Time dilation in gravitational field

[Wingeer]

Homework Statement

I recently had a test in the theory of relativity, and there was one task which I could not solve. This one has bothered me since the day I had the test.
It was a compound task about GPS-satellites.
There is no tricky calculations, or anything like that. The task aims at proving an expression.

$$\frac{{t_e }}{{\sqrt {1 - \frac{{2\gamma M}}{{c^2 r}}} }} = \frac{{t_s }}{{\sqrt{1 - \frac{{2\gamma M}}{{c^2 (r + h)}}} }}\$$

Where $${t_e }\$$ is the time on Earth's surface, $${t_s }\$$ is the time for the satellite, r is the radius from the center of Earth to the surface.

Homework Equations

$$\tau = \sqrt {1 - \frac{{2\gamma M}}{{c^2 r}}} t\$$

The Attempt at a Solution

I've tried a lot of editing these expressions, but it didn't get me anywhere. I really do not now how to prove the expression. I do imagine that there are some theory I need to figure, so that I can do something quirky with the expressions or something.

 

[Wingeer]

Oh. It looks like something happened to the LaTex-codes. Any ideas of what to do?

 

[nlightner]

Dear student,

Thank you for reaching out for help with your test question. Time dilation in a gravitational field is a fundamental concept in the theory of relativity. It refers to the idea that time moves slower in a strong gravitational field compared to a weaker one. This is due to the curvature of spacetime caused by massive objects.

In the case of GPS satellites, they are orbiting the Earth at a distance of about 20,000 km. This means they are in a weaker gravitational field compared to the Earth's surface. As a result, time moves slightly faster for the satellites compared to the surface of the Earth.

The expression you have been given is known as the gravitational time dilation formula. It relates the time experienced by an observer in a gravitational field (t) to the time experienced by an observer in a weak gravitational field (τ). In this case, the observer on the Earth's surface is experiencing time (t_e) and the observer on the satellite is experiencing time (t_s).

To prove this expression, you can use the equation you have provided:

\tau = \sqrt {1 - \frac{{2\gamma M}}{{c^2 r}}} t\

First, we need to define the variables in the equation:

- \tau is the proper time experienced by the observer in a weak gravitational field (in this case, the satellite)
- \gamma is the gravitational constant
- M is the mass of the Earth
- c is the speed of light
- r is the radius from the center of the Earth to the observer in a weak gravitational field (in this case, the satellite)

Next, we can substitute in the values for the satellite and the Earth's surface into the equation:

\tau_s = \sqrt {1 - \frac{{2\gamma M}}{{c^2 (r + h)}}} t_s\

\tau_e = \sqrt {1 - \frac{{2\gamma M}}{{c^2 r}}} t_e\

Where h is the altitude of the satellite above the Earth's surface.

Now we can rearrange the equation to get the expression you were given:

\frac{{\tau_e }}{{\tau_s }} = \frac{{t_e }}{{t_s }}\frac{\sqrt {1 - \frac{{2\gamma M}}{{c^2 (r + h)}}} }{{\sqrt {1 - \frac{{2\gamma M}}{{c^2 r}}}