While calculating $\beta_{ij}$ for the case of a Schwarzschild black hole, Hawking uses the Fourier transform of the solution of the wave equation (Particle Creation by Black Holes, S.W. Hawking, Commun. Math. Phys. 43, 199 (1975)).

So the first question I have is this: Is the reason why he prefers to work in Fourier space that in there the differential equation becomes an algebraic equation, making it easier to deal with?

The other question pertains to the mathematical machinery of the transform. The wave equation is $\Box \phi=0$, and it can be expanded in terms of the complete orthonormal set ${\{g\}}$ satisfying $\Box g=0$. In Schwarzschild spacetime we also have another additional property, i.e. $g(r,t,\theta,\phi)=\frac{1}{r}f(r,t)Y_{lm}(\theta,\phi)$, where $Y_{lm}(\theta,\phi)$ is the spherical harmonic function and $f$ satisfies the following equation $$\frac{\partial^2 f}{\partial t^2}-\frac{\partial^2f}{\partial r_*}+\left(1-\frac{2M}{r}\right)\left[\frac{l(l+1)}{r^2}+\frac{2M}{r^3}+m^2 \right]f=0 \tag{1}$$ where $r_*=r+2Mln(\frac{2M}{r}-1)$.

Now, Hawking states that Fourier transform of the incoming wave can be written as the following: $$p_{\omega^{'}lm}=\frac{1}{\sqrt{2\pi\omega^{'}}}\times\frac{1}{r}\times F_{\omega^{'}}(r)\times e^{i\omega^{'}v}\times Y_{lm}(\theta,\phi) \tag{2} $$ where $v=t+r_*$.

Going by the definition of the Fourier transform, one obtains $$p_{\omega^{'}lm}=\int_{-\infty}^{+\infty}\frac{1}{r}f(r,t)Y_{lm}(\theta,\phi)e^{-i\omega^{'}v}dv \tag{3}$$ since $dv=dt+(1-\frac{2M}{r})^{-1}dr$. I don't have any idea how to proceed any further from here, so what should I do now?

Edit: My problem is how to get the final form of $p_{\omega^{'}lm}$ and by final form I mean the equation 2.12 or 2.11

And if anyone is still confused as what one has to show it's just how one goes from $(3)$ to $(2)$.

  • $\begingroup$ Is this just the fairly standard notation convention that $F(\cdot)$ is the Fourier transform of $f(\cdot)$? If so you don't really need to "do" anything. $\endgroup$
    – alephzero
    Sep 2, 2019 at 15:57
  • $\begingroup$ @alephzero I don't think that $F$ is fourier transform of $f$ $\endgroup$
    – aitfel
    Sep 3, 2019 at 1:08
  • $\begingroup$ @aitfel Could you clarify what your question is exactly? Is it how to obtain the $p_{\omega' l m}$ as Hawking writes it? $\endgroup$
    – John Donne
    Sep 6, 2019 at 12:27
  • $\begingroup$ @JohnDonne I have written my problem at the end of the question. $\endgroup$
    – aitfel
    Sep 6, 2019 at 12:41
  • 1
    $\begingroup$ @JohnDonne Clarified in the last section of the question. $\endgroup$
    – aitfel
    Sep 6, 2019 at 16:01

1 Answer 1


Hawking considers a massless scalar field in the Schwarzschild spacetime. This satisfies the massless Klein-Gordon equation $\square \phi \equiv \nabla^a \nabla_a \phi = 0$. The expanded equation is $$-\left(1-\frac{2M}{r}\right)^{-1} \partial_t^2 \phi +\frac{1}{r^2}\partial_r \left[\left(1-\frac{2M}{r}\right) r^2 \partial_r \phi\right]+\frac{1}{r^2}\Delta_{S^2} \phi=0$$ where $\Delta_{S^2}$ is the Laplacian for the 2-sphere $S^2$. Its eigenfunctions are the spherical harmonics $Y_{lm}(\theta,\phi)$ so we can remove the angular dependence of the field by writing $\phi = \frac{1}{r}G(t,r)Y_{lm}(\theta,\varphi)$. We introduce the $1/r$ term without loss of generality because $G(t,r)$ depends on $r$, but we expect that this will simplify the equation: physically, the energy (square of the amplitude) of a wave in 3 spatial dimensions goes like $1/r^2$, so the amplitude goes like $1/r$. To simplify the equation we also introduce the usual (Eddington-Finkelstein) tortoise coordinate $r^*=r+2M\log\left(\frac{2M}{r}-1\right)$. Substituting we get $$(\partial_t^2 -\partial_{r^*}^2)G+\left(1-\frac{2M}{r}\right)\left(\frac{l(l+1)}{r^2}+\frac{2M}{r^3}\right) G=0\tag{A1}$$ Note that our notation is slightly different from that of the OP (we write $G$ instead of $f$) to be consistent with Hawking's paper, where $f$ denotes a specific set of solutions. Our equation is slightly different from OP's eq. $(1)$ in that, following Hawking, we only care about a massless scalar field. In terms of the advanced and retarded time coordinates $v = t+r^*$ and $u = t-r^*$ the equation becomes $$\partial_u \partial_v G+\left(1-\frac{2M}{r}\right)\left(\frac{l(l+1)}{r^2}+\frac{2M}{r^3}\right) G=0\tag{A2}$$

As usual in quantum field theory, we then look for a momentum-space (i.e. Fourier) decomposition of $\phi$. As such, we do not compute the Fourier transform explicitly but rather find a basis of plane-wave solutions and then express the original wave as a superposition of such solutions. (the procedure is the same as in ordinary flat-space QFT). We can then obtain the quantised field $\hat \phi$ by promoting the coefficients of the expansion to (creation and annihilation) operators.

