Reference: Moore, Thomas A., *A General Relativity Workbook*, University Science Books (2013) – Chapter 21, Problem 21.8.

Consider the metric:

We’ll have a look at what the Einstein equation has to say about gravity in a spacetime using this metric. First, we’ll find the Christoffel symbols using the method of comparing the two forms of the geodesic equation. The geodesic equation is

The following equation is formally equivalent to this (where a dot above a symbol means the derivative with respect to ):

Since the metric is diagonal and none of its components depends on or , the LHS of 2 is identically zero for or , so all Christoffel symbols with any index being or is zero.

Now look at . We get from 2, since doesn’t depend on :

Comparing with 3 we see that

Now for :

These are the only non-zero Christoffel symbols. We can get an expression for the acceleration of a particle in its rest frame by noting that

The four velocity of a particle at rest (so ) must satisfy

Plugging into 11 we get

That is, the acceleration in the direction is a constant, so this would seem to indicate a uniform gravitational field. However, let’s apply the Einstein equation and see what we get. We’ll need the Riemann tensor in order to get the Ricci tensor, so we’ll use the definition of the former:

Since all Christoffel symbols with a or index are zero, the only possibly non-zero components of are those containing only or , and due to the symmetry relations, the only independent component is

Now for the Ricci tensor. We have

Since the metric is diagonal, the upstairs components are just the reciprocals of the downstairs ones, so

and we get

where in the last line we see that can be non-zero only if and but since is diagonal, . All components of involving or indices are zero because all components of involving or indices are zero. To use the Ricci tensor in the Einstein equation, we need the upstairs version, which is

In a vacuum (assuming ) the stress-energy tensor and the Einstein equation says that

The only way this can be true is if , meaning that there is no gravitational field.

More generally, suppose that there is some fluid in the region so that . In that case

where the stress-energy scalar is

so from 38 we have

However, if we multiply 33 by and add it to 34 we get

Combining the last two equations we get

In a perfect fluid at rest, these components represent the pressure of the fluid in the and directions, so this result implies that the pressure would have to be negative, which doesn’t make sense.