0
$\begingroup$

As mentioned in Carroll's Spacetime and Geometry p. 244, a Killing vector is normal to its Killing horizon. With some help from the other forum, I could check this is true. (FYI, here the Killing horizon $\Sigma$ of a Killing vector $\chi$ is defined by a null hypersurface on which $\chi$ is null.)

But when I try to apply this general statement to a Kerr BH, something weird thing happens: in a Kerr BH, we consider a Killing vector $$\chi=\partial_t+\Omega_H\partial_\phi,$$ where $\Omega_H$ is designed to make $\chi$ to be null on the event horizon, $$\Sigma:r=r_H=M+\sqrt{M^2-a^2}.$$ So by definition $\Sigma$ is the Killing horizon of a Killing vector $\chi$. Then according to the general statement, this $\chi$ must be normal to $\Sigma$ but it doesn't look like satisfying this condition.

To be clear, note that we can write the normal vector of $\Sigma$ as $$n_\mu=\nabla_\mu(r-r_H)=(0,1,0,0).$$ But this $n$ is not parallel to $\chi$ at all. Equivalently, tangent vectors on $\Sigma$ which is orthogonal to $n$ is not orthogonal to $\chi$. This means $\chi$ is not normal to $\Sigma$...?!?!

I have no idea at this point... If you see what is going wrong here, please help me out with this nonsense!

$\endgroup$
  • $\begingroup$ Have you tried this for a Schwarzschild BH? It has the same paradox, but it's a simpler metric, so finding the source of the trouble is easier. The vector $\partial_t$ is timelike for $r>r_H$ and spacelike for $r<r_H$ and lightlike for $r=r_H$, so it is both tangent to the horizon and normal to the horizon. In contrast, the quantities $\nabla_\mu(r-r_H)$ are the components of a one-form, not a vector, and "length" of the corresponding vector $\partial_r$ is undefined (infinite) on the horizon. To make things well-defined, we need to use different coord's, and then the paradox goes away. $\endgroup$ – Chiral Anomaly Feb 7 at 23:07
1
$\begingroup$

A Killing horizon is a null hypersurface defined by the vanishing of the norm of a Killing vector $K^\mu$, that is $K^\mu K_\mu = 0$. However a vector $A^\mu$ is orthogonal to a vector $B^\mu$ if their scalar or dot product vanishes, that is if $A^\mu B_\mu = 0$. In that sense a null vector is orthogonal to itself.
I think the statement in Carroll's Spacetime and Geometry should be read in that way.

$\endgroup$

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.