Discussion:
what is really measured whit a SR measurement
beda pietanza
2021-04-07 16:31:52 UTC
a SR K frame measuring an object in K' frame gives a certain result

then if the K frame jumps to another speed while the frame K' and its object
stays still
the new measurement, of the same still object in K', changes

what the frames K are measuring are the effects of the changes that took place only in the K frame due to the changing of its absolute speed
that has logically affected the rulers and the clocks of K frame itself,

so when SR claims to do a measurement, what is really measured is not a given object solely , but the measuring tool is measured with the object measured in an inextricable mixture.

cheers
beda
Edkah Bearden
2021-04-07 16:43:48 UTC
Post by beda pietanza
a SR K frame measuring an object in K' frame gives a certain result
then if the K frame jumps to another speed while the frame K' and its
object stays still the new measurement, of the same still object in K',
changes
what the frames K are measuring are the effects of the changes that took
place only in the K frame due to the changing of its absolute speed that
has logically affected the rulers and the clocks of K frame itself,
Here's where the difference between a measurement and observation gets
visible. Relativity is rather about observations, not measurements.
However, those observation are the best you have, hence you may safely
take them as measurements. But remember, those are observations.
mitchr...@gmail.com
2021-04-07 17:52:07 UTC
We know we are in dimension but cannot measure
our frames speed moving in it... Einstein called that
his unmarked flat metric...

Mitchell Raemsch
Edkah Bearden
2021-04-07 18:37:33 UTC
We know we are in dimension but cannot measure our frames speed moving
in it... Einstein called that his unmarked flat metric...
are you an obergruppenfuhrer general of waffen, or an obersturmbannfuhrer?
Paul B. Andersen
2021-04-07 20:33:34 UTC
Post by beda pietanza
a SR K frame measuring an object in K' frame gives a certain result
then if the K frame jumps to another speed while the frame K' and its object
stays still
the new measurement, of the same still object in K', changes
what the frames K are measuring are the effects of the changes that took place only in the K frame due to the changing of its absolute speed
that has logically affected the rulers and the clocks of K frame itself,
so when SR claims to do a measurement, what is really measured is not a given object solely , but the measuring tool is measured with the object measured in an inextricable mixture.
cheers
beda
You are always incoherent, but sometimes you are even more incoherent!
--
Paul

https://paulba.no/
Cliff Hallston
2021-04-07 20:36:22 UTC
Post by beda pietanza
a SR K frame measuring an object in K' frame gives a certain result
Remedial tutorial: There is no such thing as "a SR K frame". What you are presumably trying (and failing) to say is: Given two systems K and K' of inertial coordinates, moving relative to each other with mutual speed v, an object B at rest with spatial length L in terms of K' is moving with speed v and has spatial length L*sqrt(1-v^2/c^2) in terms of K. Likewise an object A at rest with spatial length L in terms of K is moving with speed v and has spatial length L*sqrt(1-v^2/c^2) in terms of K'.

