolan splutters, “What difference does THAT make?” And
the answer is, “It matters a lot. The same two clocks may be synchronized
in one frame, but they’re not synchronized in the muon’s frame, or any
other frame in fact. This is one of the foundational observations of
special relativity.”
“But, but, but….,” fumes Pat Dolan, “I had an EXCELLENT hand!”
No, Pat, you don’t have an excellent hand. You have crap cards, all because
you have never bothered to read a book that talks about muons IN DETAIL.
Instead, you came here with a poor and slippery grasp on a simple and
common example, and you thought somehow that something SO SIMPLE had been
missed in ALL THOSE presentations that you never read.
You, Pat, are a persistent idiot, thinking you have good cards, when you’ve
never even bothered to learn how to play poker.
Post by patdolan
Nothing strange here; the labs clock is traveling at γ=2 with respect to
you so it only logs half as much elapsed time as you.
One of the lab scientists now cries out “Just a minute! That’s not how
this story goes. First of all, it is the hadron that is
Lorentz-flattened, not us. And second of all, that hadron used it’s
flattened FoR to determine that is was only 571.56 meters away from our
scintillator at the point it became a muon. Our unflattened earth FoR
clearly indicated that the hadron was actually 1143.12 meters away when
it became a muon. That’s why it took 4.4 microseconds, traveling at .866
c, to reach our scintillator. And that’s what our lab clock shows. Not
1.1 microseconds, like that dumb muon is claiming.”
Special relativity leaves us in a quandary. How much time actually did
elapse on the lab clock from the moment of the muon’s inception to the
moment of its scintillating demise? Was it 1.1 microseconds? Or was it
4.4 microseconds? It has to be one or the other, it can’t be both.
However, different elapsed times on the same lab clock can be calculated
by the muon and by the earthbound scientists with equal justification.
Special relativity is incapable of providing a unique elapsed time on the
lab clock. I challenge anyone to show that special relativity can
provide a unique elapsed time in the preceding case.
Special relativity’s greatest experimental confirmation turns out to be
it’s second greatest falsification. Gravity has been special
relativity’s greatest falsification since 1907 ( check out the author’s
brilliant post on special relativity vs. Kepler’s 3rd law ).
--
Odd Bodkin -- maker of fine toys, tools, tables
--
Odd Bodkin -- maker of fine toys, tools, tables
Odd Bodkin
2021-11-21 13:44:21 UTC
Reply
Permalink
Post by patdolan
Post by Odd Bodkin
Post by patdolan
Critical Theory comes in several flavors: Critical Law Theory, Critical
Literature Theory, Critical History Theory, Critical Race Theory. To
these we now add Critical Relativity Theory!
In the spirit of Derrida we shall deconstruct the clumsy reasoning of
special relativity and separate said reasoning from the algebraic symbols
and equations that express it, keeping in mind that mathematics is just
another form of rhetorical expression wherein falsity can be expressed
every bit as plausibly as the truth.
“There is only the text.”— J. Derrida
According to special relativity two observers in motion with respect to
each other will disagree on each other’s length. They will also disagree
on the proper flow of time. But they will always agree on the velocity
they have with respect to one another. This is exceedingly strange. How
can it be that two relative quantities, space and time, combine to
produce an absolute quantity called relative velocity? It is true that SR
does have a formula for calculating coordinate velocity; just like it has
formulas for calculating coordinate space and coordinate time. But the
Einstein velocity addition formula ONLY applies to a third object in
motion wrt a pair of FoRs. If that third object happens to be at rest
wrt one of the FoRs then Einsteinian velocity reduces to Galilean
velocity, albeit subject to the speed limit c. Relativists simply
assumed without further justification that if FoR-1 measures a velocity v
between itself and FoR-2 then FoR-2 must measure the same numerical value
v for the velocity between itself and FoR-1. Trivial? Nope. Seemingly
trivial assumptions can be monumental when constructing a theory of
motion. But a 26 year old would probably not yet have the requisite
philosophical sophistication needed to recognize this.
Einstein’s choice to make velocity strictly Galilean when calculating the
velocity between pairs of FoRs ( yes, it is a choice because it does not
follow from either the first or the second postulates ) can be
expressed mathematically as
∆x’/∆t’ = ∆x/∆t = v (1)
I now raise equation (1) to the level of a postulate and declare it to be
the third, and heretofore hidden, postulate of special relativity. In
recognizing its own structural Galileanism through this new postulate,
special relativity can finally claim to be woke.
The problem with the third postulate is that even though it is already
assumed in every equation of special relativity, it turns out to be true
only when v = c ( the second postulate ) or when v = 0. The third
postulate can be demonstrated invalid for all values of v in between.
The invalidity of the third postulate causes special relativity fall on
it’s algebraic face. Big Bang move over…Not recognizing and
acknowledging the third postulate was Einstein’s biggest blunder.
Time for some examples.
DIRK & DONO
Consider two FoRs whose x-axes are parallel and lie very close to one
another. The relative velocity between these two FoRs is .866c (γ = 2).
Dirk assumes the Lotus position at the origin of one FoR whilst Dono
assumes the fetal position at the origin of the other. Both Dirk and
Dono and their clocks are glued to the origins of their respective FoRs.
Dirk opens one eye and takes note of the meter marks on Dono’s contracted
x-axis as they whiz by. Dirk apprehends that Dono’s meter marks are
contracted to only half as long as his own meter marks. Dirk opens the
other eye and observes that Dono’s clock is ticking at only half the rate
of his own clock. Dirk begins to count Dono’s meter marks as they race
past Dirk’s position. After one year (by Dirk’s clock) of counting
Dono’s meter marks, Dirk has tallied 1.64e+16 Dono meters ( 9.5e+15 meter
marks/Ly x .866Ly x γ ). Dirk has also observed that only 0.5 years have
elapsed on Dono’s clock. Dirk now calculates what his coordinate
velocity should be according to Dono
( 1.64e+16 Dono meters ) / ( 1.58e+7 Dono seconds ) = 1.04e+9 m/s = 3.5c (2)
[ shortcut: ∆x’/∆t’ = (∆x/∆t)(γ^2) = v(γ^2) ] (3)
“Stop!” You cry, “Dirk’s and Dono’s relative velocity was already
stipulated to be an absolute .866c with respect to one another, when
measured in either FoR.”
That is true, according to (1). But remember that (1) is an arbitrary
choice made by Einstein when he built his theory. It is no more
legitimate a choice than the Dolan FoR coordinate velocity transform (3)
for determining one’s FoR coordinate velocity. It is also no less
inconsistent. For we immediately see by inspection that the Dolan FoR
coordinate velocity is always greater than the relative velocity by a
factor of γ^2. But Einstein’s choice for requiring Galilean FoR pair
velocities clangs with just as much antinomy as Dolan’s transform, as we
shall see. Special relativity’s dirty little secret is that it’s hidden
third postulate (1) destroys the theory from within.
I can see by Dono’s tightening fetal position that he still doesn’t
believe me. Very well. We shall prove SR’s mathematical inconsistency
in the next example.
SPECIAL RELATIVITY COLLAPSES UNDER THE WEIGHT OF ITS ALGEBRAIC ORIGINAL SIN
Consider a pair of FoRs whose relative velocity v is some value other
than c and other than zero. This is expressed mathematically as
v = ∆x’/∆t’ = ∆x/∆t != c (4)
and
v = ∆x’/∆t’ = ∆x/∆t != 0 (5)
The Lorentz transforms allow us to construct the FoR coordinate
velocities for pairs of FoRs
∆x’ = γ( ∆x - v∆t )
∆t’ = γ( ∆t - ∆xv/c^2 )
∆x’/∆t’ = [ γ( ∆x - v∆t ) ] / [ γ( ∆t - ∆xv/c^2 ) ] (6)
We now endeavor to solve (6) for v in hopes of demonstrating the internal
consistency of special relativity, i.e, that v = v for all pair of FoRs.
We saw how this was not the case in DIRK & DONO.
The reader will find that trying to solve for v is hopeless unless we
make the substitution
v = ∆x/∆t or v = ∆x’/∆t’
It does not matter which we use—either substitution is permitted by the
third postulate (1).
With the substitution made, we eventually arrive at the preposterous results
∆x’/∆t’ = c (7)
or
∆x’/∆t’ = 0 (8)
The derivation is left as an exercise for the reader. For those needing
help with the derivation ( I'm looking at you, Dono ) I will be happy to
provide it in another post.
The laughable results (7) and (8) directly contradict our assumptions (4)
and (5). Furthermore, the results impose the requirement that v is not
even a variable—v turns out to be a constant always equal to c or zero.
Absolutely absurd.
QED.
Algebraic relativity is thus reduced to ridiculous rubble by means
mathematical reductio ad absurdum. The root cause of special relativity’s
spectacular algebraic failure lies in the propositional calculus. I am
happy to expatiate on that subject in another post if there is interest.
Let’s do one more example—this one ripped from the headlines of experimental physics.
MUONS, SCHMUONS!
Imagine that you are a hadron in deep space minding your own business
when all of a sudden you turn to see a Lorentz-flattened earth coming at
you at a velocity of .866c. In the impending collision the first thing
to strike you is an air molecule high in the earth’s atmosphere. That
molecules knocks the stuffing out of you. What is left of you is now a
muon which means that you only have 2.2 microseconds more to live.
It turns out that the surface of the flattened earth is exactly 571.56
meters away from you, according to your own muon FoR. By a remarkable
coincidence there is a flattened scintillator and a flattened clock in a
flattened laboratory directly below you on the surface of the flattened
earth. You note the time on the flattened lab clock. Because you are a
muon, you are your own 2.2 microsecond alarm clock.
The surface of the flattened earth continues to speed towards you at
.866c. 2.2 microseconds later the flattened earth, lab and scintillator
smash into you. Just as you expire in the scintillator you note that
only 1.1 microseconds has elapsed on the lab clock.
You are an idiot. “You” the hadron cannot see the “lab clock” at t=0.
No synchronization required. The muon's clock doesn't even enter into
the argument at all. Who said that the lab clock was ever at t=0 or any
other value. E. L. A. P. S. E. D. T. I. M. E., Bodkin.
Elapsed from what moment? How does the lab clock know when to start
ticking. It isn’t anywhere near the collision of the hadron and the
atmospheric molecule. Is it supposed to get an instantaneous signal from
that event?
This is what I mean when I say you haven’t bothered to read any book about
relativity. Your “argument” is based on two and only two clocks passing
each other, both deeming the other slowed, because that’s the comic book
version of relativity. But that isn’t what relativity says. Ask yourself
why relativity talks about a *lattice* of synchronized clocks in each
frame. Why would you need the lattice?
It’s so that there is a clock NEAR the event that you’re tracking the time
of.
Quantities of elapsed time are easily determined in special relativity.
Synchronization of clocks doesn't even enter into the argument It's all
about one clock, the lab clock. One clock, two observers (the muon and
the lab scientists), TWO elapsed time. Quite brilliant on my part, n'es pas?
Tell us how synchronization has any bearing on the argument. The muon
sees the lab clock tick off 1.1 microseconds on it's trip down.
No, it doesn’t. How is it supposed to “see” that? It isn’t ANYWHERE NEAR
the lab clock. The best the muon can do is either a) monitor its own clock,
because its own clock is at ONE location in its own frame, or b) have a
LATTICE of synchronized clocks traveling along with the muon, where one of
them is far away from the muon, near the lab clock at the time the muon is
created.
If you have two events that occur at DIFFERENT PLACES in some frame of
reference, you need two DIFFERENT clocks near those events. If those clocks
are synchronized in this frame, THEN you can measure the elapsed time
between the events.
You can’t measure the elapsed time between two events that occur in
different places with ONE clock.
You, Pat Dolan, don’t understand the first thing about relativity. And it
shows.
And yet you bluster simultaneously that a) you understand relativity very
well, and b) that it is incomprehensible.
Bodkin, please declare to this forum that your argument also applies to
the scientists in the lab--especially to the scientist in the lab of Dr's
Frisch and Smith.
http://youtu.be/rbzt8gDSYIM
Frisch and Smith didn't have clocks up there alongside all the muons they
measured either. So there is NO WAY according to Bodkin that Frisch and
Smith could have possibly determined whether or not muons experienced
relativistic time dilation.
Yes they can, but not by the method you floundered around with. First of
all, these muon experiments typically measure the speed directly by having
a stack of vertically separated scintillators, so that time of flight is
readily available. Secondly, the lifetime is measured the same way it is
for radioisotopes, activity rate at different samplings. It’s actually
straightforward.

