7
$\begingroup$

I was reading about "vacuum solution" in wiki, http://en.wikipedia.org/wiki/Vacuum_solution_(general_relativity). There are some words I'm confused.

1.In general relativity, a vacuum solution is a Lorentzian manifold whose Einstein tensor vanishes identically. According to the Einstein field equation, this means that the stress–energy tensor also vanishes identically, so that no matter or non-gravitational fields are present.

2.Since $T^{ab} = 0$ in a vacuum region, it might seem that according to general relativity, vacuum regions must contain no energy. But the gravitational field can do work, so we must expect the gravitational field itself to possess energy, and it does.

It seems to be conflicted about the existence of gravitational fields. So my question is , do $T^{ab}=0$ mean no matter and non-gravitational fields?

$\endgroup$
4
$\begingroup$

That article's choice of words could certainly be improved. Basically yes, $T$ captures all the non-gravitational "stuff."


The idea of a "gravitational field" doesn't really fit in to GR. There is stress-energy $T$ everywhere, and there is a metric $g$ everywhere, and that's really all you need to define what exists.

The article is trying to say that even if spacetime is more or less empty of stress-energy, there is a potential for stuff to happen thanks to the metric being nontrivial. Really, this is nothing new -- in Newtonian gravity, two separated bodies have as a system a gravitational potential energy that doesn't really reside anywhere. Trying to localize this potential "energy" in space is more difficult/ill-defined in GR than in Newtonian gravity.

In GR you can have two isolated masses sitting in space initially at rest. The stress-energy tensor is nonzero only in the regions occupied by the masses. Exterior to the masses the homogeneous Einstein equation is the same as that for empty Minkowski space, but the solution depends on the boundary conditions imposed by the masses and is in fact not Minkowski but rather some nonlinear superposition of Schwarzschild solutions. The nontrivial nature of the metric (which one could perhaps misleadingly call a "gravitational field") means the masses will start to move toward one another.

Another example is gravitational waves. These are again solutions to the homogeneous Einstein equation (i.e. the equation in vacuum), but they are nontrivial solutions. The metric is not just Minkowski.

To see this more mathematically, one could turn to the ADM equations of motion. Take a hypersurface of constant timelike coordinate $t$. The induced $3$-metric on this surface has components $\gamma_{ij}$ and "conjugate momentum" $\pi^{ij}$. These can be written in terms of $g$ without reference to $T$. Then there are known equations for $\partial_t \gamma_{ij}$ and $\partial_t \pi^{ij}$. In particular, only specially contrived setups will have $\partial_t \gamma_{ij} = 0$.

$\endgroup$
2
$\begingroup$

The self-energy of the gravitational field is not included in the stress-energy tensor, so $T_{\mu\nu} = 0$ does not mean no gravitational field is present.

Einstein's equation is:

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

where $R_{\mu\nu}$ is the Ricci tensor and $R$ is the Ricci scalar. When $T_{\mu\nu} = 0$ we describe this as Ricci flat i.e. $R_{\mu\nu} = 0$. However Ricci flat does not mean spacetime is flat because the Ricci tensor is a contraction of the Riemann tensor that describes the curvature, and the Riemann tensor can be non-zero even when the Ricci tensor is zero.

A good example of this is the Schwarzschild metric that describes the curvature outside a spherically symmetric body e.g. (approximately) the Earth. Everywhere outside the body $T_{\mu\nu} = 0$ and therefore $R_{\mu\nu} = 0$, but of course the Earth still has a gravitational field. Another example would be a gravitational wave propagating through empty space. The gravitational wave is a curvature oscillation, so the curvature can have an (oscillating) non-zero value when $T_{\mu\nu} = 0$.

$\endgroup$
  • $\begingroup$ So, you mean the stress-energy tensor $T^{\mu\nu}$ just contains energy density of the matter, but not includes the energy of gravitational field outside of the matter? $\endgroup$ – Simon Sep 25 '14 at 8:08
  • $\begingroup$ See the Wikipedia article for exactly what goes into the stress-energy tensor. It does not include energy due to the gravitational field. $\endgroup$ – John Rennie Sep 25 '14 at 8:11
0
$\begingroup$

Any matter (or non-gravitational fields) will act as sources for gravity, so to have a "vacuum" solution, you cannot have any of

One can define $T^{ab}_{nongrav} \equiv \frac{\delta S}{\delta g_{ab}}$ for non-gravitational fields, and this will be lower bounded (at each point) by zero. (One can show that this is as good as a stress-energy tensor defined by Noether's theorem) So if the stress-energy tensor vanishes, then one must necessarily have no non-gravitational fields.

I don't have much expertise in GR, but defining energy for the gravitational field is much more complicated. An excerpt from the Wikipedia page:

"Noether's theorem applies to any system which can be described by an action principle. Noether's theorem associates conserved energies with time-translation symmetries. When the time-translation symmetry is a finite parameter continuous group, such as the Poincaré group, Noether's theorem defines a scalar conserved energy for the system in question. However, when the symmetry is an infinite parameter continuous group, the existence of a conserved energy is not guaranteed. In a similar manner, Noether's theorem associates conserved momenta with space-translations, when the symmetry group of the translations is finite-dimensional. Because General Relativity is a diffeomorphism invariant theory, it has an infinite continuous group of symmetries rather than a finite-parameter group of symmetries, and hence has the wrong group structure to guarantee a conserved energy. Noether's theorem has been extremely influential in inspiring and unifying various ideas of mass, system energy, and system momentum in General Relativity."


BTW, imagine that you have a spacetime which is a solution of Einstein's equations, and you put a very small "probe mass" somewhere. ("Probe" means that your mass probes and responds to the gravitational field while it's too weak to affect the gravitational field).

You will expect the probe mass to flow along the geodesic, right? We need a gravitational field to move the mass, and in moving the probe mass, the gravitational field does work. Only, since the spacetime is a solution of Einstein's equations, the field at every point is not varying with time.

This helps confirm that even if $T^{ab}=0$ you still have gravitational fields, after-all you have some metric, and you can find the geodesics.

$\endgroup$

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.