Alternative Ways to Realize Invariance: Lorentz Transformation

In summary: Thus, according to these arguments there is a unique way to realize invariance and that is through the Lorentz group.
  • #1
msumm21
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TL;DR Summary
Wondering if there are alternative ways to realize invariance, other than the Lorentz transformation?
The Lorentz transformation ensures different inertial observers measure the same speed of light. Are there other transformations, or other ways to setup a "space-time" that also have this property of invariance? Is the Lorentz transformation the unique solution?
 
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  • #2
Galilean relativity kind of allows it - one can broadly make the case that infinite speed is invariant in that case. More rigorously, any locally Lorentzian spacetime (i.e., what general relativity studies) respects the invariance of local measures of light speed.

If you dump the principle of relativity it might be possible to come up with other solutions. I'm not aware of any, though.
 
  • #3
A paper by Polash Pal, "Nothing but Relativity" shows that the assumption of relativity will only allow Galilean relativity (all inertial frames share a universal time) or Einseteinian relativity (constant speed of light in a vacuum).
(See https://arxiv.org/abs/physics/0302045 :
Abstract
We deduce the most general space-time transformation laws consistent with the principle of relativity. Thus, our result contains the results of both Galilean and Einsteinian relativity. The velocity addition law comes as a bi-product of this analysis. We also argue why Galilean and Einsteinian versions are the only possible embodiments of the principle of relativity. )
 
  • #4
msumm21 said:
TL;DR Summary: Wondering if there are alternative ways to realize invariance, other than the Lorentz transformation?

The Lorentz transformation ensures different inertial observers measure the same speed of light. Are there other transformations, or other ways to setup a "space-time" that also have this property of invariance? Is the Lorentz transformation the unique solution?

Flipping this around, there are various topological arguments that causality implies the Lorentz group, e.g.,

Zeeman, E. C. "Causality Implies the Lorentz Group", Journal of Mathematical Physics 5 (4): 490-493; (1964);

Nanda S., "A geometrical proof that causality implies the Lorentz group", Mathematical Proceedings of the Cambridge Philosophical Society , Volume 79 , Issue 3 , May 1976 , pp. 533 - 536.
 
  • #5
By the way are you aware that the Lorentz Transformations in Einstein's version cannot in general reduce to Galilean Transformations?
 
  • #6
stefanoquattrini said:
By the way are you aware that the Lorentz Transformations in Einstein's version cannot in general reduce to Galilean Transformations?
Yes - Galileo is a low speed short distance approximation to Lorentz. Why do you ask?
 
  • #7
Ibix said:
Yes - Galileo is a low speed short distance approximation to Lorentz. Why do you ask?
Yes, low speed and short distances indeed.. but only low-speed v/c<<1 is involved for a classic approximation to be found. There is something ill posed on assuming also short distances
Maybe you want to suggest that t'=t-vx/c2
( first order Lorentz transformation) should be chosen instead of Galilean Transformations in general?
I ask because such constraint has consequences...
 
  • #8
stefanoquattrini said:
Yes, low speed and short distances indeed.. but only low-speed v/c<<1 is involved for a classic approximation to be found. There is something ill posed on assuming also short distances
Maybe you want to suggest that t'=t-vx/c2
( first order Lorentz transformation) should be chosen instead of Galilean Transformations in general?
I ask because such constraint has consequences...
PLEASE PLEASE PLEASE use Latex in your postings. Is it so hard to write ##t'=t-\frac{vx}{c}##?

But with that said... I'm still not seeing the point you're trying to make in this post and the one above. Start with the Lorentz transformations and take ##v/c\ll 1## and we get the Galilean approximation which works well enough when that condition holds (and fortunately for us that includes just about all of humanity's daily experience throughout most of history). Conversely, when that condition does not hold (so that I cannot drop the ##\frac{vx}{c}## term) your expression is just plain incorrect - it fails to respect the observational fact of the invariance of light speed.
 
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  • #9
Nugatory said:
PLEASE PLEASE PLEASE use Latex in your postings. Is it so hard to write ##t'=t-\frac{vx}{c}##?

