4.1 The Einstein-Hilbert Action

All our fundamental theories of physics are described by action principles. Gravity is no different. Furthermore, the straight-jacket of differential geometry places enormous restrictions on the kind of actions that we can write down. These restrictions ensure that the action is something intrinsic to the metric itself, rather than depending on our choice of coordinates.

Spacetime is a manifold M, equipped with a metric of Lorentzian signature. An action is an integral over M. We know from Section 2.4.4 that we need a volume-form to integrate over a manifold. Happily, as we have seen, the metric provides a canonical volume form, which we can then multiply by any scalar function. Given that we only have the metric to play with, the simplest such (non-trivial) function is the Ricci scalar R. This motivates us to consider the wonderfully concise action

$$S=\int d^{4}x\sqrt{-g}R\text{\hspace{1em}(4.1)}$$

This is the Einstein-Hilbert action. Note that the minus sign under the square-root arises because we are in a Lorentzian spacetime: the metric has a single negative eigenvalue and so its determinant, $g=\det g_{\mu \nu },$ is negative.

As a quick sanity check, recall that the Ricci tensor takes the schematic form (3.39) $R\sim \partial \Gamma +\Gamma \Gamma$ while the Levi-Civita connection itself is $\Gamma \sim \partial g$ . This means that the Einstein-Hilbert action is second order in derivatives, just like most other actions we consider in physics.

Varying the Einstein-Hilbert Action

We would like to determine the Euler-Lagrange equations arising from the action (4.1). We do this in the usual way, by starting with some fixed metric $g_{\mu \nu }(x)$ and seeing how the action changes when we shift

$$g_{\mu \nu }(x)\to g_{\mu \nu }(x)+\delta g_{\mu \nu }(x)$$

Writing the Ricci scalar as $R=g^{\mu \nu }{R}_{\mu \nu }.$ , the Einstein-Hilbert action clearly changes as

$$\delta S=\int d^{4}x\left((\delta \sqrt{-g})g^{\mu \nu }{R}_{\mu \nu }+\sqrt{-g}(\delta g^{\mu \nu })R_{\mu \nu }+\sqrt{-g}{g}^{\mu \nu }\delta R_{\mu \nu }\right)\text{\hspace{1em}(4.2)}$$

It turns out that it’s slightly easier to think of the variation in terms of the inverse metric $\delta g^{\mu \nu }$ . This is equivalent to the variation of the metric $\delta g_{\mu \nu };$ the two are related by

$$g_{\rho \mu }{g}^{\mu \nu }={\delta }_{\rho }^{\nu }\Rightarrow (\delta g_{\rho \mu })g^{\mu \nu }+g_{\rho \mu }\delta g^{\mu \nu }=0\Rightarrow \delta g^{\mu \nu }=-g^{\mu \rho }{g}^{\nu \sigma }\delta g_{\rho \sigma }$$

The middle term in (4.2) is already proportional to $\delta g^{\mu \nu }$ . We now deal with the first and third terms in turn. We will need the following result:

Claim

The variation of $\sqrt{-g}$ is given by

$$\delta \sqrt{-g}=-\frac{1}{2}\sqrt{-g}{g}_{\mu \nu }\delta g^{\mu \nu }$$

Proof

We use the fact that any diagonalisable matrix A obeys the identity

$$\log \det A=\mathrm{tr}\log A$$

This is obviously true for diagonal matrices. (The determinant is the product of eigenvalues while the trace is the sum of eigenvalues.) But because both the determinant and the trace are invariant under conjugation, it is also true for a diagonalisable matrix. Using this, we have,

$$\frac{1}{\det A}\delta (\det A)=\mathrm{tr}(A^{-1}\delta A)$$

Applying this to the metric, we have

$$\delta \sqrt{-g}=\frac{1}{2}\frac{1}{\sqrt{-g}}(-g)g^{\mu \nu }\delta g_{\mu \nu }=\frac{1}{2}\sqrt{-g}{g}^{\mu \nu }\delta g_{\mu \nu }$$

Using $g^{\mu \nu }\delta g_{\mu \nu }=-g_{\mu \nu }\delta g^{\mu \nu }$ then gives the result.

