5.1 Linearised Theory

To proceed, we need to construct the various curvature tensors from the metric (5.1). For each, we work at linear order in h. To leading order, the inverse metric is

$$g^{\mu \nu }={\eta }^{\mu \nu }-h^{\mu \nu }$$

The Christoffel symbols are then

$${\Gamma }_{\nu \rho }^{\sigma }=\frac{1}{2}{\eta }^{\sigma \lambda }\left({\partial }_{\nu }{h}_{\lambda \rho }+{\partial }_{\rho }{h}_{\nu \lambda }-{\partial }_{\lambda }{h}_{\nu \rho }\right)\text{\hspace{1em}(5.2)}$$

The Riemann tensor is

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

The ΓΓ terms are second order in $h,$ so to linear order we have

$$\begin{array}{rl}& R_{\rho \mu \nu }^{\sigma }={\partial }_{\mu }{\Gamma }_{\nu \rho }^{\sigma }-{\partial }_{\nu }{\Gamma }_{\mu \rho }^{\sigma }\\ & =\frac{1}{2}{\eta }^{\sigma \lambda }\left({\partial }_{\mu }{\partial }_{\rho }{h}_{\nu \lambda }-{\partial }_{\mu }{\partial }_{\lambda }{h}_{\nu \rho }-{\partial }_{\nu }{\partial }_{\rho }{h}_{\mu \lambda }+{\partial }_{\nu }{\partial }_{\lambda }{h}_{\mu \rho }\right)\end{array}\text{\hspace{1em}(5.3)}$$

The Ricci tensor is then

$$R_{\mu \nu }=\frac{1}{2}\left({\partial }^{\rho }{\partial }_{\mu }{h}_{\nu \rho }+{\partial }^{\rho }{\partial }_{\nu }{h}_{\mu \rho }-\square h_{\mu \nu }-{\partial }_{\mu }{\partial }_{\nu }h\right)$$

with $h=h^{\mu }{}_{\mu }$ the trace of $h_{\mu \nu }$ and $\square ={\partial }^{\mu }{\partial }_{\mu }$ . The Ricci scalar is

$$R={\partial }^{\mu }{\partial }^{\nu }{h}_{\mu \nu }-\square h\text{\hspace{1em}(5.4)}$$

By the time we get to the Einstein tensor, we’ve amassed quite a collection of terms

$$G_{\mu \nu }=\frac{1}{2}[{\partial }^{\rho }{\partial }_{\mu }{h}_{\nu \rho }+{\partial }^{\rho }{\partial }_{\nu }{h}_{\mu \rho }-\square h_{\mu \nu }-{\partial }_{\mu }{\partial }_{\nu }h-({\partial }^{\rho }{\partial }^{\sigma }{h}_{\rho \sigma }-\square h){\eta }_{\mu \nu }]\text{\hspace{1em}(5.5)}$$

The Bianchi identity for the full Einstein tensor is ${\nabla }^{\mu }{G}_{\mu \nu }=0$ . For the linearised Einstein tensor, this reduces to

$${\partial }^{\mu }{G}_{\mu \nu }=0\text{\hspace{1em}(5.6)}$$

It’s simple to check explicitly that this is indeed obeyed by (5.5).

The Einstein equations then become the linear, but somewhat complicated, set of partial differential equations

$${\partial }^{\rho }{\partial }_{\mu }{h}_{\nu \rho }+{\partial }^{\rho }{\partial }_{\nu }{h}_{\mu \rho }-\square h_{\mu \nu }-{\partial }_{\mu }{\partial }_{\nu }h-({\partial }^{\rho }{\partial }^{\sigma }{h}_{\rho \sigma }-\square h){\eta }_{\mu \nu }=16\pi G{T}_{\mu \nu }\text{\hspace{1em}(5.7)}$$

where, for consistency, the source $T_{\mu \nu }$ must also be suitably small. The left-hand side of this equation should be viewed as a second order, linear differential operator acting on $h_{\mu \nu }$ . This is known as the Lichnerowicz operator.

