Finding the angular momentum of a Kerr black hole

[etotheipi]

[Mentor Note -- Specialized question moved to the general technical forums]

Homework Statement:: To show that $J = Ma$ for the charged Kerr metric [Wald Ch. 11 Pr. 6]
Relevant Equations:: \begin{align*} \mathrm{d}s^2 = &- \left( \frac{\Delta - a^2 \sin^2{\theta}}{\Sigma}\right) \mathrm{d}t^2 - \frac{2a\sin^2{\theta}(r^2 + a^2 - \Delta)}{\Sigma}\mathrm{d}t \mathrm{d}\phi \\

&+ \left[ \frac{(r^2 + a^2)^2 -\Delta a^2 \sin^2{\theta}}{\Sigma}\right] \sin^2{\theta} \mathrm{d}\phi^2
+ \frac{\Sigma}{\Delta} \mathrm{d}r^2 + \Sigma \mathrm{d}\theta^2 \end{align*}Since $g_{ab}$ is independent of $\phi$, this spacetime admits a Killing field $\psi^a = (\partial / \partial \phi)^a$ for which we can define a Komar integral,$$J:= \frac{1}{16\pi} \int_{\partial \Sigma} *\mathrm{d}\psi_a = \frac{1}{16\pi} \int_{\partial \Sigma} \epsilon_{abcd} \nabla^c \psi^d$$Defining a two-form $\boldsymbol{\alpha} := \epsilon_{abcd} \nabla^c \psi^d$ and then following the procedure demonstrated by Wald on pages 288-289 let's you write down $\mathrm{d}\boldsymbol{\alpha} = 3\nabla_{[l} \epsilon_{mn]cd} \nabla^c \psi^d = 2{R^e}_f \psi^f \epsilon_{elmn}$, from which it follows by Stokes that\begin{align*}
\frac{1}{16\pi} \int_{\partial \Sigma} \epsilon_{abcd} \nabla^c \psi^d &= \frac{3}{16 \pi} \int_{\Sigma} \nabla_{[l} \epsilon_{mn]cd} \nabla^c \psi^d \\
&= \frac{(-1)}{8\pi} \int_{\Sigma} R_{ab} n^a \psi^b \mathrm{d}V \\

&= (-1)\int_{\Sigma} \left(T_{ab} - \frac{1}{2} T g_{ab} \right) n^a \psi^b \mathrm{d}V

\end{align*}which follows from contracting Einstein's equation, ${G^a}_a = {R^a}_a - \frac{1}{2} R {g^a}_a = 8 \pi {T^a}_a \implies R = - 8\pi T$ which we can then use to eliminate $R$ in the original Einstein equation.

How do you actually do that integral? I can't see why converting it into a volume integral makes it easier; I would have thought we should instead consider that $\nabla^c \psi^d = g^{ce}\Gamma^d_{ef} \psi^f$ such that the two-form $\epsilon_{abcd}g^{ce}\Gamma^d_{ef} \psi^f = h \mathrm{d}\theta \wedge \mathrm{d}\phi$ for some function $h$ to be determined. Then you'd need to find the Christoffel symbols and compute\begin{align*}
\frac{1}{16\pi} \int_{\partial \Sigma} \epsilon_{abcd} g^{ce}\Gamma^d_{ef} \psi^f &= \int_0^{2\pi} \mathrm{d}\theta \int_0^{\pi} \mathrm{d}\phi \, h
\end{align*}Is that the right way to go about doing the integral? I ask because I'd prefer not to waste a lot of time doing completely the wrong thing ... thanks!

 

[PeterDonis]

How do you actually do that integral?

Since Kerr is a vacuum spacetime, that integral vanishes identically.

 

[etotheipi]

Since Kerr is a vacuum spacetime, that integral vanishes identically.

I thought the only condition on this analysis is that the spacetime is asymptotically flat, i.e. in the vacuum region we can set $R_{ab} = 0$ such that $d\boldsymbol{\alpha} = 0$.

But surely $R_{ab}$ cannot vanish everywhere in $\Sigma$? Otherwise we're just left with the case where $J=0$ and $M=0$, in which case it holds trivially...

 

[PeterDonis]

I thought the only condition on this analysis is that the spacetime is asymptotically flat

The problem you referenced (part (c) of Problem 6 in Chapter 11 of Wald) specifically asks about Kerr spacetime, not just a general asymptotically flat, stationary, axisymmetric spacetime.

surely $R_{ab}$ cannot vanish everywhere in $\Sigma$? Otherwise we're just left with the case where $J=0$ and $M=0$, in which case it holds trivially...