So far, we have managed to write the variation of the action (4.2) as

$$\delta S=\int d^{4}x\sqrt{-g}\left(R_{\mu \nu }-\frac{1}{2}R{g}_{\mu \nu }\right)\delta g^{\mu \nu }+\sqrt{-g}{g}^{\mu \nu }\delta R_{\mu \nu }$$

We now need only worry about the final term. For this, we use:

Claim

The variation of the Ricci tensor is a total derivative

$$\delta R_{\mu \nu }={\nabla }_{\rho }\delta {\Gamma }_{\mu \nu }^{\rho }-{\nabla }_{\nu }\delta {\Gamma }_{\mu \rho }^{\rho }$$

where

$$\delta {\Gamma }_{\mu \nu }^{\rho }=\frac{1}{2}{g}^{\rho \sigma }\left({\nabla }_{\mu }\delta g_{\sigma \nu }+{\nabla }_{\nu }\delta g_{\sigma \mu }-{\nabla }_{\sigma }\delta g_{\mu \nu }\right)$$

Proof

We start by looking at the variation of the Christoffel symbols, ${\Gamma }_{\mu \nu }^{\rho }.$ . First note that, although the Christoffel symbol itself is not a tensor, the variation $\delta {\Gamma }_{\mu \nu }^{\rho }$ is a tensor. This is because it is the difference of Christoffel symbols, one computed using $g_{\mu \nu }$ and the other using $g_{\mu \nu }+\delta g_{\mu \nu }$ . But the extra derivative term in the transformation of ${\Gamma }_{\mu \nu }^{\rho }$ is independent of the metric and so cancels out when we take the difference, leaving us with an object which transforms nicely as a tensor.

This is a useful observation. At any point $p\in M$ we can choose to work in normal coordinates such that ${\partial }_{\rho }{g}_{\mu \nu }=0$ and, correspondingly, ${\Gamma }_{\mu \nu }^{\rho }=0$ . Then, to linear order in the variation, the change in the Christoffel symbol evaluated at $p$ is

$$\begin{array}{rl}& \delta {\Gamma }_{\mu \nu }^{\rho }=\frac{1}{2}{g}^{\rho \sigma }({\partial }_{\mu }\delta g_{\sigma \nu }+{\partial }_{\nu }\delta g_{\sigma \mu }-{\partial }_{\sigma }\delta g_{\mu \nu })\\ & =\frac{1}{2}{g}^{\rho \sigma }({\nabla }_{\mu }\delta g_{\sigma \nu }+{\nabla }_{\nu }\delta g_{\sigma \mu }-{\nabla }_{\sigma }\delta g_{\mu \nu })\end{array}$$

where we’re at liberty to replace the partial derivatives with covariant derivatives because they differ only by the Christoffel symbols ${\Gamma }_{\mu \nu }^{\rho }$ which, in normal coordinates, vanish at $p.$ However, both the left and right-hand sides of this equation are tensors which means that although we derived this expression using normal coordinates, it must hold in any coordinate system. Moreover, the point p was arbitrary so the final expression holds generally.

Next we look at the variation of the Riemann tensor. In normal coordinates, the expression (3.39) becomes

$$R_{\rho \mu \nu }^{\sigma }={\partial }_{\mu }{\Gamma }_{\nu \rho }^{\sigma }-{\partial }_{\nu }{\Gamma }_{\mu \rho }^{\sigma }$$

and the variation is

$$\delta R_{\rho \mu \nu }^{\sigma }={\partial }_{\mu }\delta {\Gamma }_{\nu \rho }^{\sigma }-{\partial }_{\nu }\delta {\Gamma }_{\mu \rho }^{\sigma }={\nabla }_{\mu }\delta {\Gamma }_{\nu \rho }^{\sigma }-{\nabla }_{\nu }\delta {\Gamma }_{\mu \rho }^{\sigma }$$

where, as before, we replace partial derivatives with covariant derivatives as we are working in normal coordinates where the Christoffel symbols vanish. Once again, our final expression relates two tensors and must, therefore, hold in any coordinate system. Contracting indices (and working to leading order), we have

$$\delta R_{\rho \nu }={\nabla }_{\mu }\delta {\Gamma }_{\nu \rho }^{\mu }-{\nabla }_{\nu }\delta {\Gamma }_{\rho \mu }^{\mu }$$

as claimed.

