Vacuum solution question

1. Sep 7, 2013

Thinkor

The general field equation for GR is

Rab - 1/2 gab R = 8πTab

where I am setting G = 1 and c = 1.

Also, the vacuum solution is

Rab = 0

But it seems to me that this "vacuum" solution must hold even when there is matter present. Pick a point within a planet. Then excavate an infinitesimal vacuum chamber about the point. That can't affect the solution there because the local contribution to the solution is infinitesimal. Therefore, whether there is a vacuum at a point or not makes no difference, the vacuum solution still holds. Rab = 0 is a constraint on curvature that is everywhere satisfied.

So why can't the field equation then be simplified to this by replacing Rab with 0?

- 1/2 gab R = 8πTab

That makes it look a lot like Gauss's law of gravity, which does not add in a zero term representing a vacuum solution.

2$\varphi$ = 4πρ

This makes more sense to me. The solution is determined by the sources only, not by a zero valued vacuum solution.

Did Einstein glue together two equations into one for brevity? If so, why doesn't Gauss's law need another equation? And why would a vacuum solution be needed in addition to the source term in GR?

Last edited: Sep 7, 2013
2. Sep 7, 2013

phyzguy

This is not correct. You could make this argument for calculating an integral, for example. I could say "The contribution to the integral from each infinitesimal dx is infinitesimal, therefore I can ignore it and set it to zero, and the integral of any function is zero." The problem with your argument is that, while the contribution to the solution for each infinitesimal volume is infinitesimal, when I add up an infinite number of them, I get a non-zero result. Imagine taking your "inifinitesimal vacuum chamber" to be finite but small, then take the limiting procedure of taking the volume of the chamber to zero. While the volume of the chamber tends to zero, the number of chambers tends to infinity and the sum of the contributions from all of the chambers is non-zero.

3. Sep 7, 2013

WannabeNewton

In addition to phyzguy's comments, if $R_{ab} = 0$ then clearly the field equations do not resemble $\nabla^2 \varphi = 4\pi \rho$ for non-vanishing $\rho$ because $R_{ab} = 0$ trivially implies $R = 0$ which reduces your equation to $T_{ab} = 0$, the Newtonian analogue of which is Laplace's equation $\nabla^2 \varphi = 0$.

4. Sep 7, 2013

Thinkor

Yes, thanks. I failed to recognize in it the more complex context that exists here.