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!

  • $\begingroup$ 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. $\endgroup$ Feb 7, 2019 at 23:07

1 Answer 1


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.


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