Discussion:
[SR] Don't panic
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Python
2022-01-08 14:45:46 UTC
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A student takes a science exam, and a question is about the theory of
relativity.
Unfortunately, he forgot about Lorentz's transformations.
No problem, he can derive them easily from first principles.
He is then asked to give the coordinates of a space rocket by changing
the frame of reference.
Do not panic!
He knows that y'= y and z'= z.
x'= [x+sqrt(x²+y²+z²).v/c]/sqrt(1-v²/c²)
Why would he remind such an atrocious bullshit which is not
linear (so in contradiction with Newton laws of motion) while
the correct equation is far simpler?
With that, he can get by.
For To', just set To'=sqrt(x'²+y²+z²)/c
A real student in physics wouldn't use a silly equation in
contradiction with Newton's laws.
He is saved.
He is dead in the water, he failed miserably. He won't pass.

Maybe he should consider a career in medicine, standard there
are clearly lower.
Maciej Wozniak
2022-01-08 15:17:57 UTC
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Post by Python
A student takes a science exam, and a question is about the theory of
relativity.
Unfortunately, he forgot about Lorentz's transformations.
No problem, he can derive them easily from first principles.
And, of course, in the meantime in the real world forbidden
by your moronic religion TAI, using sane methods of
synchronization - keep measuring t'=t, just like all serious
clocks always did.
Python
2022-01-08 16:30:13 UTC
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Post by Python
A student takes a science exam, and a question is about the theory of
relativity.
Unfortunately, he forgot about Lorentz's transformations.
No problem, he can derive them easily from first principles.
And, of course, in the meantime in the real world ...
In the real world you are a wannabe "information engineer" with
0 knowledge in physics, maths or engineering yelling on Usenet
without convincing anyone with your idiotic rant.
Have you already learnt the difference between a square
root of 7 and sqrt(7)?
Still fighting with the basics, Woz? Sigh...
Buddy Good
2022-01-08 15:37:08 UTC
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A student takes a science exam, and a question is about the theory of
relativity. Unfortunately, he forgot about Lorentz's transformations.
He is then asked to give the coordinates of a space rocket by changing
the frame of reference. Do not panic!
BEYOND RETARDED - Hospitalized Canadian Patient Diagnosed With 'Climate
Change' 🤣🤣 https://www.bitchute.com/video/f0njKXIKXwpb/
Python
2022-01-08 18:56:32 UTC
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x'= [x+sqrt(x²+y²+z²).v/c]/sqrt(1-v²/c²)
is not linear
It's bijective, and that's enough for me.
Could you prove it is even bijective ?
I am a great humble man, you know.
No. You are an arrogant idiot.
I'm not looking for numbers in silver, or mathematical signs in gold.
This is not a proper English sentence, it is meaningless.
The coherence of nature is enough for me.
Newton's laws are part of this "coherence". Learn.
I leave scientific puritanism to others.
This is not "puritanism", it is sticking with overwhelming experimental
evidence.
Richard Hachel
2022-01-08 19:49:44 UTC
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Le 08/01/2022 à 19:56, Python a écrit :

Bon, Jean-Pierre.

Maintenant, tu arrêtes, tu n'es pas gentil du tout.

C'est mal.
Post by Python
x'= [x+sqrt(x²+y²+z²).v/c]/sqrt(1-v²/c²)
is not linear
It's bijective, and that's enough for me.
Could you prove it is even bijective ?
Not only is it bijective, but it is reciprocal if you change v to -v.

Exemple :

E(R)=(12,9,0) (E)R'=(40,9,0) v=0.8c

x'= [x+sqrt(x²+y²+z²).v/c]/sqrt(1-v²/c²)

x'= [12+sqrt(12²=9²+0²).0.8]/0.6 ---> x=40

Réciproque :

x= [x'-sqrt(x'²+y'²+z'²).v/c]/sqrt(1-v²/c²)

x= [40-sqrt(40²+9²+0²).0.8]/0.6 ---> x=12

Cette équation permet d'obtenir très facilement les coordonnées
spatiales et temporelles
dans R' d'un événement qui se passe dans R au moment du croisement O et
O'.

Evidemment, puisqu'on déclenche les montres à cet instant, on a t'=t=0.

Pour connaître l'heure de l'événement en temps local, il suffit de
poser
To=-d/c dans R, et To'=-d'/c dans R'.

Ici To=sqrt((12²+9²+0²)/c= -15 et To'=sqrt(40²+9²+0²)/c= -41

On obtient la même chose en passant par les TL(attention au signe de
To).

x'=(x-vTo)/sqrt(1-v²/c²)
y'=y
z'=z
To'=(To-xv/c²)/sqrt(1-v²/c²)

<http://news2.nemoweb.net/?DataID=***@jntp>
R.H.
--
"Mais ne nous y trompons pas. Il n'y a pas que de la violence
avec des armes. Il y a des situations de violence".
Abbé Pierre.
Python
2022-01-08 21:02:16 UTC
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...
Post by Richard Hachel
Post by Python
x'= [x+sqrt(x²+y²+z²).v/c]/sqrt(1-v²/c²)
is not linear
It's bijective, and that's enough for me.
Could you prove it is even bijective ?
Not only is it bijective, but it is reciprocal if you change v to -v.
You don't know what bijective mean? Really? So
why do you use the word?
mitchr...@gmail.com
2022-01-08 22:25:15 UTC
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Post by Python
...
Post by Richard Hachel
Post by Python
x'= [x+sqrt(x²+y²+z²).v/c]/sqrt(1-v²/c²)
is not linear
It's bijective, and that's enough for me.
Could you prove it is even bijective ?
Not only is it bijective, but it is reciprocal if you change v to -v.
You don't know what bijective mean? Really? So
why do you use the word?
What quantities are you using you moron?
Richard Hachel
2022-01-09 00:17:56 UTC
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Post by Python
...
Post by Richard Hachel
Post by Python
x'= [x+sqrt(x²+y²+z²).v/c]/sqrt(1-v²/c²)
is not linear
It's bijective, and that's enough for me.
Could you prove it is even bijective ?
Not only is it bijective, but it is reciprocal if you change v to -v.
You don't know what bijective mean? Really? So
why do you use the word?
Mais tu ne t'arrêtes donc jamais, toi?