Also, please note that in terms of any other system of inertial coordinates K", moving with speed u relative to K', the spatial length of B (at rest in K') is L*sqrt(1-u^2/c^2).
Post by beda pietanza
Then if the K frame jumps to another speed...
No, K is a specified system of inertial coordinates, it does not "jump". What you are trying (and failing) to say is: Suppose object A is initially at rest in K and is then accelerated to a different state of motion so that it comes to rest in another system of inertial coordinates K" that is moving with speed u in terms of K'.
Post by beda pietanza
while the frame K' and its object [B] stays still. The new measurement, of the same still
object in K', changes
What you are trying to say is: The spatial length of B in terms of K" is L*sqrt(1-u^2/c^2). This was always the length of B in terms of K". Of course, the length of B in terms of K is still L*sqrt(1-v^2/c^2), and the length of B in terms of K' is L. None of these has changed. All that has changed is that A was initially at rest in terms of K and now it has been accelerated so it is at rest in terms of K".
Post by beda pietanza
What the frames K are measuring...
There are not multiple frames K, there are three inertial coordinate systems under discussion, K, K', and K". Also, please note that frames do not measure things; rather, things have spatial and temporal relations in terms of specified systems of coordinates.
Post by beda pietanza
are the effects of the changes that took place only in the K frame due to the
changing of its absolute speed
The inertial coordinate system K does not change, nor do any of the others. These coordinate systems are related by certain linear transformations because they are in relative motion. An object has infinitely many speeds at any given instant, depending on the system of coordinates. The spatial extent of B does not change in terms of any inertial coordinate system when we accelerate A. This has been explained to you before. Have you forgotten?
Post by beda pietanza
that has logically affected the rulers and the clocks of K frame itself,
No, K is an inertial coordinate system, and all the rulers and clocks at rest in K do not change merely because we have accelerated object A. The only thing that changes by accelerating object A is that object A's state of motion changes, and it reaches an equilibrium configuration when it comes to rest in K" that is congruent to its configuration in terms of K when it was at rest in K.
Post by beda pietanza
so when SR claims to do a measurement...
Special relativity is a theory, it does not claim to do measurements. Special relativity asserts that inertial coordinate systems, such as K, K', and K", are related by Lorentz transformations.
Post by beda pietanza
what is really measured is not a given object solely, but the measuring tool is
measured with the object measured in an inextricable mixture.
What you are trying (and failing) to say is that objects are described in terms of inertial coordinate systems, and those systems are related by Lorentz transformations. YTour phrase "measure solely" makes no sense, like one hand clapping. Understand?
beda pietanza
2021-04-08 17:24:36 UTC
Post by beda pietanza
a SR K frame measuring an object in K' frame gives a certain result
Remedial tutorial: There is no such thing as "a SR K frame". What you are presumably trying (and failing) to say is: Given two systems K and K' of inertial coordinates, moving relative to each other with mutual speed v, an object B at rest with spatial length L in terms of K' is moving with speed v and has spatial length L*sqrt(1-v^2/c^2) in terms of K. Likewise an object A at rest with spatial length L in terms of K is moving with speed v and has spatial length L*sqrt(1-v^2/c^2) in terms of K'.
Also, please note that in terms of any other system of inertial coordinates K", moving with speed u relative to K', the spatial length of B (at rest in K') is L*sqrt(1-u^2/c^2).
Post by beda pietanza
Then if the K frame jumps to another speed...
No, K is a specified system of inertial coordinates, it does not "jump". What you are trying (and failing) to say is: Suppose object A is initially at rest in K and is then accelerated to a different state of motion so that it comes to rest in another system of inertial coordinates K" that is moving with speed u in terms of K'.
Post by beda pietanza
while the frame K' and its object [B] stays still. The new measurement, of the same still
object in K', changes
What you are trying to say is: The spatial length of B in terms of K" is L*sqrt(1-u^2/c^2). This was always the length of B in terms of K". Of course, the length of B in terms of K is still L*sqrt(1-v^2/c^2), and the length of B in terms of K' is L. None of these has changed. All that has changed is that A was initially at rest in terms of K and now it has been accelerated so it is at rest in terms of K".
Post by beda pietanza
What the frames K are measuring...
There are not multiple frames K, there are three inertial coordinate systems under discussion, K, K', and K". Also, please note that frames do not measure things; rather, things have spatial and temporal relations in terms of specified systems of coordinates.
Post by beda pietanza
are the effects of the changes that took place only in the K frame due to the
changing of its absolute speed
The inertial coordinate system K does not change, nor do any of the others. These coordinate systems are related by certain linear transformations because they are in relative motion. An object has infinitely many speeds at any given instant, depending on the system of coordinates. The spatial extent of B does not change in terms of any inertial coordinate system when we accelerate A. This has been explained to you before. Have you forgotten?
Post by beda pietanza
that has logically affected the rulers and the clocks of K frame itself,
No, K is an inertial coordinate system, and all the rulers and clocks at rest in K do not change merely because we have accelerated object A. The only thing that changes by accelerating object A is that object A's state of motion changes, and it reaches an equilibrium configuration when it comes to rest in K" that is congruent to its configuration in terms of K when it was at rest in K.
Post by beda pietanza
so when SR claims to do a measurement...
Special relativity is a theory, it does not claim to do measurements. Special relativity asserts that inertial coordinate systems, such as K, K', and K", are related by Lorentz transformations.
Post by beda pietanza
what is really measured is not a given object solely, but the measuring tool is
measured with the object measured in an inextricable mixture.
What you are trying (and failing) to say is that objects are described in terms of inertial coordinate systems, and those systems are related by Lorentz transformations. YTour phrase "measure solely" makes no sense, like one hand clapping. Understand?
beda

I don't give to SR frames any properties that you assign them, to me they just represent the positions
of objects in space and time nothing else associated to that.

