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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.
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.