# Trying to understand GR and density

1. Jun 25, 2010

### diagopod

I've been trying to get a sense of the differences b/w Newtonian gravity and GR. One thing I was curious about. I'm reasonably certain that Newtonian gravity predicts that two objects of the same mass, but different densities, exert the same gravitational force on a given object at distance R from their centers. Does GR predict that same equivalence, or does the density of the object have an impact on the gravity at distance R?

2. Jun 25, 2010

### Mentz114

In the spherically symmetric non-rotating exterior solution of the Einstein equations ( Schwarzschild metric) only the mass appears in the equations. From which we could deduce that the constitution of the mass is not relevant.

3. Jun 25, 2010

### Passionflower

I have to disagree with your conclusion. The Schwarzschild solution is a vacuum solution thus density of matter does not come into question. So there is nothing to deduce from with respect to density.

Generally speaking in GR an increase in density (and pressure) means an increase in gravitational strength.

Last edited: Jun 25, 2010
4. Jun 25, 2010

### diagopod

Thanks. Just to make sure I follow you, are you saying that if Observer A is the same distance from two bodies B and C, both with the same rest mass, but with C having twice the density of B, Observer A will experience the same attractive "force" to B and C? Or will A experience a greater attraction to C b/c its greater density translates into greater gravitational strength? Thanks again for your guidance on this.

5. Jun 26, 2010

### Phrak

I have to disagree. In the Schwrzchild metric, the value M already accounts for pressure. Density doesn't enter, at all.

6. Jun 26, 2010

### bcrowell

Staff Emeritus
What Mentz114 is referring to is Birkhoff's theorem: http://en.wikipedia.org/wiki/Birkhoff's_theorem_(relativity [Broken]) , and he's carefully stated the necessary conditions for B's theorem to apply. Based on B's theorem, we know that, for example, if a star collapses and becomes more dense, there will be no observable effect on the exterior field, provided that spherical symmetry is maintained.

If we want to generalize this to, e.g., cases without spherical symmetry, I don't know how to say anything definitive. There is no uniquely defined notion of gravitational mass in GR. For a particular definition of gravitational mass, we could ask whether that quantity is conserved. But GR doesn't allow us to state general conservation laws.

Last edited by a moderator: May 4, 2017
7. Jun 26, 2010

### JesseM

But for a massive nonrotating sphere of mass M and uniform density, the spacetime at any radius beyond the surface of the sphere has the same curvature as the Schwarzschild solution for a black hole of mass M at that radius, it's only once you hit the surface that the two metrics differ (see this thread and this one for info one what the metric inside the surface would be)

8. Jun 27, 2010

### bcrowell

Staff Emeritus
IMO this is a little bit of an oversimplification. The Schwarzschild metric is the unique spherically symmetric vacuum solution, and it has only one adjustable parameter. We simply *call* this parameter the mass. It doesn't follow from that that if we allow a spherically symmetric massive body to compress under its own gravity, its external field will remain the same. To determine that, you need three things:
-1. Birkhoff's theorem
-2. a belief that some form of conservation of mass-energy should hold
-3. the ability to match boundary conditions with a non-singular spherical body in the Newtonian limit
If you believe in those premises, then the reasoning goes like this. From Birkhoff's theorem, we know that the compression can't cause any gravitational radiation. Since it can't radiate, it can't lose any mass-energy. Based on 3, we know that the mass-energy matches up, at least in the Newtonian limit, with the parameter M appearing in the Schwarzschild metric.

It may seem like I'm just being pedantic, but there really are some nontrivial issues involved here. E.g., there are multiple ways of extending the definition of mass from the Newtonian limit to GR. If you don't fill in the steps in the reasoning, the whole thing becomes tautological: we define gravitational mass based on what the body's exterior field looks like, and therefore it's merely a matter of definition to say that all bodies with a given mass have the same exterior field.

9. Jun 27, 2010

### Phrak

It's difficult to tell where the disagreements lay, though M determines both the interior and exterior solution.

If I take a pound of stuff and shape it into a sphere it will have a different M than if I take the same stuff and compress it. The change in density goes along with the compression, as no material is perfectly incompressible. I think this is well understood by all, but the source of disagreement in how it has been talked about.

10. Jun 27, 2010

### Yuripe

If bodies B and C have the same mass but body C is twice the density of B, than body C has smaller radius then B.
So distance between A and the centers of B/C as stated is the same, but the distance from A to the surface of B is less then to the surface of C.
As gravity of the body fades with the square starting from its surface, than attractive force of body C should be less then of body B to the body A.

11. Jun 27, 2010

### atyy

Can I add to the pedantry by saying that we also don't know what R means in the original question - is it the radial parameter in the combined interior and exterior solution, or just the radial parameter from where the exterior solution takes over, or the proper distance from the centre of the interior solution?

12. Jun 27, 2010

### atyy

13. Jun 28, 2010

### diagopod

Good point. So suppose one takes a simpler case, where compression would not seem to result in additional mass-energy. Suppose, for example, one had a spherical cluster of relatively motionless asteroids, each with a rest mass of 10^5kg, and each 10 meters on average from its surrounding neighbors in the cluster. Then take another cluster, identical in number of asteroids and rest mass per asteroid, but with only 5 meters on average between each asteroid and its surrounding neighbors. Then suppose we place a billiard ball far away in space, equidistant from the centers of the two asteroid clusters. Will the ball move toward one rather than the other, or will its acceleration, if any at all, have no bias for one rather than the other?

14. Jun 28, 2010

### Yuripe

Volume increases with power of 3 with radius, and gravity decreases with power of 2 with radius. Value of gravity is determined by mass for a given volume.

Within solid body starting from its center and going outwards volume increase adds more mass and overpowers gravity's decrease with distance so overall gravity value increases toward the surface of this body. When you cross the surface and continue to move outwards volume still increases but there is no additional mass increase. This puts radius as main determinant of gravity's value.

With asteroid field same principle applies only the mass increase is smaller because of lower overall density of given volume.