# Does compressing a mass to a smaller sphere change the curvature of space?

1. May 21, 2010

### TCS

If you take a homogenously distributed spherical mass and compress it to a smaller radius while maintaining its overall total energy and momentum, will it change the curvature of space (gravity) outside of the original sphere?

According to Newton, it stays the same. However, it seems like the shape/derivatives of the curves would change in GR.

2. May 21, 2010

### Ich

If I may interpret "maintaining its overall total energy" as keeping the http://en.wikipedia.org/wiki/Komar_mass" [Broken] constant, then the gravity outside the sphere is (trivially) also unchanged.
Basically, if no energy leaks from the sphere or flows into it, the outside geometry will be unchanged.

Last edited by a moderator: May 4, 2017
3. May 21, 2010

### TCS

I must be misunderstanding something because it seem like the curature of space would have to change within the original sphere and that changing the first derivative within the sphere would affect the higher order dierviatives outside the sphere.

If I was programing a ray tracing numerical solution, I would consider the steps inside the sphere for determining a solution outside the sphere.

Last edited by a moderator: May 4, 2017
4. May 21, 2010

### Jonathan Scott

It's a standard consequence of Birkhoff's Theorem that for a spherically symmetric static solution the shape of space outside a given radius does not depend on the radial density distribution of the matter inside that radius, but only on the total effective mass/energy. This applies even if the matter is shrunk into a black hole.

The word "compressing" suggests adding to the total energy, in which case the external curvature would be affected. Another word like "shrinking" might avoid this accidental implication.

5. May 22, 2010

### litup

I think you are confusing the close in gravitational field vs the uncompressed matter field. At the original radius, space does not know the difference but there certainly will be a big difference in the curvature of space near the newer smaller radius. If you were on a mass the size of the sun and massed a hundred times your earth mass, supposing you could stand on the surface of a sun, then if you compressed that sun to something like the radius of earth you would probably weigh thousands of times what you weighed on the sun before compression. A planet orbiting at a distance that was in a year long orbit would still be in a yearlong orbit around the much smaller compressed sun so the curvature of space AT THAT RADIUS is unchanged. But you can bet your boopy there would be a much greater curvature at the new radius. That's what makes black holes.

6. May 23, 2010

### D.S.Beyer

I had this same question and was convinced that the spacetime curvature would be different. But then I read 'A Breif History of the Universe' by Hawking and he nonchalantly said that the curvature would be the same....so I had to change my mind.

I am a very visual person so I made a diagram. Below is a side view of the classic Bowling Ball analogy of spacetime curvature. The large blue circle is the same mass as the small red one. Inside each you can see a bent line which indicates the curvature of spacetime within the planet (NOTE : I have never seen a graph of the interior curvature so I was just approximating. Assuming that the center is less but not none.) Thus one can see that the spacetime deformation does not change, except to "deepen".

7. May 23, 2010

### yuiop

I think you are on the right lines here for the curvature outside the circles, but inside the curvature should continue downwards because the potential continues to decrease inside the bodies and is at its lowest at the centre. If you reverse the curvature of the bent lines within the circles you will not be far off a reasonable aproximation of what it should look like.

8. May 24, 2010

### Jonathan Scott

I agree - inside the body the potential is still getting lower but the field (gradient) is decreasing towards the centre, so the curvature is upwards, and the minimum potential occurs in the middle of an upward curving line.

9. May 24, 2010

### D.S.Beyer

Kev,

In another post of https://www.physicsforums.com/showthread.php?t=404153" you state, "If we have an ordinary massive body like the Earth, we could in principle drill a hole to the centre and measure the distance with rulers and in that case the circumference of a ring around the Earth will be less than 2Pi*r where r is the measured radius from the centre of the Earth"

Does not this infer then that the curvature of space 'bends' upward in the center of planets? Assuming gravitational length contraction aligns with spacetime curvature.
Or am I lumping to many attributes into one line?

In another post of mine I tried to determined if the graph line of time dilation was the same as the graph lines of spacetime deformations (https://www.physicsforums.com/showthread.php?t=400908"). I only received one enigmatic answer that "It was close." Are the "bowling ball" graphs of time dilation, gravitational influence, gravitational length contraction, and spacetime geodesics IDENTICAL, SIMILAR, or CLOSE at all?

But for simplicity, one at a time, does the graph of gravitational length contraction correspond to the spacetime curvature graph? (see below...Again a little diagram).

Last edited by a moderator: Apr 25, 2017
10. May 24, 2010

### Jonathan Scott

The complication with talking about what happens to space is that it depends a lot on your coordinate system. The coordinate system is just a map. If you stretch it radially, it still describes the same thing, so the relationship of the radial coordinate to ruler length in the radial direction is a matter of convention, not physics.

The most practical coordinate system is an isotropic one, where the radial and tangential ruler sizes are assumed to vary in the same way relative to coordinate space. In that case, you can say that rulers shrink with potential (approximately as 1-Gm/rc2).

Roughly speaking, in the bowling ball model, the curve the ball makes in the surface is a representation of the potential. That is, the radius of curvature points downwards outside the ball, where the field is increasing, and upwards underneath the ball, where the field is decreasing.

However, the model is pretty misleading at best, so don't think too much about it - it is more of an analogy than a realistic model.

11. May 24, 2010

### DrGreg

There are two sorts of diagram you can draw, which superficially look very similar -- a "well" or depression within an otherwise nearly-flat plane -- but which represent very different things.

The first is a diagram of gravitational potential, a "gravity well", where the slope represents "acceleration due to gravity" and height is related to time dilation.

The second represents the curvature of space (not spacetime) where distance along the curved surface represents "ruler distance" measured in space by a hovering observer, as suggested in post #9. Horizontal radius represents the Schwarzschild r coordinate. Outside of a spherically symmetric non-rotating mass it is called a Flamm's paraboloid. I don't know if it's possible to do the same thing inside a mass, but if it were, it wouldn't matter whether you drew it as a "hill" or a "hole", so you might as well draw it as a "hole" to keep the surface smooth.

(Note: for black holes, the Flamm's paraboloid stops, vertically, when you reach the event horizon.)

12. May 24, 2010

### Passionflower

To avoid confusion, the r coordinate in the Schwarzschild metric does not describe a physical radial distance.

Last edited: May 24, 2010