But what is clear through out this is that you never READ exactly how
Frisch and Smith did this experiment, and you’ve never READ how relativity
describes this phenomenon, both of which are common expositions available
in lots of books.

If you’d only do that, you’d bluster less with loaded squirt guns you’re
holding backwards.
Post by patdolan
Declare this, you checkmated chump.
Post by Odd Bodkin
The lab scientists, using relativity and the standard muon life-span,
calculate the same lab clock ticks off 4.4 seconds for the same trip.
Now take a deep breath and form an argument that addresses the facts of
the case, not imaginary t=0 stuff.
can see the clock adjacent to you, which is 572 m above the lab clock. And,
by the way, the one adjacent to you and the lab clock are not synchronized,
according to “you”, and so “you” wouldn’t dream of subtracting the readings
on the two clocks at rest in the earth frame to come up with an elapsed
time according to the lab clocks.
“What?” Pat Dolan protests, “the lab clock and the clock up by the
atmosphere are perfectly synchronized beforehand. They are still
synchronized now.” But the question then is, “In what frame were they
synchronized?” Pat Dolan splutters, “What difference does THAT make?” And
the answer is, “It matters a lot. The same two clocks may be synchronized
in one frame, but they’re not synchronized in the muon’s frame, or any
other frame in fact. This is one of the foundational observations of
special relativity.”
“But, but, but….,” fumes Pat Dolan, “I had an EXCELLENT hand!”
No, Pat, you don’t have an excellent hand. You have crap cards, all because
you have never bothered to read a book that talks about muons IN DETAIL.
Instead, you came here with a poor and slippery grasp on a simple and
common example, and you thought somehow that something SO SIMPLE had been
missed in ALL THOSE presentations that you never read.
You, Pat, are a persistent idiot, thinking you have good cards, when you’ve
never even bothered to learn how to play poker.
Post by patdolan
Nothing strange here; the labs clock is traveling at γ=2 with respect to
you so it only logs half as much elapsed time as you.
One of the lab scientists now cries out “Just a minute! That’s not how
this story goes. First of all, it is the hadron that is
Lorentz-flattened, not us. And second of all, that hadron used it’s
flattened FoR to determine that is was only 571.56 meters away from our
scintillator at the point it became a muon. Our unflattened earth FoR
clearly indicated that the hadron was actually 1143.12 meters away when
it became a muon. That’s why it took 4.4 microseconds, traveling at .866
c, to reach our scintillator. And that’s what our lab clock shows. Not
1.1 microseconds, like that dumb muon is claiming.”
Special relativity leaves us in a quandary. How much time actually did
elapse on the lab clock from the moment of the muon’s inception to the
moment of its scintillating demise? Was it 1.1 microseconds? Or was it
4.4 microseconds? It has to be one or the other, it can’t be both.
However, different elapsed times on the same lab clock can be calculated
by the muon and by the earthbound scientists with equal justification.
Special relativity is incapable of providing a unique elapsed time on the
lab clock. I challenge anyone to show that special relativity can
provide a unique elapsed time in the preceding case.
Special relativity’s greatest experimental confirmation turns out to be
it’s second greatest falsification. Gravity has been special
relativity’s greatest falsification since 1907 ( check out the author’s
brilliant post on special relativity vs. Kepler’s 3rd law ).
--
Odd Bodkin -- maker of fine toys, tools, tables
--
Odd Bodkin -- maker of fine toys, tools, tables
--
Odd Bodkin -- maker of fine toys, tools, tables
patdolan
2021-11-21 16:04:16 UTC
Reply
Permalink
Post by Odd Bodkin
Post by patdolan
Post by Odd Bodkin
Post by patdolan
Critical Theory comes in several flavors: Critical Law Theory, Critical
Literature Theory, Critical History Theory, Critical Race Theory. To
these we now add Critical Relativity Theory!
In the spirit of Derrida we shall deconstruct the clumsy reasoning of
special relativity and separate said reasoning from the algebraic symbols
and equations that express it, keeping in mind that mathematics is just
another form of rhetorical expression wherein falsity can be expressed
every bit as plausibly as the truth.
“There is only the text.”— J. Derrida
According to special relativity two observers in motion with respect to
each other will disagree on each other’s length. They will also disagree
on the proper flow of time. But they will always agree on the velocity
they have with respect to one another. This is exceedingly strange. How
can it be that two relative quantities, space and time, combine to
produce an absolute quantity called relative velocity? It is true that SR
does have a formula for calculating coordinate velocity; just like it has
formulas for calculating coordinate space and coordinate time. But the
Einstein velocity addition formula ONLY applies to a third object in
motion wrt a pair of FoRs. If that third object happens to be at rest
wrt one of the FoRs then Einsteinian velocity reduces to Galilean
velocity, albeit subject to the speed limit c. Relativists simply
assumed without further justification that if FoR-1 measures a velocity v
between itself and FoR-2 then FoR-2 must measure the same numerical value
v for the velocity between itself and FoR-1. Trivial? Nope. Seemingly
trivial assumptions can be monumental when constructing a theory of
motion. But a 26 year old would probably not yet have the requisite
philosophical sophistication needed to recognize this.
Einstein’s choice to make velocity strictly Galilean when calculating the
velocity between pairs of FoRs ( yes, it is a choice because it does not
follow from either the first or the second postulates ) can be
expressed mathematically as
∆x’/∆t’ = ∆x/∆t = v (1)
I now raise equation (1) to the level of a postulate and declare it to be
the third, and heretofore hidden, postulate of special relativity. In
recognizing its own structural Galileanism through this new postulate,
special relativity can finally claim to be woke.
The problem with the third postulate is that even though it is already
assumed in every equation of special relativity, it turns out to be true
only when v = c ( the second postulate ) or when v = 0. The third
postulate can be demonstrated invalid for all values of v in between.
The invalidity of the third postulate causes special relativity fall on
it’s algebraic face. Big Bang move over…Not recognizing and
acknowledging the third postulate was Einstein’s biggest blunder.
Time for some examples.
DIRK & DONO
Consider two FoRs whose x-axes are parallel and lie very close to one
another. The relative velocity between these two FoRs is .866c (γ = 2).
Dirk assumes the Lotus position at the origin of one FoR whilst Dono
assumes the fetal position at the origin of the other. Both Dirk and
Dono and their clocks are glued to the origins of their respective FoRs.
Dirk opens one eye and takes note of the meter marks on Dono’s contracted
x-axis as they whiz by. Dirk apprehends that Dono’s meter marks are
contracted to only half as long as his own meter marks. Dirk opens the
other eye and observes that Dono’s clock is ticking at only half the rate
of his own clock. Dirk begins to count Dono’s meter marks as they race
past Dirk’s position. After one year (by Dirk’s clock) of counting
Dono’s meter marks, Dirk has tallied 1.64e+16 Dono meters ( 9.5e+15 meter
marks/Ly x .866Ly x γ ). Dirk has also observed that only 0.5 years have
elapsed on Dono’s clock. Dirk now calculates what his coordinate
velocity should be according to Dono
( 1.64e+16 Dono meters ) / ( 1.58e+7 Dono seconds ) = 1.04e+9 m/s = 3.5c (2)
[ shortcut: ∆x’/∆t’ = (∆x/∆t)(γ^2) = v(γ^2) ] (3)
“Stop!” You cry, “Dirk’s and Dono’s relative velocity was already
stipulated to be an absolute .866c with respect to one another, when
measured in either FoR.”
That is true, according to (1). But remember that (1) is an arbitrary
choice made by Einstein when he built his theory. It is no more
legitimate a choice than the Dolan FoR coordinate velocity transform (3)
for determining one’s FoR coordinate velocity. It is also no less
inconsistent. For we immediately see by inspection that the Dolan FoR
coordinate velocity is always greater than the relative velocity by a
factor of γ^2. But Einstein’s choice for requiring Galilean FoR pair
velocities clangs with just as much antinomy as Dolan’s transform, as we
shall see. Special relativity’s dirty little secret is that it’s hidden
third postulate (1) destroys the theory from within.
I can see by Dono’s tightening fetal position that he still doesn’t
believe me. Very well. We shall prove SR’s mathematical inconsistency
in the next example.
SPECIAL RELATIVITY COLLAPSES UNDER THE WEIGHT OF ITS ALGEBRAIC ORIGINAL SIN
Consider a pair of FoRs whose relative velocity v is some value other
than c and other than zero. This is expressed mathematically as
v = ∆x’/∆t’ = ∆x/∆t != c (4)
and
v = ∆x’/∆t’ = ∆x/∆t != 0 (5)
The Lorentz transforms allow us to construct the FoR coordinate
velocities for pairs of FoRs
∆x’ = γ( ∆x - v∆t )
∆t’ = γ( ∆t - ∆xv/c^2 )
∆x’/∆t’ = [ γ( ∆x - v∆t ) ] / [ γ( ∆t - ∆xv/c^2 ) ] (6)
We now endeavor to solve (6) for v in hopes of demonstrating the internal
consistency of special relativity, i.e, that v = v for all pair of FoRs.
We saw how this was not the case in DIRK & DONO.
The reader will find that trying to solve for v is hopeless unless we
make the substitution
v = ∆x/∆t or v = ∆x’/∆t’
It does not matter which we use—either substitution is permitted by the
third postulate (1).
With the substitution made, we eventually arrive at the preposterous results
∆x’/∆t’ = c (7)
or
∆x’/∆t’ = 0 (8)
The derivation is left as an exercise for the reader. For those needing
help with the derivation ( I'm looking at you, Dono ) I will be happy to
provide it in another post.
The laughable results (7) and (8) directly contradict our assumptions (4)
and (5). Furthermore, the results impose the requirement that v is not
even a variable—v turns out to be a constant always equal to c or zero.
Absolutely absurd.
QED.
Algebraic relativity is thus reduced to ridiculous rubble by means
mathematical reductio ad absurdum. The root cause of special relativity’s
spectacular algebraic failure lies in the propositional calculus. I am
happy to expatiate on that subject in another post if there is interest.
Let’s do one more example—this one ripped from the headlines of experimental physics.
MUONS, SCHMUONS!
Imagine that you are a hadron in deep space minding your own business
when all of a sudden you turn to see a Lorentz-flattened earth coming at
you at a velocity of .866c. In the impending collision the first thing
to strike you is an air molecule high in the earth’s atmosphere. That
molecules knocks the stuffing out of you. What is left of you is now a
muon which means that you only have 2.2 microseconds more to live.
It turns out that the surface of the flattened earth is exactly 571.56
meters away from you, according to your own muon FoR. By a remarkable
coincidence there is a flattened scintillator and a flattened clock in a
flattened laboratory directly below you on the surface of the flattened
earth. You note the time on the flattened lab clock. Because you are a
muon, you are your own 2.2 microsecond alarm clock.
The surface of the flattened earth continues to speed towards you at
.866c. 2.2 microseconds later the flattened earth, lab and scintillator
smash into you. Just as you expire in the scintillator you note that
only 1.1 microseconds has elapsed on the lab clock.
You are an idiot. “You” the hadron cannot see the “lab clock” at t=0.
No synchronization required. The muon's clock doesn't even enter into
the argument at all. Who said that the lab clock was ever at t=0 or any
other value. E. L. A. P. S. E. D. T. I. M. E., Bodkin.
Elapsed from what moment? How does the lab clock know when to start
ticking. It isn’t anywhere near the collision of the hadron and the
atmospheric molecule. Is it supposed to get an instantaneous signal from
that event?
This is what I mean when I say you haven’t bothered to read any book about
relativity. Your “argument” is based on two and only two clocks passing
each other, both deeming the other slowed, because that’s the comic book
version of relativity. But that isn’t what relativity says. Ask yourself
why relativity talks about a *lattice* of synchronized clocks in each
frame. Why would you need the lattice?
It’s so that there is a clock NEAR the event that you’re tracking the time
of.
Quantities of elapsed time are easily determined in special relativity.
Synchronization of clocks doesn't even enter into the argument It's all
about one clock, the lab clock. One clock, two observers (the muon and
the lab scientists), TWO elapsed time. Quite brilliant on my part, n'es pas?
Tell us how synchronization has any bearing on the argument. The muon
sees the lab clock tick off 1.1 microseconds on it's trip down.
No, it doesn’t. How is it supposed to “see” that? It isn’t ANYWHERE NEAR
the lab clock. The best the muon can do is either a) monitor its own clock,
because its own clock is at ONE location in its own frame, or b) have a
LATTICE of synchronized clocks traveling along with the muon, where one of
them is far away from the muon, near the lab clock at the time the muon is
created.
If you have two events that occur at DIFFERENT PLACES in some frame of
reference, you need two DIFFERENT clocks near those events. If those clocks
are synchronized in this frame, THEN you can measure the elapsed time
between the events.
You can’t measure the elapsed time between two events that occur in
different places with ONE clock.
You, Pat Dolan, don’t understand the first thing about relativity. And it
shows.
And yet you bluster simultaneously that a) you understand relativity very
well, and b) that it is incomprehensible.
Bodkin, please declare to this forum that your argument also applies to
the scientists in the lab--especially to the scientist in the lab of Dr's
Frisch and Smith.
http://youtu.be/rbzt8gDSYIM
Frisch and Smith didn't have clocks up there alongside all the muons they
measured either. So there is NO WAY according to Bodkin that Frisch and
Smith could have possibly determined whether or not muons experienced
relativistic time dilation.
Yes they can, but not by the method you floundered around with. First of
all, these muon experiments typically measure the speed directly by having
a stack of vertically separated scintillators, so that time of flight is
readily available. Secondly, the lifetime is measured the same way it is
for radioisotopes, activity rate at different samplings. It’s actually
straightforward.
If F&S know the time of flight and the velocity of the muons in their lab then F&S can also know the time of flight and velocity of the earth in the muon's FoR.