R.H.
Jean-Michel Affoinez
2022-01-09 14:31:57 UTC
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Post by Python
...
Post by Richard Hachel
Post by Python
x'= [x+sqrt(x²+y²+z²).v/c]/sqrt(1-v²/c²)
is not linear
It's bijective, and that's enough for me.
Could you prove it is even bijective ?
Not only is it bijective, but it is reciprocal if you change v to -v.
You don't know what bijective mean? Really? So
why do you use the word?
Yes, is the question.
why this fool use that words.
Because : "Il est fou".
jma
Python
2022-01-09 15:12:07 UTC
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This is the best you can do when your crap is debuncked, Lengrand? Post
more nonsense using a sock puppet fake name?
So, is Euclidean geometry a valid geometry of our
universe? According to you and [snip profanity], of
course.
Still no answer, [snip profanity]?
You shouldn't ask questions that you don't understand what
they actually mean. You're only making a fool of yourself
and exposing what an idiot you are Maciej.
Richard Hachel
2022-01-09 15:23:50 UTC
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Post by Python
This is the best you can do when your crap is debuncked, Lengrand? Post
more nonsense using a sock puppet fake name?
So, is Euclidean geometry a valid geometry of our
universe? According to you and [snip profanity], of
course.
Still no answer, [snip profanity]?
You shouldn't ask questions that you don't understand what
they actually mean. You're only making a fool of yourself
and exposing what an idiot you are Maciej.
Ce qui m'étonnera toujours, c'est la connerie humaine, teintée de haine,
de violence, de censure,
d'appels à la délation et au "doute professionnel".

Cette saloperie humaine a existé.

On appelle cela la profondeur du serpent.

Des centaines de siècles attestent cela, couronnés par la sagesse des
sages.

Le serpent croit qu'il a une grosse bite (alors qu'il n'a même pas de
pattes).

Il se donne des noms, le serpent.

Il est fou.

On lui demande d'arrêter.

On lui dit : Jean-Pierre, maintenant, tu arrêtes. STOP!

Il continue.

Il est fou.

R.H.
Odd Bodkin
2022-01-09 19:43:38 UTC
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This is the best you can do when your crap is debuncked, Lengrand? Post
more nonsense using a sock puppet fake name?
So, is Euclidean geometry a valid geometry of our
universe?
Depends on the application. As any engineer should know.
According to you and your idiot gurus, of
course.
Still no answer, poor stinker?
--
Odd Bodkin -- maker of fine toys, tools, tables
Maciej Wozniak
2022-01-09 19:54:50 UTC
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Post by Odd Bodkin
This is the best you can do when your crap is debuncked, Lengrand? Post
more nonsense using a sock puppet fake name?
So, is Euclidean geometry a valid geometry of our
universe?
Depends on the application. As any engineer should know.
It's not a common engineering knowledge, and the
concept is completely alien to a relativistic moron
mumbling about his imagined Laws of Nature.
Python
2022-01-09 19:59:46 UTC
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Post by Maciej Wozniak
Post by Odd Bodkin
This is the best you can do when your crap is debuncked, Lengrand? Post
more nonsense using a sock puppet fake name?
So, is Euclidean geometry a valid geometry of our
universe?
Depends on the application. As any engineer should know.
It's not a common engineering knowledge
How really? How would you know?

(actually it is)
Maciej Wozniak
2022-01-09 20:05:57 UTC
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Post by Python
Post by Maciej Wozniak
Post by Odd Bodkin
This is the best you can do when your crap is debuncked, Lengrand? Post
more nonsense using a sock puppet fake name?
So, is Euclidean geometry a valid geometry of our
universe?
Depends on the application. As any engineer should know.
It's not a common engineering knowledge
How really? How would you know?
I see you don't have it, poor stinker. So, is this your answer
now? It's valid or not, it depends? And how about

"Neither Einstein nor anyone ever pretended that SR (or Riemann's work)
make EG invalid. How can you be so illiterate???"
Was it a lie, poor stinker? Of course it was.
Python
2022-01-09 20:15:30 UTC
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Post by Maciej Wozniak
Post by Python
Post by Maciej Wozniak
Post by Odd Bodkin
This is the best you can do when your crap is debuncked, Lengrand? Post
more nonsense using a sock puppet fake name?
So, is Euclidean geometry a valid geometry of our
universe?
Depends on the application. As any engineer should know.
It's not a common engineering knowledge
How really? How would you know?
I see you don't have it, poor stinker. So, is this your answer
now? It's valid or not, it depends? And how about
"Neither Einstein nor anyone ever pretended that SR (or Riemann's work)
make EG invalid. How can you be so illiterate???"
Was it a lie, [snip profanity]? Of course it was.
Not a lie. It's a very simple obvious fact for anyone who knows
maths. Too bad you don't and too sad you'll never learn as you
are a stubborn ignorant ranting imbecile Maciej.
Odd Bodkin
2022-01-09 20:18:41 UTC
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Post by Maciej Wozniak
Post by Odd Bodkin
This is the best you can do when your crap is debuncked, Lengrand? Post
more nonsense using a sock puppet fake name?
So, is Euclidean geometry a valid geometry of our
universe?
Depends on the application. As any engineer should know.
It's not a common engineering knowledge,
It’s not common engineering knowledge that models, including the
mathematical infrastructure, depend on application?
Post by Maciej Wozniak
and the
concept is completely alien to a relativistic moron
mumbling about his imagined Laws of Nature.
Of course it’s not alien. There’s a reason why the Bohr model of the atom
is still taught, why p=mv is still taught, why Newtonian gravitation is
still taught. What’s wrong with you?
--
Odd Bodkin -- maker of fine toys, tools, tables
Maciej Wozniak
2022-01-09 20:36:01 UTC
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Post by Odd Bodkin
Post by Maciej Wozniak
Post by Odd Bodkin
This is the best you can do when your crap is debuncked, Lengrand? Post
more nonsense using a sock puppet fake name?
So, is Euclidean geometry a valid geometry of our
universe?
Depends on the application. As any engineer should know.
It's not a common engineering knowledge,
It’s not common engineering knowledge that models, including the
mathematical infrastructure, depend on application?
It's a common knowledge of information engineering.
It's not a common engineering knowledge.
Post by Odd Bodkin
Of course it’s not alien. There’s a reason why the Bohr model of the atom
And there is a reason why your bunch of idiots insist
that the real geometry of the universe is not euclidean.
You should have listened to Poincare. He was really smart
as for an amateur.
Michael Moroney
2022-01-09 20:21:45 UTC
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Post by Maciej Wozniak
Post by Odd Bodkin
This is the best you can do when your crap is debuncked, Lengrand? Post
more nonsense using a sock puppet fake name?
So, is Euclidean geometry a valid geometry of our
universe?
Depends on the application. As any engineer should know.
It's not a common engineering knowledge, and the
concept is completely alien to a relativistic moron
mumbling about his imagined Laws of Nature.
Is Euclidean geometry a valid geometry for surveying land on earth?
Why or why not?
Maciej Wozniak
2022-01-09 20:38:01 UTC
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Post by Michael Moroney
Post by Maciej Wozniak
Post by Odd Bodkin
This is the best you can do when your crap is debuncked, Lengrand? Post
more nonsense using a sock puppet fake name?
So, is Euclidean geometry a valid geometry of our
universe?
Depends on the application. As any engineer should know.
It's not a common engineering knowledge, and the
concept is completely alien to a relativistic moron
mumbling about his imagined Laws of Nature.
Is Euclidean geometry a valid geometry for surveying land on earth?
Sure.
Post by Michael Moroney
Why or why not?
Because it's been crafted with skill, wisdom and luck.
Unlike your Shit, having nothing but luck.
Michael Moroney
2022-01-09 21:00:59 UTC
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Post by Maciej Wozniak
Post by Michael Moroney
Post by Maciej Wozniak
Post by Odd Bodkin
This is the best you can do when your crap is debuncked, Lengrand? Post
more nonsense using a sock puppet fake name?
So, is Euclidean geometry a valid geometry of our
universe?
Depends on the application. As any engineer should know.
It's not a common engineering knowledge, and the
concept is completely alien to a relativistic moron
mumbling about his imagined Laws of Nature.
Is Euclidean geometry a valid geometry for surveying land on earth?
Sure.
Post by Michael Moroney
Why or why not?
Because it's been crafted with skill, wisdom and luck.
Unlike your Shit, having nothing but luck.
Then why do the US states of Colorado and Wyoming, legislatively defined
as rectangular, have northern borders which are shorter than their
southern borders, despite having 4 right angles and east and west
borders of the same length*?