Any vacuum spacetime has $R_{ab} = 0$ everywhere, not just flat Minkowski spacetime.

Have you worked part (a) of problem 5 in that chapter of Wald? If not, I suggest doing that before tackling part (c) of problem 6. (If you have worked that problem, how did you solve it?) Both have the same apparent problem: you are asked to evaluate a global quantity ($M$ for Schwarzschild, $J$ for Kerr) in a vacuum spacetime, where any integral involving the stress-energy tensor vanishes identically, but clearly the global quantity does not vanish. So you have to find some other integral to evaluate that global quantity, one that gives a nonzero answer even in a vacuum spacetime. (Hint: stick with the surface integrals instead of trying to convert them to volume integrals.)

 

[etotheipi]

Thanks! Interesting; I do not understand this topic well at all. Let me re-read the chapter now, and then I'll attempt a solution using your hints and post it tomorrow, when I'm slightly less tired

 

[PeterDonis]

(Hint: stick with the surface integrals instead of trying to convert them to volume integrals.)

Btw, another aspect of this hint is: why would volume integrals not make sense for the Schwarzschild and Kerr vacuum spacetimes? Think about the topology of the surfaces of constant time in the maximal extensions of these spacetimes. (The recent thread of yours on electric field lines through wormholes is actually relevant here.)

 

[etotheipi]

Okay, let's see how it goes!

For part a), let's again denote by $\boldsymbol{\alpha} := \epsilon_{abcd} \nabla^c \psi^d$. We already showed that $d\boldsymbol{\alpha}$ vanishes in the asymptotic region, so let's consider a region $\Sigma'$ which is bounded by two concentric surfaces $S$ and $S'$, and write$$0 = \int_{\Sigma'} d\boldsymbol{\alpha} = \int_{S'} \boldsymbol{\alpha} - \int_S \boldsymbol{\alpha} \quad \implies \quad \int_{S'} \boldsymbol{\alpha} = \int_S \boldsymbol{\alpha}$$Thus $J$ is independent of $S$. As shown in the OP we can get to the equation
\begin{align*}
J = - \int_{\Sigma} \left( T_{ab} - \frac{1}{2} T g_{ab} \right)n^a \psi^b dV &=- \int_{\Sigma} T_{ab} n^a \psi^b dV - \frac{1}{2} \int_{\Sigma} T n_b \psi^b dV \\
&=- \int_{\Sigma} T_{ab} n^a \psi^b dV
\end{align*}which holds because $\psi^a$ is orthogonal to $n^a$, the unit normal to $\Sigma$. Now let's try to prove $J = Ma$ for the charged Kerr metric, i.e. part (c). Since $\nabla^{\mu} \psi^{\nu} = g^{\mu \rho}\Gamma^{\nu}_{\rho \sigma} \psi^{\sigma} = g^{\mu \rho}\Gamma^{\nu}_{\rho \phi}$ we can write\begin{align*}
J = \frac{1}{16\pi} \int_{S} \epsilon_{abcd} \nabla^c \psi^d &= \frac{1}{16\pi} \int_S dS_{\mu \nu} g^{\mu \rho} \Gamma^{\nu}_{\rho \phi} = \frac{1}{16\pi} \int_S dS_{tr} g^{t\rho} \Gamma^{r}_{\rho \phi}
\end{align*}We need to work out some of the inverse metric components. The $g^{rr} = \frac{\Delta}{\Sigma}$ and $g^{\theta \theta} = \frac{1}{\Sigma}$ are straightforward but for the others we can write\begin{align*}
\begin{pmatrix} g^{tt} & g^{t\phi} \\ g^{\phi t} & g^{\phi \phi} \end{pmatrix}

= \frac{1}{K} \begin{pmatrix} g_{\phi \phi} & -g_{t \phi} \\ -g_{\phi t} & g_{tt} \end{pmatrix}
\end{align*}where $K$ is given by\begin{align*}

K = g_{tt} g_{\phi \phi} - g_{\phi t} g_{t \phi} = &- \left( \frac{\Delta - a^2 \sin^2{\theta}}{\Sigma}\right)\left( \frac{(r^2 + a^2)^2 -\Delta a^2 \sin^2{\theta}}{\Sigma}\right) \sin^2{\theta} \\