Will you now deign to delight this forum with your calculation of the latter? Thank you, in advance. Also, please use an online translator to render a version of your answer in French for those Francophones looking in on our debate. Thank you again in advance.
Post by Odd Bodkin
But what is clear through out this is that you never READ exactly how
Frisch and Smith did this experiment, and you’ve never READ how relativity
describes this phenomenon, both of which are common expositions available
in lots of books.
If you’d only do that, you’d bluster less with loaded squirt guns you’re
holding backwards.
Post by patdolan
Declare this, you checkmated chump.
Post by Odd Bodkin
The lab scientists, using relativity and the standard muon life-span,
calculate the same lab clock ticks off 4.4 seconds for the same trip.
Now take a deep breath and form an argument that addresses the facts of
the case, not imaginary t=0 stuff.
can see the clock adjacent to you, which is 572 m above the lab clock. And,
by the way, the one adjacent to you and the lab clock are not synchronized,
according to “you”, and so “you” wouldn’t dream of subtracting the readings
on the two clocks at rest in the earth frame to come up with an elapsed
time according to the lab clocks.
“What?” Pat Dolan protests, “the lab clock and the clock up by the
atmosphere are perfectly synchronized beforehand. They are still
synchronized now.” But the question then is, “In what frame were they
synchronized?” Pat Dolan splutters, “What difference does THAT make?” And
the answer is, “It matters a lot. The same two clocks may be synchronized
in one frame, but they’re not synchronized in the muon’s frame, or any
other frame in fact. This is one of the foundational observations of
special relativity.”
“But, but, but….,” fumes Pat Dolan, “I had an EXCELLENT hand!”
No, Pat, you don’t have an excellent hand. You have crap cards, all because
you have never bothered to read a book that talks about muons IN DETAIL.
Instead, you came here with a poor and slippery grasp on a simple and
common example, and you thought somehow that something SO SIMPLE had been
missed in ALL THOSE presentations that you never read.
You, Pat, are a persistent idiot, thinking you have good cards, when you’ve
never even bothered to learn how to play poker.
Post by patdolan
Nothing strange here; the labs clock is traveling at γ=2 with respect to
you so it only logs half as much elapsed time as you.
One of the lab scientists now cries out “Just a minute! That’s not how
this story goes. First of all, it is the hadron that is
Lorentz-flattened, not us. And second of all, that hadron used it’s
flattened FoR to determine that is was only 571.56 meters away from our
scintillator at the point it became a muon. Our unflattened earth FoR
clearly indicated that the hadron was actually 1143.12 meters away when
it became a muon. That’s why it took 4.4 microseconds, traveling at .866
c, to reach our scintillator. And that’s what our lab clock shows. Not
1.1 microseconds, like that dumb muon is claiming.”
Special relativity leaves us in a quandary. How much time actually did
elapse on the lab clock from the moment of the muon’s inception to the
moment of its scintillating demise? Was it 1.1 microseconds? Or was it
4.4 microseconds? It has to be one or the other, it can’t be both.
However, different elapsed times on the same lab clock can be calculated
by the muon and by the earthbound scientists with equal justification.
Special relativity is incapable of providing a unique elapsed time on the
lab clock. I challenge anyone to show that special relativity can
provide a unique elapsed time in the preceding case.
Special relativity’s greatest experimental confirmation turns out to be
it’s second greatest falsification. Gravity has been special
relativity’s greatest falsification since 1907 ( check out the author’s
brilliant post on special relativity vs. Kepler’s 3rd law ).
--
Odd Bodkin -- maker of fine toys, tools, tables
--
Odd Bodkin -- maker of fine toys, tools, tables
--
Odd Bodkin -- maker of fine toys, tools, tables
Townes Olson
2021-11-21 16:57:24 UTC
Reply
Permalink
Post by patdolan
If F&S know the time of flight and the velocity of the muons in their lab then F&S can
also know the time of flight and velocity of the earth in the muon's FoR.
Will you [please teach me the] calculation of the latter?
That is one of the oldest freshman befuddlements in the book. Let S be inertial coordinates in which a particle P on the earth’s surface is at rest, and let S’ be a system in which the muon is at rest. The speed of each system in terms of the other is the same. The muon, directly approaching P at high speed, is created at event E1, which is simultaneous with event E2 of P in terms of S, and with event E3 of P in terms of S’. Say the absolute interval between E1 and E2 is 14000 meters, and the absolute interval between E1 and E3 is 600 meters. Thus E3 is 46.62 microseconds later than E2. The times of flight are just distance divided by speed. There's nothing paradoxical about this. The two absolute space-like intervals (14000 and 600) are between two different pairs of events.

