I was studying simultaneity of events and there was a line which quotes that "The relativity of simultaneity is a direct consequence of the invariance of the speed of light in all frames of reference". I wanted to ask that if the invariance of light's speed is false(I know it's not but if we just assume to be), will then the events be simultaneous to all reference frames moving uniformly relative to others? Can we prove that the time interval between the two events is equal to zero in the reference frames mathematically?



3 Answers 3


It depends. It's all a matter of what changes and what doesn't, as much as a certain symmetry in how these things (space and time coordinates) change.

You can arrange to have space-time transformations that do not make the speed of light an invariant, but do not give absolute simultaneity of events.

In standard special relativity, all the kinematical (having to do with motion) results are contained in the Lorentz transformations: $$ t'=\frac{t-vx/c^{2}}{\sqrt{1-v^{2}/c^{2}}} $$ $$ x'=\frac{x-vt}{\sqrt{1-v^{2}/c^{2}}} $$ where $\left(t,x\right)$ and $\left(t',x'\right)$ are the time and space coordinates corresponding to inertial frames moving with relative speed $v$, and I'm assuming, for simplicity, that everything takes place in one spatial direction, $x$. It's not difficult to deduce from here that if an object moves with speed $v_x$ in the unprimed reference frame (S), it will be seen as moving at speed, $$ v_{x}'=\frac{v_{x}-v}{1-vv_{x}/c^{2}} $$ in the primed reference frame (S').

Suppose an observer at S sees an object moving at speed $c$: $$ v_{x}=c $$ Then, an observer at S' will see it moving at speed: $$ v'_x=\frac{c-v}{1-vc/c^{2}}=c\frac{c-v}{c-v}=c $$ This is the famous invariance of the speed of light.

Now you can try to generalise this. Suppose Lorentz's transformations displayed above were not exactly correct. Let us assume instead, $$ t'=f\left(v\right)\left(t-g\left(v\right)x/c^{2}\right) $$ $$ x'=f\left(v\right)\left(x-cg\left(v\right)t\right) $$ for certain universal functions $f\left(v\right)$ and $g\left(v\right)$ characterising the relative motion between S and S'. Assume also that you have two events separated by time interval $\Delta t$ and space interval $\Delta x$ in frame S. You would have, $$ \triangle t'=f\left(v\right)\left(\triangle t-g\left(v\right)\triangle x/c^{2}\right) $$ $$ \triangle x'=f\left(v\right)\left(\triangle x-cg\left(v\right)\triangle t\right) $$ and, $$ v'_{x}=\frac{\triangle x'}{\triangle t}=\frac{f\left(v\right)\left(\triangle x-cg\left(v\right)\triangle t\right)}{f\left(v\right)\left(\triangle t-g\left(v\right)\triangle x/c^{2}\right)}=\frac{\frac{\triangle x}{\triangle t}-cg\left(v\right)\frac{\triangle t}{\triangle t}}{\frac{\triangle t}{\triangle t}-g\left(v\right)\frac{\triangle x}{\triangle t}/c^{2}}=\frac{v_{x}-cg\left(v\right)}{1-g\left(v\right)v_{x}/c^{2}} $$ This transformation law for velocities still gives you an invariant speed of light:

$$ v'_{x}=\frac{\triangle x'}{\triangle t}=c\frac{c-cg\left(v\right)}{c-cg\left(v\right)}=c $$

So the idea is that you would have to play with these $f$ and $g$ $v$-dependent factors to make them non-symmetrical between space and time, and you would arrive to a freakish space-time in which simultaneity of events would not hold, and yet the speed of light would not be an invariant. So yes, it is a logical possibility. It would meet all kinds of problems with homogeneity and isotropy, but that's another story.