to understand what is the working (math) mechanism of a SR frame I go like this:

in a moving SR frame the ruler is contracted to sqrt(1-v^2); v is the speed of the SR frame
the clocks ( having the time rate of sqrt(1-v^2), at two ends of the ruler, are offset so, that the the far end clock
(in the direction of the movement) is retarded of a value of v (in terms of local time of the SR frames) due to the Esynchro

with this setting any SR frame can see any other SR frame as having the ruler (whose real value is sqrt(1-v''^2)
as having the value of sqrt(1-vrel^2); where vrel is given by the formula
vrel= (v'+v'')/(1+(v'*v'')); v' and v'' are the absolute speeds of the two frames.

all the math story of SR is above, all the physical properties, associated to the SR math model, are illusory.

please don't loose time to repeat your mantras, beside the perfect math, for macroscopic objects, high speeds larger than a very
low fraction of c are forbitten, so your SR is only a math model, just that.

(physicists have to work somehow)

cheers
beda
Cliff Hallston
2021-04-08 18:30:13 UTC
Post by beda pietanza
I don't give to SR frames any properties that you assign them, to me they just represent
the positions of objects in space and time nothing else associated to that.
*Of course* they just represent the positions of objects in space and time. If you think anything I have said denies is inconsistent with this, then you have not read correctly. What you fail to grasp is that there is a unique class of coordinate systems in terms of which the laws of physics take their simplest homogeneous and isotropic form. This is shown by the fact that the equations of physics in terms of our lab frame are Lorentz invariant. That's all there is to it. You ought to be able to understand this.
Post by beda pietanza
In a moving SR frame...
Stop. You have not defined "moving" and you have not defined "SR frame". Here is the sane grown-up way of saying what you might be trying to say: "In terms of a system of inertial coordinates K that is moving at speed v in terms of the local inertial coordinates K_CMBR in which the CMBR is isotropic..."
Post by beda pietanza
... the ruler is contracted to sqrt(1-v^2); v is the speed of the SR frame
Be careful with the active phrase "is contracted to". The correct statement would be "A ruler at rest with length L in terms of K is moving at speed v and has length L*sqrt(1-v^2/c^2) in terms of K_CMBR". That is a true statement, and of course it is perfectly consistent with special relativity. In fact, special relativity is what has informed us of this fact. And it is equally true that a ruler at rest with length L in terms of K_CMBR is moving at speed v and has length L*sqrt(1-v^2/c^2) in terms of K. Agreed?
Post by beda pietanza
The clocks, having the time rate of sqrt(1-v^2), at two ends of the ruler...
You can't keep behaving like a spoiled self-indulgent child. You need to specify what you mean. Ideal clocks at rest in K advance at a rate dtau/dt = sqrt(1-v^2/c^2) where t is the time coordinate of K_CMBR. Likewise ideal clocks at rest in K_CMBR advance at a rate dtau/dt' = sqrt(1-v^2/c^2) where t' is the time coordinate of K. Agreed?
Post by beda pietanza
are offset...
Not necessarily. The synchronization of those clocks has not been specified. If you want those clocks to indicate time in such a way that the laws of physics are homogeneous and isotropic in terms of K, then they must be inertially synchronized by an isotropic process in terms of K.
Post by beda pietanza
With this setting any SR frame can see any other SR frame as having the ruler [length]
(whose real value is sqrt(1-v^2)
Correction: A ruler at rest in K has length L, then it has spatial length L*sqrt(1-v^2/c^2) in terms of any other system of inertial coordinates in which it is moving at speed u. You have not defined the word "real", but what your little brain means is "in terms of K_CMBR". In other words, what you call "v" is v_rel between K and K_CMBR.
Post by beda pietanza
as having the value of sqrt(1-vrel^2); where vrel is given by the formula
vrel= (v'+v'')/(1+(v'*v'')); v' and v'' are the absolute speeds of the two frames.
Your brain malfunctioned again. You typed "absolute speeds" when you meant "speeds in terms of K_CMBR". With that correction, yes, for any two frames, such as K and K_CMBR, the effects of length contraction and time dilation and skew of simultaneity depend on vrel between those two frames. There is nothing special about K_CMBR in this regard. Hence all your talk about absoluteness is sheer nonsense. Consider 20 inertial coordinate systems, K1, K2, ..., K20. One of those is K_CMBR, but I'm not telling you which one. Each pair of those systems is related by Lorentz transformation with parameter vrel(i,j) between the two systems Ki and Kj. This is true regardless of which of those systems in the CMBR system. Understand?