(*) Legislative definition. Surveyors later place boundary stones along
the borders and made some mistakes so in real life they zigzag a little.
I'm not talking about that.
Maciej Wozniak
2022-01-09 21:22:25 UTC
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Post by Michael Moroney
Post by Maciej Wozniak
Post by Michael Moroney
Post by Maciej Wozniak
Post by Odd Bodkin
This is the best you can do when your crap is debuncked, Lengrand? Post
more nonsense using a sock puppet fake name?
So, is Euclidean geometry a valid geometry of our
universe?
Depends on the application. As any engineer should know.
It's not a common engineering knowledge, and the
concept is completely alien to a relativistic moron
mumbling about his imagined Laws of Nature.
Is Euclidean geometry a valid geometry for surveying land on earth?
Sure.
Post by Michael Moroney
Why or why not?
Because it's been crafted with skill, wisdom and luck.
Unlike your Shit, having nothing but luck.
Then why do the US states of Colorado and Wyoming, legislatively defined
as rectangular
They are not. From wiki: "United States Congress defined the boundaries of
the new Territory of Colorado exclusively by lines of latitude and longitude,"
and they surely knew Earth isn't flat. Unlike you, it seems.
Michael Moroney
2022-01-09 22:14:13 UTC
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Post by Maciej Wozniak
Post by Michael Moroney
Post by Maciej Wozniak
Post by Michael Moroney
Post by Maciej Wozniak
Post by Odd Bodkin
This is the best you can do when your crap is debuncked, Lengrand? Post
more nonsense using a sock puppet fake name?
So, is Euclidean geometry a valid geometry of our
universe?
Depends on the application. As any engineer should know.
It's not a common engineering knowledge, and the
concept is completely alien to a relativistic moron
mumbling about his imagined Laws of Nature.
Is Euclidean geometry a valid geometry for surveying land on earth?
Sure.
Post by Michael Moroney
Why or why not?
Because it's been crafted with skill, wisdom and luck.
Unlike your Shit, having nothing but luck.
Then why do the US states of Colorado and Wyoming, legislatively defined
as rectangular
They are not. From wiki: "United States Congress defined the boundaries of
the new Territory of Colorado exclusively by lines of latitude and longitude,"
and they surely knew Earth isn't flat. Unlike you, it seems.
Yet the states have four right angles for corners.
Now that you have read the congressional definition and suddenly
realized that the earth is not flat, do you change your answer to the
Post by Maciej Wozniak
Is Euclidean geometry a valid geometry for surveying land on earth?
Maciej Wozniak
2022-01-10 06:01:51 UTC
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Post by Michael Moroney
Post by Maciej Wozniak
Post by Michael Moroney
Post by Maciej Wozniak
Post by Michael Moroney
Post by Maciej Wozniak
Post by Odd Bodkin
This is the best you can do when your crap is debuncked, Lengrand? Post
more nonsense using a sock puppet fake name?
So, is Euclidean geometry a valid geometry of our
universe?
Depends on the application. As any engineer should know.
It's not a common engineering knowledge, and the
concept is completely alien to a relativistic moron
mumbling about his imagined Laws of Nature.
Is Euclidean geometry a valid geometry for surveying land on earth?
Sure.
Post by Michael Moroney
Why or why not?
Because it's been crafted with skill, wisdom and luck.
Unlike your Shit, having nothing but luck.
Then why do the US states of Colorado and Wyoming, legislatively defined
as rectangular
They are not. From wiki: "United States Congress defined the boundaries of
the new Territory of Colorado exclusively by lines of latitude and longitude,"
and they surely knew Earth isn't flat. Unlike you, it seems.
Yet the states have four right angles for corners.
Yet their borders aren't straight lines. Neither in Euclidean
geometry nor in Riemann's idiocy. They can no way be
considered rectangles (in Riemann's there is no such shape).
Post by Michael Moroney
Now that you have read the congressional definition and suddenly
realized that the earth is not flat, do you change your answer to the
Post by Maciej Wozniak
Is Euclidean geometry a valid geometry for surveying land on earth?
Of course. Euclidean geometry deals with a sphere with
ease.
Michael Moroney
2022-01-10 07:20:38 UTC
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Post by Maciej Wozniak
Post by Michael Moroney
Post by Maciej Wozniak
Post by Michael Moroney
Post by Maciej Wozniak
Post by Michael Moroney
Post by Maciej Wozniak
Post by Odd Bodkin
This is the best you can do when your crap is debuncked, Lengrand? Post
more nonsense using a sock puppet fake name?
So, is Euclidean geometry a valid geometry of our
universe?
Depends on the application. As any engineer should know.
It's not a common engineering knowledge, and the
concept is completely alien to a relativistic moron
mumbling about his imagined Laws of Nature.
Is Euclidean geometry a valid geometry for surveying land on earth?
Sure.
Post by Michael Moroney
Why or why not?
Because it's been crafted with skill, wisdom and luck.
Unlike your Shit, having nothing but luck.
Then why do the US states of Colorado and Wyoming, legislatively defined
as rectangular
They are not. From wiki: "United States Congress defined the boundaries of
the new Territory of Colorado exclusively by lines of latitude and longitude,"
and they surely knew Earth isn't flat. Unlike you, it seems.
Yet the states have four right angles for corners.
Yet their borders aren't straight lines. Neither in Euclidean
geometry nor in Riemann's idiocy.
But they must be straight lines. Congress set up four corners at the
crossings of lines of latitude and longitude, connected them with line
segments (remember, we're using Euclidean geometry because you said so)
and they must be line segments since we only define end points, we check
the corners and sure enough, they're 90 degrees. So according to
Euclidean geometry (used because you said so) we have defined a
rectangle. Yet despite the corners being right angles somehow two
parallel sides aren't the same length.