$\square$

The upshot of these calculations is that

$$g^{\mu \nu }\delta R_{\mu \nu }={\nabla }_{\mu }{X}^{\mu }\mathrm{with}{X}^{\mu }=g^{\rho \nu }\delta {\Gamma }_{\rho \nu }^{\mu }-g^{\mu \nu }\delta {\Gamma }_{\nu \rho }^{\rho }$$

The variation of the action (4.2) can then be written as

$$\delta S=\int d^{4}x\sqrt{-g}\left[\left(R_{\mu \nu }-\frac{1}{2}R{g}_{\mu \nu }\right)\delta g^{\mu \nu }+{\nabla }_{\mu }{X}^{\mu }\right]\text{\hspace{1em}(4.3)}$$

This final term is a total derivative and, by the divergence theorem of Section 3.2.4, we ignore it. Requiring that the action is extremised, so $\delta S=0$ , we have the equations of motion

$$G_{\mu \nu }:=R_{\mu \nu }-\frac{1}{2}R{g}_{\mu \nu }=0\text{\hspace{1em}(4.4)}$$

where $G_{\mu \nu }$ is the Einstein tensor defined in Section 3.4.1. These are the Einstein field equations in the absence of any matter. In fact they simplify somewhat: if we contract (4.4) with $g^{\mu \nu }$ , we find that we must have $R=0$ . Substituting this back in, the vacuum Einstein equations are simply the requirement that the metric is Ricci flat,

$$R_{\mu \nu }=0\text{\hspace{1em}(4.5)}$$

These deceptively simple equations hold a myriad of surprises. We will meet some of the solutions as we go along, notably gravitational waves in Section 5.2 and black holes in Section 6.

Before we proceed, a small comment. We happily discarded the boundary term in (4.3), a standard practice whenever we invoke the variational principle. It turns out that there are some situations in general relativity where we should not be quite so cavalier. In such circumstances, one can be more careful by invoking the so-called Gibbons-Hawking boundary term.

4.1.1 An Aside on Dimensional Analysis

As it stands, there’s something a little fishy about the action (4.1): it doesn’t have the right dimensions. This isn’t such an issue since we have just a single term in the action and multiplying the action by a constant doesn’t change the classical equations of motion. Nonetheless, it will prove useful to get it right at this stage.

If we take the coordinates $x^{\mu }$ to have dimension of length, then the metric $g_{\mu \nu }$ is necessarily dimensionless. The Ricci scalar involves two spatial derivatives so has dimension $[R]=L^{-2}$ . Including the integration measure, the action (4.1) then has dimensions $[S]=L^2$ . However, actions should have dimensions of energy × time (it’s the same dimensions as $\hslash )$ , or $[S]=M{L}^2{T}^{-1}$ . This means that the Einstein-Hilbert action should be multiplied by a constant with the appropriate dimensions. We take

$$S=\frac{c^3}{16\pi G}\int d^{4}x\sqrt{-g}R$$

where c is the speed of light and G is Newton’s constant,

$$G\approx 6.67\times 10^{-11}{\mathrm{m}}^3{\mathrm{kg}}^{-1}{s}^{-2}$$

This factor doesn’t change the equation of motion in vacuum, but we will see in Section 4.5 that it determines the strength of the coupling between the gravitational field and matter, as we might expect.

It’s no fun carrying around a morass of fundamental constants in all our equations. For this reason, we often work in “natural units” in which various constants are set equal to 1. From now on, we will set $c=1$ . (Any other choice of $c,$ including $3\times 10^8$ 2 is simply dumb.) This means that units of length and time are equated.

