0
$\begingroup$

I apologize to ask, again and again, a question that seems to come back often, but I believe this is new.

I am trying to test the equivalence principle applied to two similar situations:

  1. Flat spacetime. An inertial observer Alice (equiped with a set of coordinates ($x$,$t$)) is watching a non-inertial observer Bob (proper time $\tau'$) uniformly accelerating toward her with a constant proper acceleration $\alpha$. At time t=0, they are separated by a distance $L$. What is the relationship between d$\tau'$ and d$t$ ? For $L=0$, Rindler says (after some work)

$d\tau' = \frac{dt}{\sqrt{1+(\frac{\alpha t}{c})^2}}$

I just wanted to make sure this is true for $L\neq0$.

  1. Non-rotating mass $M$. Bob, still a non-inertial observer with proper time $\tau'$, is stationnary with respect to the mass $M$. Alice is radially free falling toward Bob, from an initial distance $L$ between the two. I don't believe the Schwarzschild coordinates apply here, because Alice (equiped with her set of coordinates $(x,t)$) is free-falling, she is not "far away" from Bob and stationnary to him. (I do know the gravitational time dilation formula). Alice sees Bob (and an entire planet) uniformaly accelerating toward her with a proper acceleration $\alpha$. Again, what is the relationship between $dt$ and $d\tau'$ ?

Can you help me ? Please do not use $c=1$, or $G=1$. Do we agree that, in the situation 2), Bob feels a proper acceleration of

$\alpha = \frac{1}{\sqrt{1-\frac{r}{r_{s}}}}\frac{GM}{r^2}$

Using the popular convention?

Thank you !

$\endgroup$

1 Answer 1

1
$\begingroup$

In the first problem, Bob's position relative to Alice and/or relative to $x=0$ is irrelevant, since only velocity and acceleration figure into the calculation. You need to specify that Bob is at rest at $t=0$. With that additional assumption, that formula is correct.

In the second problem, Bob's proper acceleration is correct, but it isn't clear what the $(x,t)$ coordinates are supposed to be. You can't just put inertial coordinates on this spacetime, since it isn't flat. You could try to find "approximately inertial" local coordinates, but they would be either inexact or very ugly. You could also calculate quantities actually measurable by Alice, such as Bob's apparent Doppler shift and angular size as a function of Alice's proper time, and compare them to the same quantities calculated in the special relativistic problem, which would be easier but tedious. Either way the results would be similar to the special-relativistic case but not identical since there is spacetime curvature.

$\endgroup$
1
  • $\begingroup$ Thank you. I am going to think about doppler effects in the two cases and I'll come back to you. $\endgroup$ Commented Sep 29, 2020 at 22:47

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.