Also I have this map of the world. I lay it flat to ensure it really is
Euclidean geometry, and look at that! Greenland is as large as South
America! But if I look up the areas of Greenland and South America, I
get very different answers! How did that happen for this Euclidean world?
Maciej Wozniak
2022-01-10 07:28:44 UTC
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Post by Michael Moroney
Post by Michael Moroney
Post by Maciej Wozniak
Post by Michael Moroney
Post by Maciej Wozniak
Post by Michael Moroney
Post by Maciej Wozniak
Post by Odd Bodkin
This is the best you can do when your crap is debuncked, Lengrand? Post
more nonsense using a sock puppet fake name?
So, is Euclidean geometry a valid geometry of our
universe?
Depends on the application. As any engineer should know.
It's not a common engineering knowledge, and the
concept is completely alien to a relativistic moron
mumbling about his imagined Laws of Nature.
Is Euclidean geometry a valid geometry for surveying land on earth?
Sure.
Post by Michael Moroney
Why or why not?
Because it's been crafted with skill, wisdom and luck.
Unlike your Shit, having nothing but luck.
Then why do the US states of Colorado and Wyoming, legislatively defined
as rectangular
They are not. From wiki: "United States Congress defined the boundaries of
the new Territory of Colorado exclusively by lines of latitude and longitude,"
and they surely knew Earth isn't flat. Unlike you, it seems.
Yet the states have four right angles for corners.
Yet their borders aren't straight lines. Neither in Euclidean
geometry nor in Riemann's idiocy.
But they must be straight lines.
Neither Euclid, nor Riemann, nor any sane person
is sharing this opinion of yours, stupid Mike. I'm not
sharing it either.
Michael Moroney
2022-01-10 12:42:01 UTC
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Post by Maciej Wozniak
Post by Michael Moroney
Post by Michael Moroney
Post by Maciej Wozniak
Post by Michael Moroney
Post by Maciej Wozniak
Post by Michael Moroney
Post by Maciej Wozniak
Post by Odd Bodkin
This is the best you can do when your crap is debuncked, Lengrand? Post
more nonsense using a sock puppet fake name?
So, is Euclidean geometry a valid geometry of our
universe?
Depends on the application. As any engineer should know.
It's not a common engineering knowledge, and the
concept is completely alien to a relativistic moron
mumbling about his imagined Laws of Nature.
Is Euclidean geometry a valid geometry for surveying land on earth?
Sure.
Post by Michael Moroney
Why or why not?
Because it's been crafted with skill, wisdom and luck.
Unlike your Shit, having nothing but luck.
Then why do the US states of Colorado and Wyoming, legislatively defined
as rectangular
They are not. From wiki: "United States Congress defined the boundaries of
the new Territory of Colorado exclusively by lines of latitude and longitude,"
and they surely knew Earth isn't flat. Unlike you, it seems.
Yet the states have four right angles for corners.
Yet their borders aren't straight lines. Neither in Euclidean
geometry nor in Riemann's idiocy.
But they must be straight lines.
Neither Euclid, nor Riemann, nor any sane person
is sharing this opinion of yours, stupid Mike. I'm not
sharing it either.
I'm glad to see that you consider yourself outside the group of "any
sane person". It's a start.