However, different communities have different conventions when it comes to $G.$ . Relativists will typically set $G=1$ . Since we have already set $c=1$ , we have $[G]=L{M}^{-1}$ Setting $G=1$ then equates mass with length. This is useful when discussing gravitational phenomenon where the mass is often directly related to the length. For example, the Schwarzschild radius of black hole is $R_s=2GM/c^2$ which becomes simply $R_s=2M$ once we set $G=c=1$

However, if you’re interested in phenomena other than gravity, then it’s no more sensible to set $G=1$ than to set, say, the Fermi coupling for the weak force $G_F=1$ Instead, it is more useful to choose the convention where $\hslash =1$ , a choice which equates energy with inverse time (also known as frequency). With this convention, Newton’s constant has dimension $[G]=M^{-2}$ . The corresponding energy scale is known as the (reduced) Planck mass; it is given by

$$M_{\mathrm{pl}}^2=\frac{\hslash c}{8\pi G}$$

It is around $10^{18}\mathrm{GeV}$ . This is a very high energy scale, way beyond anything we have probed in experiment. This can be traced to the weakness of the gravitational force. With $c=\hslash =1$ , we can equally well write the Einstein-Hilbert action as

$$S=\frac{1}{2}{M}_{\mathrm{pl}}^2\int d^{4}x\sqrt{-g}R$$

You might be tempted to set $c=\hslash =G=1$ . This leaves us with no remaining dimensional quantities. It is typically a bad idea, not least because dimensional analysis is a powerful tool and one we do not want to lose. In these lectures, we will focus only on gravitational physics. Nonetheless, we will retain G in all equations.

4.1.2 The Cosmological Constant

We motivated the Einstein-Hilbert action as the simplest term we could write down. While it’s true that it’s the simplest term that results in interesting dynamics for the gravitational field, there is in fact a simpler term which we could add to the action. This comes from multiplying the volume form by a constant. The resulting action is

$$S=\frac{1}{16\pi G}\int d^{4}x\sqrt{-g}(R-2\Lambda )$$

Here Λ is referred to as the cosmological constant. It has dimension $[\Lambda ]=L^{-2}$ . The minus sign in the action comes from thinking of the Lagrangian as ${}^{66}T-V^{\prime }$ : the cosmological constant is like the potential energy V.

Varying the action as before now yields the Einstein equations,

$$R_{\mu \nu }-\frac{1}{2}R{g}_{\mu \nu }=-\Lambda g_{\mu \nu }$$

This time, if we contract with $g^{\mu \nu }$ , we get $R=4\Lambda$ . Substituting this back in, the vacuum Einstein equations in the presence of a cosmological constant become

$$R_{\mu \nu }=\Lambda g_{\mu \nu }$$

We will solve these shortly in Section 4.2.

Higher Derivative Terms

The Einstein-Hilbert action (with cosmological constant) is the simplest thing we can write down but it is not the only possibility, at least if we allow for higher derivative terms. For example, there are three terms that contain four derivatives of the metric,

$$S_{4-\mathrm{deriv}}=\int d^{4}x\sqrt{-g}\left(c_1{R}^2+c_2{R}_{\mu \nu }{R}^{\mu \nu }+c_3{R}_{\mu \nu \rho \sigma }{R}^{\mu \nu \rho \sigma }\right)$$

with $c_1,c_2$ and $c_3$ dimensionless constants. General choices of these constants will result in higher order equations of motion which do not have a well-defined initial value problem. Nonetheless, it turns out that one can find certain combinations of these terms which conspire to keep the equations of motion second order. This is known as Lovelock’s theorem. In $d=4$ dimensions, this combination has a rather special topological property: a generalisation of the Gauss-Bonnet theorem states that

$$\frac{1}{8{\pi }^2}{\int }_M{d}^{4}x\sqrt{g}\left(R^2-4R_{\mu \nu }{R}^{\mu \nu }+R_{\mu \nu \rho \sigma }{R}^{\mu \nu \rho \sigma }\right)=\chi (M)$$

where $\chi (M)\in \mathbf{Z}$ is the Euler character of M that we previously defined in (2.35). In Lorentzian signature, this combination of curvature terms is also a total derivative and does not affect the classical equations of motion.