The Fierz-Pauli Action

The linearised equations of motion can be derived from an action principle, first written down by Fierz and Pauli,

$$S_{FP}=\frac{1}{8\pi G}\int d^{4}x\left[-\frac{1}{4}{\partial }_{\rho }{h}_{\mu \nu }{\partial }^{\rho }{h}^{\mu \nu }+\frac{1}{2}{\partial }_{\rho }{h}_{\mu \nu }{\partial }^{\nu }{h}^{\rho \mu }+\frac{1}{4}{\partial }_{\mu }h{\partial }^{\mu }h-\frac{1}{2}{\partial }_{\nu }{h}^{\mu \nu }{\partial }_{\mu }h\right]\text{\hspace{1em}(5.8)}$$

This is the expansion of the Einstein-Hilbert action to quadratic order in h (after some integration by parts). (At linear order, the expansion of the Lagrangian is equal to the linearised Ricci scalar (5.4) which is a total derivative.)

Varying the Fierz-Pauli action, and performing some integration by parts, we have

$$\begin{array}{rl}\delta S_{FP}& =\frac{1}{8\pi G}\int d^{4}x\left[\frac{1}{2}{\partial }_{\rho }{\partial }^{\rho }{h}_{\mu \nu }-{\partial }^{\rho }{\partial }_{\nu }{h}_{\rho \mu }-\frac{1}{2}{\partial }^{\rho }{\partial }_{\rho }h{\eta }_{\mu \nu }+\frac{1}{2}{\partial }_{\nu }{\partial }_{\mu }h+\frac{1}{2}{\partial }_{\rho }{\partial }_{\sigma }{h}^{\rho \sigma }{\eta }_{\mu \nu }\right]\delta h^{\mu \nu }\\ & =\frac{1}{8\pi G}\int d^{4}x\left[-G_{\mu \nu }\delta h^{\mu \nu }\right]\end{array}\text{\hspace{1em}(5.9)}$$

We see that the Fierz-Pauli action does indeed give the vacuum Einstein equations $G_{\mu \nu }=0$ . We can then couple matter by adding $T_{\mu \nu }{h}^{\mu \nu }$ to the action.

5.1.1 Gauge Symmetry

Linearised gravity has a rather pretty gauge symmetry. This is inherited from the diffeomorphisms of the full theory. To see this, we repeat our consideration of infinitesimal diffeomorphisms from Section 4.1.3. Under an infinitesimal change of coordinates

$$x^{\mu }\to x^{\mu }-{\xi }^{\mu }(x)$$

with $\xi$ assumed to be small. The metric changes by (4.6)

$$\delta g_{\mu \nu }=({\mathcal{L}}_{\xi }g{)}_{\mu \nu }={\nabla }_{\mu }{\xi }_{\nu }+{\nabla }_{\nu }{\xi }_{\mu }$$

When the metric takes the form (5.1), this can be viewed as a transformation of the linearised field $h_{\mu \nu }$ . Because both ξ and h are small, the covariant derivative should be taken using the vanishing connection of Minkowski space. We then have

$$h_{\mu \nu }\to h_{\mu \nu }+({\mathcal{L}}_{\xi }\eta {)}_{\mu \nu }=h_{\mu \nu }+{\partial }_{\mu }{\xi }_{\nu }+{\partial }_{\nu }{\xi }_{\mu }\text{\hspace{1em}(5.10)}$$

This looks very similar to the gauge transformation of Maxwell theory, where the gauge potential shifts as $A_{\mu }\to A_{\mu }+{\partial }_{\mu }\alpha$ . Just as the electromagnetic field strength $F_{\mu \nu }=2{\partial }_{[\mu }{A}_{\nu ]}$ is gauge invariant, so is the linearised Riemann tensor $R^{\sigma }{}_{\rho \mu \nu }$

