General RelativityPhysics

On The Einstein Vacuum Equations

Consider Einstein’s gravitational field equations

(1)   \begin{equation*} \displaystyle{G_{\mu \nu }=R_{\mu \nu }-\frac{1}{2}g_{\mu \nu }R=\kappa T_{\mu \nu }} \end{equation*}

We can perform a trace reversal on this, to obtain the Ricci tensor in terms of the source term :

(2)   \begin{equation*} \displaystyle{R_{\mu \nu }=\kappa \left ( T_{\mu \nu }-\frac{1}{2}g_{\mu \nu }T \right )} \end{equation*}

In vacuum, the energy-momentum tensor field vanishes, so the vacuum equations reduce to the very simple form

(3)   \begin{equation*} \displaystyle{R_{\mu \nu }=0} \end{equation*}

which means that the vacuum, in the context of general relativity, is characterised by the fact that space-time in such regions is Ricci flat. The reverse is evidently also true – any Ricci flat region of space-time is necessarily a vacuum region.

Does that mean that vacuum is a flat space-time ? Then how does gravity exist ?

No. Ricci flatness does not imply a flat space-time. To see this, consider that gravity is geodesic deviation, and that deviation is described in terms of the full rank-4 Riemann curvature tensor. The Ricci tensor is merely the trace of Riemann, and as such does not contain the same amount of information; even if the Ricci tensor vanishes, all the other functionally independent components of the Riemann tensor may not. Therefore, Ricci flatness does not mean that we are dealing with a flat space-time, it only means that the allowable geometries are constrained in certain very specific ways.

So what does Ricci flatness mean then, geometrically speaking ?

To understand this intuitively and visually, consider an inflated balloon. It contains a certain amount of air, and no matter what we do to the balloon, that amount of air doesn’t change. You can deform and contort it, squeeze it, push it and pull it – the only thing that happens is its shape changes, but not the volume of air inside of it ( it’s just an analogy, so ignore the compressibility of air ).

The very same is true in a vacuum space-time, only now we are dealing with a small ball of test particles, instead of a balloon. If we place such a ball into free fall, and choose a time-like direction to correspond to the future, then over time the shape of our ball will change, but its volume remains conserved everywhere. Intuitively, in the real world, this is what we can visualise as tidal forces – if you let an object fall towards a massive central body, then that object will get stretched in the radial direction, and squeezed in the other directions perpendicular to it ( ref “spaghettification” ). More precisely, but not mathematically rigorous, the Ricci tensor measures the rate at which the volume of a small enough ball of test particles changes over time; that means that, if the Ricci tensor vanishes, the volume of our ball remains preserved. Of course, this does not tell us anything about what happens to the shape of the ball ( that is described by another quantity, the Weyl tensor ) – which is why there can still be gravity, even in a Ricci-flat space-time. Ricci curvature is only one specific aspect of overall curvature, it does not completely determine the geometry of space-time.

So here we have the meaning of the vacuum equations – in vacuum, small volumes ( but not shapes ) in free fall remain conserved over time. That is the defining characteristic of what distinguishes a vacuum from a region filled by energy-momentum, where neither shape nor volume are conserved.

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