I'm trying to perform a calculation to derive the Killing vectors of a spherically symmetric metric (so I use the Schwarzschild metric without loss of generality because the Birkhoff theorem tells me that it's the only static spherically symmetric metric that solves Einstein equations in vacuum).

I want to solve the equation: \begin{equation} \xi_{\alpha;\beta}+\xi_{\alpha;\beta}=0 \end{equation} I know that the metric has symmetries with respect time and the three independent rotations around x,y,z axis, so I should expect four Killing vectors that are solution of this equation. From an independent calculation I know that a generic stationary metric has a time-like killing vector: \begin{equation} \xi^{\alpha}=(1,0,0,0) \end{equation} and if i lower the index using Schwarzschild Metric (with +2 signature) I get: \begin{equation} \xi_{\alpha}=\left( -\left(1-\frac{R_{S}}{r}\right),0,0,0\right) \end{equation} Next I write down all the nine differential equations for the four component of the killing vectors, putting inside the nine non-vanishing connections of the metric. The system is not too difficult to solve, since though they are coupled, they are not all independent. In particular one on them holds: \begin{equation} \frac{\partial{\xi_{t}}}{\partial{t}}=\frac{R_{s}}{2}\left(1-\frac{R_{s}}{r}\right) \xi_{r} \end{equation} so $\xi_{r}$ must be zero because $\xi_{t}$ is independent from t. With similar considerations I also get $\xi_{\theta}=0$, so the only non vanishing components of a general killing vector are t and $\phi$. (which I could define in the beginning becuse these are the only two coordinates the metric doesn't depend on). The equations left lead to solve the two coupled equations:

\begin{equation} \frac{\partial^2{\xi_{\phi}}}{\partial{r}\partial{\theta}} =2\frac{cos(\theta)}{\sin(\theta)} \frac{\partial{{\xi}_{\phi}}}{\partial{r}} \end{equation} and \begin{equation} \frac{\partial^2{\xi_{\phi}}}{\partial{\theta}\partial{r}} =\frac{2}{r} \frac{\partial{{\xi}_{\phi}}}{\partial{\theta}} \end{equation} which leads to solve the equation:

\begin{equation} \frac{cos(\theta)}{sin(\theta)}\frac{\partial{{\xi}_{\phi}}}{\partial{r}}=\frac{1}{r} \frac{\partial{{\xi}_{\phi}}}{\partial{\theta}} \end{equation} using separation of variables I find this solution:

\begin{equation} {\xi}_{\phi}(r,\theta)= \left(A_{k}r^{k}+B_{k}\right) \left(C_{k}sin(k\theta)+D_{k}\right) \end{equation} My question is the following: is there a way to fix the constants A,B,C,D to get three independent killing vectors? I don't know if my calculations are correct, but in my opinion there's no way my solution can have three independet vectors with a single coordinate ($\phi$)"free". Hope someone can help.

  • $\begingroup$ Two of the rotation vectors have a $\theta$ component, so I think you made a mistake somewhere. $\endgroup$
    – Javier
    Apr 29, 2022 at 18:28
  • $\begingroup$ Thank you. Can you suggest me a book or a reference where I can find this calculation? I can't find someone that uses Killing equation to find killing vectors, the most popular way people use is group theory but I'm not much into it. $\endgroup$ Apr 29, 2022 at 22:50
  • $\begingroup$ I think Zee's GR book has this calculation. $\endgroup$
    – Javier
    Apr 29, 2022 at 23:38

1 Answer 1


A brute force (and ugly) derivation of the Killing fields of Schwarzschild metric