We can quickly check that the Fierz-Pauli action is invariant under the gauge symmetry (5.10). From (5.9), we have

$$\delta S_{FP}=-\frac{1}{8\pi G}\int d^{4}x2G_{\mu \nu }{\partial }^{\mu }{\xi }^{\nu }=+\frac{1}{8\pi G}\int d^{4}x2({\partial }^{\mu }{G}_{\mu \nu }){\xi }^{\nu }=0$$

where, in the second equality, we’ve integrated by parts (and discarded the boundary term) and in the third equality we’ve invoked the linearised Bianchi identity (5.6). In fact, this is just the same argument that we used to derive the Bianchi identity in Section 4.1.3, now played backwards.

When doing calculations in electromagnetism, it’s often useful to pick a gauge. One of the most commonly used is Lorentz gauge,

$${\partial }^{\mu }{A}_{\mu }=0$$

Once we impose this condition, the Maxwell equations ${\partial }^{\mu }{F}_{\mu \nu }=j_{\nu }$ reduce to the wave equations

$$\square A_{\nu }=j_{\nu }$$

We solved these equations in detail in the lectures on Electromagnetism.

We can impose a similar gauge fixing condition in linearised gravity. In this case, the analog of Lorentz gauge is called de Donder gauge

$${\partial }^{\mu }{h}_{\mu \nu }-\frac{1}{2}{\partial }_{\nu }h=0\text{\hspace{1em}(5.11)}$$

To see that this is always possible, suppose that you are handed a metric that doesn’t obey the de Donder condition but instead satisfies ${\partial }^{\mu }{h}_{\mu \nu }-\frac{1}{2}{\partial }_{\nu }h=f_{\nu }$ for some functions $f_{\nu }$ . Then do a gauge transformation (5.10). Your new gauge potential will satisfy ${\partial }^{\mu }{h}_{\mu \nu }-\frac{1}{2}{\partial }_{\nu }h+\bigsqcup {\xi }_{\nu }=f_{\nu }$ . So if you pick a gauge transformation ${\xi }_{\mu }$ that obeys $\square {\xi }_{\mu }=f_{\mu }$ then your new metric will be in de Donder gauge.

There is a version of de Donder gauge condition (5.11) that we can write down in the full non-linear theory. We won’t need it in this course, but it’s useful to know it exists. It is

$$g^{\mu \nu }{\Gamma }_{\mu \nu }^{\rho }=0\text{\hspace{1em}(5.12)}$$

This isn’t a tensor equation because the connection ${\Gamma }_{\mu \nu }^{\rho }$ is not a tensor. Indeed, if a tensor vanishes in one choice of coordinates then it vanishes for all choices while the whole point of a gauge fixing condition is to pick out a preferred choice of coordinates. If we substitute in the linearised Christoffel symbols (5.2), this reduces to the de Donder gauge condition.

The non-linear gauge condition (5.12) has a number of nice features. For example, in general the wave operator (or, on a Riemannian manifold, the Laplacian $△)$ is $\bigsqcup ={\nabla }^{\mu }{\nabla }_{\mu }=g^{\mu \nu }({\partial }_{\nu }{\partial }_{\mu }-{\Gamma }_{\nu \mu }^{\rho }{\partial }_{\rho })$ . If we fix the gauge (5.12), the annoying connection term vanishes and we simply have $\perp =g^{\mu \nu }{\partial }_{\mu }{\partial }_{\nu }$ . A similar simplification happens if we compute the covariant divergence of a one-form in this gauge: ${\nabla }^{\mu }{\omega }_{\mu }=g^{\mu \nu }{\nabla }_{\mu }{\omega }_{\nu }=$ $g^{\mu \nu }({\partial }_{\mu }{\omega }_{\nu }-{\Gamma }_{\mu \nu }^{\rho }{\omega }_{\rho })={\partial }^{\mu }{\omega }_{\mu }$