I question your odd belief that Euclid himself wouldn't consider
segments connecting two points as described to be straight line
segments. What would he say about a figure with two parallel sides of
the same length and four right angles? What would he say if a map showed
Greenland to be the same size as South America yet they weren't close in
size at all, even considering he didn't know of either place?
Maciej Wozniak
2022-01-10 13:03:14 UTC
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Post by Michael Moroney
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Post by Michael Moroney
Post by Maciej Wozniak
Post by Michael Moroney
Post by Maciej Wozniak
Post by Odd Bodkin
This is the best you can do when your crap is debuncked, Lengrand? Post
more nonsense using a sock puppet fake name?
So, is Euclidean geometry a valid geometry of our
universe?
Depends on the application. As any engineer should know.
It's not a common engineering knowledge, and the
concept is completely alien to a relativistic moron
mumbling about his imagined Laws of Nature.
Is Euclidean geometry a valid geometry for surveying land on earth?
Sure.
Post by Michael Moroney
Why or why not?
Because it's been crafted with skill, wisdom and luck.
Unlike your Shit, having nothing but luck.
Then why do the US states of Colorado and Wyoming, legislatively defined
as rectangular
They are not. From wiki: "United States Congress defined the boundaries of
the new Territory of Colorado exclusively by lines of latitude and longitude,"
and they surely knew Earth isn't flat. Unlike you, it seems.
Yet the states have four right angles for corners.
Yet their borders aren't straight lines. Neither in Euclidean
geometry nor in Riemann's idiocy.
But they must be straight lines.
Neither Euclid, nor Riemann, nor any sane person
is sharing this opinion of yours, stupid Mike. I'm not
sharing it either.
I'm glad to see that you consider yourself outside the group of "any
sane person". It's a start.
I question your odd belief that Euclid himself wouldn't consider
segments connecting two points as described to be straight line
segments.
Well, you're free to question; you're an idiot, but that was
obvious before. Latitudes are not straight even in Riemann's
nonsense (except one); in Euclidean geometry meridians are
neither.
Post by Michael Moroney
What would he say if a map showed
Greenland to be the same size as South America
Stupid Mike, Mercator's map doesn''t scale the surfaces.
Michael Moroney
2022-01-10 13:51:09 UTC
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Post by Maciej Wozniak
Post by Odd Bodkin
This is the best you can do when your crap is debuncked, Lengrand? Post
more nonsense using a sock puppet fake name?
So, is Euclidean geometry a valid geometry of our
universe?
Depends on the application. As any engineer should know.
It's not a common engineering knowledge, and the
concept is completely alien to a relativistic moron
mumbling about his imagined Laws of Nature.
Is Euclidean geometry a valid geometry for surveying land on earth?
Sure.
Post by Michael Moroney
Why or why not?
Because it's been crafted with skill, wisdom and luck.
Unlike your Shit, having nothing but luck.
Then why do the US states of Colorado and Wyoming, legislatively defined
as rectangular
They are not. From wiki: "United States Congress defined the boundaries of
the new Territory of Colorado exclusively by lines of latitude and longitude,"
and they surely knew Earth isn't flat. Unlike you, it seems.
Yet the states have four right angles for corners.
Yet their borders aren't straight lines. Neither in Euclidean
geometry nor in Riemann's idiocy.
But they must be straight lines.
Neither Euclid, nor Riemann, nor any sane person
is sharing this opinion of yours, stupid Mike. I'm not
sharing it either.
I'm glad to see that you consider yourself outside the group of "any
sane person". It's a start.
I question your odd belief that Euclid himself wouldn't consider
segments connecting two points as described to be straight line
segments.
Well, you're free to question; you're an idiot, but that was
obvious before. Latitudes are not straight even in Riemann's
nonsense (except one); in Euclidean geometry meridians are
neither.
But you said Euclidean geometry could be used for surveying. Euclidean
geometry would have the north and south borders of Colorado be the same
lengths as each other.
Post by Maciej Wozniak
Post by Michael Moroney
What would he say if a map showed
Greenland to be the same size as South America
Stupid Mike, Mercator's map doesn''t scale the surfaces.
Exactly. It tries to map a non-Euclidean space to a flat Euclidean map.
Distorting Greenland especially.

Oh look at this! I can start at the North Pole, move along the Prime
Meridian to the equator off the west coast of Africa, along the equator
to a point near the Galapagos Islands and then north to the North Pole
again and meet the Prime Meridian at a 90 degree angle. Look what I
have! A triangle with three right angles! And you say that's Euclidean!
Maciej Wozniak
2022-01-10 13:59:32 UTC
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Post by Michael Moroney
Post by Maciej Wozniak
Post by Michael Moroney
Post by Maciej Wozniak
Post by Michael Moroney
Post by Michael Moroney
Post by Maciej Wozniak
Post by Michael Moroney
Post by Maciej Wozniak
Post by Michael Moroney
Post by Maciej Wozniak
Post by Odd Bodkin
This is the best you can do when your crap is debuncked, Lengrand? Post
more nonsense using a sock puppet fake name?
So, is Euclidean geometry a valid geometry of our
universe?
Depends on the application. As any engineer should know.
It's not a common engineering knowledge, and the
concept is completely alien to a relativistic moron
mumbling about his imagined Laws of Nature.
Is Euclidean geometry a valid geometry for surveying land on earth?
Sure.
Post by Michael Moroney
Why or why not?
Because it's been crafted with skill, wisdom and luck.
Unlike your Shit, having nothing but luck.
Then why do the US states of Colorado and Wyoming, legislatively defined
as rectangular
They are not. From wiki: "United States Congress defined the boundaries of
the new Territory of Colorado exclusively by lines of latitude and longitude,"
and they surely knew Earth isn't flat. Unlike you, it seems.
Yet the states have four right angles for corners.
Yet their borders aren't straight lines. Neither in Euclidean
geometry nor in Riemann's idiocy.
But they must be straight lines.
Neither Euclid, nor Riemann, nor any sane person
is sharing this opinion of yours, stupid Mike. I'm not
sharing it either.
I'm glad to see that you consider yourself outside the group of "any
sane person". It's a start.
I question your odd belief that Euclid himself wouldn't consider
segments connecting two points as described to be straight line
segments.
Well, you're free to question; you're an idiot, but that was
obvious before. Latitudes are not straight even in Riemann's
nonsense (except one); in Euclidean geometry meridians are
neither.
But you said Euclidean geometry could be used for surveying.
Sure.
Post by Michael Moroney
Euclidean
geometry would have the north and south borders of Colorado be the same
lengths as each other.
No, stupid Mike, it wouldn't. No surprise you don't know -
you're an idiot after all.
Post by Michael Moroney
Post by Maciej Wozniak
Stupid Mike, Mercator's map doesn''t scale the surfaces.
Exactly. It tries to map a non-Euclidean space to a flat Euclidean map.
No, stupid Mike, it's mapping an Euclidean sphere (rather - a part of it)
on an Euclidean plane.
Post by Michael Moroney
Oh look at this! I can start at the North Pole, move along the Prime
Meridian to the equator off the west coast of Africa, along the equator
to a point near the Galapagos Islands and then north to the North Pole
again and meet the Prime Meridian at a 90 degree angle. Look what I
have! A triangle with three right angles!
No, stupid Mike, you don't have any triangle. You have a spherical
triangle, which no more is a triangle than a tasmanian devil is a
devil. If you knew the definition of a triangle... but who would expect
such an idiot to know any definitions.
Odd Bodkin
2022-01-10 14:05:23 UTC
Reply
Permalink
Post by Michael Moroney
Post by Maciej Wozniak
Post by Michael Moroney
Post by Michael Moroney
Post by Maciej Wozniak
Post by Michael Moroney
Post by Maciej Wozniak
Post by Michael Moroney
Post by Maciej Wozniak
Post by Odd Bodkin
This is the best you can do when your crap is debuncked, Lengrand? Post
more nonsense using a sock puppet fake name?
So, is Euclidean geometry a valid geometry of our
universe?
Depends on the application. As any engineer should know.
It's not a common engineering knowledge, and the
concept is completely alien to a relativistic moron
mumbling about his imagined Laws of Nature.
Is Euclidean geometry a valid geometry for surveying land on earth?
Sure.
Post by Michael Moroney
Why or why not?
Because it's been crafted with skill, wisdom and luck.
Unlike your Shit, having nothing but luck.
Then why do the US states of Colorado and Wyoming, legislatively defined
as rectangular
They are not. From wiki: "United States Congress defined the boundaries of
the new Territory of Colorado exclusively by lines of latitude and longitude,"
and they surely knew Earth isn't flat. Unlike you, it seems.
Yet the states have four right angles for corners.
Yet their borders aren't straight lines. Neither in Euclidean
geometry nor in Riemann's idiocy.
But they must be straight lines.
Neither Euclid, nor Riemann, nor any sane person
is sharing this opinion of yours, stupid Mike. I'm not
sharing it either.
I'm glad to see that you consider yourself outside the group of "any
sane person". It's a start.
I question your odd belief that Euclid himself wouldn't consider
segments connecting two points as described to be straight line
segments. What would he say about a figure with two parallel sides of
the same length and four right angles? What would he say if a map showed
Greenland to be the same size as South America yet they weren't close in
size at all, even considering he didn't know of either place?
In a 2D surface, a curve is straight in a Euclidean sense if, between the 2
degrees of freedom available, only one of them is exploited. For example,
on the earth, a path along a line of longitude is straight on the surface
because there is no excursion in the direction perpendicular to that line.
Another way of saying this is that, parameterizing the curve by parameter
s, and embarking from s=0 such that ds/dx>0 and ds/dy=0, then d2s/dy2=0,
meaning that there is no curvature of the path into the other degree of
freedom.

