# Curvature inside a cavity

1. Mar 1, 2008

### Xeinstein

Inside a spherical cavity centered at the earth's center, is the space-time curvature is 0 or not 0? Would the clock run more slowly?

2. Mar 1, 2008

### Staff: Mentor

I am no GR expert, but my understanding is that there is 0 curvature. Light rays would not be deflected, nor would projectiles, and clocks at different points within the cavity would run at the same rate. These clocks would run slower than clocks on the surface because photons that went out from the cavity would be gravitationally redshifted as they climed.

3. Mar 1, 2008

### Xeinstein

Is it possible that clocks run slower and curvature is zero?
In other words, wouldn't these two contradict each other?

Last edited: Mar 1, 2008
4. Mar 1, 2008

### nanobug

Zero curvature.

Relative to what observer?

Last edited: Mar 1, 2008
5. Mar 1, 2008

### Xeinstein

Inertial observer at outer space away from any mass

Last edited: Mar 1, 2008
6. Mar 1, 2008

### George Jones

Staff Emeritus
No.

See this post.

7. Mar 1, 2008

### Staff: Mentor

No. My understanding is that there is no curvature within the cavity, so different clocks within the cavity will run at the same rate. Between the cavity and the surface is a region of curvature, and clocks within the cavity will run slower than clocks at the surface.

8. Mar 1, 2008

### Xeinstein

I'm not sure if I understand you answer. What's the relationship between clock rate and curvature?
Can you tell me if clock rate depend on "gravitational potential" or curvature?
So it's possible that clocks run slower in a region curvature is zero compared with clock at infinity

9. Mar 1, 2008

### George Jones

Staff Emeritus
Times are determined by the metric $g$, and the metric is neither constant nor zero in this situation.

10. Mar 1, 2008

### George Jones

Staff Emeritus
Maybe a hand-waving argument will help.

The gravitational potential is lower at the centre of the Earth than on the surface of the Earth. Consider a photon that is fired up a tunnel starting at the centre of the Earth. As the photon rises, it gains gravitational potential energy at the expense of its intrinsic energy. The wavelength of a photon is inversely proportional to its energy, so its wavelength increases as it rises. The photon experiences a gravitational redshift, just as DaleSpam said.

This argument is only suggestive; the math tells what actually happens.

11. Mar 2, 2008

### Staff: Mentor

My GR is not strong enough to really answer your question completely. But with the caveat that I am fairly ignorant here and may very well be wrong:

My understanding is that curvature always exists, but "gravitational potential" only exists in certain solutions to the Einstein field equations. For example, you can make a gravitational potential for a non rotating sphere but not a rotating sphere. So I believe that the clock rate cannot really be said to depend on gravitational potential in general since it does not exist in general. However, in solutions where it does exist the potential depends on the curvature so anything that would depend on the potential would also depend on the curvature.

The easy way to think about clock rate dependency is to think about where you get gravitational redshift. You don't get redshift within the cavity, but you do get redshift between the cavity and the surface. So clocks within the cavity run at the same rate, but slower than clocks at the surface.

12. Mar 2, 2008

### A.T.

13. Mar 2, 2008

### Xeinstein

The curvature of outer space, away from any mass, is zero and inside the cavity is also zero. Is it true that the clock at outer space run faster than the clock inside the cavity, why it is so?

Last edited: Mar 2, 2008
14. Mar 2, 2008

### A.T.

Because there is a region with spacetime-curvature between "away from any mass" and "inside the cavity". When you move a clock from "far away" trough that region, it's rate slows down (compared to a clock "far away"). When it reaches the cavity it is already going slower by a certain ratio, than the far-away-clock. But moving it around within the cavity doesn't change that ratio anymore.

In simple terms: On the "gravitational potential". And curvature exists where the "gravitational potential" changes.