&- \left( \frac{a\sin^2{\theta}(r^2 + a^2 - \Delta)}{\Sigma} \right)^2 \\

\end{align*}which simplifies to\begin{align*}

K &= \frac{2\Delta a^2 (r^2 + a^2) \sin^4{\theta} - \Delta (r^2 + a^2) \sin^2{\theta} - \Delta a^4 \sin^6 \theta}{\Sigma^2} \\

&= \Delta \left( \frac{2a^2(r^2 + a^2) \sin^4 \theta - r^4 \sin^2 \theta - 2r^2 a^2 \sin^2 \theta - a^4 \sin^2 {\theta} - a^4 \sin^6 \theta}{r^4 + 2a^2 r^2(1-\sin^2{\theta}) + a^4(1-\sin^2{\theta})^2} \right) \\

&= -\Delta \sin^2{\theta}
\end{align*}and the two relevant Christoffel symbols needed for the integral are\begin{align*}

\Gamma^{r}_{\phi \phi} &= \frac{1}{2} g^{r\mu} (2\partial_{\phi} g_{\phi \mu} - \partial_{\mu} g_{\phi \phi}) = \frac{\Delta}{2\Sigma} (2\partial_{\phi} g_{\phi r} - \partial_{r} g_{\phi \phi}) = - \frac{\Delta}{2\Sigma} \partial_{r} g_{\phi \phi} \\

\Gamma^{r}_{t \phi} &= \frac{1}{2} g^{r\mu} (\partial_t g_{\phi \mu} + \partial_{\phi} g_{t \mu} - \partial_{\mu} g_{t \phi}) = \frac{\Delta}{2\Sigma} (\partial_t g_{\phi r} + \partial_{\phi} g_{tr} - \partial_{r} g_{t \phi}) = - \frac{\Delta}{2\Sigma} \partial_{r} g_{t \phi}

\end{align*}I'll have to try and evaluate & simplify all of that, but it's almost 3AM and I don't have the energy to do that now! So, hopefully more to follow...

 

[etotheipi]

By definition $\Sigma := r^2 + a^2 \cos^2{\theta}$ and $\Delta := r^2 + a^2 +e^2 -2Mr$, so $\partial_r \Sigma = 2r$ and $\partial_r \Delta = 2(r-M)$. Then\begin{align*}
\Gamma^{r}_{\phi \phi} &= \frac{-\Delta}{2\Sigma} \partial_{r} g_{\phi \phi} \\

&= \frac{-\Delta \sin^2{\theta}}{2\Sigma} \partial_r \left( \frac{(r^2 + a^2)^2 - \Delta a^2 \sin^2{\theta}}{\Sigma} \right) \\

&= \frac{-\Delta \sin^2{\theta}}{2\Sigma^3} \left( \Sigma[4r(r^2 + a^2) - 2(r-M)a^2 \sin^2{\theta}] - 2r[(r^2 + a^2)^2 - \Delta a^2 \sin^2{\theta}] \right) \\ \\

\Gamma^{r}_{t \phi} &= \frac{-\Delta}{2\Sigma} \partial_{r} g_{t \phi} \\

&= \frac{\Delta a \sin^2{\theta}}{2\Sigma} \partial_r \left( \frac{2Mr - e^2}{\Sigma} \right) \\

&= \frac{\Delta a \sin^2{\theta}}{2\Sigma^3} \left( 2M \Sigma - 2r[2Mr - e^2] \right)
\end{align*}It is also required to evaluate $dS_{\mu \nu} = -2n_{[\mu} r_{\nu]} \sqrt{\sigma} d^2 \xi$, where $n^{a}$ is the time-like unit vector to $\Sigma$ and $r^{a}$ is the space-like unit vector to $S$.

 

[PeterDonis]

@etotheipi I don't have time to check your math in detail right now, but the following paper by Visser is a good, if brief, reference on Kerr spacetime:

https://arxiv.org/abs/0706.0622

You might be able to check some things against what is in that paper.

