The Einstein Field Equation in a Static, Spherically Symmetric Spacetime

This will be the first of several articles which will provide, for reference, useful equations for static, spherically symmetric spacetimes. This is a common special case that is studied in General Relativity, and it has the advantage of having a general solution for the Einstein Field equation that can be expressed in closed form equations. This first article will focus on those closed form equations for that solution.

First, we should be clear about the set of spacetimes we are talking about. The two key properties of this set of spacetimes can be defined in coordinate-independent terms as follows:

(1) A static spacetime has a timelike Killing vector field which is hypersurface orthogonal.

(2) A spherically symmetric spacetime has a 3-parameter group of spacelike Killing vector fields that satisfy the properties of the Lie group SO(3), which describes rotations in 3-dimensional space.

As noted, these definitions are coordinate-independent; they define invariant geometric properties of the spacetime. However, we can make our task much simpler by finding a coordinate chart that matches up well with those properties. Without going into the fine details, we can say that such a chart will have coordinates we label as $t$, $r$, $\theta$, $\phi$, in which the line element takes the following form:

$$ds^2 = – J(r) dt^2 + \frac{1}{1 – \frac{2 m(r)}{r}} dr^2 + r^2 d\theta^2 + r^2 \sin^2 \theta d\phi^2$$

where $J(r)$ and $m(r)$ are two functions of the radial coordinate $r$ whose specific functional forms we will discover by solving the EFE.

We can see that the metric coefficients are independent of all of the coordinates except $r$ and $\theta$, and $\theta$ only appears in the angular part. That tells us that the coordinate basis vectors $\partial_t$ and $\partial_\phi$ are Killing vectors. If the function $J(r)$ is positive, then $\partial_t$ is timelike and will be the timelike Killing vector field that we know exists because of property #1 above. From the angular part of the line element, we expect that $\partial_\phi$ will be one of the three spacelike Killing vectors that we know exist because of property #2 above. (We will not explicitly find the other two spacelike Killing vectors here; the reader is encouraged to do so as an exercise.)

Also, looking at the angular part of the line element, we can see that a 2-sphere at radial coordinate $r$ will have surface area $4 \pi r^2$. This is actually not the only possible convention for the radial coordinate, but it is the one that makes the line element look the simplest, and so we will use it here. Coordinates on any spherically symmetric spacetime that use this convention are often referred to as “Schwarzschild coordinates” (even though, ironically, Schwarzschild himself did not use this convention in his original paper).

We have made one more simplification in the above line element: the form of the $dr^2$ term, putting the function $m(r)$ in the denominator in a particular way (which should look familiar to anyone who has seen the Schwarzschild vacuum solution before) instead of just having an unknown function times $dr^2$, already anticipates the form that the solution of the 0-0 component of the EFE will take. We will discuss that further below.

With our expression for the line element we can compute its Einstein tensor, which gives us the LHS of the EFE. In order to get the RHS, we need expressions for the components of the stress-energy tensor. The key observation to make here is that spherical symmetry places severe constraints on the SET components; in fact the only nonzero components will be:

$$T^t{}_t = – \rho (r)$$

$$T^r{}_r = p (r)$$

$$T^\theta{}_\theta = T^\phi{}_\phi = s (r)$$

Note the minus sign in $T^t_t$; this is because of the (-+++) metric signature we are using. We use the mixed components because no extra factors involving the metric coefficients appear in them. Also, the spherical symmetry requires that the angular components be equal, but *not* that the spatial stress be isotropic, so the $r$-$r$ component can still be different than the angular components, hence they are shown as two different functions of $r$.

For reference, we now give the mixed Christoffel symbols for the above line element, even though we will not need them for the rest of this article; only the nonzero ones are listed.

$$\Gamma^r{}_{tt} = \frac{r – 2m}{2r} \frac{dJ}{dr}$$

$$\Gamma^t{}_{rt} = \Gamma^t{}_{tr} = \frac{1}{2J} \frac{dJ}{dr}$$

$$\Gamma^r{}_{rr} = \frac{r \frac{dm}{dr} – m}{r \left( r – 2m \right)}$$

$$\Gamma^r{}_{\theta \theta} = 2m – r$$

$$\Gamma^r{}_{\phi \phi} = \left( 2m – r \right) \sin^2 \theta$$

$$\Gamma^\theta{}_{r \theta} = \Gamma^\theta{}_{\theta r} = \frac{1}{r}$$

$$\Gamma^\phi{}_{r \phi} = \Gamma^\phi{}_{\phi r} = \frac{1}{r}$$

$$\Gamma^\theta{}_{\theta \phi} = \Gamma^\theta{}_{\phi \theta} = \frac{\cos \theta}{\sin \theta}$$

$$\Gamma^\theta{}_{\phi \phi} = – \cos \theta \sin \theta$$

We now look at the EFE components. First is the 0-0 or $t$-$t$ component:

$$G^t{}_t = 8 \pi T^t{}_t = – \frac{2}{r^2} \frac{dm}{dr} = – 8 \pi \rho$$

This immediately gives

$$\frac{dm}{dr} = 4 \pi r^2 \rho (r)$$

which shows how the form of the $dr^2$ term in the metric arises; if we write that term as just a generic function of $r$, we find that we can always define a function $m(r)$ that puts it into the form given, without loss of generality. We will defer discussion of the physical meaning of the function $m(r)$ until we have looked at the rest of the EFE components.