In order to find the Fourier modes, we need to specify appropriate boundary conditions to solve the wave equation. In the "early universe" $t\to -\infty$ a massless (null) wave will live on past null infinity $\mathscr{I}^-$, defined as the asymptotic region $t\to -\infty$, $r\to +\infty$ with $v=t+r^*$ kept constant (note that as $r\to \infty$ also $r^*\to \infty$). In the "late universe" $t\to +\infty$ the wave will live on future null infinity $\mathscr{I}^+$, the region $t\to +\infty$, $r\to +\infty$ with $u=t-r^*$ kept constant. Boundary conditions appropriate to the early universe solutions require that $G$ is an incoming positive frequency solution on $\mathscr{I}^-$, so if we call $\phi = f_{\omega l m}$ a basis of solutions to the full wave equation with this boundary condition we can write, as we did before, $$f_{\omega l m} = \frac{1}{r}G_\omega(t,r)Y_{lm}(\theta,\varphi)$$ Then $G_\omega(t,r)$ satisfies $(A2)$. On $\mathscr{I}^-$, $r\to \infty$ so the equation reduces to
$$\partial_u \partial_v G_\omega(t,r)=0\,\,\,\,\,\,\,\,\,\,\,\,\, \mathrm{on}\,\,\,\,\,\mathscr{I}^-$$ This is the same as the flat-space wave equation written in light cone coordinates, and is solved by $G_\omega(t,r)=G_1(u)+G_2(v)$ with $G_1, G_2$ arbitrary. Plane wave solutions of this equation look like $e^{i\omega u}$ (outgoing) or $e^{i\omega v}$ (incoming). We care about incoming solutions, so $$G_{\omega}(t,r)\sim\frac{1}{\sqrt{2\pi \omega}}e^{i\omega v}\,\,\,\,\,\,\,\,\,\,\,\,\, \mathrm{on}\,\,\,\,\,\mathscr{I}^-$$ The normalisation is arbitrary and as in flat space we choose it so that upon quantisation the creation/annihilation operators satisfy the canonical commutation relations with the desired normalisation. (In flat space a positive frequency wave would take the form $\sim e^{i\omega t -i kx}$).

Now write $G_\omega (t,r) = \frac{1}{\sqrt{2\pi \omega}}e^{i\omega v} F_\omega(t,r)$ (on the full spacetime, not just at infinity), without loss of generality. Going back to $(A1)$ we see that we can remove the $t$ dependence of the equation by writing $G=e^{i\omega t}R(r^*)$. Then $R(r^*)$ can indeed be taken independent of $t$, as it satisfies $$-(\partial_{r^*}^2+\omega^2)R+\left(1-\frac{2M}{r}\right)\left(\frac{l(l+1)}{r^2}+\frac{2M}{r^3}\right) R=0\tag{A1}$$ Since $v=t+r^*$ note that $G_\omega (t,r) =e^{i\omega t}\frac{1}{\sqrt{2\pi \omega}}e^{i\omega r^*} F_\omega(t,r)$, so $F_\omega(t,r)$ in fact only depends on $r^*$, or equivalently on $r$. Therefore we have the expansion $$f_{\omega l m} = \frac{1}{\sqrt{2\pi \omega}}\frac{1}{r}e^{i\omega v} F_\omega(r)Y_{lm}(\theta,\varphi)$$ which is Hawking's eq. (2.11). A similar reasoning leads to eq. (2.12), the only difference being that a positive frequency outgoing solution on $\mathscr{I}^+$ takes the asymptotic form $\sim e^{i\omega u}$ on $\mathscr{I}^+$. The same steps lead to $$p_{\omega l m} = \frac{1}{\sqrt{2\pi \omega}}\frac{1}{r}e^{i\omega u} P_\omega(r)Y_{lm}(\theta,\varphi)$$ which is Hawking's eq. (2.12) for a basis of solutions $p_{\omega l m}$ which is outgoing and positive frequency on $\mathscr{I}^+$.

  • $\begingroup$ I apologize for the fact that I didn't award you the 50 point bounty well the reason: I thought that after 1 week the topmost answer automatically gets the assigned 50 points. I apologize once again it was a terrible mistake on my part if it is possible to reward you 50 points now (through moderator's help) I am happy to do so. $\endgroup$
    – aitfel
    Sep 25, 2019 at 8:33
  • $\begingroup$ @aitfel Don't worry! :) $\endgroup$
    – John Donne
    Sep 25, 2019 at 10:34

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