Nothing falls into an evaporating black hole?

The Vaidya Metric is the metric that can be used to describe the spacetime geometry of a varying mass black hole. This metric reads $$d\tau^2=\bigg(1-\dfrac{2M(\nu)}{r}\bigg)d\nu^2+2d\nu dr - r^2 d\Omega_2^2$$ For simplicity, I will assume $d\Omega_2=0$. Now, the condition for any interval to be spacelike can be read-off from this metric as $$\dfrac{dr}{d\nu}<-\dfrac{1}{2}\bigg(1-\dfrac{2M(\nu)}{r}\bigg)\tag{1}$$ Now, suppose that the trajectory of a particle (which, of course, is supposed to be either a lightlike or a timelike trajectory) connects two events--one outside the horizon and one inside the horizon. Also, let's take these events to be very close to the horizon. Then, the coordinates of these events can be taken as $$\Big(\nu, 2M(\nu)+ \delta \xi_{o}\Big)$$ and $$\Big(\nu+d\nu, 2M(\nu+d\nu)-\delta\xi_{i}\Big)$$ where $\delta\xi_{i}$ and $\delta\xi_{o}$ are infinitesimally small parameters that we can change to make the chosen events as close to the horizon as we wish.

Now, for the pair of these two events

$$\dfrac{dr}{d\nu}=\dfrac{2M(\nu+d\nu)-\delta\xi_{i}-2M(\nu)-\delta\xi_{o}}{d\nu}=2\dot{M}(\nu)-\bigg(\dfrac{\delta\xi_{i}+\delta\xi_{o}}{d\nu}\bigg)\tag{2}$$

Combining $(1)$ and $(2)$, we get that our stipulation that a particle connects the considered two events can be true only if

$$2\dot{M}(\nu)-\bigg(\dfrac{\delta\xi_{i}+\delta\xi_{o}}{d\nu}\bigg)\geq-\dfrac{1}{2}\bigg(1-\dfrac{2M(\nu)}{r}\bigg)$$

where $r$ can be taken as $2M(\nu)$ in the limit where we make $\delta\xi_{i}$ and $\delta\xi_{o}$ sufficiently small (as compared to $d\nu$). Thus, we get

$$\dot{M}(\nu)\geq 0$$

Thus, it seems that a particle can fall into the horizon only if the black hole is either not evaporating or is gaining mass. In the case of an evaporating black hole, this calculation seems to suggest that nothing (no timelike or lightlike trajectory) can connect the exterior to the interior. Is this true?

Notice that the conclusion cannot be a result of a bad choice of the coordinates because the argument depends on the value of the generally invariant interval.

I think this is quite a surprising result and thus, I think the probability is that there is some fatal flaw in the logic of the presented argument. I would like the answers to point out the same. Presented this way, it might seem like a "check my work" question but I hope this is not a completely uninteresting and off-topic homework-like check my work question that is supposed to be avoided under the "no check my work question policy".

• How do you justify taking $d \xi_i$ and $d \xi_o$ sufficiently small compared to $d \nu$? If the particle is falling at some rate, and the black hole horizon is shrinking at a different rate, doesn't there have to be a relation between these quantities? – Peter Shor Nov 27 '17 at 1:50
• @PeterShor I have updated my question to adopt a new notation which better illustrates the nature of the differentials previously written as $d\xi_{o}$ and $d\xi_{i}$. One can think of these parameters as the parameters used to traverse through the set of events near the horizon. The smaller the parameter $\delta\xi$, the nearer the event is to the horizon. These parameters determine the coordinates of the end-points of the trajectory. The relation between them and $d\nu$ is precisely that represented by the second last inequality in my post, i.e. they are connected by a non-spacelike path.... – Dvij D.C. Nov 27 '17 at 6:09
• ...I don't think there should be any other restriction on these differentials than that represented by this inequality. Now, I don't think I have an explicit justification for making the differentials $\delta \xi$ sufficiently small compared to $d\nu$ except saying that for a given $d\nu$ I can always choose to analyze what would be the condition to make two events sufficiently close to the horizon connected by a non-spacelike path in this time $d\nu$. – Dvij D.C. Nov 27 '17 at 6:18
• But, your comment got me thinking and interestingly, even if we do not make the $\delta\xi$ sufficiently small as compared to $d\nu$, we can always make them sufficiently small as compared to $2M(\nu)$--making $1-\dfrac{2M(\nu)}{r}=\mathcal{O}(\delta\xi)$. Further $\dfrac{\delta \xi_{o}+\delta \xi_{i}}{d\nu}$ is non-negative by construction. Thus, we can anyway conclude that $\dot{M}(\nu)\geq 0$. – Dvij D.C. Nov 27 '17 at 6:18

