7
$\begingroup$

After reading parts of Chapter 8 in Hobson, 'General Relativity: An introduction for Physicists,' I have a question regarding the observation on page 184 regarding the gravitational field equations in empty space. We see that in empty space, the field equations reduce to, $$ R_{\mu\nu} = 0 $$ A table is given, $$ \begin{array}{|c|c|c|} \hline \text{# of spacetime dimensions} & \text{# of field equations} & \text{# indep. components of }R_{\mu\nu\sigma\rho} \\ \hline 2 & 3 & 1 \\ 3 & 6 & 6 \\ 4 & 10 & 20 \\ \hline \end{array} $$ Then it states:

"Thus we see that in two or three dimensions the field equations in empty space guarantee that the full curvature tensor must vanish. In four dimensions, however, ... it is therefore possible to satisfy the field equations in empty space with a non-vanishing curvature tensor."

"... we conclude that it is only in four dimensions or more that gravitational fields can exist in empty space."

I find this confusing because surely empty space means NO MATTER (or energy). Now, if the $\underline{\text{curvature of spacetime}}$ is related to the $\underline{\text{matter and energy density}}$, then how is it possible that in 4+ dimensions there can be a non-zero curvature tensor? Surely regardless of the mathematics, in empty space it should still be forced zero by the fact that nothing is there?

Forgive me if my question sounds naive, for I am only covering the derivation of the gravitational field equations now for the first time.

Ref: Hobson, M. P., Efstathiou, G. P., Lasenby, A. N., 2006. General Relativity: An introduction for Physicists. Cambridge: Cambridge University Press

$\endgroup$
  • $\begingroup$ The Riemann tensor can be broken down into two parts, one of which measures tidal fields from distant matter, and one of which measures fields from matter that is present locally. The vacuum field equations only say that the latter is zero. $\endgroup$ – user4552 Feb 22 '18 at 22:22
7
$\begingroup$

I believe one analogy might help. Think about electrostatics. The electric potential $\Phi$ satisfies Poisson's equation

$$\nabla^2\Phi=-\frac{\rho}{\epsilon_0}.$$

This is one equation between functions, so this means that when we evaluate the two functions on the same point they agree. If we have one region $U\subset \mathbb{R}^3$ on which there is no electric charge, this means that the charge density restricted to $U$, namely $\rho|_{U}$ is zero. In other words: if $x\in U$ then $\rho(x)=0$.

On this region it is then clear that

$$\nabla^2\Phi = 0$$

which is Laplace's equation. The fact that on $U$ there is no charge and $\Phi$ satisfies Laplace's equation is not an argument for inexistence of an electric field on $U$. Actually there might be some charge density elsewhere producing a field.

One common example: a localized charge density $\rho$ meaning that $\rho$ vanishes outside some large enough ball. One field will be produced and will be nonzero outside that ball, even though there is no charge elsewhere.

Now let's talk General Relativity. Einstein's field equations states that

$$R_{\mu\nu}-\frac{1}{2}g_{\mu\nu}R=\kappa T_{\mu\nu}.$$

What can happen perfectly well is that $T_{\mu\nu}$ is non-zero in some regions and zero in others like in the example above.

Think for example of one localized matter distribution: one isolated planet or star.

In these cases you can think like that: you have points on which $T_{\mu\nu}\neq 0$ and points on which $T_{\mu\nu}=0$ and there has to be a solution compatible with these two conditions. So the $T_{\mu\nu}\neq 0$ region will give you one metric tensor, which must be compatible with the solution on $T_{\mu\nu}=0$ and thus you have the matter influence on that region without matter. In other words, matter in one region produces spacetime curvature which can affect regions without matter perfectly well.

Of course the influence of matter in regions without matter doesn't extends indefinitely. If you have one isolated system for example, in regions close to it, without matter there is spacetime curvature produced by the system, but far away this influence sould become negligible and the spacetime curvature should go to zero. This ends up leading to the definition of assymptotically flat spacetimes.

$\endgroup$
11
$\begingroup$

Surely regardless of the mathematics, in empty space it should still be forced zero by the fact that nothing is there?

The moon orbits the Earth despite the fact that it is, for all intents and purposes, in a vacuum (i.e. the local matter/energy density is zero).

The point is that matter and energy in one localized region produce spacetime curvature elsewhere - even if the energy density in those other regions vanishes. You know this already - otherwise gravitational fields wouldn't exist in vacuum - but the book makes the point that the field equations imply that this is only possible for spacetimes with $d\geq 4$.

$\endgroup$
  • $\begingroup$ Yet I can imagine Newtonian gravity in 2 spatial dimensions with $F = G'\frac{Mm}{r}$ which satisfies the appropriate version of Gauss's Law. $\endgroup$ – JEB Feb 24 '18 at 5:34
  • $\begingroup$ @JEB Yes. It's interesting that the Einstein equations do not reduce to a Newtonian limit in 2+1 dimensional spacetime. $\endgroup$ – J. Murray Feb 24 '18 at 6:05
  • $\begingroup$ There is a good discussion of a more suitable theory for 3D gravity here: lss.fnal.gov/archive/other/fprint-93-54.pdf $\endgroup$ – J. Murray Feb 24 '18 at 6:26

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.