# Deriving the EFE

In light of the recent (very incredible!) discovery of gravitational waves, I thought it would be fun and interesting to describe a derivation of the (basically) fundamental result of general relativity: the Einstein Field Equations (EFEs). Note that this was written in the plural form; this is because it’s really a collection of 10 partial differential equations, written in the form of a single equation using tensors. (Why ten equations? Well, general relativity makes heavy (this is a pun) use of the metric tensor. This is a four-by-four matrix, and is also symmetric, giving us ten independent coefficients.) The post assumes knowledge of differential geometry.

Before deriving the field equations, we will talk about some physics. The principle of stationary action says that varying the action of a system gives us the equations of motion for that system. Let’s be a little more precise. Suppose $\mathscr{L}$ is the Lagrangian of a physical system. The principle of least action says that if $S=\int^{t_2}_{t_1}\mathscr{L}dt$ is the action of the mechanical system, then $\delta S = 0$. You can check that in a classical conservative system (when $\mathscr{L}$ is the kinetic energy minus the potential energy) that this yields Newton’s second law.

Let $\mathscr{M}$ be the spacetime (a smooth, connected Lorentzian manifold with a metric, which, by the Lorentzian condition, means that it has signature, for eg., +++-) For general relativity, the gravitational part of the action is, in natural units:

$\displaystyle S=\frac{1}{16 \pi}\int_\mathscr{M} R\sqrt{-g}d^4x$

Here $g=\det(g_{\mu \nu})$, and $R = g^{\mu \nu}R_{\mu\nu}$. This is called the Einstein-Hilbert action. The whole action is, therefore, simply given by

$\displaystyle S = \int_\mathscr{M}\left(\frac{1}{16\pi}R + \mathscr{L}_\mathrm{matter}\right)\sqrt{-g}d^4x$

Let’s note here that the energy-momentum tensor is $\displaystyle T_{\mu\nu} = \frac{-2}{\sqrt{-g}}\frac{\delta\sqrt{-g}\mathscr{L}_\mathrm{matter}}{\delta g^{\mu\nu}}$.

The principle of least action tells us that $\delta S = 0$. What is $\delta S$? First, let’s consider $\displaystyle\frac{\delta(\sqrt{-g}R)}{\delta g^{\mu\nu}}$. This is simply given by $\displaystyle\sqrt{-g}\frac{\delta R}{\delta g^{\mu\nu}} + R\frac{\delta \sqrt{-g}}{\delta g^{\mu\nu}}$. We will expand each term separately. Since $R = g^{\mu\nu}R_{\mu\nu}$, varying this gives $R_{\mu\nu} + g^{\mu\nu}\frac{\delta R_{\mu\nu}}{\delta g^{\mu\nu}}$; but the latter term vanishes, so this is just $R_{\mu\nu}$.

What about $\displaystyle\frac{\delta \sqrt{-g}}{\delta g^{\mu\nu}}$? We will use Jacobi’s determinant formula, $\delta g = gg^{\mu\nu}\delta g_{\mu\nu}$. Now, $\displaystyle\delta\sqrt{-g} = -\frac{1}{2\sqrt{g}}\delta g$ (basic calculus), and hence, $\displaystyle\frac{\delta\sqrt{-g}}{\delta g^{\mu\nu}} = \frac{1}{2}\sqrt{-g}g^{\mu\nu}\frac{\delta g_{\mu\nu}}{\delta g^{\mu\nu}}$. Since $g^{\mu\nu}\delta g_{\mu\nu} = -g_{\mu\nu}\delta g^{\mu\nu}$, this is equivalent to $-\displaystyle\frac{1}{2}\sqrt{-g}g_{\mu\nu}$.

Now, let’s go back to the general action. The principle of stationary action tells us that:

$\displaystyle \frac{\delta R}{\delta g^{\mu\nu}} + \frac{R}{\sqrt{-g}}\frac{\delta\sqrt{-g}}{\delta g^{\mu\nu}} = \frac{-16\pi}{\sqrt{-g}}\frac{\delta(-\sqrt{g}\mathscr{L}_\mathrm{matter})}{\delta g^{\mu\nu}}$

This is obtained by setting the terms inside the expression for $\delta S$ equal to $0$, and then multiplying the terms by $\displaystyle \frac{\sqrt{-g}}{\sqrt{-g}}\frac{\delta g^{\mu\nu}}{\delta g^{\mu\nu}}$. Using the above equations for $\displaystyle \frac{\delta R}{\delta g^{\mu\nu}}$ and $\displaystyle \frac{\delta\sqrt{-g}}{\delta g^{\mu\nu}}$, we end up with the famed Einstein equations:

$\displaystyle R_{\mu\nu} - \frac{1}{2}Rg_{\mu\nu} = 8\pi T_{\mu\nu}$

(For example, in a vacuum, if we expand $R_{\mu\nu}$ out in terms of the Christoffel symbols, which are derivatives of the metric tensor, we end up with a second-order partial differential equation.) Let’s be honest, though – this was more of a mathematical derivation than a physical derivation. In the next post, we’ll describe a physical derivation of the Einstein field equations, and show that the equation indeed reduces to the Poisson equation for the Newtonian potential.

Have fun,
SKD