As in any field theory, higher derivative terms in the action only become relevant for fast varying fields. In General Relativity, they are unimportant for all observed physical phenomena and we will not discuss them further in this course.

4.1.3 Diffeomorphisms Revisited

Here’s a simple question: how many degrees of freedom are there in the metric $g_{\mu \nu }?$ Since this is a symmetric $4\times 4$ matrix, our first guess is $\frac{1}{2}\times 4\times 5=10$

However, not all the components of the metric $g_{\mu \nu }$ are physical. Two metrics which are related by a change of coordinates, $x^{\mu }\to {\tilde{x}}^{\mu }(x)$ describe the same physical spacetime. This means that there is a redundancy in any given representation of the metric, which removes precisely 4 of the 10 degrees of freedom, leaving just 6 behind.

Mathematically, this redundancy is implemented by diffeomorphisms. (We defined diffeomorphisms in Section 2.1.3.) Given a diffeomorphism, $\varphi :M\to M$ , we can use this to map all fields, including the metric, on M to a new set of fields on M. The end result is physically indistinguishable from where we started: it describes the same system, but in different coordinates. Such diffeomorphisms are analogous to the gauge symmetries that are familiar in Maxwell and Yang-Mills theory.

Let’s look more closely at the implication of these diffeomorphisms for the path integral. We’ll consider a diffeomorphism that takes a point with coordinates $x^{\mu }$ to a nearby point with coordinates

$$x^{\mu }\to {\tilde{x}}^{\mu }(x)=x^{\mu }+\delta x^{\mu }$$

We could view this either as an “active change”, in which one point with coordinates $x^{\mu }$ is mapped to another point with coordinates $x^{\mu }+\delta x^{\mu }$ , or as a “passive” change, in which we use two different coordinate charts to label the same point. Ultimately, the two views lead to the same place. We’ll adopt the passive perspective here, simply because we have a lot of experience of changing coordinates. Later we’ll revert to the active picture.

We can think of the change of coordinates as being generated by an infinitesimal vector field $X$

$$\delta x^{\mu }=-X^{\mu }(x)$$

The metric transforms as

$$g_{\mu \nu }(x)\to {\tilde{g}}_{\mu \nu }(\tilde{x})=\frac{\partial x^{\rho }}{\partial {\tilde{x}}^{\mu }}\frac{\partial x^{\sigma }}{\partial {\tilde{x}}^{\nu }}{g}_{\rho \sigma }(x)$$

With our change of coordinate ${\tilde{x}}^{\mu }=x^{\mu }-X^{\mu }(x)$ , with infinitesimal $X^{\mu }$ , we can invert the Jacobian matrix to get

$$\frac{\partial {\tilde{x}}^{\mu }}{\partial x^{\rho }}={\delta }_{\rho }^{\mu }-{\partial }_{\rho }{X}^{\mu }\Rightarrow \frac{\partial x^{\rho }}{\partial {\tilde{x}}^{\mu }}={\delta }_{\mu }^{\rho }+{\partial }_{\mu }{X}^{\rho }$$

where the inverse holds to leading order in the variation X. Continuing to work infinitesimally, we then have

$$\begin{array}{rl}& {\tilde{g}}_{\mu \nu }(\tilde{x})=({\delta }_{\mu }^{\rho }+{\partial }_{\mu }{X}^{\rho })({\delta }_{\nu }^{\sigma }+{\partial }_{\nu }{X}^{\sigma })g_{\rho \sigma }(x)\\ & =g_{\mu \nu }(x)+g_{\mu \rho }(x){\partial }_{\nu }{X}^{\rho }+g_{\nu \rho }(x){\partial }_{\mu }{X}^{\rho }\end{array}$$