Of course, for each particle, when it is a given proper time from collision, in terms of it's own rest frame coordinates the spatial distance to the other particle is the same.
patdolan
2021-11-22 00:07:11 UTC
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Permalink
Post by patdolan
If F&S know the time of flight and the velocity of the muons in their lab then F&S can
also know the time of flight and velocity of the earth in the muon's FoR.
Will you [please teach me the] calculation of the latter?
That is one of the oldest freshman befuddlements in the book. Let S be inertial coordinates in which a particle P on the earth’s surface is at rest, and let S’ be a system in which the muon is at rest.
The speed of each system in terms of the other is the same.
Prove it! The deconstruction of this very statement is the whole point of the original post.

The muon, directly approaching P at high speed, is created at event E1, which is simultaneous with event E2 of P in terms of S, and with event E3 of P in terms of S’. Say the absolute interval between E1 and E2 is 14000 meters, and the absolute interval between E1 and E3 is 600 meters. Thus E3 is 46.62 microseconds later than E2. The times of flight are just distance divided by speed. There's nothing paradoxical about this. The two absolute space-like intervals (14000 and 600) are between two different pairs of events.

What does all this even mean??? Are you sure you know, Townes? Use some equations and more words. 14000 m??? 600 m??? 46.62 microseconds??? From where does all this come???

And try to remember that Pat Dolan is holding court in this forum at the moment. That means you are in the Big Leagues for the time being, Townes. You need to bring your A game.
Of course, for each particle, when it is a given proper time from collision, in terms of it's own rest frame coordinates the spatial distance to the other particle is the same.
Townes Olson
2021-11-22 01:29:35 UTC
Reply
Permalink
Post by Townes Olson
Post by patdolan
If F&S know the time of flight and the velocity of the muons in their lab then F&S can
also know the time of flight and velocity of the earth in the muon's FoR.
Will you [please teach me the] calculation of the latter?
Let S be inertial coordinates in which a particle P on the earth’s surface is at rest,
and let S’ be a system in which the muon is at rest. The muon, directly approaching
P at high speed, is created at event E1, which is simultaneous with event E2 of P in
terms of S, and with event E3 of P in terms of S’... There's nothing paradoxical about this.
The two absolute space-like intervals are between two different pairs of events.
What does all this even mean??? Use some equations and more words.
Sure. Let E4 be the event where the muon collides with particle P on earth's surface, and let v denote the mutual speed between the muon and P, and in terms of S let D denote the distance traveled by the muon from E1 to E4, so its time of flight is D/v. In terms of S' (in which the muon is at rest) the distance traveled by the earth particle P from E3 to E4 is D sqrt(1-v^2/c^2) and the time of flight is (D/v) sqrt(1-v^2/c^2).
patdolan
2021-11-22 01:36:31 UTC
Reply
Permalink
Post by Townes Olson
Post by patdolan
If F&S know the time of flight and the velocity of the muons in their lab then F&S can
also know the time of flight and velocity of the earth in the muon's FoR.
Will you [please teach me the] calculation of the latter?
Let S be inertial coordinates in which a particle P on the earth’s surface is at rest,
and let S’ be a system in which the muon is at rest. The muon, directly approaching
P at high speed, is created at event E1, which is simultaneous with event E2 of P in
terms of S, and with event E3 of P in terms of S’... There's nothing paradoxical about this.
The two absolute space-like intervals are between two different pairs of events.
What does all this even mean??? Use some equations and more words.
Sure. Let E4 be the event where the muon collides with particle P on earth's surface, and let v denote the mutual speed between the muon and P, and in terms of S let D denote the distance traveled by the muon from E1 to E4, so its time of flight is D/v. In terms of S' (in which the muon is at rest) the distance traveled by the earth particle P from E3 to E4 is D sqrt(1-v^2/c^2) and the time of flight is (D/v) sqrt(1-v^2/c^2).
I ask for clarification on E1, E2 and E3. Instead I get E4!

So Bodkin...is it Mitch & Townes, Townes & Mitch?
Townes Olson
2021-11-22 01:44:59 UTC
Reply
Permalink
Post by Townes Olson
Post by patdolan
If F&S know the time of flight and the velocity of the muons in their lab then F&S can
also know the time of flight and velocity of the earth in the muon's FoR.
Will you [please teach me the] calculation of the latter?
Let S be inertial coordinates in which a particle P on the earth’s surface is at rest,
and let S’ be a system in which the muon is at rest. The muon, directly approaching
P at high speed, is created at event E1, which is simultaneous with event E2 of P in
terms of S, and with event E3 of P in terms of S’... There's nothing paradoxical about this.
The two absolute space-like intervals are between two different pairs of events.
What does all this even mean??? Use some equations and more words.
Sure. Let E4 be the event where the muon collides with particle P on earth's surface, and let v denote the mutual speed between the muon and P, and in terms of S let D denote the distance traveled by the muon from E1 to E4, so its time of flight is D/v. In terms of S' (in which the muon is at rest) the distance traveled by the earth particle P from E3 to E4 is D sqrt(1-v^2/c^2) and the time of flight is (D/v) sqrt(1-v^2/c^2).
I ask for clarification on E1, E2 and E3. Instead I get E4!
What is it that you don't understand about those events? They are explicitly defined above. Again, the muon is created at E1 and it collides with the earth particle P at E4. Event E2 of P is simultaneous with E1 in terms of S, and event E3 of P is simultaneous with E1 in terms of S'. Do you understand this?
Odd Bodkin
2021-11-22 02:07:11 UTC
Reply
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Post by Townes Olson
Post by Townes Olson
Post by patdolan
If F&S know the time of flight and the velocity of the muons in
their lab then F&S can
also know the time of flight and
velocity of the earth in the muon's FoR.
Will you [please teach me the] calculation of the latter?
Let S be inertial coordinates in which a particle P on the earth’s surface is at rest,
and let S’ be a system in which the muon is at rest. The muon, directly approaching
P at high speed, is created at event E1, which is simultaneous with event E2 of P in
terms of S, and with event E3 of P in terms of S’... There's nothing
paradoxical about this.
The two absolute space-like intervals are between two different pairs of events.
What does all this even mean??? Use some equations and more words.
Sure. Let E4 be the event where the muon collides with particle P on
earth's surface, and let v denote the mutual speed between the muon and
P, and in terms of S let D denote the distance traveled by the muon
from E1 to E4, so its time of flight is D/v. In terms of S' (in which
the muon is at rest) the distance traveled by the earth particle P from
E3 to E4 is D sqrt(1-v^2/c^2) and the time of flight is (D/v) sqrt(1-v^2/c^2).
I ask for clarification on E1, E2 and E3. Instead I get E4!
What is it that you don't understand about those events? They are
explicitly defined above. Again, the muon is created at E1 and it
collides with the earth particle P at E4. Event E2 of P is simultaneous
with E1 in terms of S, and event E3 of P is simultaneous with E1 in terms
of S'. Do you understand this?
He’s walking you through something standard, Pat.
--
Odd Bodkin — Maker of fine toys, tools, tables
patdolan
2021-11-22 02:29:53 UTC
Reply
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Post by Odd Bodkin
Post by Townes Olson
Post by patdolan
If F&S know the time of flight and the velocity of the muons in
their lab then F&S can
also know the time of flight and
velocity of the earth in the muon's FoR.
Will you [please teach me the] calculation of the latter?
Let S be inertial coordinates in which a particle P on the earth’s surface is at rest,
and let S’ be a system in which the muon is at rest. The muon, directly approaching
P at high speed, is created at event E1, which is simultaneous with event E2 of P in
terms of S, and with event E3 of P in terms of S’... There's nothing
paradoxical about this.
The two absolute space-like intervals are between two different pairs of events.
What does all this even mean??? Use some equations and more words.
Sure. Let E4 be the event where the muon collides with particle P on
earth's surface, and let v denote the mutual speed between the muon and
P, and in terms of S let D denote the distance traveled by the muon
from E1 to E4, so its time of flight is D/v. In terms of S' (in which
the muon is at rest) the distance traveled by the earth particle P from
E3 to E4 is D sqrt(1-v^2/c^2) and the time of flight is (D/v) sqrt(1-v^2/c^2).
I ask for clarification on E1, E2 and E3. Instead I get E4!
What is it that you don't understand about those events? They are
explicitly defined above. Again, the muon is created at E1 and it
collides with the earth particle P at E4. Event E2 of P is simultaneous
with E1 in terms of S, and event E3 of P is simultaneous with E1 in terms
of S'. Do you understand this?
He’s walking you through something standard, Pat.
I know. But it's like he doesn't understand that the very assumption he relies on, namely ∆x'/∆t' = ∆x/∆t, is what we are debating. Use your influence to get him off my thread until he gets a clue. Naturally I would like to use young Townes for more cannon fodder; but only after he understands the topography of this battlefield.
Post by Odd Bodkin
--
Odd Bodkin — Maker of fine toys, tools, tables
Townes Olson
2021-11-22 02:50:18 UTC
Reply
Permalink
Post by Townes Olson
Post by patdolan
If F&S know the time of flight and the velocity of the muons in
their lab then F&S can
also know the time of flight and
velocity of the earth in the muon's FoR.
Will you [please teach me the] calculation of the latter?
Let S be inertial coordinates in which a particle P on the earth’s surface is at rest,
and let S’ be a system in which the muon is at rest. The muon, directly approaching
P at high speed, is created at event E1, which is simultaneous with event E2 of P in
terms of S, and with event E3 of P in terms of S’... There's nothing
paradoxical about this.
The two absolute space-like intervals are between two different pairs of events.
What does all this even mean??? Use some equations and more words.
Sure. Let E4 be the event where the muon collides with particle P on
earth's surface, and let v denote the mutual speed between the muon and
P, and in terms of S let D denote the distance traveled by the muon
from E1 to E4, so its time of flight is D/v. In terms of S' (in which
the muon is at rest) the distance traveled by the earth particle P from
E3 to E4 is D sqrt(1-v^2/c^2) and the time of flight is (D/v) sqrt(1-v^2/c^2).
I ask for clarification on E1, E2 and E3. Instead I get E4!
What is it that you don't understand about those events? They are
explicitly defined above. Again, the muon is created at E1 and it
collides with the earth particle P at E4. Event E2 of P is simultaneous
with E1 in terms of S, and event E3 of P is simultaneous with E1 in terms
of S'. Do you understand this?
The very assumption you rely on, namely ∆x'/∆t' = ∆x/∆t, is what we are debating.
That's not the assumption, that's the conclusion of the calculation. In tiny baby steps: In terms of S the events E1, E2, E3, and E4 of the scenario you stipulated have the (x,t) coordinates (0,0), (D,0), (D,vD), and (D,D/v) respectively. These are stipulated in your question, given the speed and time of flight of the muon in terms of S. Given this, you asked for the speed and time of flight of P in terms of S', which is easily computed as above. What part of this do you consider to be debatable?
... get him off my thread...
I don't think that's going to work, because the answer is here for all to see, so even if you had the power to ban me, you can't un-ring the bell. You would have to somehow get my post deleted from the servers, but that would be nearly impossible. I';m afraid the jig is up.
Thank you in advance.
You're welcome.
patdolan
2021-11-22 03:30:20 UTC
Reply
Permalink
Post by Townes Olson
Post by patdolan
If F&S know the time of flight and the velocity of the muons in
their lab then F&S can
also know the time of flight and
velocity of the earth in the muon's FoR.
Will you [please teach me the] calculation of the latter?
Let S be inertial coordinates in which a particle P on the earth’s surface is at rest,
and let S’ be a system in which the muon is at rest. The muon, directly approaching
P at high speed, is created at event E1, which is simultaneous with event E2 of P in
terms of S, and with event E3 of P in terms of S’... There's nothing
paradoxical about this.
The two absolute space-like intervals are between two different pairs of events.
What does all this even mean??? Use some equations and more words.
Sure. Let E4 be the event where the muon collides with particle P on
earth's surface, and let v denote the mutual speed between the muon and
P, and in terms of S let D denote the distance traveled by the muon
from E1 to E4, so its time of flight is D/v. In terms of S' (in which
the muon is at rest) the distance traveled by the earth particle P from
E3 to E4 is D sqrt(1-v^2/c^2) and the time of flight is (D/v) sqrt(1-v^2/c^2).
I ask for clarification on E1, E2 and E3. Instead I get E4!
What is it that you don't understand about those events? They are
explicitly defined above. Again, the muon is created at E1 and it
collides with the earth particle P at E4. Event E2 of P is simultaneous
with E1 in terms of S, and event E3 of P is simultaneous with E1 in terms
of S'. Do you understand this?
The very assumption you rely on, namely ∆x'/∆t' = ∆x/∆t, is what we are debating.
That's not the assumption, that's the conclusion of the calculation. In tiny baby steps: In terms of S the events E1, E2, E3, and E4 of the scenario you stipulated have the (x,t) coordinates (0,0), (D,0), (D,vD), and (D,D/v) respectively. These are stipulated in your question, given the speed and time of flight of the muon in terms of S. Given this, you asked for the speed and time of flight of P in terms of S', which is easily computed as above. What part of this do you consider to be debatable?
Lemesee...Dirk foundered on ALGEBRAIC RELATIVITY'S ORIGINAL SIN...and Bodkin cried uncle on MUONS, SCHMUONS....so Townes, my boy, why don't you have a go at DIRK & DONO and see what you can make of that as yet unaddressed proof of my troika. In the spirit Derrida, try to address the text that you find--resist casting my examples in your own E1, E2, etc. gibberish. Now run along my boy and come back when you think you have something to show us.
... get him off my thread...
I don't think that's going to work, because the answer is here for all to see, so even if you had the power to ban me, you can't un-ring the bell. You would have to somehow get my post deleted from the servers, but that would be nearly impossible. I';m afraid the jig is up.
Thank you in advance.
You're welcome.
Dono.
2021-11-22 03:39:35 UTC
Reply
Permalink
snip cretinisms<
Pattycakes,