Now, Maciej is apparently having difficulty understanding that the surface
of a sphere is in fact a 2D space. His common sense tells him, no, no, no,
it obviously occupies 3D. No, Maciej, the surface does not because there
would be 3 degrees of freedom and hence it would be a 3D space, not a 2D
space, and it’s expanse would be a volume and not a surface area.
--
Odd Bodkin -- maker of fine toys, tools, tables
Odd Bodkin
2022-01-10 14:32:28 UTC
Reply
Permalink
Post by Odd Bodkin
Post by Michael Moroney
Post by Maciej Wozniak
Post by Michael Moroney
Post by Michael Moroney
Post by Maciej Wozniak
Post by Michael Moroney
Post by Maciej Wozniak
Post by Michael Moroney
Post by Maciej Wozniak
Post by Odd Bodkin
This is the best you can do when your crap is debuncked, Lengrand? Post
more nonsense using a sock puppet fake name?
So, is Euclidean geometry a valid geometry of our
universe?
Depends on the application. As any engineer should know.
It's not a common engineering knowledge, and the
concept is completely alien to a relativistic moron
mumbling about his imagined Laws of Nature.
Is Euclidean geometry a valid geometry for surveying land on earth?
Sure.
Post by Michael Moroney
Why or why not?
Because it's been crafted with skill, wisdom and luck.
Unlike your Shit, having nothing but luck.
Then why do the US states of Colorado and Wyoming, legislatively defined
as rectangular
They are not. From wiki: "United States Congress defined the boundaries of
the new Territory of Colorado exclusively by lines of latitude and longitude,"
and they surely knew Earth isn't flat. Unlike you, it seems.
Yet the states have four right angles for corners.
Yet their borders aren't straight lines. Neither in Euclidean
geometry nor in Riemann's idiocy.
But they must be straight lines.
Neither Euclid, nor Riemann, nor any sane person
is sharing this opinion of yours, stupid Mike. I'm not
sharing it either.
I'm glad to see that you consider yourself outside the group of "any
sane person". It's a start.
I question your odd belief that Euclid himself wouldn't consider
segments connecting two points as described to be straight line
segments. What would he say about a figure with two parallel sides of
the same length and four right angles? What would he say if a map showed
Greenland to be the same size as South America yet they weren't close in
size at all, even considering he didn't know of either place?
In a 2D surface, a curve is straight in a Euclidean sense if, between the 2
degrees of freedom available, only one of them is exploited. For example,
on the earth, a path along a line of longitude is straight on the surface
because
Because an idiot woodworker has invented some "jargon" and insists.
Heck no. I’m referring to mathematics from the 16th century. I didn’t
invent it.

