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I am having problems taking the time it takes for a light signal from an event to reach an observer into account: For instance, if we have two observers $A$ and $B$ who synchronize their clocks when they pass each other at the origin $O$ and $A$ then emits a light pulse at a time $t$ as measured on their clock, we have the event:

$$X_{1}^\mu=\left(\frac{t}{c},0,0,0\right) \implies X_{1}'^\mu=\left(\frac{\gamma t}{c},-\frac{\beta \gamma t}{c},0,0\right)$$

However, if I'm correct, then $X_{1}'^{\mu}$ is not the time that $B$ observes the signal, it is merely the time which the event takes place in the reference frame at which $B$ is at rest. Therefore we have the actual time that $X_{1}$ is observed in $B$ as:

$$t_{1}'=\gamma t + \frac{\beta \gamma t}{c^{2}}=t\left(\gamma+\frac{\beta\gamma}{c^{2}}\right)$$

Therefore, if $B$ emits a light signal in response $2t$ after they receive the light signal, we have:

$$X_{2}'^{\mu}=\left(\frac{t}{c}\left(2+\gamma + \frac{\beta \gamma}{c^{2}}\right),0,0,0\right) \\\implies X_{2}^\mu = \left(\frac{\gamma\left(2+\gamma+\frac{\beta \gamma}{c^{2}}\right)t}{c},\frac{\beta\gamma\left(2+\gamma+\frac{\beta \gamma}{c^{2}}\right)t}{c},0,0\right)$$

And therefore, the time that $A$ receives the signal would be:

$$t_{2}=\gamma\left(2+\gamma+\frac{\beta\gamma}{c^{2}}\right)t+\frac{\beta\gamma\left(2+\gamma+\frac{\beta\gamma}{c^{2}}\right)}{c^{2}}$$

However, this expression doesn't look right to me as it doesn't simplify and has factors of $c^{2}$, but I'm not sure where I've erred.

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is not the time that B observes the signal, it is merely the time which the event takes place in the reference frame at which B is at rest.

But that's what observe means in SR. From the Wikipedia article "Observer (special relativity)":

Physicists use the term "observer" as shorthand for a specific reference frame from which a set of objects or events is being measured. Speaking of an observer in special relativity is not specifically hypothesizing an individual person who is experiencing events, but rather it is a particular mathematical context which objects and events are to be evaluated from. The effects of special relativity occur whether or not there is a sentient being within the inertial reference frame to witness them.


Therefore we have the actual time that X1 is observed in B as:

No, the time coordinate of event 1 in the primed frame is the 'actual' time the event is observed by B:

$$t'_1 = \gamma t$$

The event that a photon from event 1 arrives at the origin of B's frame is an entirely different event (let's call it event 2) and the time coordinate of that event is not the 'actual' time event 1 is observed by B:

$$t'_2 = t'_1 + \beta \gamma t$$

The point is that seeing (or photographing) and observing, in special relativity, are not synonymous. For further reading, see "Photography in Special Relativity"

In Special Relativity there is a very specific meaning to the word observer. An observer is one of an infinite collection, in space and time, of robots whose sole task in life is to record the time and location of each detected event. When we say that an observation was made we simply mean that the space and time coordinates of an event have been recorded.

A photograph is a record of all of the photons that were received at the focal point of the camera at the time the photograph was taken.

These are fundamentally different processes and they give us very different information about the universe.

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  • $\begingroup$ Thank you for your answer; I'm a little confused as to why you have: $$t'_{2} = t_{1}'+\beta\gamma t$$ Rather than $$t'_{2}=t_{1}'+\frac{\beta\gamma t}{c^{2}}$$ I thought that the distance the photon had to travel (at speed $c$) was $\frac{\beta \gamma t}{c}$? $\endgroup$ Commented Mar 31, 2014 at 16:00
  • $\begingroup$ @Shaktal, if $v$ is the relative speed of the primed frame in the unprimed frame, the spatial coordinate in the primed frame for event 1 is $x'_1 = -\gamma vt$. The time required for a photon to propagate from $x'_1$ to the primed origin is then $\Delta t' = \frac{\gamma vt}{c} = \beta \gamma t$ $\endgroup$ Commented Mar 31, 2014 at 16:29

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