# General Relativity - Circular Orbit around Earth

## Homework Statement

(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 ## The Attempt at a Solution

Part(a)
[/B]
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)..

TSny
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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##

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

Does ##dx^2 + dy^2 +dz^2 = 0## for two neighboring points on the orbit?
They are at the same radius, so yes.

TSny
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They are at the same radius, so yes.
For general motion in the xy plane (z = 0), how would you express ##dx^2+dy^2## in polar coordinates ##(r, \theta)##?

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

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

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}##.

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

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

TSny
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Right.

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

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

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

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

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

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

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

TSny
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Looks good.

Looks good.

I am stuck on part (b) though..

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

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

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

TSny
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...and ##ds^2 = ##?

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

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##?

TSny
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For part (c), is ##dr^2 = c^2 dt^2##?
No. ##dr/dt## for the radially traveling light pulse is just a "coordinate speed of light" and need not equal c.

But no matter what the coordinate speed of light, what is ##ds## for any two neighboring space-time events on the worldline of a light pulse?

No. ##dr/dt## for a light pulse is just a "coordinate speed of light" and need not equal c.

But no matter what the coordinate speed of light, what is ##ds## for any two neighboring space-time events on the worldline of a light pulse?

Time-like, no rest frame. so ##ds = 0##.

TSny
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It's neither spacelike or timelike. It's "lightlike". Yes, ds = 0.

Last edited:
Yes.
So does this mean the equation is simply:

$$0 = -\left( 1 - \frac{2GM}{c^2R} \right) c^2 dt^2 + \left( 1 + \frac{2GM}{c^2r} \right) R^2 d\phi^2$$
$$0 = -\left( 1 - \frac{2GM}{c^2R} \right) c^2 dt^2$$

TSny
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The light is traveling radially. So, what is ##dx^2 + dy^2 + dz^2## in polar coordinates?

But, you are interested in the proper time. How is ##ds## related to ##d\tau##?

The light is traveling radially. So, what is ##dx^2 + dy^2 + dz^2## in polar coordinates?

But, you are interested in the proper time. How is ##ds## related to ##d\tau##?

Since it is only moving radially, ##d\phi = 0## so ##dx^2 + dy^2 = dr^2##.

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

$$ds^2 = -c^2 ~ d\tau^2 = 0$$

TSny
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$$ds^2 = -\left( 1 - \frac{2GM}{c^2r} \right) c^2 dt^2 + \left( 1 + \frac{2GM}{c^2r} \right) dr^2$$
OK, but you don't need the metric for part (c).

You know ##ds^2 = -c^2d\tau^2##. So, for an increment along the light path, what is the change in proper time ##d\tau##?

OK, but you don't need the metric for part (c).

You know ##ds^2 = -c^2d\tau^2##. So, for an increment along the light path, what is the change in proper time ##d\tau##?

I don't understand why it isn't simply ## ds^2 = -c^2 ~ d\tau^2 = 0 ##

TSny
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That's correct.

That's correct.
So the result doesn't really say anything other than:

$$0 = -\left( 1 - \frac{2GM}{c^2R} \right) c^2 dt^2$$

which means either ##dt = 0## which is not possible since ##dt = P##. Therefore it implies that ##R = \frac{2GM}{c^2}##, which happens to be the schwarzschild radius.