beda pietanza
2021-04-09 10:51:00 UTC
Post by beda pietanza
I don't give to SR frames any properties that you assign them, to me they just represent
the positions of objects in space and time nothing else associated to that.
*Of course* they just represent the positions of objects in space and time. If you think anything I have said denies is inconsistent with this, then you have not read correctly. What you fail to grasp is that there is a unique class of coordinate systems in terms of which the laws of physics take their simplest homogeneous and isotropic form. This is shown by the fact that the equations of physics in terms of our lab frame are Lorentz invariant. That's all there is to it. You ought to be able to understand this.
Post by beda pietanza
In a moving SR frame...
Stop. You have not defined "moving" and you have not defined "SR frame". Here is the sane grown-up way of saying what you might be trying to say: "In terms of a system of inertial coordinates K that is moving at speed v in terms of the local inertial coordinates K_CMBR in which the CMBR is isotropic..."
Post by beda pietanza
... the ruler is contracted to sqrt(1-v^2); v is the speed of the SR frame
Be careful with the active phrase "is contracted to". The correct statement would be "A ruler at rest with length L in terms of K is moving at speed v and has length L*sqrt(1-v^2/c^2) in terms of K_CMBR". That is a true statement, and of course it is perfectly consistent with special relativity. In fact, special relativity is what has informed us of this fact. And it is equally true that a ruler at rest with length L in terms of K_CMBR is moving at speed v and has length L*sqrt(1-v^2/c^2) in terms of K. Agreed?
Post by beda pietanza
The clocks, having the time rate of sqrt(1-v^2), at two ends of the ruler...
You can't keep behaving like a spoiled self-indulgent child. You need to specify what you mean. Ideal clocks at rest in K advance at a rate dtau/dt = sqrt(1-v^2/c^2) where t is the time coordinate of K_CMBR. Likewise ideal clocks at rest in K_CMBR advance at a rate dtau/dt' = sqrt(1-v^2/c^2) where t' is the time coordinate of K. Agreed?
Post by beda pietanza
are offset...
Not necessarily. The synchronization of those clocks has not been specified. If you want those clocks to indicate time in such a way that the laws of physics are homogeneous and isotropic in terms of K, then they must be inertially synchronized by an isotropic process in terms of K.
Post by beda pietanza
With this setting any SR frame can see any other SR frame as having the ruler [length]
(whose real value is sqrt(1-v^2)
Correction: A ruler at rest in K has length L, then it has spatial length L*sqrt(1-v^2/c^2) in terms of any other system of inertial coordinates in which it is moving at speed u. You have not defined the word "real", but what your little brain means is "in terms of K_CMBR". In other words, what you call "v" is v_rel between K and K_CMBR.
Post by beda pietanza
as having the value of sqrt(1-vrel^2); where vrel is given by the formula
vrel= (v'+v'')/(1+(v'*v'')); v' and v'' are the absolute speeds of the two frames.
Your brain malfunctioned again. You typed "absolute speeds" when you meant "speeds in terms of K_CMBR". With that correction, yes, for any two frames, such as K and K_CMBR, the effects of length contraction and time dilation and skew of simultaneity depend on vrel between those two frames. There is nothing special about K_CMBR in this regard. Hence all your talk about absoluteness is sheer nonsense. Consider 20 inertial coordinate systems, K1, K2, ..., K20. One of those is K_CMBR, but I'm not telling you which one. Each pair of those systems is related by Lorentz transformation with parameter vrel(i,j) between the two systems Ki and Kj. This is true regardless of which of those systems in the CMBR system. Understand?
beda
source at rest versus the CMBR has the property of sending a light isotropically in space speed wise and frequency wise
a receiver at rest versus the CMBR has the property of receiving light isotropically without changing the arriving frequencies
versus a point at rest in the CMBR there is a natural symmetry between two objects moving in opposite directions at same speeds
all these properties are correlated to the a CMBR frame whose clocks are absolutely synchronized.

you claim that these properties are shared by all SR frames, that is not true, because in all the frames (different from the CMBR frame)
the clocks are offset differently in each frame, (with he same procedure) but the result is not an absolute synchrony.

