Co-rotating electric potential for KN solution

[etotheipi]

I am trying to work out the co-rotating electric potential $\Phi = \xi^{\mu} A_{\mu}$ for the KN solution. First it's necessary to prove that the hypersurfaces $r = r_{\pm}$ are Killing horizons $\mathcal{N}_{\pm}$ of a Killing field of the form $\xi = k + \Omega_H m$ for some Killing fields $k,m$ and constant $\Omega_H$. It is possible to work out the normal vectors to $\mathcal{N}_{\pm}$, for some arbitrary function $f_{\pm}$, \begin{align*}
l_{\pm} = f_{\pm} g^{\mu r}\big{|}_{\mathcal{N}_{\pm}} \partial_{\mu} = f_{\pm} g^{rr}\big{|}_{\mathcal{N}_{\pm}} \partial_r = f_{\pm} \frac{\Delta}{\Sigma} \big{|}_{\mathcal{N}_{\pm}} \partial_r = f_{\pm} \left( \frac{r_{\pm}^2 - 2Mr_{\pm} + (a^2 + e^2)}{r_{\pm}^2 + a^2 \cos^2{\theta}}\right) \partial_r
\end{align*}where $\Delta = (r-r_{+})(r-r_{-})$ and $r_{\pm} = M \pm \sqrt{M^2 - (a^2 + e^2)}$. We can check the $\mathcal{N}_{\pm}$ really are null because $l_{\pm}^2 \propto g_{rr} \big{|}_{\Delta = 0} = 0$.

In order for $\mathcal{N}_{\pm}$ to be a Killing horizon of $\xi$, it's required that $\xi \big{|}_{\mathcal{N}_{\pm}} \propto l_{\pm}$. I can't see exactly how to come up with the form of $\xi$. I'd be grateful if someone could offer some hints, or tell me if I've gone wrong already. Thanks!

 

[PeterDonis]

I can't see exactly how to come up with the form of $\xi$.

Do you have any ideas on what the Killing fields $k$ and $m$ might be? (Hint: there aren't that many to choose from since the spacetime is stationary and axisymmetric, with no other symmetries.)

 

[etotheipi]

Do you have any ideas on what the Killing fields $k$ and $m$ might be? (Hint: there aren't that many to choose from since the spacetime is stationary and axisymmetric, with no other symmetries.)

I presumed that they are $k = \partial_t$ and $m = \partial_{\phi}$, but in that case we require $(\partial_t + \Omega_H \partial_{\phi}) \big{|}_{\mathcal{N}_{\pm}} \propto l_{\pm}$ where $l_{\pm}$ is as above, but since $l_{\pm} \propto \partial_r$ I can't see what to write for $\Omega_H$ that would make it work

 

[PeterDonis]

I presumed that they are $k = \partial_t$ and $m = \partial_{\phi}$

Yes.

in that case we require $(\partial_t + \Omega_H \partial_{\phi}) \big{|}_{\mathcal{N}_{\pm}} \propto l_{\pm}$ where $l_{\pm}$ is as above, but since $l_{\pm} \propto \partial_r$ I can't see what to write for $\Omega_H$ that would make it work

Isn't $\partial_r$ tangent to the horizon? (Remember the horizon is a null surface so a vector that is normal to the horizon is also tangent to it.) And given that, wouldn't $\partial_r$ also be parallel to any null vector on the horizon that, heuristically, "goes in the same direction"?

 

[PeterDonis]

wouldn't $\partial_r$ also be parallel to any null vector on the horizon that, heuristically, "goes in the same direction"?

For a simpler example of the same sort of thing, consider the vector fields usually labeled as $\partial_t$ and $\partial_r$ in Schwarzschild spacetime. (Strictly speaking, we can't use Schwarzschild coordinates to evaluate these on the horizon, but the vector fields themselves are perfectly well defined on the horizon, and we can use, say, Kruskal coordinates to analyze them if we need coordinates.) On the horizon, these vector fields are parallel, even though they are also orthogonal to each other; this works because the vector fields are both null on the horizon and "go in the same direction".

 

[etotheipi]

Oh yeah! I'd forgotten that null hyper-surfaces have that strange property that normal vectors are also tangent vectors. This is a bit of stab in the dark, but in that case what about, with $k = \partial_t$ and $\Omega_H = 0$, $$
\cancel{\xi_{\pm} = \left( \frac{r_{\pm}^2 - 2Mr_{\pm} + (a^2 + e^2)}{r_{\pm}^2 + a^2 \cos^2{\theta}}\right) k}$$

 

[PeterDonis]

with $k = \partial_t$ and $\Omega_H = 0$?

This can't possibly be right, since $\partial_t$ is spacelike on the horizon.

Another hint: what do you think $\Omega_H$ represents?

 

[PeterDonis]

This is a bit of stab in the dark

You shouldn't need to stab. You know what the vectors $k$ and $m$ are; all you need to do is find a linear combination of them that is null on the horizon.

