# GR Vacuum solutions

1. Sep 8, 2010

### TrickyDicky

I would like to have a conceptually better understanding of GR solutions in the absence of matter, and a little difficulty I usually run into is that seemingly since in these solutions we have both the Einstein and stress-energy tensors equal to zero (which is what it means to have no matter-energy there) we are ignoring the curvature produced by any mass, however, when applying the solution to the Mercury precession problem or the bending of light by the sun problem, we are actually introducing the mass of the sun to solve them, and this to me seems a bit contradictory with the premise that there is no matter in the manifold under consideration.
Is there a simple way to explain away this false contradiction?
My own way to explain this is that actually in the Schwartzschild metric we have the expresion (1-2m/r) which seems to imply that we are substracting(that is close to ignoring) the curvature produced by the spherical symmetric mass from the total spacetime curvature, but I'm not sure if this is really so or even has any logic, it just helps me understand conceptually the premise of this vacuum solution.

2. Sep 8, 2010

### tom.stoer

The Schwarzschild solution has two patches matching at the Schwarzschild radius. The outer solution is the familiar vacuum solution, whereas there is an inner solution which is NOT a vacuum metric and wich therefore differs from the well-known inner vacuum solution used for a black hole. Instead a spherical symmetric, non-rotating, incompressible fluid is used which leads to a regular solution w/o singularity at r=0.

3. Sep 8, 2010

### TrickyDicky

Thanks but you answer has nothing to do whatsoever with the point of my question, nothing to do with the inner solution or black hole, I am referring to the outer solution only and the apparent incongruence of using the sun's mass in a solution whose premise is the absence of mass.

4. Sep 8, 2010

### Staff: Mentor

Conceptually this is similar to EM, where charge is the source of the field, but vacuum solutions exist. Specifically, in EM the charge is proportional to the divergence of the E-field, but a specific region of vacuum may have a field due to charge located elsewhere. Similarly, in GR the stress energy tensor is proportional to the Ricci curvature and the curvature scalar, but a specific region of vacuum may have a Riemann curvature due to stress-energy located elsewhere.

5. Sep 8, 2010

### tom.stoer

That's not correct; the outer solution IS a vacuum solution (vacuum outside the sun) generated by a mass confined to the sun's volume (it is not a global vacuum with T=0 everywhere).

Look at classical electrodynamics: the Coulomb potential is a vacuum solution of the Maxwell equation valid outside the charge distribution = "in vacuum".

6. Sep 8, 2010

### bcrowell

Staff Emeritus
Tom, maybe I'm misunderstanding you, but this seems incorrect to me. It's true that in the case of an object that's not a black hole, you would have to have a separate equation for the interior of the object, and this would be a non-vacuum solution without a singularity. However, the surface where this is glued onto the exterior vacuum spacetime is not at the Schwarzschild radius. In the case you're talking about, the Schwarzschild radius is less than the radius of the object, and nothing physically happens at the Schwarzschild radius. I think your answer would be correct if you just changed "Schwarzschild radius" to "radius of the object." Maybe that was what you intended, but you wrote "Schwarzschild radius" by mistake?

To put this in less technical language, I think the issue boils down to distinguishing between two types of curvature. There is tidal curvature, which can be caused by mass outside that volume of space, and there's a different type of curvature (measured by the Einstein tensor) that can only be caused by mass that exists inside that volume of space. TrickyDicky, if you have access to either the Feynman lectures or Penrose's The Road to Reality, both of those have good nontechnical discussions of this distinction. My own explanation is here: http://www.lightandmatter.com/html_books/genrel/ch05/ch05.html#Section5.1 [Broken]

-Ben

Last edited by a moderator: May 4, 2017
7. Sep 9, 2010

### tom.stoer

Ben,

Yes, you are absolutely right. Patching of outer = vacuum and inner = non-vacuum solution obviously is done at the physical radius of the object, not at the Schwarzschild radius, which is irrelevant in that case. Sorry for the confusion!