if you deny this you are fooling yourself

cheers
beda
Cliff Hallston
2021-04-09 14:35:09 UTC
Source at rest versus the CMBR has the property of sending a light isotropically...
Grown-up statement: The laws of physics (including but not limited to the laws governing the propagation of light) are isotropic and homogeneous in terms of inertial coordinates in which the CMBR is isotropic (not coincidentally!). That is a true statement, but your mistake is failing to realize that the laws of physics are also homogeneous and isotropic in terms of *every other* local system of inertial coordinates as well. This is established conclusively by tests of Lorentz invariance to all speeds.
you claim that these properties are shared by all SR frames, that is not true,
because in all the frames (different from the CMBR frame) the clocks are offset
differently in each frame, (with he same procedure)...
Your statement is self-contradictory and an obvious non-sequitur. Yes, the same procedure for establishing the coordinates is used in every frame, and yes this results in different systems of coordinates for the different frames. They differ both in their space coordinates and their time coordinates. The relationships between these systems are Lorentz transformations, which are perfectly symmetrical and reciprocal. The laws of physics take exactly the same homogeneous and isotropic form in terms of each of these systems, regardless of whether the system is at rest in terms of the frame in which the CMBR is isotropic. This is Lorentz invariance.

Again, you are free to hypocritically assign the label "absolute" to the frame in which your left foot (or the CMBR isotropy or flying pink elephants or an undetectable ether) is at rest, but that is pointless, and it does not contradict local Lorentz invariance.
beda pietanza
2021-04-09 21:30:57 UTC
Source at rest versus the CMBR has the property of sending a light isotropically...
Grown-up statement: The laws of physics (including but not limited to the laws governing the propagation of light) are isotropic and homogeneous in terms of inertial coordinates in which the CMBR is isotropic (not coincidentally!). That is a true statement, but your mistake is failing to realize that the laws of physics are also homogeneous and isotropic in terms of *every other* local system of inertial coordinates as well. This is established conclusively by tests of Lorentz invariance to all speeds.
you claim that these properties are shared by all SR frames, that is not true,
because in all the frames (different from the CMBR frame) the clocks are offset
differently in each frame, (with he same procedure)...
Your statement is self-contradictory and an obvious non-sequitur. Yes, the same procedure for establishing the coordinates is used in every frame, and yes this results in different systems of coordinates for the different frames. They differ both in their space coordinates and their time coordinates. The relationships between these systems are Lorentz transformations, which are perfectly symmetrical and reciprocal. The laws of physics take exactly the same homogeneous and isotropic form in terms of each of these systems, regardless of whether the system is at rest in terms of the frame in which the CMBR is isotropic. This is Lorentz invariance.
Again, you are free to hypocritically assign the label "absolute" to the frame in which your left foot (or the CMBR isotropy or flying pink elephants or an undetectable ether) is at rest, but that is pointless, and it does not contradict local Lorentz invariance.
beda
the same Esynchro procedure leads to a peculiarity on a frame at rest in the CMBR; the clocks of that frame are
absolutely synchronized, in all the other frames they are not

no macroscopic bodies traveling at high speeds, no Lorentz invariance at any speed.

I understand it is hard to accept it for you, but that is the reality of nature
your Lorentz invariance is pure math, the real Lorentz contraction and the Esynchro gives you an illusion of invariance at low speeds,
you extend that illusion at high speeds, you must cope with the fact that in not so,

if we cannot make an experiment with macroscopic bodies at very high speeds, this implies logically that the
Lorentz invariance is illusory.

Remains the humble Lorentz contraction (to be taken with prudence) that parrots something alike
with much little pretences

