I am trying to work through a proof of causality in special relativity using the Lorentz transformations, but there is one assumption that is necessary for the proof that I don't see as correct.

The question I am working on specifically asks: "In frame S, event A causes event B and therefore occurs before event B. Show that in any frame, event A always occurs before event B." I have understood the proof except for one assumption: "The fastest way information from A can reach B is at the speed of light, so Δx/Δt < c". Why is it not $Δx/Δt ≤ c$? I know that if I use this latter assumption, then the proof doesn't really work, as it results in the possibility that the causal event in the S frame could look like a simultaneous event in the S' frame, but I don't see why my assumption is wrong.

I came up with a simple thought experiment: say event A is the firing of a photon at a target one lightsecond away; event B is the reception of that photon one second later. This gives Δt=1 s, Δx=3*10^8 m, and so Δx/Δt = c. If I plug this in, however, another observer would measure Δt'=0, so they would see the two events happen simultaneously, which breaks (or at least muddies) causality.

Where did I go wrong in my thinking?

EDIT: here is a photo of the proof (it comes from the Cambridge Physics IB textbook):

enter image description here

I understand all the math and substitutions, but I think that the inequality should be Δx/Δt ≤ c, which would lead to a final conclusion of Δt'≥0.

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    $\begingroup$ It should also work with Δx/Δt ≤ c. Can you provide a link to the material you are using or some useful resource? Is this of any help? physics.stackexchange.com/q/100448/226902 physics.stackexchange.com/q/495364/226902 $\endgroup$
    – Quillo
    Apr 23 at 15:23
  • $\begingroup$ EDIT: I posted a photo of the proof. My concern is with the implications of getting Δt'≥0, as it would lead to a causal event in the S frame being a simultaneous event in the S' frame. I know from other examples involving lightning strikes and trains etc. that this isn't unheard of in SR, but I'm not sure if it is conceptually applicable here. $\endgroup$ Apr 23 at 15:57
  • $\begingroup$ Proper time of a photon is zero. In other words: in the photon's frame all events happen at the same time. $\endgroup$
    – Kurt G.
    Apr 23 at 16:08
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    $\begingroup$ @KurtG I don’t think “the photon’s frame” is well-defined. $\endgroup$
    – robphy
    Apr 23 at 17:01
  • $\begingroup$ @robphy Sure. It lacks a dimension. However we formulate it OP should probably restrict those investigations to frames that have relative speeds less than $c$. $\endgroup$
    – Kurt G.
    Apr 23 at 17:08

1 Answer 1


You are correct that information can travel at the speed of light. In the notation of the proof above: $\frac{\Delta x}{\Delta t} \leq c$. We can rewrite the proof allowing for this

\begin{align} \Delta t' &= \gamma\Delta t\left(1-\frac{v\Delta x}{c^2\Delta t}\right) \\ &\geq \gamma\Delta t\left(1-\frac{vc}{c^2}\right) \\ &\geq \gamma\Delta t\left(1-\frac{v}{c}\right) \end{align}

In the last line, I've only canceled a "$c$", I did not set $v$ equal to $c$.

I think the confusion comes from differentiating $v$ and $\frac{\Delta x}{\Delta t}$. The latter is a ratio of space and time intervals between two events in the frame $S$. If the two events happen to be connected by a null (photon) trajectory $\frac{\Delta x}{\Delta t} = c$. However $v$ is the relative velocity of frame $S'$ with respect to $S$. As both frames correspond to timelike observers, neither can travel at the speed of light: $v<c$. Now $v$ can be arbitrarily close to $c$, but not equal to it. So in the last inequality above, the right hand side can get arbitrarily close to zero, but not equal to it. Another way of saying this is $\Delta t' > 0$ (the $\geq$ has become a strict inequality: $>$).

  • $\begingroup$ Hmm okay, I think I may see it clearly now. I'll have to continue my studying to fully understand this idea of "timelike", but I do think I've heard that an observer can never travel at the speed of light because of the increase to an infinite mass (although maybe I'm not using the right thinking here). I do think that there is a confusion in the proof, though: going from line 3 to the first part of 4, the only change is from Δx/Δt to c, but this is accompanied by a > sign, indicating to me that they've subbed in Δx/Δt>c. Then it looks like they do set v=c in the second part of line 4. $\endgroup$ Apr 23 at 16:34
  • $\begingroup$ @ACommonScholar "timelike" is lingo for observers traveling slower than the speed of light (really it's talking about the magnitude of vectors in spacetime). The idea of increasing relativistic mass is pretty dated. Today I think most physicists prefer to say an infinite amount of energy is required to move a massive body at the speed of light. $\endgroup$
    – Aiden
    Apr 23 at 16:45
  • $\begingroup$ Ah okay, that makes sense. So it seems like v>c by definition if they are timelike observers. Would there be a way to have non-timlike observers? For instance, in my proposed thought experiment, could we have an observer moving with the photon, in which case the relative velocity of their frame of reference is = c? I know that, at this point, I may be asking questions which require a more complex knowledge of relativity to appreciate -- I've only just begun my study, and I know there's a lot I have yet to learn. If a simple, discursive explanation isn't possible, I understand! $\endgroup$ Apr 23 at 16:52
  • $\begingroup$ I've edited my answer to address some of your earlier questions. In relativity observers are restricted to travel slower than the speed of light locally. That being said, there are all sorts of systems where there is apparent super-luminal travel, i.e. some object with mass can beat light to a destination using a wormhole or a "warp drive" of some sort. Most of these spacetimes require some form of exotic energy and really need the full machinery of general relativity examine properly. $\endgroup$
    – Aiden
    Apr 23 at 17:21
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    $\begingroup$ I really appreciate it! Given what we've discussed, the proof makes sense and is crystal clear to me now (although I'm concerned about what appears to me to be a pretty big error in a textbook like this, though it's not the first time...). I'll continue my studying to learn more about this idea of timelike observers, but for now I understand that, as far as the questions I'm dealing with are concerned, v < c. Thanks a lot! $\endgroup$ Apr 23 at 17:29

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