I have this silly doubt in my head and it's bugging me for a real long time now. Let us consider the Galilean transformation $x=x'+vt$ for two frames measuring coordinates $x$ and $x'$. For simplicity, I'll call these frames YOU(measuring $x'$) and ME (measuring $x$).Now, there are two ways to look at this(I'll mention both so that the reader gets the drift-changing one concept of Galilean relativity gives Lorentz transforms)-

(1) I (i.e. frame ME) sees the origin of YOU at a distance of $vt$ after a time $t$ ( the regular stuff about origins coinciding at $t=0$ holds of course). Now, there is an event, $E$, which occurs at a coordinate $x'$ in you frame. So,

(Distance of E from ME)=(Distance of E from YOU) + (Distance of YOU from ME)

So $x=x'+vt$, the Galilean transformation law.

Now, in SR, we just let go of the idea that all frames measure the same length. Since it is possible to show the existence of time dilation and length contraction in SR from purely physical reasoning (i.e. without invoking Lorentz transforms), we may write

(Distance of E from ME)=(Distance of E from YOU 'as seen by ME') + (Distance of YOU from ME)

And we get $x=x'/\gamma+vt$, using the result of lorentz contraction for the 'length' $x'$ YOU measured. And we have derived the lorentz transformation rule.

Now, the problem is here-

(2) The way I am now trying to look at this is to imagine the following. Suppose YOU measures a coordinate $x'$, and communicates his measurement to me. At this instant, YOU is at a distance $vt$ from me. If YOU could communicate this INSTANTLY, ME could geometrically add the information about distances he knows to get his coordinate- $x=x'+vt$. Thus, Galilean transformations can sort of be ascribed to this 'instantaneous' communication of measurements.

Now, if we account for the finite speed at which a signal can be transmitted, so that NO instantaneous communication is possible, can a similar line of reasoning as above lead to Lorentz Transforms? Or are Lorentz transformations much more fundamental than that? I have a feeling that there is a very trivial point I have overlooked and gotten myself in this mess.

I tried to work it out (not too diligently, I admit), but did not get anywhere close. So is my guess that the Lorentz transformations are simply a correction induced in SR due to finiteness of signal speed incorrect? This is annoying. Any help would be appreciated.


1 Answer 1


Ok, so as far as I understand, there are two parts to answering your question:

  1. It is true that Lorentz transformation can be thought of as corrections to the Galilean transformation due to the finiteness of the speed of causality. In the sense that if you take the limit in which $c$ goes to $\infty$ then the Lorentz transformations reduce to Galilean transformations.

  2. I don't think a direct method you propose of communicating with the YOU frame can give us Lorentz transformation--at least not trivially. Because, trivially, the communication lag would be incorporated as $x=x'+v\bigg(t+\dfrac{t}{c}\bigg)$. Further, this approach is flawed in its very origin. See, an observer in SR means a whole array of measuring instruments spread over all the spacetime points. So, for ME frame to communicate with YOU frame, it would take no time at all. There is an instrument which is a part of the ME frame in the very vicinity of the origin of the YOU frame.

  • $\begingroup$ Thanks a lot @Dvij!..I was missing the fact that for 2 frames to 'communicate', no distance needs to be traveled by the signal, as you rightly pointed out. I was kinda hopeful because this sort of reasoning gave the correct answer for galilean relativity. Is there maybe a flaw in that too? $\endgroup$
    – GRrocks
    Dec 19, 2017 at 10:13
  • $\begingroup$ @GRrocks Yes, this reasoning is flawed on its very basis due to the simple fact that you realize that communication between frames is instantaneous no matter whether the speed of causality is finite or infinite. The reason it works out seems to be purely coincidental. Somewhat mathematically speaking, Lorentz transformation is not the only transformation that is mathematically conceivable that reduces to Galilean transformation in $c$ goes to zero limit. There could be many. For example, $x'=x-v(t+\dfrac{t}{c})$. $\endgroup$
    – user87745
    Dec 19, 2017 at 13:04
  • $\begingroup$ The logic you use works, in a mathematical way, because this clearly wrong transformation reduces to Galilean transformation in the concerned limit. I could write another crazy transformation that does the same in the same limit and I could formulate a corresponding verbal description of the crazy formula and argue that since it works for Galilean limit it should work for the cases in which $c$ is finite. Clearly, as you can see, this is not logical. Hope this helps. $\endgroup$
    – user87745
    Dec 19, 2017 at 13:04

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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