Killing vectors of Schwarzschild metric

Question

Exterior Schwarzschild metric

$$ ds^2=\Big(1-\frac{2M}{r}\Big)dt^2- \Big(1-\frac{2M}{r}\Big)^{-1}dr^2-r^2 d\theta^2-r^2\sin^2\theta d\phi^2 $$

I knew that there are two Killing vectors associated with the Schwarzschild metric, $K^{(1)}=(1, 0, 0, 0)$ and $K^{(2)}=(0, 0, 0, 1)$.

But, in an article written that there are four Killing vectors in Schwarzschild metric. $$ K^{(1)}_\mu=\Big(1-\frac {2M}{r}\Big)\delta^t_\mu $$

$$ K^{(2)}_\mu=r^2\sin^2\theta \delta^\phi_\mu $$

$$ K^{(3)}_\mu=r^2(\sin \phi\delta^\theta_\mu+\sin\theta\cos\theta\cos\phi\delta^\phi_\mu) $$

$$ K^{(4)}_\mu=r^2(\cos \phi\delta^\theta_\mu-\sin\theta\cos\theta\sin\phi\delta^\phi_\mu) $$

corresponding to time translations and infinitesimal spatial rotations.

I did not understand how are Killing vectors defined? Why are they found differently from classical GR books? What is the method to obtain the Killing vectors of arbitrary metric?

If the metric does not depend on $x^k$ coordinate $K^\mu=\delta^\mu_k$ is a Killing vector. The covariant components which are $ K_\mu=g_{\mu\nu}K^\nu=g_{\mu\nu}\delta^\nu_k $. $$ K^{(1)}_\mu=g_{\mu\nu}\delta^\nu_k=g_{k\nu}\delta^\nu_\mu=g_{tt}\delta^t_\mu $$

$$ K^{(2)}_\mu=g_{\mu\nu}\delta^\nu_k=g_{k\nu}\delta^\nu_\mu=g_{\phi\phi}\delta^\phi_\mu $$

This is understandable. But how are obtained the second and third Killing vectors? If they are rotation associated vectors around which axis rotation carried out?

I found the solutions for $\xi_\theta$, $\xi_\phi$ from the Killing equations for the metric of the $2D$ sphere.

But, how they are picking the arbitrary values for $A$, $B$, $C$ is unclear.

Answer 1

Any vector field that satisfies $$ \mathcal{L}_v g_{\mu \nu} = 0 $$ with $\mathcal{L}$ a Lie derivative is a Killing field. Infinitesimal diffeomorphisms generated by it preserve the components of the metric tensor, so they form a group of isometries, much like $ISO(3,1)$ in Minkowski space.

Expanding the definition of the Lie derivative, $$ \mathcal{L}_v g_{\mu \nu} = v^{\sigma} \partial_{\sigma} g_{\mu \nu} + g_{\sigma \nu} \partial_{\mu} v^{\sigma} + g_{\mu \sigma} \partial_{\nu} v^{\sigma}. $$

This is already a tensor, by definition of the Lie derivative. We can, however, substitute the partial derivative for covariant derivative. In the normal coordinates, $\Gamma_{\mu \nu}^{\sigma}$ all vanish, so we'll get the same answer. Moreover, since both objects (with partial and with covariant derivatives) are tensors, and they are equal in one coordinate frame, they must be equal in all frames. Therefore,

$$ \mathcal{L}_v g_{\mu \nu} = v^{\sigma} \nabla_{\sigma} g_{\mu \nu} + g_{\sigma \nu} \nabla_{\mu} v^{\sigma} + g_{\mu \sigma} \nabla_{\nu} v^{\sigma} = \nabla_{\mu} v_{\nu} + \nabla_{\nu} v_{\mu} = 0. $$

That is the equation that a Killing vector field must satisfy. Now all you have to do is check that your fields satisfy this equation.

Answer 2

Killing vectors are just infinitessimal motions that keave the metric unchanged. In other words isometries. The four that seem to puzzle you are time translation (with an added normalization compared to your original one) and the three isometries of the 3-sphere $r=constant$ corresponding to rotating about the $x$, $y$, and $z$ axes. The three rotations are not linearly independent of course, and from them you can get isometries from rotation about any axis.