1. Jun 10, 2013

### WannabeNewton

Hello there. I have to do a calculation (problem 5 of chapter 12 in Wald) verifying the super-radiance of electromagnetic waves incident on Kerr black holes and have a few preliminary questions.

As background: on pages 328-329 of Wald, there is a discussion of super-radiance achieved by Klein Gordon fields incident on Kerr black holes. Using the standard $(t,r,\theta,\phi)$ coordinates for the Kerr space-time, we consider a scalar wave of the form $\phi = \text{Re}[\phi_{0}(r,\theta)e^{-i\omega t}e^{im\phi}]$ with $0 < \omega < m\Omega_{H}$ where $m, \omega$ are constants and the constant $\Omega_{H}$ is related to the killing field tangent to the null geodesic generators of the event horizon by $\chi^{a} = (\partial_{t})^{a} + \Omega_{H}(\partial_{\phi})^{a}$; it physically represents the "angular velocity" of the event horizon.

We then construct the conserved energy-current $J_{a} = -T_{ab}\xi^{b}$ (here $\xi^{a} = (\partial_t)^{a}$ is the time-like killing field) and calculate the time averaged flux for the scalar wave incident on the horizon i.e. we calculate $-\left \langle J_{a}\chi^{a} \right \rangle = \left \langle T_{ab}\chi^{a}\xi^{b} \right \rangle$. The energy-momentum tensor for the Klein Gordon field is given by $T_{ab} = \nabla_{a}\phi \nabla_{b}\phi - \frac{1}{2}g_{ab}(\nabla_{c}\phi \nabla^{c}\phi + \mathfrak{m}^{2}\phi)$ but $g_{ab}\xi^{a}\chi^{b} = 0$ on the horizon so we just end up with the expression $\left \langle (\xi^{a}\nabla_{a}\phi) (\chi^{b}\nabla_{b}\phi) \right \rangle$. Plugging in the above form for the scalar wave into this expression yields, after a very simple calculation, $\left \langle (\xi^{a}\nabla_{a}\phi) (\chi^{b}\nabla_{b}\phi) \right \rangle = \frac{1}{2}\omega(\omega - m\Omega_{H})\left | \phi_{0}(r,\theta) \right |$ so the energy flux through the horizon is negative i.e. the reflected wave carries back greater energy than the incident wave. This phenomena is called super-radiance.

I must now show this also holds for electromagnetic waves. My main question is: what form of an electromagnetic wave do I consider? Would I consider a simple wave solution analogous to the scalar wave above i.e. $A_{a} = \text{Re}[(A_0)_{a}(r,\theta)e^{-i\omega t}e^{im \phi}]$?

Secondly, would this wave fall under the regime of the geometrical optics approximation? That is, would the space-time derivatives of $(A_0)_{a}(r,\theta)$ be negligible? The reason I ask is, Wald never specifies the nature of the amplitude of the Klein Gordon wave above but in that case it isn't an issue since the derivatives of the amplitude never come into the expression for the time averaged flux ($\xi^{\mu}\nabla_{\mu}\phi = \partial_{t}\phi$ and $\chi^{\mu}\nabla_{\mu}\phi = \partial_{t}\phi + \Omega_{H}\partial_{\phi}\phi$). However, for an electromagnetic wave, the time averaged flux becomes $\left \langle T_{ab}\chi^{a}\xi^{b} \right \rangle = \left \langle F_{ac}F_{b}{}{}^{c}\chi^{a}\xi^{b} \right \rangle$ and since $F_{ab} = 2\nabla_{[a}A_{b]}$ this expression looks like it can potentially contain space-time derivatives of $(A_0)_{a}(r,\theta)$. As such, I would like to know if the geometrical optics approximation is valid since I can then drop any terms involving space-time derivatives of $(A_0)_{a}(r,\theta)$ that could potentially show up in the time averaged flux, saving me a lot of calculation.

It is quite possible that in the end space-time derivatives of $(A_0)_{a}(r,\theta)$ don't show up at all in the expression for the time averaged flux once I manage to figure out how to use Wald's hint (which says to first prove that $\mathcal{L}_{X}F_{ab} = -2\nabla_{[a}(F_{b]c}X^{c})$ for any vector field $X^{a}$, which is very easy to prove, and then says to use this to relate $F_{ab}\xi^{b}$ to $F_{ab}\chi^{b}$, which I am still working on) but I just want to have it clarified beforehand (so that it is less daunting from the start of course!) and more importantly I want to make sure that the form $A_{a} = \text{Re}[(A_0)_{a}(r,\theta)e^{-i\omega t}e^{im \phi}]$ is actually the correct kind of wave solution to use for this problem as I am unsure of this as well.