 

[etotheipi]

Defining the 3-surface $\Sigma$ by $\Phi(x^{\alpha}) = 0$ where $\Phi(x^{\alpha}) := t - c_1$ we have $\partial_{t} \Phi = 1$ and thus $n_a = (n_t, 0,0,0)$ where\begin{align*}

n_t = \frac{-1}{|g^{\mu \nu} \partial_{\mu} \Phi \partial_{\nu} \Phi|^{1/2}} &= \frac{-1}{| \frac{g_{\phi \phi}}{-\Delta \sin^2{\theta}}|^{1/2}} \\

&= -\sqrt{\Sigma \Delta} \left( (r^2 + a^2)^2 - \Delta a \sin^2{\theta} \right)^{-1/2}

\end{align*}$\Sigma$ is described by co-ordinates $y^{\alpha} = r, \theta, \phi$. Defining the 2-surface $S$ by $\psi(y^{\alpha}) = 0$ where $\psi(y^{\alpha}) := r - c_2$ we have $\partial_{r} \psi = 1$ and thus $r_a = (0, r_r, 0, 0)$ where\begin{align*}

r_r = \frac{1}{|g^{\mu \nu} \partial_{\mu} \psi \partial_{\nu} \psi|^{1/2}} = \frac{1}{|g^{rr}|^{1/2}} = \sqrt{\frac{\Sigma}{\Delta}}

\end{align*}The induced metric on $S$, which is described by co-ordinates $\xi^{\alpha} = \theta, \phi$, is$$\sigma_{AB} = g_{\alpha \beta} \left( \frac{\partial y^{\alpha}}{\partial \xi^A} \right) \left( \frac{\partial y^{\beta}}{\partial \xi^B} \right)$$which is diagonal and has components\begin{align*}
\sigma_{\theta \theta} &= g_{\theta \theta} = \Sigma \\
\sigma_{\phi \phi} &= g_{\phi \phi} = \sin^2{\theta} \left( \frac{(r^2 + a^2)^2 - \Delta a^2 \sin^2{\theta}}{\Sigma} \right) \\
\end{align*}thus the determinant of the induced metric is just$$\mathrm{det}(\sigma_{AB}) := \sigma = \sigma_{\theta \theta} \sigma_{\phi \phi} = \sin^2{\theta} \left( (r^2 + a^2)^2 -\Delta a^2 \sin^2{\theta} \right)$$Finally we can write down the integral. It is, given $dS_{tr} = -n_t r_r \sqrt{\sigma} d^2\xi$\begin{align*}J &= \frac{1}{16\pi} \int_S dS_{tr} g^{t\rho} \Gamma^{r}_{\rho \phi} \\

&= \frac{1}{16\pi} \int_0^{\pi} d\theta \int_0^{2\pi} d\phi \Sigma \sin{\theta} \left[ g^{tt} \Gamma^{r}_{t\phi} + g^{t\phi} \Gamma^{r}_{\phi \phi} \right] \\

&= \frac{1}{8} \int_0^{\pi} d\theta \Sigma \sin{\theta} \left[ g^{tt} \Gamma^{r}_{t\phi} + g^{t\phi} \Gamma^{r}_{\phi \phi} \right]

\end{align*}where the Christoffel symbols as are in post #8, and also $g^{tt} = g_{\phi \phi}/(-\Delta \sin^2{\theta})$ and $g^{t \phi} = -g_{t \phi}/(-\Delta \sin^2{\theta})$.

 

[etotheipi]

I gave up trying to integrate it exactly, so it's time to get a little bit cavalier. Let's just take $c_1 \rightarrow \infty$, and now state a few limits, firstly:\begin{align*}

\lim_{r\rightarrow \infty} n_t &= - \lim_{r \rightarrow \infty} \left( r^4 + \mathcal{O}(r^3) \right)^{1/2} \left( r^4 + \mathcal{O}(r^3) \right)^{-1/2} = -1 \\

\lim_{r\rightarrow \infty} r_r &= \lim_{r \rightarrow \infty} \left( \frac{r^2 + \mathcal{O}(r) }{r^2 + \mathcal{O}(r) }\right)^{1/2} = 1 \\

\lim_{r\rightarrow \infty} \Sigma &= r^2\end{align*}which is good, because that's is agreement with the last line of #10. Some more limits:\begin{align*}

\lim_{r \rightarrow \infty} \Gamma^{r}_{\phi \phi} &= \lim_{r \rightarrow \infty} \frac{-\Delta \sin^2{\theta}}{2\Sigma^3} \left( \Sigma[4r(r^2 + a^2) - 2(r-M)a^2 \sin^2{\theta}] - 2r[(r^2 + a^2)^2 - \Delta a^2 \sin^2{\theta}] \right) \\

&= \frac{-\sin^2 \theta}{2} \lim_{r \rightarrow \infty} \left( \frac{\Delta}{\Sigma^3} \right) \lim_{r\rightarrow \infty} \left( 2r^5 + \mathcal{O}(r^4) \right) \\