  • $\begingroup$ The functions $f$ and $g$ can't truly be independent, though. It's quite easy to show that $$f(v) = \frac{1}{\sqrt{1-g(v)^2}}$$ (use $\Lambda\Lambda^{-1}=1$). Furthermore, unless you plan on changing the definition of velocity, you can easily show that $g(v) = v$. (Observe how the origin of $S'$ is seen to move according to someone in $S$. Since by definition $S'$ is moving with a velocity $v$ with respect to $S$, you should be able to show that $g(v)=v$. So if you impose physical constraints -- and if you want to use linear transformations -- you get back the Lorentz Transformations! $\endgroup$
    – Philip
    Apr 15, 2021 at 11:20
  • 1
    $\begingroup$ @Philip, you're absolutely right. The point I wanted to illustrate is that non-absolute character of simultaneity of events is not a consequence of $c$ being an invariant alone. The other principle that you must invoke is precisely the one you mention. Namely, $$ \Lambda^{T}\eta\Lambda=\eta $$ An overarching principle of symmetry that strongly shapes the argument. $\endgroup$
    – joigus
    Apr 15, 2021 at 11:36

It depends on how you make it the case that the speed of light is not invariant. You could assume that photon is an ever-so-slightly massive particle and then its speed not being invariant would not have any impact on anything.

Assuming homogeneity of space and time, isotropy of space, and the principle of relativity (i.e., the principle that experiments done in inertial frames should be indistinguishable and a frame that moves at a uniform velocity w.r.t. an inertial frame is also an inertial frame), you arrive at the conclusion that the coordinate transformations among inertial frames ought to take the form

\begin{align} x'&=\frac{x-vt}{\sqrt{1-Kv^2}}\\ t'&=\frac{t-Kvx}{\sqrt{1-Kv^2}} \end{align}

where $v$ is the velocity of the primed frame w.r.t. the unprimed frame and $K$ is a non-negative universal constant that falls out of the formalism. See, Nothing but Relativity by Palash B. Pal. As you can see, these transformations imply that there is an invariant speed $\frac{1}{\sqrt{K}}$. Thus, if the parameter $K=0$ then we have $t'=t$ and simultaneity will be absolute. But, if $K\neq 0$ then simultaneity will be relative. In other words, the symmetries of homogeneity of space and time, isotropy of space, and relativity among inertial frames allow for both Galilean and Einsteinian relativity. If we do not find a finite invariant speed in our experiments then it means that the universe is such that Galilean relativity is correct and simultaneity will be absolute. Otherwise, Einsteinian relativity is correct and simultaneity will be relative.

I will not go into an actual derivation (in line with the theme of the answer ;)) but simply point out that it also follows that massless particles must travel at the invariant speed whatever it might be, finite or infinite. Thus, to test as to whether Galilean relativity is correct or Einsteinian relativity, it would suffice to find a massless particle and test as to whether its speed (which ought to be trivially invariant if it is an infinite speed and which ought to be non-trivially invariant if it is a finite speed) is finite or infinite.

Thus, as I mentioned earlier, light could have been a very light massive particle and it would have no effect on which relativity is actual as long as there were some other massless particles that we could have found to be traveling at an invariant finite speed.


That’s what was implicitly assumed since Galileo and Newton, before Maxwell and Einstein noticed that the Galilean transform used in the Newtonian formalism couldn’t be completely coherent with the motion of light (described by Maxwell’s equations).

In order to account for the finite and invariant value of the speed of light, the relativistic formalism introduces the Lorentz transformation which generalizes Galileo’s change of referential transformation.

One way to fall back on the Newtonian formulation (absolute time, composition of speed by simple addition without corrective term, etc.) is to pretend that the value of the speed of light is infinite.

If you pretend that c is infinite, then all the corrective terms which appear in the Lorentz transformation vanish (they vary like v/c) and the Lorentz transformation becomes Galileo’s transformation again.

$\beta = \frac{v}{c}$

$\gamma = \frac{1}{\sqrt{1 - \beta^2}}$

$t' = \gamma \left(t - \frac{vx}{c^2}\right)$

$x' = \gamma(x - vt)$

$y' = y$

$z' = z$


$x' = x - vt$

$y' = y$

$z' = y$

$t' = t$


In which $t' = t$ no matter the relative speed of the two referentials.

  • $\begingroup$ Why does the fact that c isn't invariant imply that c is infinite? $\endgroup$ Apr 15, 2021 at 9:11
  • $\begingroup$ I’m not saying it’s the only solution. I was merely saying it’s one trivial solution. In other words when c tends towards an infinite value, the Lorentz transformation tends towards the Galilean transformation. $\endgroup$ Apr 15, 2021 at 16:07

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