You are still struggling with basic concepts. Give that you are an old fart, you are guaranteed to never getting it. In other words, you were born an imbecile and your only consolation is that you are dying one.
Townes Olson
2021-11-22 04:09:04 UTC
Reply
Permalink
Post by patdolan
If F&S know the time of flight and the velocity of the muons in their lab
then F&S can also know the time of flight and velocity of the earth in the
muon's FoR. Will you [please teach me the] calculation of the latter?
The muon is created at E1 and it collides with the earth particle P at E4. Event E2 of P is simultaneous with E1 in terms of S, and event E3 of P is simultaneous with E1 in terms of S'. In terms of S let D denote the distance traveled by the muon from E1 to E4, so its time of flight is D/v. Therefore, in terms of S, the events E1, E2, E3, and E4 have the (x,t) coordinates (0,0), (D,0), (D,vD), and (D,D/v) respectively. Thus in terms of S' (in which the muon is at rest) the distance traveled by the earth particle P from E3 to E4 is D sqrt(1-v^2/c^2) and the time of flight is (D/v) sqrt(1-v^2/c^2).

What part of this do you consider to be debatable?
Post by patdolan
Lemesee... [followed by crazed gibberish]
So, you can't find anything debatable in the answer to your question. Excellent.
Post by patdolan
Thank you in advance.
You're welcome.
Odd Bodkin
2021-11-22 12:12:45 UTC
Reply
Permalink
Post by patdolan
Post by Townes Olson
Post by patdolan
If F&S know the time of flight and the velocity of the muons in
their lab then F&S can
also know the time of flight and
velocity of the earth in the muon's FoR.
Will you [please teach me the] calculation of the latter?
Let S be inertial coordinates in which a particle P on the
earth’s surface is at rest,
and let S’ be a system in which the muon is at rest. The muon,
directly approaching
P at high speed, is created at
event E1, which is simultaneous with event E2 of P in
terms of S, and with event E3 of P in terms of S’... There's nothing
paradoxical about this.
The two absolute space-like intervals are between two different pairs of events.
What does all this even mean??? Use some equations and more words.
Sure. Let E4 be the event where the muon collides with particle P on
earth's surface, and let v denote the mutual speed between the muon and
P, and in terms of S let D denote the distance traveled by the muon
from E1 to E4, so its time of flight is D/v. In terms of S' (in which
the muon is at rest) the distance traveled by the earth particle P from
E3 to E4 is D sqrt(1-v^2/c^2) and the time of flight is (D/v) sqrt(1-v^2/c^2).
I ask for clarification on E1, E2 and E3. Instead I get E4!
What is it that you don't understand about those events? They are
explicitly defined above. Again, the muon is created at E1 and it
collides with the earth particle P at E4. Event E2 of P is simultaneous
with E1 in terms of S, and event E3 of P is simultaneous with E1 in terms
of S'. Do you understand this?
The very assumption you rely on, namely ∆x'/∆t' = ∆x/∆t, is what we are debating.
That's not the assumption, that's the conclusion of the calculation. In
tiny baby steps: In terms of S the events E1, E2, E3, and E4 of the
scenario you stipulated have the (x,t) coordinates (0,0), (D,0), (D,vD),
and (D,D/v) respectively. These are stipulated in your question, given
the speed and time of flight of the muon in terms of S. Given this, you
asked for the speed and time of flight of P in terms of S', which is
easily computed as above. What part of this do you consider to be debatable?
Lemesee...Dirk foundered on ALGEBRAIC RELATIVITY'S ORIGINAL SIN...and
Bodkin cried uncle on MUONS, SCHMUONS....
Oh, bullshit. You showed you didn’t know how the experiment was actually
done, and you showed you didn’t know how to do the Lorentz transforms to
find out what the lab clock time would read, which is unambiguous.

The more you cajole, the more pathetic you look.

Why not just read a book? Need a recommendation?
Post by patdolan
so Townes, my boy, why don't you have a go at DIRK & DONO and see what
you can make of that as yet unaddressed proof of my troika. In the
spirit Derrida, try to address the text that you find--resist casting my
examples in your own E1, E2, etc. gibberish. Now run along my boy and
come back when you think you have something to show us.
... get him off my thread...
I don't think that's going to work, because the answer is here for all
to see, so even if you had the power to ban me, you can't un-ring the
bell. You would have to somehow get my post deleted from the servers,
but that would be nearly impossible. I';m afraid the jig is up.
Thank you in advance.
You're welcome.
--
Odd Bodkin -- maker of fine toys, tools, tables
Odd Bodkin
2021-11-22 12:10:49 UTC
Reply
Permalink
Post by patdolan
Post by Odd Bodkin
Post by Townes Olson
Post by patdolan
If F&S know the time of flight and the velocity of the muons in
their lab then F&S can
also know the time of flight and
velocity of the earth in the muon's FoR.
Will you [please teach me the] calculation of the latter?
Let S be inertial coordinates in which a particle P on the earth’s
surface is at rest,
and let S’ be a system in which the muon is at rest. The muon, directly approaching
P at high speed, is created at event E1, which is simultaneous with event E2 of P in
terms of S, and with event E3 of P in terms of S’... There's nothing
paradoxical about this.
The two absolute space-like intervals are between two different pairs of events.
What does all this even mean??? Use some equations and more words.
Sure. Let E4 be the event where the muon collides with particle P on
earth's surface, and let v denote the mutual speed between the muon and
P, and in terms of S let D denote the distance traveled by the muon
from E1 to E4, so its time of flight is D/v. In terms of S' (in which
the muon is at rest) the distance traveled by the earth particle P from
E3 to E4 is D sqrt(1-v^2/c^2) and the time of flight is (D/v) sqrt(1-v^2/c^2).
I ask for clarification on E1, E2 and E3. Instead I get E4!
What is it that you don't understand about those events? They are
explicitly defined above. Again, the muon is created at E1 and it
collides with the earth particle P at E4. Event E2 of P is simultaneous
with E1 in terms of S, and event E3 of P is simultaneous with E1 in terms
of S'. Do you understand this?
He’s walking you through something standard, Pat.
I know. But it's like he doesn't understand that the very assumption he
relies on, namely ∆x'/∆t' = ∆x/∆t, is what we are debating.
This is easy to demonstrate if you start with Lorentz transforms.

