1
$\begingroup$

I have a question about the derivation of Lorentz transformations. My syllabus starts the proof in the following way: We have two inertial reference frames, S and S', that coincide at their origins at t=t'=0. S' is moving in the positive x-direction relative to S with velocity $v$. At t=0, an observer at the origin of S shines a light beam in the positive x-direction. We obtain $x-ct=x'-ct'$. This is still true if we consider a more general form, $x-ct=\lambda(x'-ct')$, where $\lambda$ is yet to be determined. So far, so good. Now they say that this observer simultaneously also shines a light beam in the negative x-direction, which yields to $x+ct=x'+ct'$. However, the syllabus now says that we would get a different general form $x+ct=\mu (x'+ct')$, instead of using the same $\lambda$ again.

How is this possible? If we found $\lambda$, then it should make sense that that same $\lambda$ would work in the negative direction. Could someone explain to me why we're working with two unknowns here?

In case it helps, this is the proof as written in the syllabus (it's in Dutch). It's basically the "From physical principles" derivation found at the wikipedia page: https://en.wikipedia.org/wiki/Derivations_of_the_Lorentz_transformations

enter image description here

enter image description here

$\endgroup$
  • 2
    $\begingroup$ Could you cite the textbook? I am guessing that the author is working towards the notion of spatial isotropy: that the thought experiment looks the same whatever direction in space the boost may happen in. Whilst thoroughly reasonable, and altogether usual in this kind of discussion, it is nonetheless an assumption that has to be explicitly made. So is there a part where the author discusses the assertion that $\lambda = \mu$ is justified by spatial isotropy, or something like this? $\endgroup$ – WetSavannaAnimal Jan 6 '17 at 12:05
  • $\begingroup$ @WetSavannaAnimalakaRodVance I can't cite the textbook, because it is written in Dutch. It's actually a syllabus. I will make a screenshot of it and include it in my original post. $\endgroup$ – Sha Vuklia Jan 6 '17 at 12:21
  • $\begingroup$ Ah. There are Dutch speakers here, but I'm not one of them. Check that there isn't some discussion of spatial isotropy coming up in your text. $\endgroup$ – WetSavannaAnimal Jan 6 '17 at 12:23
  • $\begingroup$ There isn't. They don't even get back to the original $\lambda$ and $\mu$, since they've defined new symbols, $a$ and $b$. $\endgroup$ – Sha Vuklia Jan 6 '17 at 12:24
  • 1
    $\begingroup$ Title? Author? Page? $\endgroup$ – Qmechanic Jan 6 '17 at 12:44
1
$\begingroup$

I'm fairly sure I can see what the issue is (I say this tentatively since I've used Google translate and the likeness of many of the unknown words to those of a language I do read for my day job (German)).

In a general setting, the two constants $\mu$ and $\lambda$ could indeed be different. This would be saying that the thought experiment would give different Lorentz factors depending on the direction of the relative motion, or, to put it another way that space is somehow not isotropic and behaves differently depending on the direction of the boost. We generally don't make such a complicated assumption in relativity.

However, the author specializes his/ her discussion in the paragraph beginning "Nu gebruiken we het relativiteitsprincipe ...", where it is stated that the length of a rod at rest in frame $S$ as seen from the relatively moving frame $S^\prime$ is the same length as it would be seen from frame $S$ if the rod were at rest with respect to $S^\prime$ instead.

This is effectively what forces $\mu = \lambda$ and asserts that the experiment works the same way whether the motion should be in the positive or negative direction - i.e. it asserts spatial isotropy.

I actually disagree that one should call this the "Relativity Principle". Mostly that name is reserved for Galileo's principle that it is impossible to detect uniform motion of a frame $A$ relative to any other $B$ if one makes measurements confined to frame $A$ alone, or similar statements that the laws of physics are unaffected by such relative uniform motion. It would be conceivable to have a physics wherein space were anisotropic but Galileo's principle nonetheless would hold. Rather, that paragraph is a particular statement about spatial isotropy.

There is indeed a special name for the assertion that the transformation from $A$ to $B$ is the same as the transformation from $B$ to $A$ with the sign of the velocity reversed: it's called relativistic reciprocity and I discuss it further in my answer here.

That's not to criticize the author's work - it looks altogether sound and the style seems very concise and readable. But it is a little confusing, as your valid question shows, to raise the possibility of a different $\mu$ and $\lambda$ without explicitly saying why (i.e. without saying we're allowing for potential anisotropy) and then explicitly saying where the specialization is made.

$\endgroup$
  • 1
    $\begingroup$ Wow, thanks for the effort! In the mean time, I've read the derivation on wikipedia, and I had also already watched the derivation on Khan academy, and I've noticed there are different styles and flavours, yet the essence is the same. I'm glad to hear that there is indeed something 'going on' by using $\mu$ and $\lambda$, but the actual reason for this distinction goes a little bit too far for my current course, so I will just accept the fact that spatial isotropy wasn't assumed at the beginning of the proof, and that they just did it as they did it. Thanks again for your efforts! $\endgroup$ – Sha Vuklia Jan 6 '17 at 13:18
  • 1
    $\begingroup$ @ShaVuklia You're very welcome. I was intrigued by the difference and have thought about these issues a great deal - there are many papers around where anisotropy is allowed for but these are fairly advanced / "pedantic" and usually you would say straight up in a first course (particularly since the author describes his / her chapter as "Eenvoudige") that we assume that they are the same because we assume the effect is independent of the boost's direction. $\endgroup$ – WetSavannaAnimal Jan 6 '17 at 13:26

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.