Now, what might be true is that the only mathematics and mathematics lingo
you know is what you learned in high school about Ancient Greek geometry,
and so that’s good enough for you.
Do you also insist that Earth surface is a plane? You can! You would be
in a good company.
Post by Odd Bodkin
Now, Maciej is apparently having difficulty understanding that the surface
of a sphere is in fact a 2D space.
Now, Odd is apparently having difficulty understanding that what
he announces as "a fact" is just his totally unsupported assertion.
--
Odd Bodkin -- maker of fine toys, tools, tables
Maciej Wozniak
2022-01-10 14:48:36 UTC
Reply
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Post by Odd Bodkin
Post by Odd Bodkin
Post by Michael Moroney
Post by Maciej Wozniak
Post by Michael Moroney
Post by Michael Moroney
Post by Maciej Wozniak
Post by Michael Moroney
Post by Maciej Wozniak
Post by Michael Moroney
Post by Maciej Wozniak
Post by Odd Bodkin
This is the best you can do when your crap is debuncked, Lengrand? Post
more nonsense using a sock puppet fake name?
So, is Euclidean geometry a valid geometry of our
universe?
Depends on the application. As any engineer should know.
It's not a common engineering knowledge, and the
concept is completely alien to a relativistic moron
mumbling about his imagined Laws of Nature.
Is Euclidean geometry a valid geometry for surveying land on earth?
Sure.
Post by Michael Moroney
Why or why not?
Because it's been crafted with skill, wisdom and luck.
Unlike your Shit, having nothing but luck.
Then why do the US states of Colorado and Wyoming, legislatively defined
as rectangular
They are not. From wiki: "United States Congress defined the boundaries of
the new Territory of Colorado exclusively by lines of latitude and longitude,"
and they surely knew Earth isn't flat. Unlike you, it seems.
Yet the states have four right angles for corners.
Yet their borders aren't straight lines. Neither in Euclidean
geometry nor in Riemann's idiocy.
But they must be straight lines.
Neither Euclid, nor Riemann, nor any sane person
is sharing this opinion of yours, stupid Mike. I'm not
sharing it either.
I'm glad to see that you consider yourself outside the group of "any
sane person". It's a start.
I question your odd belief that Euclid himself wouldn't consider
segments connecting two points as described to be straight line
segments. What would he say about a figure with two parallel sides of
the same length and four right angles? What would he say if a map showed
Greenland to be the same size as South America yet they weren't close in
size at all, even considering he didn't know of either place?
In a 2D surface, a curve is straight in a Euclidean sense if, between the 2
degrees of freedom available, only one of them is exploited. For example,
on the earth, a path along a line of longitude is straight on the surface
because
Because an idiot woodworker has invented some "jargon" and insists.
Heck no. I’m referring to mathematics from the 16th century.
Heck no. You're referring to your own personal fantasies.
In the mathematics of XVI century (and in the modern mathematics
as well) - a line is a straight line in Euclidean sense when it matches
Euclid's axioms of a straight line. Meridians don't, so they are not
straight lines in Euclidean sense. Sorry, poor halfbrain.
Michael Moroney
2022-01-10 14:59:22 UTC
Reply
Permalink
Do you also insist that Earth surface is a plane? You can! You would be
in a good company.
Apparently you consider the surface of a sphere to be a plane, since you
say Euclidean geometry can be used.
Post by Odd Bodkin
Now, Maciej is apparently having difficulty understanding that the surface
of a sphere is in fact a 2D space.
Exactly. The fact it's curved in a third dimension is what makes it
non-Euclidean, as Colorado and the triangle with 3 right angles demonstrate.

Surveying is two dimensional although topographic info (the third
dimension) is often an add-on plus work is necessary to get the plot on
a single elevation. On a small scale it is treated as flat (Euclidean)
plane space but on a larger scale, "corrections" are initially
necessary, and at an even larger scale different techniques are used as
the space becomes increasingly non-Euclidean. My father was a surveyor,
he told me about something like that when I was young, although I don't
remember the terms he used (not "Euclidean").

It's fairly easy to visualize the problem of projecting the surface of a
sphere/almost-a-sphere (earth) to a flat map. Much harder to visualize
doing the same for the universe's 3D space to a "flat" 3D space we
understand at a small scale.
Maciej Wozniak
2022-01-10 15:15:29 UTC
Reply
Permalink
Post by Michael Moroney
Do you also insist that Earth surface is a plane? You can! You would be
in a good company.
Apparently you consider the surface of a sphere to be a plane, since you
say Euclidean geometry can be used.
Stupid Mike, have you really never heard of Euclidean 3d
geometry?
Post by Michael Moroney
Exactly. The fact it's curved in a third dimension is what makes it
non-Euclidean,
No, stupid Mike. It simply make it curved.
Michael Moroney
2022-01-10 15:29:35 UTC
Reply
Permalink
Post by Maciej Wozniak
Post by Michael Moroney
Do you also insist that Earth surface is a plane? You can! You would be
in a good company.
Apparently you consider the surface of a sphere to be a plane, since you
say Euclidean geometry can be used.
Stupid Mike, have you really never heard of Euclidean 3d
geometry?
I'm discussing 2d surveying of the surface of the earth.
Post by Maciej Wozniak
Post by Michael Moroney
Exactly. The fact it's curved in a third dimension is what makes it
non-Euclidean,
No, stupid Mike. It simply make it curved.
Not in 2d plotting.

Townes Olson
2022-01-08 21:17:38 UTC
Reply
Permalink
x'= [x+sqrt(x²+y²+z²).v/c]/sqrt(1-v²/c²)
That equation is valid only for events that are light-like separated from the origin, meaning that c²t² = x²+y²+z². In general we have tau^2 = c²t² - x² - y² - z², so to cover time-like intervals your equation would have to be
x'= [x + sqrt(x²+y²+z²+tau^2)v/c] / sqrt(1-v²/c²) .
where tau is the proper time along the interval. Writing the transformation equation that way is not advantageous.
Maciej Wozniak
2022-01-08 21:28:05 UTC
Reply
Permalink
x'= [x+sqrt(x²+y²+z²).v/c]/sqrt(1-v²/c²)
That equation is valid only for events that are light-like separated from the origin, meaning that c²t² = x²+y²+z². In general we have tau^2 = c²t² - x² - y² - z², so to cover time-like intervals your equation would have to be
x'= [x + sqrt(x²+y²+z²+tau^2)v/c] / sqrt(1-v²/c²) .
where tau is the proper time along the interval. Writing the transformation equation that way is not advantageous.
In the meantime in the real world, however, forbidden
by your moronic religion TAI keep measuring t'=t, just
like all serious clocks always did.
Richard Hachel
2022-01-09 00:13:27 UTC
Reply
Permalink
Post by Townes Olson
x'= [x+sqrt(x²+y²+z²).v/c]/sqrt(1-v²/c²)
That equation is valid only for events that are light-like separated from the
origin, meaning that c²t² = x²+y²+z². In general we have tau^2 = c²t² -
x² - y² - z², so to cover time-like intervals your equation would have to be
x'= [x + sqrt(x²+y²+z²+tau^2)v/c] / sqrt(1-v²/c²) .
where tau is the proper time along the interval. Writing the transformation
equation that way is not advantageous.
Ah! Un posteur qui réfléchit et qui ne vient pas faire du pugilat.
Je ne dis pas qu'il n'y en a pas sur ce site, car il s'en trouve. Mais il
n'y en a pas des dizaines.
Donc félicitations, monsieur.

Ah! A poster who thinks and does not come to fight.
I'm not saying there aren't any on this site, because there are. But there
aren't dozens of them.
So congratulations, sir.

Je ne sais pas à quoi correspond votre équation, je vous donne donc la
mienne puisque vous voulez aller plus loin et parler en fonction de t.

Je précise bien en fonction de t. C'est à dire le moment où
l'observateur perçoit l'événement.

Je différencie le temps local To par rapport au temps propre t de la
montre placée en O. Ce n'est pas la même chose.

