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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 positive 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, $\left \langle (\xi^{a}\nabla_{a}\phi) (\chi^{b}\nabla_{b}\phi) \right \rangle = \frac{1}{2}\omega(\omega - m\Omega_{H})\left | \phi_{0}(r,\theta) \right |^{2}$ 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.

I'm still trying 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 still can't figure out) but I just want to have it clarified beforehand (so that it is less daunting from the start of course!) and [I]more importantly[/I] 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. Thanks in advance!

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    $\begingroup$ very interesting question! $\endgroup$
    – lurscher
    Jun 12, 2013 at 16:20
  • $\begingroup$ i did a somewhat related question a while ago: physics.stackexchange.com/q/46870/955 There are so much questions open regarding Hawking radiation for non-schwarzschild BH on semi-classical gravity. I'm not sure if people still work on these subjects more than as an exercise to eventually settle in other more professionally rewarding problems in string theory $\endgroup$
    – lurscher
    Jun 13, 2013 at 1:37
  • $\begingroup$ btw, not directly related questions, but hopefully will have your head scratching as well: physics.stackexchange.com/q/47668/955 physics.stackexchange.com/q/48511/955 $\endgroup$
    – lurscher
    Jun 13, 2013 at 1:45
  • $\begingroup$ I'm not sure how the hint helps. You can use $A_\mu=A^0_\mu e^{-i\omega t-m\phi}$, where $A^0_\mu=(0,A(r),0,0)$ (radial wave). Plugging this into the definition of $F$ and then into the expression, $F_{ac}F_b^c\chi^a\xi^b$ yields an expression proportional to $(\omega-m\Omega)$ as needed. $\endgroup$
    – sobol
    Mar 14, 2021 at 2:42
  • $\begingroup$ I know this is an old question, but I might provide my own thought, athough not very successful. We might want to apply Stokes' theorem to $\int_S\mathscr L_wF_{ab}=-\oint_LF_{ab}l^aw^b$ with $S$ a 2-surface bounded by $L$. Choose $L$ such that one part of it is the radial coordinate line from a point in horizon to the infinity, and the remaining part shall be a "good" curve such that the curve integral along it vanishes. If we can find such $L$, we can calculate $F_{ab}w^b$ on horizon in terms of $\int_S\mathscr L_wF_{ab}$. But I failed to find such $L$... $\endgroup$ Oct 12, 2022 at 8:27

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