If you don’t want to start there, then it follows from linearity
requirements.
Post by patdolan
Use your influence to get him off my thread until he gets a clue.
Naturally I would like to use young Townes for more cannon fodder; but
only after he understands the topography of this battlefield.
Post by Odd Bodkin
--
Odd Bodkin — Maker of fine toys, tools, tables
--
Odd Bodkin -- maker of fine toys, tools, tables
patdolan
2021-11-22 15:30:02 UTC
Reply
Permalink
Post by Odd Bodkin
Post by Odd Bodkin
Post by Townes Olson
Post by patdolan
If F&S know the time of flight and the velocity of the muons in
their lab then F&S can
also know the time of flight and
velocity of the earth in the muon's FoR.
Will you [please teach me the] calculation of the latter?
Let S be inertial coordinates in which a particle P on the earth’s
surface is at rest,
and let S’ be a system in which the muon is at rest. The muon, directly approaching
P at high speed, is created at event E1, which is simultaneous with event E2 of P in
terms of S, and with event E3 of P in terms of S’... There's nothing
paradoxical about this.
The two absolute space-like intervals are between two different pairs of events.
What does all this even mean??? Use some equations and more words.
Sure. Let E4 be the event where the muon collides with particle P on
earth's surface, and let v denote the mutual speed between the muon and
P, and in terms of S let D denote the distance traveled by the muon
from E1 to E4, so its time of flight is D/v. In terms of S' (in which
the muon is at rest) the distance traveled by the earth particle P from
E3 to E4 is D sqrt(1-v^2/c^2) and the time of flight is (D/v) sqrt(1-v^2/c^2).
I ask for clarification on E1, E2 and E3. Instead I get E4!
What is it that you don't understand about those events? They are
explicitly defined above. Again, the muon is created at E1 and it
collides with the earth particle P at E4. Event E2 of P is simultaneous
with E1 in terms of S, and event E3 of P is simultaneous with E1 in terms
of S'. Do you understand this?
He’s walking you through something standard, Pat.
I know. But it's like he doesn't understand that the very assumption he
relies on, namely ∆x'/∆t' = ∆x/∆t, is what we are debating.
This is easy to demonstrate if you start with Lorentz transforms.
Wow! What an admission, Bodkin. You and Dirk now have an issue between you. Just like Long Ears and Paul A#2.

This will be fun to watch. Relativist trying to cancel relativist.
Post by Odd Bodkin
If you don’t want to start there, then it follows from linearity
requirements.
Use your influence to get him off my thread until he gets a clue.
Naturally I would like to use young Townes for more cannon fodder; but
only after he understands the topography of this battlefield.
Post by Odd Bodkin
--
Odd Bodkin — Maker of fine toys, tools, tables
--
Odd Bodkin -- maker of fine toys, tools, tables
Odd Bodkin
2021-11-22 15:40:46 UTC
Reply
Permalink
Post by patdolan
Post by Odd Bodkin
Post by Odd Bodkin
Post by Townes Olson
Post by patdolan
If F&S know the time of flight and the velocity of the muons in
their lab then F&S can
also know the time of flight and
velocity of the earth in the muon's FoR.
Will you [please teach me the] calculation of the latter?
Let S be inertial coordinates in which a particle P on the earth’s
surface is at rest,
and let S’ be a system in which the muon is at rest. The muon,
directly approaching
P at high speed, is created at
event E1, which is simultaneous with event E2 of P in
terms of S, and with event E3 of P in terms of S’... There's nothing
paradoxical about this.
The two absolute space-like intervals are between two different pairs of events.
What does all this even mean??? Use some equations and more words.
Sure. Let E4 be the event where the muon collides with particle P on
earth's surface, and let v denote the mutual speed between the muon and
P, and in terms of S let D denote the distance traveled by the muon
from E1 to E4, so its time of flight is D/v. In terms of S' (in which
the muon is at rest) the distance traveled by the earth particle P from
E3 to E4 is D sqrt(1-v^2/c^2) and the time of flight is (D/v) sqrt(1-v^2/c^2).
I ask for clarification on E1, E2 and E3. Instead I get E4!
What is it that you don't understand about those events? They are
explicitly defined above. Again, the muon is created at E1 and it
collides with the earth particle P at E4. Event E2 of P is simultaneous
with E1 in terms of S, and event E3 of P is simultaneous with E1 in terms
of S'. Do you understand this?
He’s walking you through something standard, Pat.
I know. But it's like he doesn't understand that the very assumption he
relies on, namely ∆x'/∆t' = ∆x/∆t, is what we are debating.
This is easy to demonstrate if you start with Lorentz transforms.
Wow! What an admission, Bodkin.
Why? Why do you think this is a problem?
Post by patdolan
You and Dirk now have an issue between you. Just like Long Ears and Paul A#2.
This will be fun to watch. Relativist trying to cancel relativist.
Well, it’s funny that this is what you’re trying to do, rather than asking
about relativity. So what are you interested in, relativity? Or trying to
generate squabbles on a backwater of the internet? (And if the latter, WHY
would anybody do that?)
Post by patdolan
Post by Odd Bodkin
If you don’t want to start there, then it follows from linearity
requirements.
See? Not a problem.
Post by patdolan
Post by Odd Bodkin
Use your influence to get him off my thread until he gets a clue.
Naturally I would like to use young Townes for more cannon fodder; but
only after he understands the topography of this battlefield.
Post by Odd Bodkin
--
Odd Bodkin — Maker of fine toys, tools, tables
--
Odd Bodkin -- maker of fine toys, tools, tables
--
Odd Bodkin -- maker of fine toys, tools, tables
Odd Bodkin
2021-11-22 15:53:40 UTC
Reply
Permalink
Post by Odd Bodkin
Post by patdolan
Post by Odd Bodkin
Post by Odd Bodkin
Post by Townes Olson
Post by patdolan
If F&S know the time of flight and the velocity of the muons in
their lab then F&S can
also know the time of flight and
velocity of the earth in the muon's FoR.
Will you [please teach me the] calculation of the latter?
Let S be inertial coordinates in which a particle P on the earth’s
surface is at rest,
and let S’ be a system in which the muon is at rest. The muon,
directly approaching
P at high speed, is created at
event E1, which is simultaneous with event E2 of P in
terms of S, and with event E3 of P in terms of S’... There's nothing
paradoxical about this.
The two absolute space-like intervals are between two different pairs of events.
What does all this even mean??? Use some equations and more words.
Sure. Let E4 be the event where the muon collides with particle P on
earth's surface, and let v denote the mutual speed between the muon and
P, and in terms of S let D denote the distance traveled by the muon
from E1 to E4, so its time of flight is D/v. In terms of S' (in which
the muon is at rest) the distance traveled by the earth particle P from
E3 to E4 is D sqrt(1-v^2/c^2) and the time of flight is (D/v) sqrt(1-v^2/c^2).
I ask for clarification on E1, E2 and E3. Instead I get E4!
What is it that you don't understand about those events? They are
explicitly defined above. Again, the muon is created at E1 and it
collides with the earth particle P at E4. Event E2 of P is simultaneous
with E1 in terms of S, and event E3 of P is simultaneous with E1 in terms
of S'. Do you understand this?
He’s walking you through something standard, Pat.
I know. But it's like he doesn't understand that the very assumption he
relies on, namely ∆x'/∆t' = ∆x/∆t, is what we are debating.
This is easy to demonstrate if you start with Lorentz transforms.
Wow! What an admission, Bodkin.
Why? Why do you think this is a problem?
Post by patdolan
You and Dirk now have an issue between you. Just like Long Ears and Paul A#2.
This will be fun to watch. Relativist trying to cancel relativist.
Well, it’s funny that this is what you’re trying to do, rather than asking
about relativity. So what are you interested in, relativity? Or trying to
generate squabbles on a backwater of the internet? (And if the latter, WHY
would anybody do that?)
Post by patdolan
Post by Odd Bodkin
If you don’t want to start there, then it follows from linearity
requirements.
See? Not a problem.
Post by patdolan
Post by Odd Bodkin
Use your influence to get him off my thread until he gets a clue.
Naturally I would like to use young Townes for more cannon fodder; but
only after he understands the topography of this battlefield.
Post by Odd Bodkin
--
Odd Bodkin — Maker of fine toys, tools, tables
--
Odd Bodkin -- maker of fine toys, tools, tables
You know, Pat, it’s a bit of a circus, you first floundering around, not
knowing what relativity predicts for a ground clock in a muon decay
experiment; then floundering around, not knowing how the experimental
measurement with muons was actually made; then asking for and walking away
from someone else’s attempt to teach you how to use Lorentz transforms with
a handful of events in different frames; then spluttering that there might
be some problem with the symmetry of relative speeds in two frames, when
there is no such problem; then pretending that it’s all just a game to
produce infighting on the internet. And all, the whole, proclaiming
yourself king of the science hill.

Good lord, man, it’s no wonder you haven’t learned a thing. You can’t stay
focused long enough to read a few pages in a book that might actually
explain it to you.

