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Assume there are two points in spacetime $a=(t,x,y,z)$ and $a'=(t',x',y',z')$. Let's say that the first one is in the origin of spacetime i.e. $a=(0,0,0,0)$. The point $a'$ has two possibilities

  1. $a'=(0,10cm,0,0)$ i.e. it's $10cm$ right of $a$
  2. $a'=(1ns,1cm,0,0)$ i.e. it's $1cm$ right of $a$ but $1ns$ ahead of it as well.

Someone want to send a signal from $a$ to $a'$. Is there a way that these two points(the two $a'$) receive the signal instantly either in space or time?

What I am trying to understand is that if someone can consider those points($a,a'$) to be identical either in space or in time.

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2 Answers 2

Calculate the spacetime interval $$\Delta s^2 = -c^2\Delta t^2 + \Delta x^2 + \Delta y^2 + \Delta z^2\text{.}$$ In this sign convention, $$\begin{cases} \Delta s^2 < 0\text{,}& \mbox{timelike separation}\\ \Delta s^2 = 0\text{,}& \mbox{lightlike separation}\\ \Delta s^2 > 0\text{,}& \mbox{spacelike separation}\\ \end{cases}$$

Someone want to send a signal from $a$ to $a′$. Is there a way that these two points(the two a′) receive the signal instantly either in space or time?

If the events are timelike-separated, there exists an inertial frame in which the events are at the same spatial location ('identical in space'?). Intuitively, if the spatial distance between them, $\sqrt{\Delta x^2 + \Delta y^2 + \Delta z^2}$ divided by the elapsed time $\Delta t$ is less than $c$, then it is possible for a subluminal signal to go between them at this speed $v$. Hence you can pick an inertial frame comoving with that subluminal signal, in which the signal will be stationary... and hence those two events at the same spatial location.

Similarly, if the events are spacelike-separated, there exists an inertial frame in which the events are simultaneous ('identical in time'?). Relative to the old inertial frame, it will have speed $c^2/v$, where $v$ is as above.

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For (1), the event $a'$ is co-located in time, i.e., simultaneous with event $a$ and thus the associated interval is space-like (the distance through space is greater than the distance through time in any frame).

There is no world line (time-like curve) that includes the events $a$ and $a'$ so, the event $a'$ cannot be the event "receive signal initiated at event $a$."

For (2), the temporal distance is $10^{-9} s \cdot 3 \times 10^{10} \frac{cm}{s} = 30\ cm$. Since the temporal distance is larger than the $1 \ cm$ spatial distance, the associated interval is time-like and thus, there is a frame of reference (the rest frame) in which the two events are co-located in space.

However, I think your notion of "sending a signal from $a$ to either $a'$" needs refinement. You should be thinking in terms of world lines and the event $a$ as "the event that a signal is sent" and event $a'$ as "the event that the signal is received". Thinking of it this way, it's clear that (1) cannot be an $a'$.

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