The Schwarzschild metric is \begin{equation} ds^2 = -\left(1-\frac{R_{\text{S}}}{r}\right) \text{d} t^2 + \left(1-\frac{R_{\text{S}}}{r}\right)^{-1} \text{d} r^2 + r^2 (\text{d} \theta^2 + \sin^2\theta \,\text{d} \phi^2). \end{equation} The Christoffel symbols are \begin{align*} \Gamma^t_{tr} &= \frac{R_{\text{S}}}{2r(r-R_{\text{S}})}, & \Gamma^r_{tt} &= \frac{R_{\text{S}}}{2r^3}(r-R_{\text{S}}), & \Gamma^r_{rr} &= \frac{-R_{\text{S}}}{2r(r-R_{\text{S}})}, \\ \Gamma^\theta_{r\theta} &= \frac{1}{r}, & \Gamma^r_{\theta\theta} &= -(r-R_{\text{S}}), & \Gamma^\phi_{r\phi} &= \frac{1}{r}, \\ \Gamma^r_{\phi\phi} &= -(r-R_{\text{S}}) \sin^2 \theta, & \Gamma^\theta_{\phi\phi} &= -\sin\theta \cos\theta, & \Gamma^\phi_{\theta\phi} &= \frac{\cos\theta}{\sin\theta}. \end{align*} The Killing equations are \begin{align} \tag{1a} &K_{t,t} - \Gamma^r_{tt} K_r = 0, \\ \tag{1b} &K_{t,r} + K_{r,t} - 2 \Gamma^t_{tr} K_t = 0, \\ \tag{1c} &K_{t,\theta} + K_{\theta,t} = 0, \\ \tag{1d} &K_{t,\phi} + K_{\phi,t} = 0, \\ \tag{1e} &K_{r,r} - \Gamma^r_{rr} K_r = 0, \\ \tag{1f} &K_{r,\theta} + K_{\theta,r} - 2 \Gamma^\theta_{r\theta} K_\theta = 0, \\ \tag{1g} &K_{r,\phi} + K_{\phi,r} - 2 \Gamma^\phi_{r\phi} K_\phi = 0, \\ \tag{1h} &K_{\theta,\theta} - \Gamma^r_{\theta\theta} K_r = 0, \\ \tag{1i} &K_{\theta,\phi} + K_{\phi,\theta} - 2 \Gamma^\phi_{\theta\phi} K_\phi = 0, \\ \tag{1j} &K_{\phi,\phi} - \Gamma^r_{\phi\phi} K_r - \Gamma^\theta_{\phi\phi} K_\theta = 0. \end{align}
From ($1$e) we have \begin{equation}\tag{2} K_r = T(t,\theta,\phi) \left( \frac{r}{r-R_{\text{S}}} \right)^{1/2}. \end{equation} Differentiating ($1$b) with respect to $t$ and substituting ($1$a), ($1$e) and ($2$) into the result, we obtain \begin{equation}\tag{3} \left( \frac{\partial}{\partial r} \Gamma^r_{tt} + 3 \Gamma^r_{tt} \Gamma^r_{rr} \right) T(t,\theta,\phi) + \frac{\partial^2}{\partial t^2} T(t,\theta,\phi) = 0. \end{equation} Since \begin{equation} \frac{\partial}{\partial r} \Gamma^r_{tt} + 3 \Gamma^r_{tt} \Gamma^r_{rr} = \frac{R_{\text{S}}}{r^3}\left( -1 + \frac{3R_{\text{S}}}{4r} \right) \end{equation} is a function of $r$ only, and is not identically zero, ($3$) holds only when $T(t,\theta,\phi) \equiv 0$. Then we have $K_r \equiv 0$. So we can simplify the Killing equations to \begin{align} \tag{4a} &K_{t,t} = 0, \\ \tag{4b} &K_{t,r} = 2 \Gamma^t_{tr} K_t, \\ \tag{4c} &K_{t,\theta} + K_{\theta,t} = 0, \\ \tag{4d} &K_{t,\phi} + K_{\phi,t} = 0, \\ \tag{4e} &K_{\theta,r} = 2 \Gamma^\theta_{r\theta} K_\theta, \\ \tag{4f} &K_{\phi,r} = 2 \Gamma^\phi_{r\phi} K_\phi, \\ \tag{4g} &K_{\theta,\theta} = 0, \\ \tag{4h} &K_{\theta,\phi} + K_{\phi,\theta} = 2 \Gamma^\phi_{\theta\phi} K_\phi, \\ \tag{4i} &K_{\phi,\phi} = \Gamma^\theta_{\phi\phi} K_\theta. \end{align} From ($4$a), ($4$b), ($4$e), ($4$f) and ($4$g) we have \begin{align} K_t &= A(\theta,\phi) \left( 