I don't know what your equation is, so I'm giving you mine since you want
to go further and speak in terms of t.

I am precise in terms of t. That is to say the moment when the observer
perceives the event.

I differentiate the local time To compared to the proper time t of the
watch placed in O. It is not the same thing.

L'équation qui prend en compte t est la suivante :
x'=[x+({sqrt(x²+y²+z²)/c}-t).v/c]/sqrt(1-v²/c²)

Sa réciproque est :
x=[x'-({sqrt(x'²+y²+z²)/c}-t).v/c]/sqrt(1-v²/c²)


The equation which takes into account t is the following:
x'=[x+({sqrt(x²+y²+z²)/c}-t)v/c]/sqrt(1-v²/c²)

Its reciprocal is:
x = [x'- ({sqrt(x'²+y²+z²)/c}-t').v/c]/sqrt(1-v²/c²)

Prenons un exemple simple. Une fusée passe à proximité de la terre à
vitesse v=0.8c.
On lui assigne l'axe Ox.
Dans le référentiel terrestre une explosion cosmique à lieu (c'est à
dire est VISIBLE) au moment où les deux observateurs O et O' se croisent.
x=12ly, y=9ly z=0
On a alors l'événement cosmique E suivant E=(12,9,0,-15,0) selon les
coordonnées (x,y,z,To,t).
Plusieurs sur ce forum ont alors chercher à savoir (et y sont parvenus)
ce qui se passait dans le référentiel de la fusée. Je les en remercie.
Ils ont trouvé E=(40,9,0,-41,0) pour la fusée.
Ils ont donc reçu mes félicitations.

Let's take a simple example. A rocket passes near the earth at speed v =
0.8c.
We assign it the axis Ox.
In the terrestrial frame of reference a cosmic explosion takes place (ie
is VISIBLE) at the moment when the two observers O and O 'cross. x=12ly
y=9ly z=0
We then have the following cosmic event E E = (12,9,0,-15,0) according to
the coordinates (x, y, z, To, t).
Several on this forum then tried to find out (and succeeded in doing so)
what was happening in the rocket's repository. I thank them for that.
They found E=(40,9,0,-41,0) for the rocket.
So they received my congratulations.

Mais il faut aller plus loin, vous avez raison. Car ici je n'ai employé
que l'équation :
x'= [x+sqrt(x²+y²+z²).v/c]/sqrt(1-v²/c²)
Cette équation n'est valable que pour tous les observateurs qui se
croisent en O à t=0.

Mais imaginons, que cet événement n'ai pas lieu au moment où la fusée
passe, mais six ans plus tard.
Il va donc me falloir employer une équation qui contient t (le moment où
la terre perçoit l'explosion cosmique) pendant que la fusée a
progressé, elle, en mouvement galiléen régulier de 0.8c.
Posons t=6 (six ans).
L'équation devient :
x'=[x+({sqrt(x²+y²+z²)/c}-t).v/c]/sqrt(1-v²/c²)
On a alors dans le référentiel terrestre E=(12,9,0,-9,6)
et dans le référentiel de la fusée E=(32,9,0,-31, 2.24154)

But we have to go further, you are right. Because here I only used the
equation:
x'= [x+sqrt(x²+y²+z²).v/c]/sqrt(1-v²/c²)
This equation is only valid for all the observers who intersect at O
​​at t=0.

But imagine, that this event does not take place when the rocket passes,
but six years later.
So I will have to use an equation which contains t (the moment when the
earth perceives the cosmic explosion) while the rocket has progressed in
regular Galilean motion of 0.8c.
Let t=6 (six years).
The equation becomes:
x '= [x+({sqrt(x²+y²+z²)/c}-t)v/c]/sqrt(1-v²/c²)
We then have in the terrestrial frame of reference E = (12,9,0, -9,6)
and in the frame of the rocket E = (32,9,0, -31, 2.24154)

Vérifions maintenant que la réciproque est correcte (sinon c'est
absurde).
Now let's check that the converse is correct (otherwise it's absurd).

x=[x'-({sqrt(x'²+y²+z²)/c}-t').v/c]/sqrt(1-v²/c²)

x=[32-({sqrt(32²+9²+0²)/1}-2.24154)*0.8]/0.6
x=[32-(33.24154-2.24154)*0.8]/0.6
x=[32-(31*0.8)]/0.6
x=12

R.H.
Townes Olson
2022-01-09 00:36:51 UTC
Reply
Permalink
Post by Townes Olson
x'= [x+sqrt(x²+y²+z²).v/c]/sqrt(1-v²/c²)
That equation is valid only for events that are light-like separated from the
origin, meaning that c²t² = x²+y²+z². In general we have tau^2 = c²t² -
x² - y² - z², so to cover time-like intervals your equation would have to be
x'= [x + sqrt(x²+y²+z²+tau^2)v/c] / sqrt(1-v²/c²) .
where tau is the proper time along the interval. Writing the transformation
equation that way is not advantageous.
I don't know what your equation is...
It's the space component of the Lorentz transformation, written in a rather silly way, by replacing the time coordinate, t, with the expression sqrt(x²+y²+z²+tau^2). Again, writing the equation that way is not advantageous.
x'=[x+({sqrt(x²+y²+z²)/c}-t)v/c]/sqrt(1-v²/c²)
No, that is correct only if x=y=z=0, which is not generally the case. The simplest way of expressing the Lorentz transformation, in units with c=1, is just x'=(x-vt)g, t'=(t-vx)g where g=1/sqrt(1-v^2). That is all there is to it.
Richard Hachel
2022-01-09 01:15:43 UTC
Reply
Permalink
Post by Townes Olson
Post by Richard Hachel
x'=[x+({sqrt(x²+y²+z²)/c}-t)v/c]/sqrt(1-v²/c²)
No
LOL.
Post by Townes Olson
that is correct only if x=y=z=0,
re-LOL
Post by Townes Olson
which is not generally the case. The simplest way of expressing the Lorentz
transformation, in units with c=1, is just x'=(x-vt)g, t'=(t-vx)g where
g=1/sqrt(1-v^2). That is all there is to it.
No.

You do not read what I write correctly.

My rating is this in this case:

x'=γ(x-v.To)
y'=y
z'=z
To'=γ(To-xv/c²)

N.B. Attention To and To' are negative. Here To=-15 et and To'=-41.
To2=-9(-15+6) and To2'=-31

R.H.
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