And speaking of infighting and your claim to intellectual superiority, you
could have that discussion with Ken “I’m gifted” Seto or John “King of the
Science Hill” Armisted or Maciej “world’s best logician” Wozniak or Ed “I’m
an Analyst” Lake. Let’s see if they acknowledge your mightiness and KO
count.
--
Odd Bodkin -- maker of fine toys, tools, tables
patdolan
2021-11-22 15:59:19 UTC
Reply
Permalink
Post by Odd Bodkin
Post by patdolan
Post by Odd Bodkin
Post by Odd Bodkin
Post by Townes Olson
Post by patdolan
If F&S know the time of flight and the velocity of the muons in
their lab then F&S can
also know the time of flight and
velocity of the earth in the muon's FoR.
Will you [please teach me the] calculation of the latter?
Let S be inertial coordinates in which a particle P on the earth’s
surface is at rest,
and let S’ be a system in which the muon is at rest. The muon,
directly approaching
P at high speed, is created at
event E1, which is simultaneous with event E2 of P in
terms of S, and with event E3 of P in terms of S’... There's nothing
paradoxical about this.
The two absolute space-like intervals are between two different pairs of events.
What does all this even mean??? Use some equations and more words.
Sure. Let E4 be the event where the muon collides with particle P on
earth's surface, and let v denote the mutual speed between the muon and
P, and in terms of S let D denote the distance traveled by the muon
from E1 to E4, so its time of flight is D/v. In terms of S' (in which
the muon is at rest) the distance traveled by the earth particle P from
E3 to E4 is D sqrt(1-v^2/c^2) and the time of flight is (D/v) sqrt(1-v^2/c^2).
I ask for clarification on E1, E2 and E3. Instead I get E4!
What is it that you don't understand about those events? They are
explicitly defined above. Again, the muon is created at E1 and it
collides with the earth particle P at E4. Event E2 of P is simultaneous
with E1 in terms of S, and event E3 of P is simultaneous with E1 in terms
of S'. Do you understand this?
He’s walking you through something standard, Pat.
I know. But it's like he doesn't understand that the very assumption he
relies on, namely ∆x'/∆t' = ∆x/∆t, is what we are debating.
This is easy to demonstrate if you start with Lorentz transforms.
Wow! What an admission, Bodkin.
Why? Why do you think this is a problem?
Post by patdolan
You and Dirk now have an issue between you. Just like Long Ears and Paul A#2.
This will be fun to watch. Relativist trying to cancel relativist.
--
Post by Odd Bodkin
Well, it’s funny that this is what you’re trying to do, rather than asking
about relativity. So what are you interested in, relativity? Or trying to
generate squabbles on a backwater of the internet? (And if the latter, WHY
would anybody do that?)
Post by patdolan
Post by Odd Bodkin
If you don’t want to start there, then it follows from linearity
requirements.
See? Not a problem.
"So when you write
∆x’/∆t’ = ∆x/∆t = v ,
you clearly have no idea what you are talking about. " -- Dirk Vdm

