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I know variations of these have probably been asked numerous times before, but I'm having trouble with this specific scenario.

Imagine the classic Train Paradox, except instead of lighting strikes we have an observer at the centre of the train shooting laser pulses towards the rear (Event $e_1$) and front of the train (Event $e_2$). Train is moving from left to right at a relativistic velocity $v$.

For an observer on the station, the light pulse travelling towards the rear has to travel a much lesser distance since the train is moving towards it. Let this distance be $0.5-vt$.

Obviously, station observer, who has a moving reference frame, sees the $e_1$ first.

Let us place another man at the back of the train, since he is at rest with the train, light has to travel $0.5$ (exactly half the length of the train) to reach him.

But according to the station observer for whom light has to travel only $0.5-vt$, the light reaches the man before it actually reaches him, in his own reference frame. How is the moving observer able to see an event before it even happened in the rest frame?

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  • $\begingroup$ Let us continue this discussion in chat. $\endgroup$ – xkvt Apr 26 '17 at 10:32
  • $\begingroup$ I don't pretend to understand your question (partly because the pronouns are confusing and I can't always figure out who "him" is), but no matter how it's interpreted, I don't get why you think anyone ever has knowledge of anything before it happens. A given light beam hits the eye of a given observer. That's a single event . At that event, the light beam hits the observer, and at that event the observer becomes aware of the light beam. In any frame whatsoever, that single event happens, of course, at a single time. $\endgroup$ – WillO Apr 26 '17 at 16:26
  • $\begingroup$ I'll reword my issue : a) Station observer (SO) rapidly travels to the back of the train and as such, perceives the event $e1$ immediately after it happens in the rest frame, by being closer to the back -> in Train observers (TO) frame b) back of the train rapidly moves forward towards the light beam, which strikes it around the quarter mark (if the train is at 0.99C) and reaches SO after making the return-> SO frame .These two versions are mathematically equivalent, but I fail to see how they're physically equivalent $\endgroup$ – xkvt Apr 26 '17 at 16:54
  • $\begingroup$ What does "the light reaches the man before it actually reaches him" mean? There is an event $E_1$ at which the light reaches the man. In any given frame, there is a unique time associated to $E_1$. I can't figure out what you think the problem is. $\endgroup$ – WillO Apr 26 '17 at 17:33
  • $\begingroup$ I apologize, agree that it was poorly worded. what do you think about the two frames conveying a different physical meaning? $\endgroup$ – xkvt Apr 26 '17 at 17:36
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The description you have provided of what happens and who perceives what is perfect. Your question is how can it happen that the platform observer (Sam) perceives that an event $A$ happened $before$ it has happened in the rest frame?

Now, the point is that there is no set-up in which we can meaningfully talk about whether Sam has observed an event before it has happened in the rest frame or after it has happened in the rest frame. All we can talk about is which event happens first and which later in one particular frame. We can talk about whether the LASER beam hits the front first or the back first in the rest frame. We can talk about whether the LASER beam hits the front first or the back first according to the platform frame. But we can't talk whether Sam observers the beam hitting the back before or after it has happened in the rest frame. There is no defined sense to this question in our current way of describing Physics.

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  • $\begingroup$ "Your question is how can it happen that the platform observer (Sam) perceives that an event happened before the rest frame observer (Dean) perceives it to have happened?" Actually no, i was not talking about who perceives it first. $\endgroup$ – xkvt Apr 26 '17 at 14:29
  • $\begingroup$ i was asking about sam perceving the event before it even happened in dean's frame, for dean to perceive it will obviously take twice the time $\endgroup$ – xkvt Apr 26 '17 at 14:30
  • $\begingroup$ Ok. I have edited my response accordingly. Have a look. $\endgroup$ – Feynmans Out for Grumpy Cat Apr 26 '17 at 14:40
  • $\begingroup$ Sam definitely percieves the event only after it happened in dean's frame, otherwise he would be seeing into the future! but here's two versions to consider: a) sam rapidly travels to the back of the train and as such, perceives the event immediately after it happens in the rest frame, by being closer to the back or b) back of the train rapidly moves forward towards the light beam, which strikes it around the quarter mark (if the train is at 0.99C), but since we kept the observer stationary, it takes some time for sam to percieve the event. Which makes both cases equivalent $\endgroup$ – xkvt Apr 26 '17 at 15:06
  • $\begingroup$ There is definitely something wrong with my two scenarios being equivalent but i can't quite pinpoint what it is $\endgroup$ – xkvt Apr 26 '17 at 15:09
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I'll admit that I can't quite pinpoint the issues in the question... partly because I think there are some possible confusions and misconceptions in the setup. I think @Mockingbird highlighted some of the misconceptions.

So, I offer a spacetime diagram on rotated graph paper (so we can visualize the tickmarks) to help clarify the situation.

Consider a train (whose rest length $L_{0}=10$) moving to the right with
velocity $v=\frac{AB}{OA}=(3/5)$, so that $\gamma=\frac{1}{\sqrt{1-v^2}}=\frac{OA}{OB}=(5/4)$.

Einstein train - spacetime diagram on rotated graph paper

At meeting event O [which both Station and Train assign (x=0,t=0)], light-signals are emitted.

Note: Station observes the Train to have length $L_{obs}=\frac{L_0}{\gamma}=\frac{10}{(\frac{5}{4})}=8$ (length contraction).
Thus, the Station says "the back of the train is half-a-train away: $OH=4$ units",
whereas the Train says "the back of the train is half-a-train away: $OH_0=5$ units".
(Note that Train says
$H_0=(x=-5,t=0)$ is simultaneous with event $O$, but
$H=(x=-5,t=3)$ is NOT simultaneous with event $O$ [although Station says H is simultaneous with O] (relativity of simultaneity).)

The rear-directed signal arrives at the back of the train at event $e_1$.

  • The Station says $e_1$ has spatial coordinate $x_{1,Station}=(-0.5L_{obs})+vt_{1,Station}$,
    where $t_{1,Station}=-\displaystyle\frac{x_{1,Station}}{c}$. So, $x_{1,Station}(1+v)=(-0.5\frac{L_0}{\gamma})$.
    Thus, $x_{1,Station}=\displaystyle\frac{(-0.5\frac{(10)}{\frac{5}{4}})}{1+\frac{3}{5}}=-2.5$ and so, $t_{1,Station}=2.5$.
    Station says $e_1=(x=-2.5,t=2.5)$.

  • The Train says $e_1$ has spatial coordinate $x_{1,Train}=(-0.5L_{0})$,
    where $t_{1,Train}=-\displaystyle\frac{x_{1,Train}}{c}$. So, $x_{1,Train}=(-0.5(10))=-5$ and so $t_{1,Train}=5$.
    Train says $e_1=(x=-5,t=5)$, which is 2 clock ticks after event $H$.

So, I hope this will clear up the confusion.

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protected by Qmechanic Apr 27 '17 at 17:31

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