Reference: Moore, Thomas A., A General Relativity Workbook, University Science Books (2013) – Chapter 23; Box 23.1.
We’ll return now to general relativity, and build up to a derivation of the Schwarzschild metric. As a quick review, the problem is to find a solution to the Einstein equation in the form
where is the Ricci tensor, itself a contraction of the Riemann tensor, is the stress-energy tensor and is the stress-energy scalar.
In a practical problem, will be given, and the problem is to determine the metric from the Ricci tensor. The Ricci tensor is specified in terms of Christoffel symbols, which are in turn defined in terms of the metric and its derivatives, so the Einstein equation becomes a system of coupled, non-linear partial differential equations in the components of the metric tensor.
A good starting point is to look at spacetime around a source with spherical symmetry. We can picture this spacetime as a set of nested spherical shells, on the surface of which the usual 2-d spherical metric applies:
This gives us several of the metric components already, in that
We can set up the coordinates on each shell such that any line with fixed values of and is perpendicular to all the surfaces. This means that the basis vectors and are perpendicular to the third spatial basis vector , so that . With spatial symmetry, there should be no difference in the way the metric treats motions in different directions of or , so we’d expect the terms , , and to all be zero, which gives us four more (well, eight, actually, since ) metric components.
We’re left with , and . If is a time coordinate, we must have and likewise, if is a spatial coordinate, then . We can, in fact, eliminate by making a coordinate transformation as follows:
where is some function of the original and coordinates (unknown at present). We can, in principle, always determine so that and then use as our new time coordinate. [Note that we can’t use the symmetry argument to claim that , since in a spherically symmetric situation, it does make a difference whether you are travelling in the plus or minus direction, so it isn’t necessarily so that in all cases.]
Take the differential of this equation to get
With the deductions above, our original metric equation is
Substituting 9 into the first two terms on the RHS, we get
We can now set the coefficient of to zero to get
Assuming this partial differential equation for can be solved (which we won’t be able to do a priori, since we don’t know or , but in principle, the equation can be solved), it is always possible to find a time coordinate such that , so we might as well use that time coordinate from the start. Relabelling this time coordinate from back to , the spherically symmetric metric is then
We therefore have only two metric components that need to be found by solving the Einstein equation 1, which we’ll get to in the next post.