&= \frac{-\sin^2 \theta}{2} r^{-4} \cdot 2r^5 = -r \sin^2{\theta} \\
\lim_{r \rightarrow \infty} \Gamma^{r}_{t \phi} &= \lim_{r \rightarrow \infty} \frac{\Delta a \sin^2{\theta}}{2\Sigma^3} \left( 2M \Sigma - 2r[2Mr - e^2] \right) \\

&= \frac{a \sin^2 \theta}{2} \lim_{r\rightarrow \infty} \left( \frac{\Delta}{\Sigma^3} \right) \lim_{r \rightarrow \infty} \left( -2Mr^2 + \mathcal{O}(r) \right) \\

&= \frac{a \sin^2 \theta}{2} r^{-4} \cdot -2Mr^2 = \frac{-Ma\sin^2{\theta}}{r^2}

\end{align*}... and some more limits:\begin{align*}

\lim_{r \rightarrow \infty} g^{tt} &= \lim_{r \rightarrow \infty} \frac{\left[ \frac{(r^2 + a^2)^2 -\Delta a^2 \sin^2{\theta}}{\Sigma}\right] \sin^2{\theta}}{-\Delta \sin^2{\theta}} = \lim_{r \rightarrow \infty} \frac{r^4 + \mathcal{O}(r^2)}{-\Delta \Sigma} = -1 \\

\lim_{r \rightarrow \infty} g^{t\phi} &= \lim_{r \rightarrow \infty} \frac{\left[ \frac{a\sin^2{\theta}(r^2 + a^2 - \Delta)}{\Sigma} \right]}{-\Delta \sin^2 \theta} \\

&= \lim_{r \rightarrow \infty} \left( \frac{-a}{\Sigma \Delta } \right) \lim_{r \rightarrow \infty} (2Mr + \mathcal{O}(1)) = \frac{-2Ma}{r^3}
\end{align*}Let's re-write the integral in this regime,\begin{align*}J = \frac{1}{8} \int_0^{\pi} r^2 \sin{\theta} \left( \frac{3Ma \sin^2 \theta}{r^2} \right) &= \frac{3Ma}{8} \int_0^{\pi} d\theta \sin^3 \theta \\

&= \frac{4}{3} \left( \frac{3Ma}{8} \right) = \frac{Ma}{2}\end{align*}That's actually devastating, so close but a factor of 2 out...

Btw, another aspect of this hint is: why would volume integrals not make sense for the Schwarzschild and Kerr vacuum spacetimes? Think about the topology of the surfaces of constant time in the maximal extensions of these spacetimes. (The recent thread of yours on electric field lines through wormholes is actually relevant here.)

I had a think about this too. I think it's because of the $\Sigma = 0$ singularity within the domain of integration of the volume integral that ruins the calculation. Is that along the right lines?

 

[PeterDonis]

I had a think about this too. I think it's because of the $\Sigma = 0$ singularity within the domain of integration of the volume integral that ruins the calculation. Is that along the right lines?

Not really. The $\Sigma = 0$ singularity is at $r = 0$ and $\theta = \pi / 2$ (i.e., the "ring singularity"). But this is not even in the domain of integration of the volume integral, which is over a spacelike hypersurface of constant coordinate time in Boyer-Lindquist coordinates (or Schwarzschild coordinates for the case of a static, non-rotating hole). Can you see why not?

 

[PeterDonis]

so close but a factor of 2 out...

You might check your computations of the inverse metric and Christoffel symbols against this reference:

https://arxiv.org/pdf/0904.4184.pdf

Section 2.14 gives results for Kerr spacetime.

 

[PeterDonis]

so close but a factor of 2 out...

Your derivation seems to be assuming that the covariant derivative of $\psi^a$ only has $t$, $r$ components. I'm not sure that's correct. I think the covariant derivative of $\psi^a$ will have a component in the $\theta$ direction as well. That will add more inverse metric/connection coefficient terms to the sum in your integral, which might help remove the discrepancy.