2. Jun 11, 2013

### WannabeNewton

Aside from the original questions above, I still have no idea how to use Wald's hint i.e. I have no idea what relation I can get between $F_{ab}\xi^{b}$ and $F_{ab}\chi^{b}$, using $\mathcal{L}_{X}F_{ab} = -2\nabla_{[a}(F_{b]c}X^{c})$ for any vector field $X^{a}$, that will be of any use in the calculation of the time averaged flux $\left \langle F_{ac}F_{b}{}{}^{c}\chi^{a}\xi^{b} \right \rangle$.

3. Jun 11, 2013

### 49ers2013Champ

Two words: Peter Donis.

Way above my intelligence, but I bet Mr. Donis can help:)

4. Jun 14, 2013

### bahamagreen

5. Jun 14, 2013

### WannabeNewton

Hi baha! Thanks for the link, the paper is interesting in and of itself however the workings of that paper are actually much more complicated than what I have to do for this problem. I'm afraid it won't help here because for this problem I only have to work with EM waves and am 99% sure I just have to work with a plane wave of the form $A_{a} = (A_0)_{a}(r,\theta)e^{i(-\omega t + m\phi)}$, analogous to the plane Klein Gordon wave Wald uses. Since Wald uses a simple plane KG wave of the form $\phi = \phi_0(r,\theta)e^{i(-\omega t + m\phi)}$ to prove superradiance in the regime $0 < \omega < m\Omega_h$, and in the vaguely worded statement of the problem he says to show superradiance in that same regime for EM waves, I think he just wants the reader to use an EM plane wave of an analogous form (i.e. $\phi\rightarrow A_{a},\phi_{0}(r,\theta)\rightarrow (A_0)_{a}(r,\theta)$). Unfortunately, as noted in the previous sentence, the problem statement is vague with regards to what he means exactly when he says EM waves so I could be wrong.

Nevertheless, it should work for the analogous plane EM wave. If only I can figure out how to use his hint! I just don't get how a relationship coming from $\mathcal{L}_{\chi}F_{ab} = -2\nabla_{[a}(F_{b]c}\chi^{c})$ and/or $\mathcal{L}_{\xi}F_{ab} = -2\nabla_{[a}(F_{b]c}\xi^{c})$, which have a lie derivative on one side and covariant derivatives on the other, could possibly be of any help in relating $F_{ab}\xi^{b}$ and $F_{ab}\chi^{b}$ for use in the calculation of $\left \langle F_{ac}F_{b}{}{}^{c}\chi^{a}\xi^{b} \right \rangle$, which has no derivatives of the quantities $F_{ab}$, $\xi^{a}$, and $\chi^{a}$ at all. I hope he isn't trolling the reader again with his hints!

Last edited: Jun 14, 2013
6. Jun 16, 2013

### bahamagreen

The geometrical optics approximation is appropriate when the wavelength and energy are small compared to the measuring device and the energy sensitivity of that device, right?

Is the Kerr BH playing a measurement role...?

7. Jun 16, 2013

### WannabeNewton

It applies when the space-time scale of variations of the electromagnetic field is much smaller than that of the curvature (Wald page 71). I know how the curvature varies near a Kerr black hole but I have no upper bound on the amplitude of the EM wave for this problem (Wald never specifies anything). Nevertheless, it isn't a major issue because if I have to do some extra calculation involving derivatives of the amplitude I'm fine with that :)! My main issue right now is figuring out that dang hint; I seriously have no clue how to use that hint in the calculation of the time-averaged Maxwell energy current. Thanks for responding.

8. Jun 16, 2013

### Staff: Mentor

Not really my strong suit, unfortunately. Using the same plane wave ansatz for an EM wave as for a scalar Klein-Gordon wave seems reasonable to me, but I haven't worked the problem in detail, so I can't say if the geometric optics approximation is still generally applicable in the EM wave case.

9. Jun 16, 2013

### WannabeNewton

Thanks for responding Peter! It is reassuring that you also find it reasonable to consider a plane wave ansatz for the EM wave analogous to the one for the KG wave that Wald uses. The geometrical optics approximation inquisition is more of a background question I should say; it shouldn't affect the final answer itself, at least I'm hoping. I just need to figure out that hint.