 

[etotheipi]

Another hint: what do you think $\Omega_H$ represents?

I know that this is the angular velocity of the orbits of $\xi$ relative to the orbits of $\partial_t$.

You shouldn't need to stab. You know what the vectors $k$ and $m$ are; all you need to do is find a linear combination of them that is null on the horizon.

If I denote\begin{align*}

h(r_{\pm}) := g^{rr}\big{|}_{\mathcal{N}_{\pm}} = \frac{r_{\pm}^2 - 2Mr_{\pm} + (a^2 + e^2)}{r_{\pm}^2 + a^2 \cos^2{\theta}}

\end{align*}then for $\partial_t + \Omega_H \partial_{\phi}$ to be null on the horizon it has to be orthogonal to $l_{\pm}$, so I would have thought that we have to constrain\begin{align*}

h(r_{\pm}) g_{tr} + \Omega_H h(r_{\pm}) g_{\phi r} \overset{!}{=} 0\end{align*}the problem I'm having is that $g_{tr}$ and $g_{\phi r}$ just vanish, so this relationship doesn't help... I am a little confused

 

[PeterDonis]

I know that this is the angular velocity of the orbits of $\xi$ relative to the orbits of $\partial_t$.

Yes. Physically it also represents what is sometimes called "the angular velocity of the horizon". If you consider the possible angular velocities for timelike orbits around the hole, these angular velocities get more and more constrained as you approach the horizon; in the limit, the only possible angular velocity is $\Omega_H$.

However, there is another way of looking at $\Omega_H$, by considering the following: what is the angular momentum of a light ray on the horizon whose worldline is an orbit of $\xi$? (Note that "angular momentum" here just means the constant of geodesic motion $L = k^a \psi_a$, where $k^a$ is the tangent vector to the worldline and $\psi_a$ is the axial KVF.)

for $\partial_t + \Omega_H \partial_{\phi}$ to be null on the horizon it has to be orthogonal to $l_{\pm}$

This is true, but there's also a simpler way to compute the null condition: just compute the norm of $\xi$. You know its components, just plug them into the metric and set $ds = 0$. (Note that the applicable metric coefficients for this are $g_{tt}$, $g_{t \phi}$, and $g_{\phi \phi}$ since $\xi$ only has $t$ and $\phi$ components.)

 

[etotheipi]

Ah, awesome, that makes sense!\begin{align*}
g_{\mu \nu} \xi^{\mu} \xi^{\nu} &= g_{tt} \xi^t \xi^t + g_{\phi \phi} \xi^{\phi} \xi^{\phi} + 2g_{t \phi} \xi^{t} \xi^{\phi}\\ \\

&= \frac{-(\Delta - a^2 \sin^2{\theta})}{\Sigma} + \Omega_H^2 \sin^2{\theta} \left( \frac{(r^2 + a^2)^2 - \Delta a^2 \sin^2{\theta}}{\Sigma} \right) -2a \Omega_H \sin^2{\theta} \frac{(r^2 + a^2 - \Delta)}{\Sigma} \overset{!}{=} 0
\end{align*}which gives\begin{align*}

\Omega_H = \frac{a\sin^2{\theta} [r^2 + a^2 - \Delta] \pm \sqrt{a^2 \sin^4{\theta} [r^2 + a^2 - \Delta]^2 + \sin^2{\theta}(\Delta - a^2 \sin^2{\theta})([r^2 +a^2]^2 - \Delta a^2 \sin^2{\theta})}}{\sin^2{\theta} [(r^2 + a^2)^2 - \Delta a^2 \sin^2{\theta}]}\end{align*}I'm sure that will simplify a little bit...

However, there is another way of looking at $\Omega_H$, by considering the following: what is the angular momentum of a light ray on the horizon whose worldline is an orbit of $\xi$? (Note that "angular momentum" here just means the constant of geodesic motion $L = k^a \psi_a$, where $k^a$ is the tangent vector to the worldline and $\psi_a$ is the axial KVF.)

This would be:
\begin{align*}

L = \frac{\xi^a \psi_a}{(-\xi^b \xi_b)^{1/2}}\end{align*}but since $\psi_a = (\partial / \partial {\phi})_a$, we'll just have $\xi^a \psi_a = \Omega_H$, I think. So in that case $L \propto \Omega_H$, which is sort of what we expect (e.g. in analogy to the classical formula)!

The next part of the question is to show that $\Phi$ is constant on the horizon, and then to choose a gauge where $\Phi = 0$ at infinity and show that $\Phi_H = Qr_{+}/ (r_+^2 + a^2)$. I think I have a decent idea about how to do this, so I'll post an update when I make progress!

Thanks so much

 

[etotheipi]

I think I'm a little too tired to finish this off today ... but in any case I simplified$$\sqrt{a^2 \sin^4{\theta} [r^2 + a^2 - \Delta]^2 + \sin^2{\theta}(\Delta - a^2 \sin^2{\theta})([r^2 +a^2]^2 - \Delta a^2 \sin^2{\theta})} = -\Sigma \sqrt{\Delta} \sin{\theta}$$which makes $\Omega_H$ a little nicer.

