# I'm confused about the number of Killing vectors in Schwarzschild metric

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: $$$$\xi_{\alpha;\beta}+\xi_{\alpha;\beta}=0$$$$ I know that the metric has symmetries with respect time and the three independent rotations around x,y,z axis, so I should expect four Killing vectors that are solution of this equation. From an independent calculation I know that a generic stationary metric has a time-like killing vector: $$$$\xi^{\alpha}=(1,0,0,0)$$$$ and if i lower the index using Schwarzschild Metric (with $$(-,+,+,+)$$ signature) I get: $$$$\xi_{\alpha}=\left( -\left(1-\frac{R_{S}}{r}\right),0,0,0\right)$$$$ Next I write down all the nine differential equations for the four component of the killing vectors, putting inside the nine non-vanishing connections of the metric. The system is not too difficult to solve, since though they are coupled, they are not all independent. In particular one on them holds: $$$$\frac{\partial{\xi_{t}}}{\partial{t}}=\frac{R_{s}}{2}\left(1-\frac{R_{s}}{r}\right) \xi_{r}$$$$ so $$\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:

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

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

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

• Two of the rotation vectors have a $\theta$ component, so I think you made a mistake somewhere. Commented Apr 29, 2022 at 18:28
• 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. Commented Apr 29, 2022 at 22:50
• I think Zee's GR book has this calculation. Commented Apr 29, 2022 at 23:38

The Schwarzschild metric is $$$$ds^2 = -\left(1-\frac{R_{\text{S}}}{r}\right) \text{d} t^2 + \left(1-\frac{R_{\text{S}}}{r}\right)^{-1} \text{d} r^2 + r^2 (\text{d} \theta^2 + \sin^2\theta \,\text{d} \phi^2).$$$$ The Christoffel symbols are \begin{align*} \Gamma^t_{tr} &= \frac{R_{\text{S}}}{2r(r-R_{\text{S}})}, & \Gamma^r_{tt} &= \frac{R_{\text{S}}}{2r^3}(r-R_{\text{S}}), & \Gamma^r_{rr} &= \frac{-R_{\text{S}}}{2r(r-R_{\text{S}})}, \\ \Gamma^\theta_{r\theta} &= \frac{1}{r}, & \Gamma^r_{\theta\theta} &= -(r-R_{\text{S}}), & \Gamma^\phi_{r\phi} &= \frac{1}{r}, \\ \Gamma^r_{\phi\phi} &= -(r-R_{\text{S}}) \sin^2 \theta, & \Gamma^\theta_{\phi\phi} &= -\sin\theta \cos\theta, & \Gamma^\phi_{\theta\phi} &= \frac{\cos\theta}{\sin\theta}. \end{align*} The Killing equations are \begin{align} \tag{1a} &K_{t,t} - \Gamma^r_{tt} K_r = 0, \\ \tag{1b} &K_{t,r} + K_{r,t} - 2 \Gamma^t_{tr} K_t = 0, \\ \tag{1c} &K_{t,\theta} + K_{\theta,t} = 0, \\ \tag{1d} &K_{t,\phi} + K_{\phi,t} = 0, \\ \tag{1e} &K_{r,r} - \Gamma^r_{rr} K_r = 0, \\ \tag{1f} &K_{r,\theta} + K_{\theta,r} - 2 \Gamma^\theta_{r\theta} K_\theta = 0, \\ \tag{1g} &K_{r,\phi} + K_{\phi,r} - 2 \Gamma^\phi_{r\phi} K_\phi = 0, \\ \tag{1h} &K_{\theta,\theta} - \Gamma^r_{\theta\theta} K_r = 0, \\ \tag{1i} &K_{\theta,\phi} + K_{\phi,\theta} - 2 \Gamma^\phi_{\theta\phi} K_\phi = 0, \\ \tag{1j} &K_{\phi,\phi} - \Gamma^r_{\phi\phi} K_r - \Gamma^\theta_{\phi\phi} K_\theta = 0. \end{align}
From ($$1$$e) we have $$$$\tag{2} K_r = T(t,\theta,\phi) \left( \frac{r}{r-R_{\text{S}}} \right)^{1/2}.$$$$ Differentiating ($$1$$b) with respect to $$t$$ and substituting ($$1$$a), ($$1$$e) and ($$2$$) into the result, we obtain $$$$\tag{3} \left( \frac{\partial}{\partial r} \Gamma^r_{tt} + 3 \Gamma^r_{tt} \Gamma^r_{rr} \right) T(t,\theta,\phi) + \frac{\partial^2}{\partial t^2} T(t,\theta,\phi) = 0.$$$$ Since $$$$\frac{\partial}{\partial r} \Gamma^r_{tt} + 3 \Gamma^r_{tt} \Gamma^r_{rr} = \frac{R_{\text{S}}}{r^3}\left( -1 + \frac{3R_{\text{S}}}{4r} \right)$$$$ is a function of $$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$$ are constants. In summary, we have \begin{align} K_t &= A \left( 1 - \frac{R_{\text{S}}}{r} \right), \\ K_r &= 0, \\ K_\theta &= (-D \sin\phi + E \cos\phi) \, r^2, \\ K_\phi &= [-\sin\theta \cos\theta (D \cos\phi + E \sin\phi) + F \sin^2\theta] \, r^2. \end{align} The general solution of the Killing equations of the Schwarzschild metric will be $$$$\begin{split} K &= g^{\mu\nu} K_\mu \partial_\nu \\ &= g^{tt} K_t \frac{\partial}{\partial t} + g^{\theta\theta} K_\theta \frac{\partial}{\partial \theta} + g^{\phi\phi} K_\phi \frac{\partial}{\partial \phi} \\ &= - A \frac{\partial}{\partial t} +(-D \sin\phi + E \cos\phi) \frac{\partial}{\partial \theta} + [ - \cot\theta (D \cos\phi + E \sin\phi) + F ] \frac{\partial}{\partial \phi} \\ &= - A L_{(0)} + D L_{(1)} + E L_{(2)} + F L_{(3)}, \end{split}$$$$ where \begin{align} L_{(0)} &= \frac{\partial}{\partial t}, \\ L_{(1)} &= -\sin\phi \frac{\partial}{\partial \theta} - \cot\theta \cos\phi \frac{\partial}{\partial \phi}, \\ L_{(2)} &= \cos\phi \frac{\partial}{\partial \theta} - \cot\theta \sin\phi \frac{\partial}{\partial \phi}, \\ L_{(3)} &= \frac{\partial}{\partial \phi} \end{align} form a basis of the Lie algebra of the Killing fields of Schwarzschild metric. $$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.