General Relativity - Circular Orbit around Earth

[unscientific]

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

Homework Equations

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]

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?

 

[unscientific]

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

They are at the same radius, so yes.

 

[TSny]

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 $(r, \theta)$?

 

[unscientific]

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]

See equation 6 here: http://mathworld.wolfram.com/PolarCoordinates.html
But if you want to use $\phi$ instead of $\theta$, that's ok.

 

[unscientific]

See equation 6 here: http://mathworld.wolfram.com/PolarCoordinates.html

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]

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

 

[unscientific]

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

r = R

 

[TSny]

Right.

 

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

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

 

[TSny]

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

 

[unscientific]

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]

Looks good.

 

[unscientific]

Looks good.

I am stuck on part (b) though..

 

[TSny]

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

 

[unscientific]

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]

...and $ds^2 =$?

 

[unscientific]

...and $ds^2 =$?

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

 

[unscientific]

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

 

[TSny]

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?

 

[unscientific]

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]

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

 

[unscientific]

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]

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 $d\tau$?

 

[unscientific]

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]

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

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 $d\tau$?

 

[unscientific]

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]

That's correct.

 

[unscientific]

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.