Dirk is een betere logicus dan jij, Bodkin. Dirk kent de algebraïsche catastrofe die jouw bekentenis is. Dat is waarom hij het vanaf het begin heeft vermeden.
Post by Odd Bodkin
Post by patdolan
Post by Odd Bodkin
Use your influence to get him off my thread until he gets a clue.
Naturally I would like to use young Townes for more cannon fodder; but
only after he understands the topography of this battlefield.
Post by Odd Bodkin
--
Odd Bodkin — Maker of fine toys, tools, tables
--
Odd Bodkin -- maker of fine toys, tools, tables
--
Odd Bodkin -- maker of fine toys, tools, tables
Odd Bodkin
2021-11-22 16:13:19 UTC
Reply
Permalink
Post by patdolan
Post by Odd Bodkin
Post by patdolan
Post by Odd Bodkin
Post by Odd Bodkin
Post by Townes Olson
Post by patdolan
If F&S know the time of flight and the velocity of the muons in
their lab then F&S can
also know the time of flight and
velocity of the earth in the muon's FoR.
Will you [please teach me the] calculation of the latter?
Let S be inertial coordinates in which a particle P on the earth’s
surface is at rest,
and let S’ be a system in which the muon is at rest. The muon,
directly approaching
P at high speed, is created at
event E1, which is simultaneous with event E2 of P in
terms of S, and with event E3 of P in terms of S’... There's nothing
paradoxical about this.
The two absolute space-like intervals are between two different pairs of events.
What does all this even mean??? Use some equations and more words.
Sure. Let E4 be the event where the muon collides with particle P on
earth's surface, and let v denote the mutual speed between the muon and
P, and in terms of S let D denote the distance traveled by the muon
from E1 to E4, so its time of flight is D/v. In terms of S' (in which
the muon is at rest) the distance traveled by the earth particle P from
E3 to E4 is D sqrt(1-v^2/c^2) and the time of flight is (D/v) sqrt(1-v^2/c^2).
I ask for clarification on E1, E2 and E3. Instead I get E4!
What is it that you don't understand about those events? They are
explicitly defined above. Again, the muon is created at E1 and it
collides with the earth particle P at E4. Event E2 of P is simultaneous
with E1 in terms of S, and event E3 of P is simultaneous with E1 in terms
of S'. Do you understand this?
He’s walking you through something standard, Pat.
I know. But it's like he doesn't understand that the very assumption he
relies on, namely ∆x'/∆t' = ∆x/∆t, is what we are debating.
This is easy to demonstrate if you start with Lorentz transforms.
Wow! What an admission, Bodkin.
Why? Why do you think this is a problem?
Post by patdolan
You and Dirk now have an issue between you. Just like Long Ears and Paul A#2.
This will be fun to watch. Relativist trying to cancel relativist.
--
Post by Odd Bodkin
Well, it’s funny that this is what you’re trying to do, rather than asking
about relativity. So what are you interested in, relativity? Or trying to
generate squabbles on a backwater of the internet? (And if the latter, WHY
would anybody do that?)
Post by patdolan
Post by Odd Bodkin
If you don’t want to start there, then it follows from linearity
requirements.
See? Not a problem.
"So when you write
∆x’/∆t’ = ∆x/∆t = v ,
you clearly have no idea what you are talking about. " -- Dirk Vdm
I don’t know why you think this is in conflict with anything I said. You
clearly don’t have no idea what you’re talking about.
Post by patdolan
Dirk is een betere logicus dan jij, Bodkin. Dirk kent de algebraïsche
catastrofe die jouw bekentenis is. Dat is waarom hij het vanaf het begin heeft vermeden.
I have no interest in conversing in Dutch. If you think there is an
algebraic catastrophe, out with it.
Post by patdolan
Post by Odd Bodkin
Post by patdolan
Post by Odd Bodkin
Use your influence to get him off my thread until he gets a clue.
Naturally I would like to use young Townes for more cannon fodder; but
only after he understands the topography of this battlefield.
Post by Odd Bodkin
--
Odd Bodkin — Maker of fine toys, tools, tables
--
Odd Bodkin -- maker of fine toys, tools, tables
--
Odd Bodkin -- maker of fine toys, tools, tables
--
Odd Bodkin -- maker of fine toys, tools, tables
Kip Foh
2021-11-22 15:41:48 UTC
Reply
Permalink
Post by patdolan
Wow! What an admission, Bodkin. You and Dirk now have an issue between
you. Just like Long Ears and Paul A#2.
This will be fun to watch. Relativist trying to cancel relativist.
not true, Kamala Harris Was the 1st *woman_PRESIDENT* For Two Hours While
Biden Got His Colonoscopy.
Odd Bodkin
2021-11-21 21:03:33 UTC
Reply
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Post by patdolan
Post by Odd Bodkin
Post by patdolan
Post by Odd Bodkin
Post by patdolan
Critical Theory comes in several flavors: Critical Law Theory, Critical
Literature Theory, Critical History Theory, Critical Race Theory. To
these we now add Critical Relativity Theory!
In the spirit of Derrida we shall deconstruct the clumsy reasoning of
special relativity and separate said reasoning from the algebraic symbols
and equations that express it, keeping in mind that mathematics is just
another form of rhetorical expression wherein falsity can be expressed
every bit as plausibly as the truth.
“There is only the text.”— J. Derrida
According to special relativity two observers in motion with respect to
each other will disagree on each other’s length. They will also disagree
on the proper flow of time. But they will always agree on the velocity
they have with respect to one another. This is exceedingly strange. How
can it be that two relative quantities, space and time, combine to
produce an absolute quantity called relative velocity? It is true that SR
does have a formula for calculating coordinate velocity; just like it has
formulas for calculating coordinate space and coordinate time. But the
Einstein velocity addition formula ONLY applies to a third object in
motion wrt a pair of FoRs. If that third object happens to be at rest
wrt one of the FoRs then Einsteinian velocity reduces to Galilean
velocity, albeit subject to the speed limit c. Relativists simply
assumed without further justification that if FoR-1 measures a velocity v
between itself and FoR-2 then FoR-2 must measure the same numerical value
v for the velocity between itself and FoR-1. Trivial? Nope. Seemingly
trivial assumptions can be monumental when constructing a theory of
motion. But a 26 year old would probably not yet have the requisite
philosophical sophistication needed to recognize this.
Einstein’s choice to make velocity strictly Galilean when calculating the
velocity between pairs of FoRs ( yes, it is a choice because it does not
follow from either the first or the second postulates ) can be
expressed mathematically as
∆x’/∆t’ = ∆x/∆t = v (1)
I now raise equation (1) to the level of a postulate and declare it to be
the third, and heretofore hidden, postulate of special relativity. In
recognizing its own structural Galileanism through this new postulate,
special relativity can finally claim to be woke.
The problem with the third postulate is that even though it is already
assumed in every equation of special relativity, it turns out to be true
only when v = c ( the second postulate ) or when v = 0. The third
postulate can be demonstrated invalid for all values of v in between.
The invalidity of the third postulate causes special relativity fall on
it’s algebraic face. Big Bang move over…Not recognizing and
acknowledging the third postulate was Einstein’s biggest blunder.
Time for some examples.
DIRK & DONO
Consider two FoRs whose x-axes are parallel and lie very close to one
another. The relative velocity between these two FoRs is .866c (γ = 2).
Dirk assumes the Lotus position at the origin of one FoR whilst Dono
assumes the fetal position at the origin of the other. Both Dirk and
Dono and their clocks are glued to the origins of their respective FoRs.
Dirk opens one eye and takes note of the meter marks on Dono’s contracted
x-axis as they whiz by. Dirk apprehends that Dono’s meter marks are
contracted to only half as long as his own meter marks. Dirk opens the
other eye and observes that Dono’s clock is ticking at only half the rate
of his own clock. Dirk begins to count Dono’s meter marks as they race
past Dirk’s position. After one year (by Dirk’s clock) of counting
Dono’s meter marks, Dirk has tallied 1.64e+16 Dono meters ( 9.5e+15 meter
marks/Ly x .866Ly x γ ). Dirk has also observed that only 0.5 years have
elapsed on Dono’s clock. Dirk now calculates what his coordinate
velocity should be according to Dono
( 1.64e+16 Dono meters ) / ( 1.58e+7 Dono seconds ) = 1.04e+9 m/s = 3.5c (2)
[ shortcut: ∆x’/∆t’ = (∆x/∆t)(γ^2) = v(γ^2) ] (3)
“Stop!” You cry, “Dirk’s and Dono’s relative velocity was already
stipulated to be an absolute .866c with respect to one another, when
measured in either FoR.”
That is true, according to (1). But remember that (1) is an arbitrary
choice made by Einstein when he built his theory. It is no more
legitimate a choice than the Dolan FoR coordinate velocity transform (3)
for determining one’s FoR coordinate velocity. It is also no less
inconsistent. For we immediately see by inspection that the Dolan FoR
coordinate velocity is always greater than the relative velocity by a
factor of γ^2. But Einstein’s choice for requiring Galilean FoR pair
velocities clangs with just as much antinomy as Dolan’s transform, as we
shall see. Special relativity’s dirty little secret is that it’s hidden
third postulate (1) destroys the theory from within.
I can see by Dono’s tightening fetal position that he still doesn’t
believe me. Very well. We shall prove SR’s mathematical inconsistency
in the next example.
SPECIAL RELATIVITY COLLAPSES UNDER THE WEIGHT OF ITS ALGEBRAIC ORIGINAL SIN
Consider a pair of FoRs whose relative velocity v is some value other
than c and other than zero. This is expressed mathematically as
v = ∆x’/∆t’ = ∆x/∆t != c (4)
and
v = ∆x’/∆t’ = ∆x/∆t != 0 (5)
The Lorentz transforms allow us to construct the FoR coordinate
velocities for pairs of FoRs
∆x’ = γ( ∆x - v∆t )
∆t’ = γ( ∆t - ∆xv/c^2 )
∆x’/∆t’ = [ γ( ∆x - v∆t ) ] / [ γ( ∆t - ∆xv/c^2 ) ] (6)
We now endeavor to solve (6) for v in hopes of demonstrating the internal
consistency of special relativity, i.e, that v = v for all pair of FoRs.
We saw how this was not the case in DIRK & DONO.
The reader will find that trying to solve for v is hopeless unless we
make the substitution
v = ∆x/∆t or v = ∆x’/∆t’
It does not matter which we use—either substitution is permitted by the
third postulate (1).
With the substitution made, we eventually arrive at the preposterous results
∆x’/∆t’ = c (7)
or
∆x’/∆t’ = 0 (8)
The derivation is left as an exercise for the reader. For those needing
help with the derivation ( I'm looking at you, Dono ) I will be happy to
provide it in another post.
The laughable results (7) and (8) directly contradict our assumptions (4)
and (5). Furthermore, the results impose the requirement that v is not
even a variable—v turns out to be a constant always equal to c or zero.
Absolutely absurd.
QED.
Algebraic relativity is thus reduced to ridiculous rubble by means
mathematical reductio ad absurdum. The root cause of special relativity’s
spectacular algebraic failure lies in the propositional calculus. I am
happy to expatiate on that subject in another post if there is interest.
Let’s do one more example—this one ripped from the headlines of
experimental physics.
MUONS, SCHMUONS!
Imagine that you are a hadron in deep space minding your own business
when all of a sudden you turn to see a Lorentz-flattened earth coming at
you at a velocity of .866c. In the impending collision the first thing
to strike you is an air molecule high in the earth’s atmosphere. That
molecules knocks the stuffing out of you. What is left of you is now a
muon which means that you only have 2.2 microseconds more to live.
It turns out that the surface of the flattened earth is exactly 571.56
meters away from you, according to your own muon FoR. By a remarkable
coincidence there is a flattened scintillator and a flattened clock in a
flattened laboratory directly below you on the surface of the flattened
earth. You note the time on the flattened lab clock. Because you are a
muon, you are your own 2.2 microsecond alarm clock.
The surface of the flattened earth continues to speed towards you at
.866c. 2.2 microseconds later the flattened earth, lab and scintillator
smash into you. Just as you expire in the scintillator you note that
only 1.1 microseconds has elapsed on the lab clock.
You are an idiot. “You” the hadron cannot see the “lab clock” at t=0.
No synchronization required. The muon's clock doesn't even enter into
the argument at all. Who said that the lab clock was ever at t=0 or any
other value. E. L. A. P. S. E. D. T. I. M. E., Bodkin.
Elapsed from what moment? How does the lab clock know when to start
ticking. It isn’t anywhere near the collision of the hadron and the
atmospheric molecule. Is it supposed to get an instantaneous signal from
that event?
This is what I mean when I say you haven’t bothered to read any book about
relativity. Your “argument” is based on two and only two clocks passing
each other, both deeming the other slowed, because that’s the comic book
version of relativity. But that isn’t what relativity says. Ask yourself
why relativity talks about a *lattice* of synchronized clocks in each
frame. Why would you need the lattice?
It’s so that there is a clock NEAR the event that you’re tracking the time
of.
Quantities of elapsed time are easily determined in special relativity.
Synchronization of clocks doesn't even enter into the argument It's all
about one clock, the lab clock. One clock, two observers (the muon and
the lab scientists), TWO elapsed time. Quite brilliant on my part, n'es pas?
Tell us how synchronization has any bearing on the argument. The muon
sees the lab clock tick off 1.1 microseconds on it's trip down.
No, it doesn’t. How is it supposed to “see” that? It isn’t ANYWHERE NEAR
the lab clock. The best the muon can do is either a) monitor its own clock,
because its own clock is at ONE location in its own frame, or b) have a
LATTICE of synchronized clocks traveling along with the muon, where one of
them is far away from the muon, near the lab clock at the time the muon is
created.
If you have two events that occur at DIFFERENT PLACES in some frame of
reference, you need two DIFFERENT clocks near those events. If those clocks
are synchronized in this frame, THEN you can measure the elapsed time
between the events.
You can’t measure the elapsed time between two events that occur in
different places with ONE clock.
You, Pat Dolan, don’t understand the first thing about relativity. And it
shows.
And yet you bluster simultaneously that a) you understand relativity very
well, and b) that it is incomprehensible.
Bodkin, please declare to this forum that your argument also applies to
the scientists in the lab--especially to the scientist in the lab of Dr's
Frisch and Smith.
http://youtu.be/rbzt8gDSYIM
Frisch and Smith didn't have clocks up there alongside all the muons they
measured either. So there is NO WAY according to Bodkin that Frisch and
Smith could have possibly determined whether or not muons experienced
relativistic time dilation.
Yes they can, but not by the method you floundered around with. First of
all, these muon experiments typically measure the speed directly by having
a stack of vertically separated scintillators, so that time of flight is
readily available. Secondly, the lifetime is measured the same way it is
for radioisotopes, activity rate at different samplings. It’s actually
straightforward.
If F&S know the time of flight and the velocity of the muons in their lab
then F&S can also know the time of flight and velocity of the earth in the muon's FoR.
That depends on what you mean “know”. If you mean, what does relativity
CALCULATE these to be, then yes, that’s certainly doable. You’ve in fact
already listed those results. If you want to be sure, of course, it’s a
mystery to me why you don’t read any of the copious presentations of this
very thing in many textbooks. Why you need to have explanations REPRINTED
here, is frankly beyond me, other than you just trying to get people to do
things you’re too lazy to do yourself.
Post by patdolan
Will you now deign to delight this forum with your calculation of the
latter? Thank you, in advance. Also, please use an online translator to
render a version of your answer in French for those Francophones looking
in on our debate. Thank you again in advance.
Post by Odd Bodkin
But what is clear through out this is that you never READ exactly how
Frisch and Smith did this experiment, and you’ve never READ how relativity
describes this phenomenon, both of which are common expositions available
in lots of books.
If you’d only do that, you’d bluster less with loaded squirt guns you’re
holding backwards.
Post by patdolan
Declare this, you checkmated chump.
Post by Odd Bodkin
The lab scientists, using relativity and the standard muon life-span,
calculate the same lab clock ticks off 4.4 seconds for the same trip.
Now take a deep breath and form an argument that addresses the facts of
the case, not imaginary t=0 stuff.
can see the clock adjacent to you, which is 572 m above the lab clock. And,
by the way, the one adjacent to you and the lab clock are not synchronized,
according to “you”, and so “you” wouldn’t dream of subtracting the readings
on the two clocks at rest in the earth frame to come up with an elapsed
time according to the lab clocks.
“What?” Pat Dolan protests, “the lab clock and the clock up by the
atmosphere are perfectly synchronized beforehand. They are still
synchronized now.” But the question then is, “In what frame were they
synchronized?” Pat Dolan splutters, “What difference does THAT make?” And
the answer is, “It matters a lot. The same two clocks may be synchronized
in one frame, but they’re not synchronized in the muon’s frame, or any
other frame in fact. This is one of the foundational observations of
special relativity.”
“But, but, but….,” fumes Pat Dolan, “I had an EXCELLENT hand!”
No, Pat, you don’t have an excellent hand. You have crap cards, all because
you have never bothered to read a book that talks about muons IN DETAIL.
Instead, you came here with a poor and slippery grasp on a simple and
common example, and you thought somehow that something SO SIMPLE had been
missed in ALL THOSE presentations that you never read.
You, Pat, are a persistent idiot, thinking you have good cards, when you’ve
never even bothered to learn how to play poker.
Post by patdolan
Nothing strange here; the labs clock is traveling at γ=2 with respect to
you so it only logs half as much elapsed time as you.
One of the lab scientists now cries out “Just a minute! That’s not how
this story goes. First of all, it is the hadron that is
Lorentz-flattened, not us. And second of all, that hadron used it’s
flattened FoR to determine that is was only 571.56 meters away from our
scintillator at the point it became a muon. Our unflattened earth FoR
clearly indicated that the hadron was actually 1143.12 meters away when
it became a muon. That’s why it took 4.4 microseconds, traveling at .866
c, to reach our scintillator. And that’s what our lab clock shows. Not
1.1 microseconds, like that dumb muon is claiming.”
Special relativity leaves us in a quandary. How much time actually did
elapse on the lab clock from the moment of the muon’s inception to the
moment of its scintillating demise? Was it 1.1 microseconds? Or was it
4.4 microseconds? It has to be one or the other, it can’t be both.
However, different elapsed times on the same lab clock can be calculated
by the muon and by the earthbound scientists with equal justification.
Special relativity is incapable of providing a unique elapsed time on the
lab clock. I challenge anyone to show that special relativity can
provide a unique elapsed time in the preceding case.
Special relativity’s greatest experimental confirmation turns out to be
it’s second greatest falsification. Gravity has been special
relativity’s greatest falsification since 1907 ( check out the author’s
brilliant post on special relativity vs. Kepler’s 3rd law ).
--
Odd Bodkin -- maker of fine toys, tools, tables
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Odd Bodkin -- maker of fine toys, tools, tables
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Odd Bodkin -- maker of fine toys, tools, tables
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Odd Bodkin -- maker of fine toys, tools, tables
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