Does the Vacuum Solution in General Relativity Hold in the Presence of Matter?

[Thinkor]

The general field equation for GR is

R_ab - 1/2 g_ab R = 8πT_ab

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

Also, the vacuum solution is

R_ab = 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. R_ab = 0 is a constraint on curvature that is everywhere satisfied.

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

- 1/2 g_ab R = 8πT_ab

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[itex]\varphi[/itex] = 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?

 

[phyzguy]

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.

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.

 

[WannabeNewton]

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

- 1/2 g_ab R = 8πT_ab

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 = 4πρ

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##.

 

[Thinkor]

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.

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

 

[sardar]

I appreciate your curiosity and critical thinking in regards to the vacuum solution in general relativity (GR). First, let me clarify that the vacuum solution in GR refers to a spacetime that is completely devoid of matter or energy, not just a small vacuum chamber within a planet. The vacuum solution, as you correctly stated, is given by the equation Rab = 0, which means that the curvature of spacetime is zero in the absence of any matter or energy.

Now, to address your question about simplifying the field equation by replacing Rab with 0, this is not possible because the field equation is a set of differential equations that describe how matter and energy interact with spacetime to determine its curvature. In other words, the field equation is not just about the vacuum solution, but it also includes the effects of matter and energy on the curvature of spacetime. So, simply replacing Rab with 0 would not accurately reflect the full picture of how matter and energy affect spacetime.

Furthermore, the field equation in GR is more complex than Gauss's law of gravity because it describes the curvature of spacetime, which is a four-dimensional concept, while Gauss's law only describes the gravitational force between two point masses in three-dimensional space. Therefore, the two equations cannot be compared in the way you suggested.

In addition, the vacuum solution is needed in GR because it is a fundamental aspect of the theory. In fact, the vacuum solution was one of the key predictions of GR that was later confirmed by observations. It is also important to note that the vacuum solution does not mean that there is no gravitational field in empty space, but rather that the curvature of spacetime is zero. This has important implications for the behavior of light and other particles in empty space.

In conclusion, the field equation in GR is a comprehensive and accurate description of how matter and energy interact with spacetime to determine its curvature. The vacuum solution is a fundamental aspect of this equation and cannot be simply replaced or ignored. As scientists, we must always strive to understand and explain the complexities of nature, rather than simplifying them for the sake of brevity.