1 - \frac{R_{\text{S}}}{r} \right), \\ K_\theta &= B(t,\phi) \,r^2, \\ K_\phi &= C(t,\theta,\phi) \,r^2. \end{align} Substituting these results into ($4$c) and ($4$d), we obtain \begin{align} \frac{\partial}{\partial\theta} A(\theta,\phi) \left( 1 - \frac{R_{\text{S}}}{r} \right) + \frac{\partial}{\partial t} B(t,\phi) \,r^2 = 0, \\ \frac{\partial}{\partial\phi} A(\theta,\phi) \left( 1 - \frac{R_{\text{S}}}{r} \right) + \frac{\partial}{\partial t} C(t,\theta,\phi) \,r^2 = 0. \end{align} These equations hold only when $A$ is a constant and $B$ and $C$ are independent of $t$. So we have \begin{align} K_t &= A \left( 1 - \frac{R_{\text{S}}}{r} \right), \\ K_\theta &= B(\phi) \,r^2, \\ K_\phi &= C(\theta,\phi) \,r^2. \end{align}
Substituting $K_\theta$ and $K_\phi$ into ($4$h) and ($4$i), we obtain \begin{align} \frac{\partial B}{\partial\phi} + \frac{\partial C}{\partial\theta} &= 2 \frac{\cos\theta}{\sin\theta} \,C, \\ \frac{\partial C}{\partial\phi} &= -\sin\theta \cos\theta \,B. \end{align} We can easily solve these PDEs to get \begin{align} B(\phi) &= -D \sin\phi + E \cos\phi, \\ C(\theta,\phi) &= -\sin\theta \cos\theta (D \cos\phi + E \sin\phi) + F \sin^2\theta, \end{align} where $D,E,F$ are constants. In summary, we have \begin{align} K_t &= A \left( 1 - \frac{R_{\text{S}}}{r} \right), \\ K_r &= 0, \\ K_\theta &= (-D \sin\phi + E \cos\phi) \, r^2, \\ K_\phi &= [-\sin\theta \cos\theta (D \cos\phi + E \sin\phi) + F \sin^2\theta] \, r^2. \end{align} The general solution of the Killing equations of the Schwarzschild metric will be \begin{equation} \begin{split} K &= g^{\mu\nu} K_\mu \partial_\nu \\ &= g^{tt} K_t \frac{\partial}{\partial t} + g^{\theta\theta} K_\theta \frac{\partial}{\partial \theta} + g^{\phi\phi} K_\phi \frac{\partial}{\partial \phi} \\ &= - A \frac{\partial}{\partial t} +(-D \sin\phi + E \cos\phi) \frac{\partial}{\partial \theta} + [ - \cot\theta (D \cos\phi + E \sin\phi) + F ] \frac{\partial}{\partial \phi} \\ &= - A L_{(0)} + D L_{(1)} + E L_{(2)} + F L_{(3)}, \end{split} \end{equation} where \begin{align} L_{(0)} &= \frac{\partial}{\partial t}, \\ L_{(1)} &= -\sin\phi \frac{\partial}{\partial \theta} - \cot\theta \cos\phi \frac{\partial}{\partial \phi}, \\ L_{(2)} &= \cos\phi \frac{\partial}{\partial \theta} - \cot\theta \sin\phi \frac{\partial}{\partial \phi}, \\ L_{(3)} &= \frac{\partial}{\partial \phi} \end{align} form a basis of the Lie algebra of the Killing fields of Schwarzschild metric. $L_{(0)}$ is a timelike Killing vector field that is orthogonal to a foliation of spacelike hypersurfaces, representing a static spacetime. $L_{(1)}, L_{(2)}, L_{(3)}$ are Killing fields of a 2-sphere, representing a spacetime with a SO(3) symmetry.

  • $\begingroup$ very beautiful solution $\endgroup$
    – lewton
    Dec 29, 2023 at 8:01

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