Recently a physicist I know said he was interested in the following question: "would one observer following Kruskal-Szekeres coordinates in the presence of a Schwarzschild black hole (i.e., whose comoving coordinates are K-S) experience Hawking radiation?".

Now, if I interpreted this correctly, what he wants to know is: given a Schwarzschild spacetime $(M,g)$, we consider Kruskal-Szekeres coordinates, defined in terms of the usual $(t,r,\theta,\phi)$ coordinates by

$$T=\left(\dfrac{r}{2GM}-1\right)^{1/2}e^{r/4GM}\sinh\dfrac{t}{4GM},\quad X=\left(\dfrac{r}{2GM}-1\right)^{1/2}e^{r/4GM}\cosh\dfrac{t}{4GM}$$

on the exterior region $(r > 2GM)$ and $$T=\left(1-\dfrac{r}{2GM}\right)^{1/2}e^{r/4GM}\cosh\dfrac{t}{4GM},\quad X=\left(1-\dfrac{r}{2GM}\right)^{1/2}e^{r/4GM}\sinh\dfrac{t}{4GM}$$

for the interior region $(r<2GM)$.

Then we consider one observer, whose worldline is one of the coordinate lines of the coordinate function $T$. Namely, we fix $X,\theta,\phi$ and consider the curve $\gamma : \mathbb{R}\to M$ in coordinates given by

$$T\circ\gamma(\tau)=\tau,\quad X\circ\gamma(\tau)=X_0,\quad \theta\circ \gamma(\tau)=\theta_0,\quad \phi\circ\gamma(\tau)=\phi_0.$$

We thus consider one detector following this worldline and we ask ourselves if it detects Hawking radiation. I believe this is the precise statement of the problem he has in mind. I believe the whole point to ask this question is that in these coordinates there is no apparent singularity corresponding to the event horizon.

My question here is: has this been discussed on the literature? Is there any reference on which this is discussed? If there is, where I can find it?


1 Answer 1


As you correctly state, the Kruskal-Szekeres coordinates are regular at the horizon, where nothing special happens, as expected from the equivalence principle. The associated quantum states is a vacuum state, and no particles will be detected at the horizon by a local detector.

On the contrary, an observer far away from the black hole, for which the traditional Schwarzschild coordinates are more suited, will see a flux of radiation coming from the black hole.

More mathematically, consider for simplicity a scalar field in two dimensions, as we can suppress the angular variables.

$$ S[\phi]=\frac{1}{2} \int d^2 x \, \sqrt{-g} \, g^{\mu \nu} \phi_{, \mu}\phi_{, \nu} $$

Since the action is conformally invariant we can write the solution to the scalar equation of motion in terms of different coordinates systems. First of all, we choose a frame that is singular on the horizon, for instance the lightcone coordinates $(\tilde{u},\tilde{v})$ obtained from $(t,r)$ trough the tortoise coordinate:

$$ \tilde{u}=t-r^* \qquad \tilde{v}=t+r^* \qquad dr=\left( 1-\frac{2M}{r} \right) dr^* $$

Secondly, we pick a frame that is regular everywhere apart from the central singularity, such as lightcone Kruskal-Szekeres $(u,v)$ defined by

$$ u=-4 M e^{-\frac{\tilde{u}}{4M}} \qquad v = 4M e^{\frac{\tilde{v}}{4M}} $$

An observer at rest located far away will associate particles with positive frequency modes $\Omega$ with respect to the time coordinate t. The expansion of the scalar fields will be (ignoring the left moving part for simplicity):

$$ \hat{\phi}= \int_{0}^{\infty} \frac{d\Omega}{(2\pi^{1/2})} \frac{1}{\sqrt{2\Omega}} \left[e^{-i\Omega \tilde{u}} \hat{b}_{\Omega}^-+e^{i\Omega \tilde{u}} \hat{b}_{\Omega}^+\right] $$

where the annihilation operator defines the Boulware vacuum:

$$ \hat{b}_{\Omega}^- |0_B> =0 $$

Therefore this vacuum contains no particles from the point of view of the far away observer. Nevertheless this vacuum it's not physical since it's singular on the horizon and this would require an infinite amount of energy to actually prepare such a state. For the Kruskal coordinates:

$$ \hat{\phi}= \int_{0}^{\infty} \frac{d\omega}{(2\pi^{1/2})} \frac{1}{\sqrt{2\omega}} \left[e^{-i\omega u} \hat{a}_{\omega}^-+e^{i\omega u} \hat{a}_{\omega}^+ \right] $$

$$ \hat{a}_{\omega}^- |0_K> =0 $$

The Kruskal vacuum (also known as Hartle-Hawking vacuum) is well behaved on the horizon and suitable to be a physical state. Notice that it contains particle from the point of view of the far away observer, indeed on can relate the two set of creation-annihilation operators:

$$ \hat{b}_{\Omega}^-= \int_{0}^{\infty} d\omega [\alpha_{\Omega \omega} \hat{a}_{\omega}^- -\beta_{\Omega \omega} \hat{a}_{\omega}^+] \,\, $$

via the Bogolyubov coefficients, resulting in $\hat{b}_{\Omega}^- |0_K> \neq 0$. In summary, a far away observer will see a spectrum of scalar particles coming from the black hole: $$ \langle \hat{N}_{\Omega} \rangle \equiv <0_K|\hat{b}_{\Omega}^+ \hat{b}_{\Omega}^- |0_K>= \int d \omega |\beta_{\Omega \omega}|^2 $$

$$ n_{\Omega}=\frac{\langle \hat{N}_{\Omega} \rangle }{V}= \frac{1}{e^{\frac{2\pi \Omega}{\kappa}}-1} \qquad T_{BH}=\frac{\kappa}{2 \pi} = \frac{1}{8 \pi M} $$

for a finite volume $V$. In particular this is a black body spectrum with temperature $T_{BH}$.

A reference: Introduction to Quantum Effects in Gravity


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.