But with that said... I'm still not seeing the point you're trying to make in this post and the one above. Start with the Lorentz transformations and take ##v/c\ll 1## and we get the Galilean approximation which works well enough when that condition holds (and fortunately for us that includes just about all of humanity's daily experience throughout most of history). Conversely, when that condition does not hold (so that I cannot drop the ##\frac{vx}{c}## term) your expression is just plain incorrect - it fails to respect the observational fact of the invariance of light speed.

Which expression are you talking about is incorrect?? ## t'=t-\frac{vx}{c}##?
 
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  • #10
stefanoquattrini said:
Which expression are you talking about is incorrect?? ## t'=t-\frac{vx}{c}##?
Yes. It's not even dimensionally correct.
 
  • #11
you wrote it like that not me, i wrote this one t'=t-vx/c^2 , pay attention!!! it is c2
 
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  • #12
stefanoquattrini said:
Yes, low speed and short distances indeed.. but only low-speed v/c<<1 is involved for a classic approximation to be found.
Sort of. If ##v\ll c## then ##\gamma\approx 1##, but if you have any ##x## coordinate (edit: or, more precisely, any coordinate difference) that is large enough that ##vx/c^2## is comparable to the resolution of your time measurement you will notice synchronisation problems. The so-called Andromeda Paradox is an example.
stefanoquattrini said:
There is something ill posed on assuming also short distances
Presumably you mean because ##vx/c^2## is not dimensionless and therefore "small" depends on your unit choice. As noted above, yes, in this case "small" depends on how precise your clocks are. As @Nugatory has noted, for almost all of human history it has been negligible for any experiment we could do.
stefanoquattrini said:
Maybe you want to suggest that t'=t-vx/c2
( first order Lorentz transformation) should be chosen instead of Galilean Transformations in general?
Of course it would have consequences - it's not compatible with the principle of relativity. The point is only that there are limits to how large an area you can use the Galilean transforms over before you will start seeing measurable errors, not solely limits to relative velocities.
 
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  • #13
Nugatory said:
Yes. It's not even dimensionally correct.
you wrote it like that not me, i wrote this one t'=t-vx/c2 , pay attention!!! it is c 2
 
  • #14
stefanoquattrini said:
you wrote it like that not me, i wrote this one t'=t-vx/c2 , pay attention!!! it is c 2
Ah - so you did - sorry about that, and stuff like this is why we insist on using Latex
 
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  • #15
stefanoquattrini said:
By the way are you aware that the Lorentz Transformations in Einstein's version cannot in general reduce to Galilean Transformations?
Reduce not, but you can calculate the following limits to get the GT:

## \lim_{c \rightarrow \infty}{[\gamma (x-vt)]} = x-vt ##

## \lim_{c \rightarrow \infty}{[\gamma (t-vx/c^2)]} = t ##

Source 1:
https://www.wolframalpha.com/input?i=Limit[(x+-+v+t)/Sqrt[1+-+v^2/c^2],+c+->+Infinity]
Source 2:
https://www.wolframalpha.com/input?i=Limit[(t+-+(v+x)/c^2)/Sqrt[1+-+v^2/c^2],+c+->+Infinity]

Here, ##c## is of course not meant to be the well-known physical constant, but a mathematical variable.

P.S.
You can find a link to the LaTeX Guide on the left side under the input field. Important is the paragraph "Delimiting your LaTeX code".
 
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  • #16
Sagittarius A-Star said:
Reduce not, but you can calculate the following limits to get the GT:

## \lim_{c \rightarrow \infty}{[\gamma (x-vt)]} = x-vt ##

## \lim_{c \rightarrow \infty}{[\gamma (t-vx/c^2)]} = t ##

Source 1:
https://www.wolframalpha.com/input?i=Limit[(x+-+v+t)/Sqrt[1+-+v^2/c^2],+c+->+Infinity]
Source 2:
https://www.wolframalpha.com/input?i=Limit[(t+-+(v+x)/c^2)/Sqrt[1+-+v^2/c^2],+c+->+Infinity]