Now we need to show that $\Phi = \xi^{\mu} A_{\mu}$ is constant on the horizon. The Lie derivative agrees with the covariant derivative on functions, so we just have\begin{align*}

\mathcal{L}_{\xi} \Phi = \xi^{\mu} \nabla_{\mu} (\xi^{\nu} A_{\nu}) = \xi^{\nu} \xi^{\mu} \nabla_{\mu} A_{\nu} + A_{\nu} \xi^{\mu} \nabla_{\mu} \xi^{\nu} \overset{!}{=} 0

\end{align*}Making use of $\nabla^{\mu} \nabla_{\mu} A_{\nu}= {R^{\alpha}}_{\nu} A_{\alpha}$ in vacuo, Killing's equation and the result that $R_{\mu \nu} \xi^{\mu} \xi^{\nu} = 0$ on the Killing horizon probably gives the result, but I feel like I'm about to fall asleep so I'll check tomorrow!

 

[PeterDonis]

I simplified

You should be able to simplify further by noting that on the horizon, $\Delta = 0$. Also, you can plug in $r = r_+$. However, I'm not sure that your equation is correct, because when I plug in those things I don't seem to get what I already know is the right answer for $\Omega_H$. I'll try to check things by another route when I have a chance.

 

[PeterDonis]

This would be:
\begin{align*}

L = \frac{\xi^a \psi_a}{(-\xi^b \xi_b)^{1/2}}\end{align*}

No. Remember we are on the horizon, so $\xi^a$ is null, so the denominator of the formula you wrote here is zero, hence the formula is not well-defined.

The angular momentum is just $L = \xi^a \psi_a$, i.e., the numerator of your formula.

we'll just have $\xi^a \psi_a = \Omega_H$, I think.

No. At least, that's not the answer I was expecting. The answer I was expecting is zero, since the vector field $\xi^a = t^a + \Omega_H \psi^a$ on the horizon is an instance of a ZAMO (zero angular momentum observer) vector field, so it should have $L = 0$.

I'll try to check all this by another route when I have a chance.

 

[PeterDonis]

Btw, I'll post two useful references for this general topic; one is the Visser paper that I linked to in a recent thread on Kerr spacetime:

https://arxiv.org/pdf/0706.0622.pdf

The other is a paper on ZAMO congruences:

https://arxiv.org/pdf/1408.6316.pdf

 

[etotheipi]

Yeah, I'm making a ridiculous number of mistakes this evening and I really wouldn't trust very many of the results I've written down. I'll check things over tomorrow

 

[PeterDonis]

I'll try to check things by another route when I have a chance.

I'll try to check all this by another route when I have a chance.

I have done some checking, helped by the references I gave, and while I won't post all the details here in order not to spoil whatever calculations others are doing, I'll post some brief pointers:

(1) Everywhere outside the horizon, one can define a "ZAMO" 4-velocity field $u = \beta \left( \partial_t + \omega \partial_\phi \right)$ (using the notation of the second paper I referenced), which is timelike, so $g_{ab} u^a u^b = -1$, and which has $L = 0$, so $g_{ab} u^a \psi^b = 0$, where $\psi^b$ is the axial KVF. If it seems strange that $L = 0$ when $u$ has a nonzero $\phi$ component, remember the cross term $g_{t \phi}$ in the metric; or, geometrically speaking, remember that the "timelike at infinity" KVF $\partial_t$ and the axial KVF are not orthogonal. The ZAMO 4-velocity field tells you which direction in spacetime is orthogonal to the axial KVF; $\omega$ gives the angular velocity of that direction in spacetime. We can use the $L = 0$ condition to derive an equation for $\omega$.

(2) We can take the limit of the ZAMO 4-velocity field as $r \rightarrow r_+$, i.e., as the horizon is approached. Even though Boyer-Lindquist coordinates are singular there, the coordinate singularity does not affect our ability to take this limit (since the coordinate singularity is in $g_{rr}$ and that metric coefficient doesn't come into play). This gives a well-defined value of $\omega$ at $r = r_+$.

(3) We can obtain the same value of $\omega$ at $r = r_+$ if we take the same vector field $u$ that we defined above and assume it is null on the horizon (which will seem odd since we assumed it was timelike outside the horizon), and use that to obtain an equation for $\omega$ (note that $\beta$ drops out of this equation). Alternately, we can just compute the norm of $u$ using our definition above, plugging in $r = r_+$ and the value of $\omega$ we got from (2), and verify that the norm turns out to be zero. Either way, this means $u$ is our desired null KVF on the horizon, and $\omega$ is our desired angular velocity. It also means that $L = 0$ for $u$ on the horizon, i.e., even though the null worldlines that generate the horizon are "rotating" at $\Omega_H$, they still have zero angular momentum. This is one of the weirdnesses of Kerr spacetime.