cheers
beda
Cliff Hallston
2021-04-10 00:15:49 UTC
The same Esynchro procedure leads to a peculiarity on a frame at rest in the CMBR;
the clocks of that frame are absolutely synchronized, in all the other frames they are not
But the absurd circularity of your statement is self-evident. You simply *define* the phrase "absolute synchronized" to mean "synchronized in terms of the inertial coordinates in which the CMBR is isotropic". With that definition, your assertion is an empty tautology. Physics is not about empty tautologies, nor is rational adult thinking. The point is that the exact same procedure for establishing inertial coordinates in any frame leads to coordinate systems that are perfectly symmetrical, and the relationships between them are perfectly reciprocal. This remarkable and immensely useful symmetry of nature is called the principle of relativity and Lorentz invariance.
no macroscopic bodies traveling at high speeds, no Lorentz invariance at any speed.
Your brain malfunctioned again, because you passive-aggressively referred to "speed" without specifying your coordinates. What you meant to say is that in our universe most macroscopic bodies in any particular region have speeds in terms of the local CMBR isotropic frame that are fairly small fractions of the speed of light, which is really just another way of saying that matter is distributed in the universe on the largest scale in a way that is roughly homogeneous and isotropic in terms of the same coordinates in which the CMBR is isotropic. This is hardly surprising, and it certainly does not contradict the manifest local Lorentz invariance of all physical laws.
If we cannot make an experiment with macroscopic bodies at very high speeds,
this implies logically that the Lorentz invariance is illusory.
The fallacies of your statement have already been explained. Lorentz invariance does not entail any claim that it should be easy to change the state of a large mass from 0 to 0.9999c in terms of any given system of inertial coordinates. To the contrary, special relativity implies that the amount of work required to accelerate an object of mass m to speed v is mc^2[1/sqrt(1-v^2/c^2) - 1], which approaches infinity as v approaches c. This does not represent a violation of special relativity, it is a *confirmation* of special relativity. Now do you understand?
Gary Harnagel
2021-04-10 04:41:17 UTC
Post by beda pietanza
the same Esynchro procedure leads to a peculiarity on a frame at rest in the CMBR; the clocks of that frame are
absolutely synchronized, in all the other frames they are not
"The clocks of that frame"? What does THAT mean? All clocks are in ALL frames.
Post by beda pietanza
no macroscopic bodies traveling at high speeds, no Lorentz invariance at any speed.
Define "macroscopic bodies." "Macroscopic" is a relative term. Compared to what?
Post by beda pietanza
I understand it is hard to accept it for you, but that is the reality of nature
Assertion without evidence.
Post by beda pietanza
your Lorentz invariance is pure math,
Funny. Einstein derived the Lorentz transform based on real-world phenomena. The speed
of light under various conditions supports his derivation.
Post by beda pietanza
the real Lorentz contraction and the Esynchro gives you an illusion of invariance at low speeds,
you extend that illusion at high speeds, you must cope with the fact that in not so,
This is another assertion without evidence. Actually, the Galilean transform gives the illusion
of invariance at low speeds, but it fails at high speeds. Today, however, we have instruments
that detect the failure of the GT at LOW speeds:

https://www.nist.gov/news-events/news/2010/09/nist-pair-aluminum-atomic-clocks-reveal-einsteins-relativity-personal-scale
Post by beda pietanza
if we cannot make an experiment with macroscopic bodies at very high speeds, this implies logically that the
Lorentz invariance is illusory.
Non sequitur. No logic there.
Post by beda pietanza
Remains the humble Lorentz contraction (to be taken with prudence) that parrots something alike
with much little pretences
cheers
beda
You seem to be with one enwrapped in pretense.
Sylvia Else
2021-04-10 09:55:02 UTC
Post by beda pietanza
a SR K frame measuring an object in K' frame gives a certain result
then if the K frame jumps to another speed while the frame K' and its object
stays still
the new measurement, of the same still object in K', changes
what the frames K are measuring are the effects of the changes that took place only in the K frame due to the changing of its absolute speed
that has logically affected the rulers and the clocks of K frame itself,
so when SR claims to do a measurement, what is really measured is not a given object solely , but the measuring tool is measured with the object measured in an inextricable mixture.
cheers
beda
I don't know why you feel the need to make it so complicated.

When I do a measurement of length, I use a metre rule that is not moving
relative to me. When I do a measurement of time, I use a clock that is
not moving relative to me.

I don't engage in all sorts of needless mental or mathematical contortions.

Sylvia.
Maciej Wozniak
2021-04-10 10:20:43 UTC
Post by Sylvia Else
Post by beda pietanza
a SR K frame measuring an object in K' frame gives a certain result
then if the K frame jumps to another speed while the frame K' and its object
stays still
the new measurement, of the same still object in K', changes
what the frames K are measuring are the effects of the changes that took place only in the K frame due to the changing of its absolute speed
that has logically affected the rulers and the clocks of K frame itself,
so when SR claims to do a measurement, what is really measured is not a given object solely , but the measuring tool is measured with the object measured in an inextricable mixture.
cheers
beda
I don't know why you feel the need to make it so complicated.
When I do a measurement of length, I use a metre rule that is not moving
relative to me.
And what is the longest metre rule You've seen, lady?
Post by Sylvia Else
When I do a measurement of time, I use a clock that is
not moving relative to me.
So, when You're riding a bike, the clock on
a church tower is ignored by You? Really?