 

[PeterDonis]

the two relevant Christoffel symbols needed for the integral

Are these really the only two? The overall sum that appears in the covariant derivative is $g^{\mu \rho} \Gamma^\nu{}_{\rho \phi}$. Here are all of the possible non-vanishing combinations:

$$
g^{t t} \Gamma^{r}_{t \phi}
$$

$$
g^{t t} \Gamma^{\theta}_{t \phi}
$$

$$
g^{\phi t} \Gamma^{r}_{t \phi}
$$

$$
g^{\phi t} \Gamma^{\theta}_{t \phi}
$$

$$
g^{r r} \Gamma^{t}_{r \phi}
$$

$$
g^{r r} \Gamma^{\phi}_{r \phi}
$$

$$
g^{\theta \theta} \Gamma^{t}_{\theta \phi}
$$

$$
g^{\theta \theta} \Gamma^{\phi}_{\theta \phi}
$$

$$
g^{t \phi} \Gamma^{r}_{\phi \phi}
$$

$$
g^{t \phi} \Gamma^{\theta}_{\phi \phi}
$$

$$
g^{\phi \phi} \Gamma^{r}_{\phi \phi}
$$

$$
g^{\phi \phi} \Gamma^{\theta}_{\phi \phi}
$$

(Because of the contraction with $\epsilon_{abcd}$, the $\mu$ and $\nu$ indexes cannot be the same, but it doesn't look like that eliminates any combinations.)

I don't see why any of the above would not be included in the sum that appears inside the integral.

 

[PeterDonis]

I don't see why any of the above would not be included in the sum that appears inside the integral.

Actually, on looking over the full derivation, I do see why some of those are excluded; the actual sum, once the normal to the 2-surface is taken into account, becomes $g^{t \rho} \Gamma^r_{\rho \phi} - g^{r \rho} \Gamma^t_{\rho \phi}$ (the minus sign is because of the sign flip in $\epsilon_{abcd}$ when two indexes are swapped). The terms shown in post #10 are from the first of the two terms above; but the second also should be included, as far as I can see, and gives $- g^{rr} \Gamma^t_{r \phi}$, which gives

$$
- \frac{\Delta}{\Sigma} \frac{M a \sin^2 \theta \left[ a^2 \cos^2 \theta \left( a^2 - r^2 \right) - r^2 \left( a^2 + 3 r^2 \right) \right]}{\Sigma^2 \Delta} = \frac{M a \sin^2 \theta \left[ a^2 \cos^2 \theta \left( r^2 - a^2 \right) + r^2 \left( a^2 + 3 r^2 \right) \right]}{\Sigma^3}
$$

In the limit $r \rightarrow \infty$, this becomes (noting that the leading order factor of $r$ is $3 r^4$ in the numerator and $r^6$ in the denominator)

$$
\frac{3 M a \sin^2 \theta}{r^2}
$$

which looks like it's just what is needed to fix the discrepancy.

 

[etotheipi]

You are a legend! Amazing. The mistake was back in post #7 where I said $dS_{\mu \nu} g^{\mu \rho} \Gamma^{\nu}_{\rho \phi} = dS_{tr} g^{t\rho} \Gamma^{r}_{\rho \phi}$ instead of the correct $dS_{\mu \nu} g^{\mu \rho} \Gamma^{\nu}_{\rho \phi} = dS_{tr} g^{t \rho} \Gamma^{r}_{\rho \phi} + dS_{rt} g^{r \rho} \Gamma^{t}_{\rho \phi}$. Checking quickly that it works,
\begin{align*}

\frac{1}{16\pi} \int_S dS_{\mu \nu} g^{\mu \rho} \Gamma^{\nu}_{\rho \phi} = \frac{1}{16\pi} \int_S \left( dS_{tr} g^{t \rho} \Gamma^{r}_{\rho \phi} + dS_{rt} g^{r \rho} \Gamma^{t}_{\rho \phi} \right)

\end{align*}then $dS_{tr} = -2n_{[t} r_{r]} \sqrt{\sigma} d^2 \xi = -n_t r_r \sqrt{\sigma} d^2 \xi = \sqrt{\sigma} d^2 \xi$ whilst also $dS_{rt} = -2n_{[r} r_{t]} \sqrt{\sigma} d^2 \xi = n_t r_r \sqrt{\sigma} d^2 \xi = - \sqrt{\sigma} d^2 \xi$. The second term is the one I didn't include in my working, so just checking that including it gives the result:\begin{align*}

\mathrm{2^{\mathrm{nd}} \, term} = \frac{1}{8} \int_0^{\pi} d\theta \, r^2 \sin{\theta} (-g^{r r} \Gamma^{t}_{r \phi}) &= \frac{3Ma}{8} \int_0^{\pi} d\theta \, \sin^3{\theta} = \frac{Ma}{2}
\end{align*}and when added to the first term gives $Ma$. Neat! Thanks