# General Relativity - Circular Orbit around Earth

• unscientific
Recall that the correct expression for the light pulse is ds = 0, but ##dt \neq 0## and ##dr \neq 0##.But you just need to find ##\Delta \tau = \int d\tau## for the total round trip of the photon.In summary, we discussed finding the proper time in the rest frame of a particle, in the laboratory frame, and in a photon traveling from point A to point B in time P. We derived equations for the proper time in each frame and discussed the relationship between proper time and coordinate time for a light pulse traveling radially. We also found that the result for the proper time in the photon implies a relationship between the radius and the Schwar

## 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)..

unscientific said:
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?

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

unscientific said:
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)##?

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

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

TSny said:
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}##.

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

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

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 ##

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

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

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

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

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

Looks good.

TSny said:
Looks good.

I am stuck on part (b) though..

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

TSny said:
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}##.

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

TSny said:
...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##?

unscientific said:
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?

TSny said:
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##.

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

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TSny said:
Yes.
So does this mean the equation is simply:

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

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##?

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

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

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

TSny said:
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 ##

That's correct.

TSny said:
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.

Recall that the correct expression for the light pulse is $$ds^2 = -\left( 1 - \frac{2GM}{c^2r} \right) c^2 dt^2 + \left( 1 + \frac{2GM}{c^2r} \right) dr^2$$

##ds = 0##, but ##dt \neq 0## and ##dr \neq 0##.

But you just need to find ##\Delta \tau = \int d\tau## for the total round trip of the light. So, you don't need to worry about ##dt## or ##dr##.

TSny said:
Recall that the correct expression for the light pulse is $$ds^2 = -\left( 1 - \frac{2GM}{c^2r} \right) c^2 dt^2 + \left( 1 + \frac{2GM}{c^2r} \right) dr^2$$

##ds = 0##, but ##dt \neq 0## and ##dr \neq 0##.

But you just need to find ##\Delta \tau = \int d\tau## for the total round trip of the light. So, you don't need to worry about ##dt## or ##dr##.

Yeah, but ##d\tau = 0## so ##\int d\tau = 0##. So from the light's perspective, no time has passed at all. Does this even make sense?

unscientific said:
Yeah, but ##d\tau = 0## so ##\int d\tau = 0##. So from the light's perspective, no time has passed at all.
That's right. Of course, no physical observer can travel at the speed of light. But, in a thought experiment, you can imagine the observer moving at 0.99999...c (add as many 9's as you wish). The time elapsed for this observer will approach zero. It's the ultimate "twin paradox" scenario where the traveling twin would not age at all!
Does this even make sense?
It's strange, but it's a natural consequence of SR and GR.

unscientific