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1. Nov 1, 2014

Kyrios

1. The problem statement, all variables and given/known data
An observer is orbiting at a radius r = 3GM, $\theta = \frac{\pi}{2}$ and $\phi = \omega t$ where w is constant.
The observer sends a photon around the circular orbit in the positive $\phi$ direction. What is the proper time $\Delta \tau$ for the photon to complete one orbit and return to the observer?

2. Relevant equations
Schwarzschild line element where dr =0 and $d\theta$ =0.

3. The attempt at a solution

From the line element we have
$$\left(\frac{d\tau}{dt}\right)^2 = 1 - \frac{2GM}{r} - r^2 \left(\frac{d\phi}{dt}\right)^2$$
$$\left(\frac{d\tau}{dt}\right)^2 = 1 - \frac{2GM}{r} - r^2 \omega^2$$

I was trying to use $\omega^2 = \frac{GM}{r^3}$ but that just gives

$$\left(\frac{d\tau}{dt}\right)^2 = 1 - \frac{2GM}{r} - r^2 \frac{GM}{r^3}$$
$$\left(\frac{d\tau}{dt}\right)^2 = 1 - \frac{2GM}{r} - \frac{GM}{r}$$
$$\left(\frac{d\tau}{dt}\right)^2 = 1 - \frac{3GM}{3GM}$$
as r = 3GM. This gives zero and I'm not really sure what to do with it. Have I gone wrong somewhere? What should I do with the w?

2. Nov 1, 2014

Orodruin

Staff Emeritus
Light follows light-like (naturally) geodesics. These have the defining property of ds=0 so it shoukd come as no surprise that it has length zero.

This makes me suspect that the problem wants you to compute the proper time for an observer at fixed spatial coordinates before the light signal comes back from the opposite direction.

3. Nov 1, 2014

Kyrios

I have already worked out the proper time for a stationary observer, $\Delta \tau = 6 \pi GM$. The question specifically states that the observer is not at fixed coordinates, but moving with $\phi = \omega t$ and emitting a photon while orbiting.

4. Nov 1, 2014

Orodruin

Staff Emeritus
I see, so note that the observer according to the problem statement is not necessarily moving on a geodesic. When you put $\omega=0$ you should regain your result for the stationary observer.

5. Nov 1, 2014

Kyrios

How can I simply assume that $\omega = 0$? It says that w is a constant. Since the previous question was for the case of the stationary observer, I'm quite sure that this one isn't after the same answer. Although I cannot see any way around it

6. Nov 1, 2014

Orodruin

Staff Emeritus
You cannot. What I said was that, when you have solved your problem, letting $\omega = 0$ should give you the result you had before. Your answer to the current problem should depend on $\omega$, which you can treat as an arbitrary constant (but with $\omega^2 < 3GM/r^3$ or you will have an observer following a non-time-like curve - as you found out for the limiting case already).

Edit: It is not necessary to solve the problem, but note that, regardless of the value of $\omega$, the observer will have a non-zero proper acceleration.