The Boyer-Lindquist coordinate coordinate of the Kerr Solution is $$ ds^2=\left(1-\frac{2Mr}{\Sigma}\right)dt^2+\frac{4Mar\sin^2\theta}{\Sigma}dtd\phi - \frac{\Sigma}{\Delta}dr^2-\Sigma d\theta^2-\left(r^2+a^2+\frac{2Ma^2r\sin^2\theta}{\Sigma}\right)\sin^2\theta d\phi^2 $$

Let $$ K=\frac{\partial}{\partial t}=(1,0,0,0) $$

I am asking if it is possible for the vector field $K$ to be hypersurface orthogonal.

The answer is that it is possible if and only if $J=Ma=0$, i.e. the cross term in the metric, $g_{t\phi}$, vanishes.

I can figure it out only by the following (incorrect) argument.

Assuming we want $K$ to be orthogonal to the surface $t=$ constant, the let $(t_0,r,\theta,\phi)$ be a curve in that surface, thus $V=(0,\dot r,\dot\theta,\dot\phi)$ is a tangent vector in that space. To make $K$ orthogonal, we must have $$ \langle K,V\rangle=0 $$ which is $$ \dot\phi g_{t\phi}=0 $$

so we must have $g_{t\phi}=0$.

But I think the above argument is incorrect, since we only proved that $K$ cannot be orthogonal to the surface $t=$ constant, there maybe other possibilities.

Then I tried to use the Frobenius Theorem, saying that $K$ is HSO if and only if $$ K_{[a}\nabla_bK_{c]}=0 $$

We have $$ K_a=K^bg_{ab}=g_{at} $$

Then equivalently we must show $$ g_{at}\nabla_bg_{ct}+g_{bt}\nabla_cg_{at}+g_{ct}\nabla_ag_{bt}=0 $$

But I don't know how to move on.

Furthermore, if $\nabla$ is Levi-Civita, i.e. it is metric compatible, the above equation holds naturally, no matter whether $g_{\phi t}=0$.

Could anyone point out to me how to deduce that $g_{t\phi}=0$ from this approach?


1 Answer 1


Your equivalence is not an equivalence. From the way you write, I think that your mistake comes from a problem with index notation. For the Levi-Civita connection, we have $$\nabla_a g_{\mu\nu} = 0. \tag{1}$$ But this does not mean that the $a$-derivative of the $\mu\nu$ component of $g$ is $0$. It means that the $a\mu\nu$ component of $\nabla g$ is $0$. Thus from (1) we cannot conclude that $$\nabla_b g_{ct} \overset{?}{=} 0 \tag{2}.$$ since in this equality $t$ is not an "index to be filled in", but is shorthand for $g_{c\mu} t^\mu$. We see that (2) holds iff $t^\mu$ is covariantly constant. Not surprising, since it's the definition of $t_c$. It is easy to become confused by this when doing or presenting concrete calculations.

Thus you really have to compute $$K_{[a} \nabla_b K_{c]}$$ and check when it vanishes. However since the Frobenius theorem has a formulation in terms of differential forms, one can use coordinate-independent, index-free differential forms language to solve your problem, and this could be faster. First, we have $$K_{[a} \nabla_b K_{c]} = (K \wedge dK)_{abc}.$$ (You may object that on the left we have the covariant derivative and on the right the exterior, and the former involves the metric but the latter does not. But it is easy to check that the Christoffel symbols cancel on the left, and we get the formula for $d$ on 1-forms. This and $\nabla_a f = (df)_a$ for scalars implies that for an arbitrary antisymmetric tensor $T_{\mu\nu\cdots}$, we have $\nabla_{[a} T_{\mu\nu\cdots]} = (dT)_{a\mu\nu\cdots}$. )

We have $$K_a = (1- \frac{2Mr}{\Sigma}) dt + \frac{2J\sin^2\theta}{\Sigma} d\phi.$$ I have not done the calculations myself from this point, but unless $J = 0$, $\Sigma$ depends on $\theta$, so probably we will get a $d\theta \wedge d\phi$ term in $dK$, that will not cancel.


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