Next, the 1-1 or $r$-$r$ component:

$$G^r{}_r = 8 \pi T^r{}_r = \frac{1}{r^3 J} \left( r \left( r – 2m \right) \frac{dJ}{dr} – 2 m J \right) = 8 \pi p$$

This leads quickly to an equation for J which is most conveniently written as follows:

$$\frac{1}{2J} \frac{dJ}{dr} = \frac{m + 4 \pi r^3 p}{r \left( r – 2m \right)}$$

We would normally look next at the angular components of the EFE; but rather than evaluate them directly (they are both the same, again by spherical symmetry), it turns out to be easier to evaluate the covariant divergence of the SET instead, specifically its $r$-component, to get an equation including both radial and tangential stress. The general equation is:

$$\nabla_a T^a{}_r = \partial_a T^a{}_r + \Gamma^a{}_{ab} T^b{}_r – \Gamma^b{}_{ar} T^a{}_b = 0$$

which gives

$$\partial_r T^r{}_r = – \left( \Gamma^t{}_{tr} + \Gamma^r{}_{rr} + \Gamma^\theta{}_{\theta r} + \Gamma^\phi{}_{\phi r} \right) T^r{}_r + \Gamma^t{}_{tr} T^t{}_t + \Gamma^r{}_{rr} T^r{}_r + \Gamma^\theta{}_{\theta r} T^\theta{}_\theta + \Gamma^\phi{}_{\phi r} T^\phi{}_\phi$$

This simplifies, after cancelling terms, substituting, and some algebra, to:

$$\frac{dp}{dr} = – \left( \rho + p \right) \frac{1}{2J} \frac{dJ}{dr} – \frac{2}{r} \left( p – s \right)$$

which is recognizable as a generalized form of the Tolman-Oppenheimer-Volkoff equation, i.e., this equation describes hydrostatic equilibrium, how the matter present is supported against its own gravity.

It is worth noting at this point that we have five unknown functions, $J$, $m$, $\rho$, $p$, and $s$, but we only have three equations (since there are no other nonzero components of the Einstein Tensor). That means, of course, that there are many possible solutions that satisfy our conditions of staticity and spherical symmetry. To specify a particular solution, we need to specify at least two of our unknown functions: the obvious ones to specify are $\rho$ and $s$. We will also find that it is helpful to impose other constraints, such as energy conditions, to filter out solutions that are not physically realistic.

Finally, a few comments on physical interpretation. As you have no doubt guessed, the function $m(r)$ can be physically interpreted as the mass inside radial coordinate $r$. Specifically, we have

$$
m(r) = \int_0^r 4 \pi \rho(r’) (r’)^2 dr’
$$

The interesting thing to note here is that this integral only includes $\rho$, the density; it does not include the pressure or tangential stress. This might seem contrary to the fairly common statement that, in GR, pressure is a source of gravity. This apparent discrepancy is resolved in my previous series of articles on “Does Gravity Gravitate”.

The function $J(r)$ (or more precisely its square root) has the obvious interpretation of being the “redshift factor” for an observer who is static at radial coordinate $r$. That is, light emitted by such an observer will be redshifted by this factor when received by a static observer at infinity. Also, if the static observer at $r$ compares his clock rate with that of a static observer at infinity, for example by exchanging round-trip light signals, the static observer at $r$’s clock will be seen to run slow by the factor $\sqrt{J(r)}$.

As a final note, we look at the special case where $\rho = p = s = 0$, i.e., the vacuum case, with no stress-energy present anywhere. For this case, we have $dm / dr = 0$, which means $m$ is a constant, independent of $r$, which we will call $M$ for clarity. Using this in the equation for $J$, we find that $J = 1 – 2M / r$. In other words, we have shown that any solution of the EFE that is a vacuum, static, spherically symmetric spacetime must be Schwarzschild spacetime. Of course we already know this because of Birkhoff’s theorem, but it provides a useful check on our math above.

In the next article in this series, we will look at Maxwell’s Equations in a static, spherically symmetric spacetime.