Back in our linearised world, de Donder gauge greatly simplifies the Einstein equation (5.7), which now become

$$\square h_{\mu \nu }-\frac{1}{2}\square h{\eta }_{\mu \nu }=-16\pi G{T}_{\mu \nu }\text{\hspace{1em}(5.13)}$$

It is useful to define

$${\bar{h}}_{\mu \nu }=h_{\mu \nu }-\frac{1}{2}h{\eta }_{\mu \nu }$$

Taking the trace of both sides gives $\bar{h}={\eta }^{\mu \nu }{\bar{h}}_{\mu \nu }=-h$ so, given ${\bar{h}}_{\mu \nu }$ we can trivially reconstruct $h_{\mu \nu }$ as

$$h_{\mu \nu }={\bar{h}}_{\mu \nu }-\frac{1}{2}\bar{h}{\eta }_{\mu \nu }\text{\hspace{1em}(5.14)}$$

Written in terms of ${\bar{h}}_{\mu \nu },$ the linearised Einstein equations in de Donder gauge (5.13) then reduce once again to a bunch of wave equations

$$\square {\bar{h}}_{\mu \nu }=-16\pi G{T}_{\mu \nu }\text{\hspace{1em}(5.15)}$$

and we can simply import the solutions from electromagnetism to learn something about gravity. We’ll look at some examples shortly.

5.1.2 The Newtonian Limit

Under certain circumstances, the linearised equations of general relativity reduce to the familiar Newtonian theory of gravity. These circumstances occur when we have a low-density, slowly moving distribution of matter.

For simplicity, we’ll look at a stationary matter configuration. This means that we take

$$T_{00}=\rho (\mathbf{x})$$

with the other components vanishing. Since nothing depends on time, we can replace the wave operator by the Laplacian in ${\mathbf{R}}^3:\Theta =-{\partial }_t^2+{\nabla }^2={\nabla }^2$ . The Einstein equations are then simply

$${\nabla }^2{\bar{h}}_{00}=-16\pi G\rho (\mathbf{x})\mathrm{and}{\nabla }^2{\bar{h}}_{0i}={\nabla }^2{\bar{h}}_{ij}=0$$

With suitable boundary conditions, the solutions to these equations are

$${\bar{h}}_{00}=-4\Phi (\mathbf{x})\mathrm{and}{\bar{h}}_{0i}={\bar{h}}_{ij}=0\text{\hspace{1em}(5.16)}$$

where the field Φ is identified with the Newtonian gravitational potential, obeying (0.1)

$${\nabla }^2\Phi =4\pi G\rho$$

Translating this back to $h_{\mu \nu }$ using (5.14), we use $\bar{h}=+4\Phi$ to find

$$h_{00}=-2\Phi ,h_{ij}=-2\Phi {\delta }_{ij},h_{0i}=0$$

Putting this back into the full metric $g_{\mu \nu }={\eta }_{\mu \nu }+h_{\mu \nu }$ , we have

$$d{s}^2=-(1+2\Phi )d{t}^2+(1-2\Phi )d\mathbf{x}\cdot d\mathbf{x}$$

If we take a $\Phi =-GM/r$ as expected for a point mass, we find that this coincides with the leading expansion of the Schwarzschild metric (4.8). (The $g_{00}$ term turns out to be exact; the $g_{ij}$ term is the leading order Taylor expansion of $(1+2\Phi {)}^{-1}.$ )

Way back in Section 1.2, we gave a naive, intuitive discussion of curved spacetime. There we already anticipated that the Newtonian potential Φ would appear in the $g_{00}$ component of the metric (1.26). However, in solving the Einstein equations, we learn that this is necessarily accompanied by an appearance of $\Phi$ in the $g_{ij}$ component. Ultimately, this is the reason for the factor of 2 discrepancy between the Newtonian and relativistic predictions for light bending that we met in Section 1.3

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