Mathematical proof of causality in special relativity

Question

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):

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.

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: $>$).