Reference: Moore, Thomas A., *A General Relativity Workbook*, University Science Books (2013) – Chapter 21; Box 21.1.

We can now start looking at a derivation of the Einstein equation, which is the generalization of Newton’s formula for the gravitational force. In Newtonian theory, gravity is an attractive, conservative, inverse-square force so (apart from the sign) it is mathematically identical to the electrostatic force, which means we can write a differential form of Newton’s gravitational theory using Gauss’s law. That is

where is the gravitational field, is the mass density and is the gravitational constant. The minus sign occurs because gravity is attractive, whereas the electric force for like charges is repulsive. Because the force is conservative, can be written as the gradient of a potential so an alternative form of the equation is

or

The derivation of the Einstein equation is, like so many derivations in relativity, based on a plausibility argument. We want to find a tensor equation that generalizes Newton’s equation, and we want this tensor equation to reduce to Newton’s equation in the weak field limit.

First, the generalization of Newtonian mass density is the stress-energy tensor we can try replacing the RHS of 3 by where is a scalar constant. Since the RHS is now a rank-2 tensor, the LHS must also be a rank-2 tensor, so we must have an equation like

where the form of needs to be determined. To do this, think about what we want the theory to do. The idea behind general relativity is that the energy density in a region of space should determine the curvature of the space in that region. The Riemann tensor and the metric tensor describe the curvature of space-time, so it makes sense that could depend on these two tensors.

Suppose we try to express solely in terms of the Riemann tensor. What other constraints can we impose to narrow things down? First, since the Riemann tensor is rank 4 and is rank 2, we’ll need to contract the Riemann tensor to get rid of 2 of its indices. One candidate is the Ricci tensor, defined as the contraction of the Riemann tensor over its first and third indices:

To use , we need to raise both its indices, so we get

where in the third line, we’ve swapped the dummy indices and . Since is symmetric, must also be symmetric, but since , this condition is satisfied.

From conservation of energy and momentum, we know that , so we must also have (since is a constant). This is where we run into a snag. The condition must apply everywhere, in every reference frame, so it must apply the origin of a locally inertial frame (LIF). In a LIF, the Riemann tensor reduces to

The Ricci tensor is then

In a LIF, the first derivatives of are all zero (by definition of the LIF), so

In a LIF, the total derivative reduces to the ordinary derivative so we get

The indexes in the first term can be relabelled by swapping with and with to give

which is the negative of the third term (since and the order of the partial derivatives doesn’t matter), so these two terms cancel and we’re left with

The only way this can be identically zero is if we could swap with in the first term and have it equal the negative of the second term. However, if we try this, we get

The partial derivatives match up with those in the second term, but the product of the three metric tensors doesn’t, so in general this isn’t zero, meaning that

Thus setting won’t work, and we’ll need to try something else.