The method is based on the following fact (see for example [1])

If the electromagnetic field $F_{ab}$ satisfies Maxwell's equations
\begin{equation*}
\nabla_{[a}F_{bc]}= 0 \quad\text{and}\quad \nabla^a F_{ab} =0
\end{equation*}
and there is a spinor $\psi$ such that
\begin{equation}\label{eq:20161115b}
(\nabla_{\mu} + i \sqrt{4 \pi} \not F \gamma_{\mu} ) \psi = 0
\end{equation}
and $i \bar \psi \gamma^{\mu} \psi$ is time-like

then the Einstein equations are satisfied as well: \begin{equation*} R_{\mu\nu} - \frac{1}{2} R g_{\mu\nu} = 8 \pi T_{\mu\nu} \end{equation*}

A spinor $\psi$ satisfying equation \eqref{eq:20161115b} is called a Killing spinor.
Definitions and conventions about the formulae above can be found at the bottom of this post.
then the Einstein equations are satisfied as well: \begin{equation*} R_{\mu\nu} - \frac{1}{2} R g_{\mu\nu} = 8 \pi T_{\mu\nu} \end{equation*}

I take the following ansatz for the metric and electromagnetic field of an extreme black hole \begin{align*} ds^2 &= - f(x)^2 dt^2 + g(x)^2 ( dx_1^2 + dx_2^2 + dx_3^2)\\ A_{\mu} dx^{\mu} &= \frac{1}{\sqrt{4 \pi}} h(x) dt, \quad\text{with}\quad F = dA \end{align*} The functions $f(x)$, $g(x)$ and $h(x)$ do not depend on $t$. Although all calculations can be easily performed on a computer after choosing explicit gamma matrices, it gives more insight to work by hand. I take the following dual tetrad: $\theta^t = f(x) dt$ and $\theta^i = g(x) dx^i$. The components of the connection are then \begin{align*} \Gamma_{tti}&= \frac{1}{f g}\partial_i f\\ \Gamma_{ijk}&= \frac{1}{g^2} \left( - \delta_{ij} \partial_k g + \delta_{ik} \partial_j g\right) \end{align*} thus \begin{equation*} \not\Gamma_t = -\frac{1}{2} \frac{1}{f g}\partial_i f\ \gamma^{ti} \\ \end{equation*} and \begin{equation*} \not\Gamma_i = \frac{1}{2} \frac{1}{g^2} \partial_k g\ \gamma^{ik} \end{equation*} also \begin{equation*} F = \frac{1}{\sqrt{4 \pi}} \partial_i h \ dx^i dt = \frac{1}{\sqrt{4 \pi}} \partial_i h \ g^{-1} f^{-1}\ \theta^i \theta^t \end{equation*} thus \begin{equation*} \not F = \frac{2}{4} \frac{1}{\sqrt{4 \pi}} \partial_i h \ g^{-1} f^{-1} \gamma^{it} \end{equation*} The $\mu = t$ component of equation \eqref{eq:20161115b} then gives \begin{equation*} \left[ -\frac{1}{2} \frac{1}{fg} \partial_i f \ \gamma^{ti} + \frac{i}{2} \partial_i h \ g^{-1} f^{-1} \gamma^{it} \gamma_t \right]\psi = 0 \end{equation*} A solution of this equation is given by $h=f$ and $\psi$ satisfying $i \gamma_t \psi = - \psi$. The $\mu = i$ component of equation \eqref{eq:20161115b} gives \begin{equation*} \left[e_i +\frac{1}{2} \frac{1}{g^2} \partial_k g\ \gamma^{ik}+ \frac{i}{2} \partial_k h \ g^{-1} f^{-1} \gamma^{kt} \gamma_i \right]\psi = 0 \end{equation*} Using $i \gamma_t \psi = - \psi$, this equation is equivalent with \begin{equation*} \left[e_i +\frac{1}{2} \frac{1}{g^2} \partial_k g\ \gamma^{ik} - \frac{1}{2} \partial_i h \ g^{-1} f^{-1} - \frac{1}{2} \partial_k h \ g^{-1} f^{-1} \gamma^{ki} \right]\psi = 0 \end{equation*} This is solved by $e_i(\psi) -\dfrac{1}{2} \partial_i h \ g^{-1} f^{-1}\psi =0$ and $\dfrac{1}{2} \dfrac{1}{g^2} \partial_k\ g + \dfrac{1}{2} \partial_k h \ g^{-1} f^{-1}=0$. Thus $g = f^{-1}$. Writing $\psi = k(x) \epsilon$, with $\epsilon$ a constant spinor, we find $k = f ^{1/2}$. Finally, the Maxwell equation $\dfrac{1}{\sqrt{-g}} \partial_{\mu} (\sqrt{-g} F^{\mu\nu} ) =0$ gives $\partial_i ( f^{-2} \partial_i f) =0$, thus $f^{-1}$ is a harmonic function.

All in all, if $U(x_1,x_2,x_3)$ is a harmonic function, then $f = U^{-1}$, $g = U$, $h = U^{-1}$ is a solution of the Einstein-Maxwell equations with Killing spinor $\psi = U^{-1/2} \epsilon$ where $i \gamma_0 \epsilon = - \epsilon$. The special case $U(x_1,x_2,x_3) = 1+ \frac{m}{r}$ with $r^2 = x_1^2 + x_2^2 + x_3^3$ is the Reissner–NordstrÃ¶m solution with mass equal to the charge. The case with general $U$ is called the Majumdar-Papapetrou solution. Notice that the equations we solved above were all linear. This is much easier than solving the non-linear Einstein equations.

Further reading

This method is related to supersymmetry. The equation \eqref{eq:20161115b} expresses that the supersymmetry variation of the gravitino is zero, and therefore that the solution is supersymmetric. The method can also be generalized to higher dimensions. For example, in [2], the authors classified all supersymmetric solution of (some version of) the five-dimensional Einstein-Maxwell equations. They based their classification on an equation analogous to \eqref{eq:20161115b}.

References

- Black hole solutions in string theory, Maeda and Nozawa, 2011. hep-th/1104.1849
- All supersymmetric solutions of minimal supergravity in five dimensions, Gauntlett, Gutowski, Hull, Pakis and Reall, 2002. hep-th/0209114

Definitions and conventions

- $\gamma^a$ are the gamma matrices that satisfy $\{ \gamma^a, \gamma^b\} = 2 \eta^{ab}$ with $\eta^{ab}=\text{diagonal}(-1,1,1,1)$
- $\gamma^{ab} = \frac{1}{2} (\gamma^a \gamma^b - \gamma^b \gamma^a)$
- The covariant derivative of a spinor is \begin{equation*} (\nabla_{\mu} \psi)^{\alpha} = \partial_{\mu} \psi^{\alpha} + \Gamma^{\alpha}_{\mu\beta} \psi^{\beta} \end{equation*} with $\Gamma^{\alpha}_{\mu\beta} = - \frac{1}{4} \Gamma_{\mu ab} (\gamma^{ab})^{\alpha}_{\ \ \beta}$.
- $\Gamma_{\mu ab}$ are the components of the spin connection where $\omega^a_{\ \ b} = \Gamma^a_{\mu b} dx^{\mu}$ and $\Gamma_{\mu ab} = \eta{bc}\Gamma^c_{\mu a}$.

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