Doesn't this transformation also preserve speed of light?:

$B$ is moving with speed $v$ relative to $A$ in the $+x$ direction. $B$ passes $A$ at $t=0$, and $A$ shines a torch at that moment.

Let $(x', t')$, the coordinates of the light beam in $B$'s frame, be related to the $(x,t)$ coordinates in $A$'s frame by these transformations:

$$x'=x-vt =ct-vt =ct\left(1-\frac{v}{c}\right)$$


As we can see, this transformation gives $\frac{x'}{t'}=c$. So the speed of light is the same in $B$'s frame. This transformation tweaks the time but lets distance be the same as given by Galilean transformation.

What issues does a transformation like this cause? And what makes 'Lorentz transformation' the right way to go to account for the constancy of speed of light?

  • $\begingroup$ The Lorentz transformation is global if it's a space-time independent constant matrix. The invariance of physical laws with respect to a global Lorentz transformation is just the principal of the invariance of the speed of light in special relativity. $\endgroup$ – Cinaed Simson Jul 15 at 9:08
  • $\begingroup$ How would $y$ and $z$ transform? What would be the speed of a light beam that wasn’t traveling in the $x$-direction? $\endgroup$ – G. Smith Jul 15 at 16:22

Consider observer $C$, who is traveling at a speed relative $-v$ in the $x$-direction relative to $B$. By your logic, the coordinates $x''$ and $t''$ measured by $C$ should be $$ x'' = x' - (-v) t' = (x - vt) + v (t - v/c) = x - v^2/c \\ t'' = t'( 1 - (-v)/c) = t ( 1 - v/c)(1 + v/c) = t \left( 1 - \frac{v^2}{c^2} \right) $$ But $C$'s coordinates should be the same as $A$'s coordinates, i.e., $x'' = x$ and $t'' = t$. This is a contradiction.

The other way to look at it is that the inverse transformation law from $B$ to $A$, according to your equations, is $$ t = \frac{t'}{1 - v/c} \qquad x = x' + \frac{t'}{1 - v/c} $$ But this means that the transformation laws to get between reference frames are different between these two frames. This means that the principle of relativity is broken; the transformation laws should have substantially the same form in all reference frames.

  • $\begingroup$ Is Lorentz transformation the only way to achieve 'transformation laws having same form in all reference frames'? Do we know that? $\endgroup$ – Ryder Rude Jul 16 at 3:21
  • $\begingroup$ @RyderRude: It probably is under some additional mathematical assumptions, but I'm not sure what those are; I'd have to think about it some more. One way to approach this is via k-calculus, which starts from the assumptions of (I think) linearity of transformed proper times and the equivalence of frames and builds up the Lorentz transformations from there. $\endgroup$ – Michael Seifert Jul 16 at 11:55

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