# Special relativity: when can we say in our good own right that something happened before something else?

I am studying "Introducing Einstein's Relativity" (D'Inverno, Oxford University Press) and I am trying to understand Fig.2.12, page 23. There it goes: Observer B sends:

• a light ray towards Q at time t(R) which bounces back at t(U)

• a light ray towards P at time t(S) which bounces back at t(V)

I'd say that B "sees" Q before P because t(U) < t(V).

The book says explicitly that B "sees" Q before P because t(R) < t(S).

I am willing to believe the book, but can anyone explain to me why t(U) < t(V) isn't a good argument?

What further troubles me is that on the same page 23 the book places Einstein's image of the train with the two light bulbs. My understanding of that experiment is that indeed the moment of light reception is relevant to convince B that the bulb at the head of the carriage was switched on before the one in the back of the carriage.

• Very briefly: the word "see" is not being used literally. B knows that light travels at a finite speed, and can calculate the time and position of P and Q. – Javier Mar 24 '17 at 11:51
• I don't know about you, but I usually "quote" concepts that I am not "using literally". Having agreed on that, what is your answer to my question? (which is purposedly not quoted) – Marco Faustinelli Mar 24 '17 at 12:03
• Calm down, I meant that the book is not using the word "see" literally. It's not that B sees Q before P with their eyes (though that is true), it's that in the B frame Q happens before P. – Javier Mar 24 '17 at 12:50
• Can you manage to elaborate an answer? – Marco Faustinelli Mar 24 '17 at 20:25

The text says "By symmetry RU = SV and so these events are equidistant according to B. However, the signal RQ was sent before the signal SP and so B concludes that the event Q took place well before P. Hence, events that A judges to be simultaneous, B judges not to be simultaneous. Similarly, A maintains that P, O, and Q occurred simultaneously, whereas B maintains that they occurred in the order Q, then O, and then P."

Note that t(R) < t(S) is equivalent to t(U) < t(V).

Since we are told that RU=SV (where R, U, S, and V are events on Observer-B's worldline), then t(U)-t(R)=t(V)-t(S)... call this common time-interval T.

Then t(R)=t(U)-T and t(S)=t(V)-T.

So, t(R) < t(S) implies t(U)-T < t(V)-T, which implies t(U) < t(V).

You can run the argument in reverse to show that t(U) < t(V) implies t(R) < t(S).

So, it seems to me that the author could have used either inequality.

For the bulbs in the train, it seems like it is a practical consideration that the reception events convince observer-B. (Presumably, Observer-B could have done radar measurements [with transmissions and receptions] before the meeting at event O to conclude that he is in the middle of the train.)

• I appreciate your words and I subscribe your geometrical conclusions. However we are both second-guessing professor D'Inverno. All this while Einstein (afaik) says nothing about emitted light and builds his conclusions of the train experiment only reasoning about received light. – Marco Faustinelli Mar 24 '17 at 14:48
• I agree with you that B needs a radar to be able to say he's in the middle of the carriage. I am actually about to post another question right on that. This question is strictly about timing of events. – Marco Faustinelli Mar 24 '17 at 14:54
• Often spacetime diagrams are drawn optimally to encode many aspects of the situation. The radar measurements establishes that P and Q are equidistant from these observers. Note that the midtime clock reading is what the observer uses to assign a time to Q. I think Einstein focused on received light because his original experiment involved lightning strikes. The problem could have been "when should I send light signals to turn on the bulbs simultaneously in A's frame?" – robphy Mar 24 '17 at 15:28
• I am reflecting after reading your words (and I am thankful for them). It looks like, if B has a radar, then Q and P happen to him halfway RU and SV respectively. If B is only receiving light from P and Q, he can only measure the time interval between them, but cannot place them precisely in his own timeframe. Which is nonsense, so I must be missing something :-) – Marco Faustinelli Mar 24 '17 at 15:50

In a sense, you're correct that observer B sees Q before P because the light ray from Q arrives before the one from P. This is the way things work most of the time, after all: we see things by receiving the light from them, so it makes sense that we judge time ordering from the order of reception.

The problem is that in relativity we tend to use the word "see" in another way. We assume that the observers have a good knowledge of electromagnetism and therefore they know that light travels at a finite speed; therefore, they can correct for this.

Let me give you an example. Suppose we look at the sky, and we observe that star A goes supernova, and one hour later we observe star B going supernova (I didn't say this would be a realistic example!). It seems to us like star A went supernova first. But if we know that star A is 1 light year away while star B is 100 light years away, it's clear that in fact star B went supernova first - it's just that its light took longer to reach us. So in the literal meaning of "see" we see star A exploding first, but in the physics-meaning of "see" we see star B exploding first.

Getting back to your question, in the physics version of "seeing", B sees Q before P. Maybe a better way of putting it would be to say that in the frame in which B is at rest, Q happens before P. But I disagree with the book in that it's because $t(R) < t(S)$; I say it's because $t(S) < t(U)$. See my modification of the picture: I've drawn in blue the lines of simultaneity for B; these are events which are simultaneous in B's frame (but B doesn't see them as simultaneous in the colloquial sense of the word). If $t$ is B's time coordinate, these are lines of constant $t$. You can see the according to B, Q happens at $t(S)$ while P happens at $t(U)$, and clearly $t(S)<t(U)$.