P.S.
You can find a link to the LaTeX Guide on the left side under the input field. Important is the paragraph "Delimiting your LaTeX code".
What you wrote is obvious and well-accepted. The issue stays in the classical limit v<<c which is not fulfilled in general
 
  • #17
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  • #18
Ibix said:
Sort of. If ##v\ll c## then ##\gamma\approx 1##, but if you have any ##x## coordinate (edit: or, more precisely, any coordinate difference) that is large enough that ##vx/c^2## is comparable to the resolution of your time measurement you will notice synchronisation problems. The so-called Andromeda Paradox is an example.
That is an issue because in general, at low speeds, there cannot be differences in the counting of clocks.
Once the clocks have been set in sync while overlapping there is no reason for them be out of sync of that quantity when they are at a distance.

Ibix said:
Presumably you mean because ##vx/c^2## is not dimensionless and therefore "small" depends on your unit choice. As noted above, yes, in this case "small" depends on how precise your clocks are. As @Nugatory has noted, for almost all of human history it has been negligible for any experiment we could do.
That term is simply a time. You should know that it is also derivable as a result of the Sagnac effect (from SE one gets double the value due to the difference in the arrival times of the two counterpropagating beams) . That is maybe the only way to derive it and measure it (1913). That was measured also in rectilinear motion (fiber optics experiment).

Ibix said:
Of course it would have consequences - it's not compatible with the principle of relativity. The point is only that there are limits to how large an area you can use the Galilean transforms over before you will start seeing measurable errors, not solely limits to relative velocities.
the t'=t involves a timestamp sent at infinite speed connecting A and B,

the error that one would make by using a signal at the speed of light is ##\frac{H}{c}+\frac{vx}{c^2}##

by considering that the synchronization already occurred when the clocks were at 0 distance
we have ## t'=t-\frac{vx}{c^2}## ,
where ##\frac{vx}{c^2}## is simply the light-time required to reach a moving target initially separated by x from the source.
 
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  • #19
stefanoquattrini said:
Once the clocks have been set in sync while overlapping there is no reason for them be out of sync of that quantity when they are at a distance.
Sure there is - the Lorentz transforms say that two flocks of clocks in relative motion will be out of sync except in at most one place due to the relativity of simultaneity, and the offsets will drift due to time dilation. That never goes away no matter how low you make your speeds - it just becomes negligible for most purposes, but not all. Low speeds just mean that you have to have a really widely separated flock of clocks, or very precise clocks, for the difference to be measurable.
stefanoquattrini said:
That term is simply a time.
Yes - it's the time in the unprimed frame when a clock at rest in the primed frame reads zero and is at position ##x##. In other words, it's the difference in zeroing of the clocks, and it grows without bound. For any non-zero ##v## there is an ##x## where that quantity is very, very large.
stefanoquattrini said:
You should know that it is also derivable as a result of the Sagnac effect (from SE one gets double the value due to the difference in the arrival times of the two counterpropagating beams) . That is maybe the only way to derive it and measure it (1913)
The way I read this, you seem to be saying that the only way to check if two clocks at the same location both read zero at the same time is to do an experiment with the Sagnac effect. That's obviously absurd. Perhaps you could explain what you actually mean.
stefanoquattrini said:
the t'=t involves a timestamp sent at infinite speed connecting A and B,

the error that one would make by using a signal at the speed of light is ##\frac{H}{c}+\frac{vx}{c^2}##

by considering that the synchronization already occurred when the clocks were at 0 distance
we have ## t'=t-\frac{vx}{c^2}## ,
where ##\frac{vx}{c^2}## is simply the light-time required to reach a moving target initially separated by x from the source.
I can't work out what you are arguing here. Infinite speed signals are not possible; you haven't defined ##H##; and I thought you agreed that ##t'=t-vx/c^2## was neither the Lorentz nor Galilean transform, but merely a restriction on the range of validity of the Galilean transforms.
 
  • #20
@stefanoquattrini I am finding it very difficult to understand the point you are making. Please post a professional scientific reference that can explain your point.
 
  • #21
After moderator review, we will leave this thread closed. The recent posts by @stefanoquattrini are not based on the professional scientific literature, and do not merit further discussion here.
 

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