# General Relativity - Circular Orbit around Earth

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1. Feb 1, 2015

### unscientific

1. The problem statement, all variables and given/known data

(a) Find the proper time in the rest frame of particle
(b) Find the proper time in the laboratory frame
(c) Find the proper time in a photon that travels from A to B in time P

2. Relevant equations

3. The attempt at a solution

Part(a)

The metric is given by:

$$ds^2 = -\left( 1 - \frac{2GM}{c^2r} \right) c^2 dt^2 + \left( 1 + \frac{2GM}{c^2r} \right) dr^2$$

Circular orbit implies that $dr^2 = 0$, so
$$ds^2 = c^2 d\tau^2 = -\left( 1 - \frac{2GM}{c^2R} \right) c^2 dt^2$$
$$\left( \frac{d\tau}{dt} \right)^2 = \left( 1 - \frac{2GM}{c^2R} \right)$$
$$\frac{d\tau}{dt} = \sqrt { \left( 1 - \frac{2GM}{c^2R} \right) }$$

Since the time between event A and B is $dt = P$, the time experienced in the rest frame must be $d\tau = \sqrt { \left( 1 - \frac{2GM}{c^2R} \right) } P$?

I'm not sure how to approach parts (b) and (c)..

2. Feb 1, 2015

### TSny

Does $dx^2 + dy^2 +dz^2 = 0$ for two neighboring points on the orbit?

3. Feb 1, 2015

### unscientific

They are at the same radius, so yes.

4. Feb 1, 2015

### TSny

For general motion in the xy plane (z = 0), how would you express $dx^2+dy^2$ in polar coordinates $(r, \theta)$?

5. Feb 1, 2015

### unscientific

$$r^2sin^2 \theta d\phi$$

6. Feb 1, 2015

7. Feb 1, 2015

### unscientific

Oh I must have confused the r's. It should be:

$$ds^2 = -\left( 1 - \frac{2GM}{c^2r} \right) c^2 dt^2 + \left( 1 + \frac{2GM}{c^2r} \right) R^2 d\phi^2$$

Since for $z=0$ we let $\theta = \frac{\pi}{2}$.

8. Feb 1, 2015

### TSny

On the circular orbit of radius R, what is the value of little r?

9. Feb 1, 2015

### unscientific

r = R

10. Feb 1, 2015

### TSny

Right.

11. Feb 1, 2015

### unscientific

$$ds^2 = -\left( 1 - \frac{2GM}{c^2R} \right) c^2 dt^2 + \left( 1 + \frac{2GM}{c^2r} \right) R^2 d\phi^2$$

Using $d\phi = \omega dt = \frac{2\pi}{P} dt$

$$ds^2 = -\left( 1 - \frac{2GM}{c^2R} \right) c^2 dt^2 + \left( 1 + \frac{2GM}{c^2 R} \right) R^2 (\frac{2\pi}{P})^2 dt^2$$

12. Feb 1, 2015

### TSny

Not quite. Does $R^2P^2dt^2$ have the correct dimensions of distance squared?

13. Feb 1, 2015

### unscientific

Using $d\phi = \omega dt = \frac{2\pi}{P} dt$,

$$ds^2 = -\left( 1 - \frac{2GM}{c^2R} \right) c^2 dt^2 + \left( 1 + \frac{2GM}{c^2 R} \right) R^2 (\frac{2\pi}{P})^2 dt^2$$
$$-c^2 d\tau^2 = -\left( 1 - \frac{2GM}{c^2R} \right) c^2 dt^2 + \left( 1 + \frac{2GM}{c^2 R} \right) R^2 (\frac{2\pi}{P})^2 dt^2$$
$$\left( \frac{d\tau}{dt} \right)^2 = \left( 1 - \frac{2GM}{c^2R} \right) - \left( 1 + \frac{2GM}{c^2 R} \right) R^2 \left(\frac{2\pi}{Pc}\right)^2$$

14. Feb 1, 2015

### TSny

Looks good.

15. Feb 1, 2015

### unscientific

I am stuck on part (b) though..

16. Feb 1, 2015

### TSny

What is $dx^2 + dy^2 + dz^2$ for a fixed point on the orbit?

17. Feb 1, 2015

### unscientific

For a fixed point, it is $0$, so $\frac{d\tau}{dt} = 1 - \frac{2GM}{c^2R}$.

18. Feb 1, 2015

### TSny

...and $ds^2 =$?

19. Feb 1, 2015

### unscientific

Solving, we get $\frac{d\tau}{dt} = 1 - \frac{2GM}{c^2R}$.

20. Feb 1, 2015

### unscientific

For part (c), since beam of light travels at speed $c$ in all frames, is $dr^2 = c^2 dt^2 = c^2 P^2$?