In this particular metric: $$ds^2=\frac{1}{z}(-dt^2 +dx^2+dy^2)+zdz^2$$ we have the three Killings $$\xi_1=\partial_t$$ $$\xi_2=\partial_x$$ $$\xi_3=\partial_y$$and we also have $$\xi_4=x \partial_y-y \partial_x$$ $$\xi_5=x\partial_t-t\partial_x$$ $$\xi_6=y\partial_t-t\partial_y$$

However, in the Schwarzschild metric, we of course have $$\xi_1=\partial_t$$ $$\xi_2=\partial_\phi$$ Why don't we have $$\xi_3=t \partial_\phi-\phi \partial_t$$


Your metric contains the piece $-\mathrm dt^2 +\mathrm dx^2+\mathrm dy^2$ which is invariant under rotations in the $(t,x,y)$ plane. Thus, the generators of these rotations are conserved.

The Schwarzschild metric, $$ \mathrm ds^2\sim f(r)\mathrm dt^2+g(r,\theta)\mathrm d\phi^2+\cdots $$ is not invariant in the $(t,\phi)$ plane. Thus, there is no conservation law associated to these rotations, and no Killing field.

  • $\begingroup$ Oh, I see. Thanks. Do you know where I could find information on how to obtain the killings og the Friedman-Lemaitre-Robertson-Walker metric? I've been looking everywhere, and can't seem to find anything. $\endgroup$ – Francisco Ardevol Martinez Feb 7 '17 at 20:17

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.