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I am a beginner of general relativity. I read and follow the notation in Andrew Steane's book, Relativity Made Relatively Easy, chap 9.2. Suppose there is an rigid accelerating frame with constant proper acceleration with respect to inertial frame then the motion of grids of the frame is a set of hyperbolas. There is a relationship between inertial coordinate and Rindler frame: (Suppose $c=1$)

$$ \begin{cases} t=h\sinh(\theta)\\ x=h\cosh(\theta) \end{cases} $$ , where $(t, x)$ are inertial frame coordinates and $(h, \theta)$ are the hyperbolic coordinates. The coordinate lines are hyperbola. $\theta$ is a measure of local time (not local proper time).

Now there is a light signal sent from $h_0$ to $h_1$ then and there is a mirror at $h_1$ and once the signal arrives at $h_1$ it is immediately reflected back to $h_0.$

enter image description here

The calculations of both trips are:

$$ \begin{cases} (h_0\sinh(\theta_0),h_0\cosh(\theta_0))+(\Delta t,\Delta t)=(h_1\sinh(\theta),h_1\cosh(\theta))\\ (h_1\sinh(\theta),h_1\cosh(\theta))+(\Delta t,-\Delta t)=(h_0\sinh(\theta_1),h_0\cosh(\theta_1)) \end{cases} $$

I got two relationships:

$$ e^{\theta-\theta_0}=\frac{h_1}{h_0}\rightarrow\theta-\theta_0=\ln(\frac{h_1}{h_0})\\ e^{\theta_1-\theta}=\frac{h_1}{h_0}\rightarrow\theta_1-\theta=\ln(\frac{h_1}{h_0}) $$

So the round-trip time is independent when the light is sent and the time is the same in both direction.

Now if there are three observers with equal distance in the accelerating frame ($h_1-h_0=h_2-h_1$), and they send light to the middle observer at their 12:00pm according to their local clock

enter image description here

Then I got a conclusion from the above calculation that the middle observer does not get synchronous signals, because

$$ \theta_2-\theta_1=\ln(\frac{h_2}{h_1}) \neq\theta_1-\theta_0=\ln(\frac{h_1}{h_0}) $$

which also means the clocks at $h_0$ and $h_2$ are not synchronous. However, according to my intuition, I know the grids are rigid, speed of light is a constant, and those two observers remain the same distance with the middle one and according to the diagram, those local clocks should be synchronous so the middle observer should receive synchronous signals in this experiment .

So what did I do it wrong here?

Thanks for your patience

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1 Answer 1

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Now if there are three observers with equal distance in the accelerating frame ($h_1-h_0=h_2-h_1$), and they send light to the middle observer

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Then I got a conclusion from the above calculation that the middle observer does not get synchronous signals

That is correct, if you use the condition $h_1-h_0=h_2-h_1$ then the middle observer does not get synchronous signals. For the middle observer to get synchronous signals you need to use $h_1/h_0 = h_2/h_1$.

This is because of the metric. In the Rindler frame the metric is $ds^2=-h^2 d\theta^2 + dh^2$. So what this result shows is not so much an issue of simultaneity as much as an issue that in this frame ruler distances and radar distances are not the same. Your $h$ coordinate gives a ruler distance, but your experiment is based on radar distances.

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  • $\begingroup$ I got it. Thanks for your reply $\endgroup$
    – Hsu Bill
    May 30 at 9:37

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