Reading this paper which is itself an exposition of Parikh and Wilczek's paper, I get to a point where I fail to be able to follow the calculation. Now this is undoubtably because my calculational skills have been affected by decades of atrophy, so I wonder if someone can help. The paper computes the tunneling transmission coefficient for a particle (part of a particle/antiparticle pair) created just inside the horizon. The WKB transmission coefficient is given by $$T = exp({-\frac{2}{\hbar} Im(S)})$$ where "Im" is the imaginary part, and the action, $S$, is evaluated over the classically forbidden region. Using Painleve-Gullstrand coordinates for the black hole, the paper derives, fairly straightforwardly, $$Im(S) = Im \int_{2M}^{2(M-\omega)}{dr \int_{0}^{\omega}d\omega' {\frac{1}{1-\sqrt{\frac{2(M-\omega')}{r}}}}}$$ (eq 41). $\omega$ is the energy carried out by the tunneling particle. Now the next step is where I get stuck. The following equation (42) suggests that the r integration has been performed to get from (41)->(42). Here's what happens when I try to do it:

Presumably the contour integral being referred to is in the complexified r variable. The way the "r" appears isn't very nice, so we make a substitution $z=r^{\frac{1}{2}}$, giving $$Im \int_{2M}^{2(M-\omega)}dz{\frac{2z^2}{z-\sqrt{2(M-\omega')}}}$$. The talk of deforming the contour "in the E plane" suggests we make the energy slightly imaginary - add $i\delta$ to $\omega$ where $\delta$ is small. The expression $\sqrt{2(M-\omega+i\delta)}$ can then be expanded in powers of delta, leaving $\sqrt{2(M-\omega)}+i\delta$ (after the expansion I've redefined delta to take out the constant factors - this doesn't matter because we're going to contour-integrate round the pole anyway, so moving the pole up and down a bit makes no difference). So the z-plane looks like the first diagram here, where the large X is the displaced pole. We ultimately want to evaluate the integral between the two points on the real axis.

enter image description here

Well, the one thing I can do is integrate round the green contour shown in the second figure. The answer is just $2\pi i$ times the residue at the simple pole, i.e. in this case $$2\pi i\cdot\lim_{z \to {\sqrt{2(M-\omega')}+i\delta}}(z-\sqrt{2(M-\omega')}+i\delta)\cdot \frac{2z^2}{(z-\sqrt{2(M-\omega')}+i\delta} $$, $$=2\pi i\cdot2\cdot(\sqrt{2(M-\omega')}+i\delta)^2$$ $$=8\pi i \cdot(M-\omega')$$ approx.

This looks similar to what I want, namely $4\pi i \cdot(M-\omega')$ in order to get equation (42) (apart from a factor of 2). However

(1) The next step would be to relate the closed contour integral to the integral along the real axis. Unfortunately, this relies on the integrand vanishing for large $|z|$ in the positive half plane, but this appears not to be the case here

(2) And anyway we want the integral between $\sqrt{2(M-\omega)}$ and $\sqrt{2M}$, so how would I deal with the integral along other parts on the real axis (i.e. outside of the classically forbidden region)?

Any hints would be welcome (or an alternative way of computing the tunneling transmission coefficient).


The (finite) imaginary part comes solely from the singularity at $r = 2(M-\omega^\prime)$ in the integration over $r$ away from this singularity the integral is finite and real.

Performing a change of variables: $r = 2(M-\omega^\prime) + u$, since only the singularity contributes to the imaginary part , we can approximate the integrand by by his singular part in the vicinity of u = 0 :

$\frac{1}{1-\sqrt{\frac{2(M-\omega^\prime) }{r}}} \approx - 4(M-\omega^\prime) \frac{1}{u}$

Thus, the integral over $r$:

$I= 4(M-\omega^\prime) \mathrm{Im}\int_{-2\omega^\prime}^{(\omega-\omega^\prime)} \frac{du}{u}$.

Using the relation

$\frac{1}{u} = P.V.(\frac{1}{u} ) + \pi i \delta(u)$.

(P.V. denotes the Cauchy principal value. We get:

$I= 4(M-\omega^\prime) $

  • $\begingroup$ Thank you so much. That was really helpful - I was going in completely the wrong direction! $\endgroup$ – twistor59 Nov 27 '12 at 19:09

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