# Gravitational time dilation and escape velocity

1. Jun 17, 2013

### Shaw

The time dilation caused by gravity on the surface of a planet is equal to the time dilation for an object moving at the planet's escape velocity in space. This can be proved using the Schwarzschild metric. GR doesn't explain why this is true. It seems to be an odd coincidence.

2. Jun 17, 2013

### Staff: Mentor

Meaning, I assume, in free space, far away from all gravitating bodies?

Yes, given the assumption I made above.

Why should GR have to explain it?

AFAIK, yes, it is.

3. Jun 17, 2013

### Mentz114

It's a conspiracy. No matter how hard you try nothing can ever be seen to reach the speed of light.

I tried to do this in GR by using a spacetime with two static sources separated by distance a. I found the geodesic of a test body released in the plane of symmetry at distance R from the axis on which the sources lie. The maximum velicity reached in the limit as R->∞ and a -> 0 is c, but only for an infinitessimal time as the body crosses the axis. I wrote this up ( amateurishly) in my blog.

4. Jun 17, 2013

### Staff: Mentor

I don't understand this comment. Isn't a proof that it is true the same as a proof why it is true. I.e. It is true because it must logically be true if the premises are true, as shown by the proof.

I just don't get what you are looking for. What is the difference between a proof that something is true and an explanation why it is true?

5. Jun 17, 2013

### Mentz114

It's a conspiracy. No matter how hard you try nothing can ever be seen to reach the speed of light.

I tried to do this in GR by using a spacetime with two static sources separated by distance a. I found the geodesic of a test body released in the plane of symmetry at distance R from the axis on which the sources lie. The maximum velicity reached in the limit as r->∞ and a -> 0 is c, but only for an infinitessimal time as the body crosses the axis. I wrote this up (amateurishly) in my blog.

I can't see how to undelete a blog entry so I'll put the url here

http://www.blatword.co.uk/space-time/GR-osc-2.pdf

This double post is from an attempt to edit ...

6. Jun 17, 2013

### WannabeNewton

If you approximate the Earth as a uniform spherical mass, drill a hole through the surface, and fall in you will execute simple harmonic motion. You can use this to show that the time it takes to complete one period (i.e. return to the point from where you jumped) is exactly equal to the time it takes for a low orbit satellite to circle the Earth once. As far as I know this is simply a nice coincidence. The point is sometimes coincidental results are just that and nothing more.

7. Jun 17, 2013

### George Jones

Staff Emeritus
Isn't this just Kepler's third law?

8. Jun 17, 2013

### WannabeNewton

I can't say I immediately see the connection; my textbook just makes it out to be a cool coincidence. Can it be used to explain the equality? That would be nice.

9. Jun 17, 2013

### Staff: Mentor

No, it isn't. The trajectory of the person jumping into the tunnel is just a degenerate case of an elliptical orbit.

10. Jun 17, 2013

### Bill_K

No it isn't. The "ellipse" in this example has two apogees.

11. Jun 17, 2013

### Staff: Mentor

Oops--good point. Now I have to think up another reason why it isn't a coincidence. Let's see...

<furrows brow>

Ah, yes, I can't remember where I first came across this argument, but it's pretty straightforward: consider the centripetal acceleration of the person in the circular orbit. (Of course I'm using Newtonian terminology here.) It has a constant magnitude (since the orbital altitude is constant), but its direction changes; relative to the original direction, the one where the circular orbiting person was just passing the person jumping into the tunnel, the circular orbital acceleration is more and more "sideways", until, at the point 90 degrees around the orbit, the acceleration is exactly "sideways", perpendicular to the original direction.

The key question is, how much sideways? Since the orbit is circular, the component in the original direction is just $cos \phi$, where $\phi$ is the angular coordinate going around the orbit. But $cos \phi$ is just $r / R$, where $R$ is the radius of the Earth and $r$ is the radial distance to the chord passing through the Earth, perpendicular to the original direction, that intersects the circular orbit (i.e., the Earth's surface) at angle $\phi$ from the original direction.

But $r / R$ is also the ratio of the acceleration of the person who jumps into the tunnel, when he is at radius $r$, to his starting acceleration, at radius $R$. (This is, of course, only true on the assumption of a spherical Earth of uniform density.) So basic geometry and trigonometry ensures that both periods will be the same, because the motions of both objects, projected onto the radial line going through the tunnel, are identical--same starting position, velocity, and acceleration, therefore same trajectory.

12. Jun 18, 2013

### A.T.

I think he is asking about the connection between the two distinct phenomena in this context: gravitational time dilation and motion time dilation.

Here an idea based on the the Equivalence Principle: Consider a rotating frame of reference, where you have artificial gravity.

- In the inertial frame the time dilation at the circumference is motion time dilation, due to tangential motion.

- In the rotating frame it is gravitational time dilation due to the centrifugal gravity.

But it is the same time dilation relative to the center, just explained differently. And the escape velocity in the rotating frame (to get from the circumference to the center) equals the tangential velocity at the circumference in the inertial frame.

13. Jun 18, 2013

### Staff: Mentor

Yes, I understand that. But my point is a general one. Any time that you have a proof of X you also have an explanation about why X. Specifically, X is true because the premises of the proof logically imply X, as shown by the proof.

The OP acknowledges that the equality between gravitational and motional time dilation can be proven, but claims that GR doesn't explain why that is true. I don't think that claim makes sense because the proof is an explanation why.

14. Jun 18, 2013

### Bill_K

In the Schwarzschild solution, grr gtt = 1. If I ask, "Why is grr gtt = 1?", what I mean is, Does it follow from some general, more intuitive principle, or is it just an inexplicable peculiarity of the Schwarzschild metric.

15. Jun 18, 2013

### Staff: Mentor

Wasn't there a thread about this recently? Not here, but over in "Classical" or "General"?

16. Jun 18, 2013

### A.T.

Yes, that's how I understand this type of question, that the OP asked. Maybe the OPs question can be answered based on the Equivalence Principle, by applying coordinate transformations to convert gravitational time dilation to motion time dilation. I gave a very simple example in post #12, and wonder if it could be generalized.

17. Jun 18, 2013

### Staff: Mentor

You prove that grr gtt = 1 by assuming a form of the metric with specific properties (spherical, static, vacuum) and solving the EFE for that form of the metric. Therefore, grr gtt = 1 because that is logically implied by the EFE and the assumed form of the metric.

If you make different assumptions or solve different equations then you get different results, so the reason "why" is the premises of the proof. I think what the OP is really asking for is something like what is the minimal set of assumptions or the most general assumptions that lead to the result in question. Because once you have a proof you have "why". Or perhaps the OP is asking if similar results occur in other spacetimes under different assumptions.

18. Jun 18, 2013

### WannabeNewton

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