Meanwhile, we can Taylor expand the left-hand side

$${\tilde{g}}_{\mu \nu }(\tilde{x})={\tilde{g}}_{\mu \nu }(x+\delta x)={\tilde{g}}_{\mu \nu }(x)-X^{\lambda }{\partial }_{\lambda }{\tilde{g}}_{\mu \nu }(x)$$

Comparing the the different metrics at the same point $x,$ we find that the metric undergoes the infinitesimal change

$$\delta g_{\mu \nu }(x)={\tilde{g}}_{\mu \nu }(x)-g_{\mu \nu }(x)=X^{\lambda }{\partial }_{\lambda }{g}_{\mu \nu }+g_{\mu \rho }{\partial }_{\nu }{X}^{\rho }+g_{\nu \rho }{\partial }_{\mu }{X}^{\rho }\text{\hspace{1em}(4.6)}$$

But this is something we’ve seen before: it is the Lie derivative of the metric. In other words, if we act with an infinitesimal diffeomorphism along $X$ , the metric changes as

$$\delta g_{\mu \nu }=({\mathcal{L}}_{X}g{)}_{\mu \nu }$$

This makes sense: it’s like the leading term in a Taylor expansion along X.

In fact, we can also massage (4.6) into a slightly different form. We lower the index on $X^{\rho }$ in the last two $\partial X^{\rho }$ terms by taking the metric inside the derivative. This results in two further terms in which the derivative hits the metric, and these must be cancelled of. We’re left with

$$\delta g_{\mu \nu }={\partial }_{\mu }{X}_{\nu }+{\partial }_{\nu }{X}_{\mu }+X^{\rho }\left({\partial }_{\rho }{g}_{\mu \nu }-{\partial }_{\mu }{g}_{\rho \nu }-{\partial }_{\nu }{g}_{\mu \rho }\right)$$

But the terms in the brackets are the Christoffel symbols, $2g_{\rho \sigma }{\Gamma }_{\mu \nu }^{\sigma }$ . We learn that the infinitesimal change in the metric can be written as

$$\delta g_{\mu \nu }={\nabla }_{\mu }{X}_{\nu }+{\nabla }_{\nu }{X}_{\mu }\text{\hspace{1em}(4.7)}$$

Let’s now see what this means for the path integral. Under a general change of the metric, the Einstein-Hilbert action changes as (4.3)

$$\delta S=\int d^{4}x\sqrt{-g}{G}^{\mu \nu }\delta g_{\mu \nu }$$

where we have discarded the boundary term. Insisting that $\delta S=0$ for any variation $\delta g_{\mu \nu }$ gives the equation of motion $G^{\mu \nu }=0$ . In contrast, symmetries of the action are those variations $\delta g_{\mu \nu }$ for which $\delta S=0$ for any choice of metric. Since diffeomorphisms are (gauge) symmetries, we know that the action is invariant under changes of the form (4.7). Using the fact that $G_{\mu \nu }$ is symmetric, we must have

$$\delta S=2\int d^{4}x\sqrt{-g}{G}^{\mu \nu }{\nabla }_{\mu }{X}_{\nu }=0\mathrm{forall}{X}_{\mu }(x)$$

After integrating by parts, we find that this results in something familiar: the Bianchi identity

$${\nabla }_{\mu }{G}^{\mu \nu }=0$$

We already know that the Bianchi identity holds from our work in Section 3.4, but the derivation there was a little fiddly. Here we learn that, from the path integral perspective, the Bianchi identity is a result of diffeomorphism invariance.

In fact it makes sense that the two are connected. Naively, the Einstein equation $G_{\mu \nu }=0$ comprises ten independent equations. But, as we’ve seen, diffeomorphism invariance means that there aren’t ten independent components of the metric, so one might worry that the Einstein equations are overdetermined. Happily, diffeomorphisms also ensure that not all the Einstein equations are independent either; they are related by the four Bianchi constraints. We see that, in fact, the Einstein equations give only six independent conditions on the six independent degrees of freedom in the metric.

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