1 Answer

Let us start with considering instead the Schwarzschild black hole: as you may know, the maximally extended coordinate system is the Kruskal-Szekeres one: $$ds^2 = \frac{32M}{r}e^{-r/2M}dU\,dV - r^2d\Omega^2,$$ which describes not only the exterior, event horizon, and interior of the black hole, but also a parallell exterior region, and the event horizon and interior of a white hole, inaccessible to timelike and null observers. The Vadiya metric corresponds more directly to Eddington-Finkelstein coordinates, which can be written in ingoing coordinates, $$ds^2 = \left(1 - \frac{2M}{r}\right)dv^2 - 2dv\,dr - r^2d\Omega^2,$$ or outgoing ones, $$ds^2 = \left(1 - \frac{2M}{r}\right)du^2 + 2du\,dr - r^2d\Omega^2.$$ Both of these allow us chart across an event horizon, but while the ingoing coordinates chart across the black hole horizon, the outgoing ones chart across the white hole horizon! Perhaps you can see where this is going?

Now, by letting $M = M(v)$ the ingoing Eddington-Finkelstein coordinates become the ingoing Vadiya metric. And similarly, by letting $M = M(u)$ the outgoing Eddington-Finkelstein coordinates become the outgoing Vadiya metric. Notice that the line element you have written down is the outgoing Vadiya. That is to say, your metric already describes an "un-crossable" event horizon even if the radiation is set to zero. That it evaporates will not change this.

To clarify: the radiation emitted from an outgoing Vadiya hole corresponds to radiation classically crossing the event horizon, and thus demands a white hole structure. Meanwhile, the Hawking radiation is a quantum phenomenon that introduces radiation even from a black hole, and cannot be accurately described by the outgoing Vadiya metric (for further confirmation see e.g. this (relatively recent) article, where the authors use the ingoing Vadiya to analyze Hawking radiation around a dynamical black hole)

But, wait a minute. Does that mean that if $M(u)_{,u} > 0$ in the outoing Vadiya metric, we can actually access the white hole? Since the white hole is inaccessible to timelike and null observers, we may suspect a contradiction. Indeed, straightforward calculation shows that the only non-zero Ricci components are given by \begin{align} R_{uu} &= -2\frac{M(u)_{,u}}{r^2}, & R_{vv} &= 2\frac{M(v)_{,v}}{r^2}, \end{align} respectively. Since $g^{uu} = 0 = g^{vv}$, it follows that the only non-zero components of the stress energy tensor are \begin{align} T_{uu} &= -2\kappa\frac{M(u)_{,u}}{r^2}, & T_{vv} &= 2\kappa\frac{M(v)_{,v}}{r^2}, \end{align} where we let $\kappa$ denote the proportionality constant in the Einstein Field Equations. Thus, from the null energy condition (which guarantees that a null observer observes non-negative energy density) we must have $M(u)_{,u} \leq 0$ and $M(v)_{,v} \geq 0$. Concluding we remark that this means that the ingoing Vadiya necessarily describes an absorbing body, while the outoing Vadiya necessarily describes an emitting body.

• Thanks for your elaborate answer. I understand how it resolves my question but the part that I don't understand is the application of the null-energy condition to the ingoing metric. It seems to imply that the mass of a black hole can never decrease--which, of course, is a classically appropriate assertion. But the ingoing metric can be taken as a metric describing a variable mass black hole whose mass varies due to quantum mechanical effects, right? In that case, how does the null-energy condition's implication survive? – Dvij D.C. Nov 27 '17 at 11:44
• @Dvij Honestly, the details of Hawking radiation is currently beyond my expertise, but it is likely that one cannot use the Vadiya metric to attempt a calculation of the backreaction. Traditionally, I believe one assumes that Hawking radiation has no effect on the mass parameter, so as to retain the timelike Killing vector. I cannot say what happens once this assumption is abandoned, and do not have the expertise to evaluate papers for recommendation. Perhaps someone else does; you could try opening a new question. – Erik Jörgenfelt Nov 27 '17 at 15:27
• Okay, yes, I also think that the Hawking calculation is done assuming that the mass parameter is constant. But, I was thinking that if we don't require the explicit formula for the temperature but we just assume that due to quantum mechanical effects, the BH spherically symmetrically radiates then we can probably use the Vaidya metric. I will post a separate question regarding this specific point. – Dvij D.C. Nov 27 '17 at 15:57
• A point that I recently came to appreciate--which I think further confuses me: The $\nu$ of the ingoing Vaidya is, by construction, a coordinate that is a constant of motion for an ingoing photon. This, in some sense, makes it completely absurd to treat it as a temporal coordinate with the usual sense of the word. – Dvij D.C. Dec 4 '17 at 5:49
• Another thing: Can you provide a reference where I can better understand why outgoing Vaidya describes a white hole? The reason I am confused is that outgoing Vaidya doesn't seem to be the metric one gets when one reverses the time direction in the Schwarzschild metric and then produces the ingoing Vaidya for that reversed metric. If such were the case then it would be pretty clear to me that outgoing Vaidya is just a white whole. Thanks! – Dvij D.C. Dec 4 '17 at 5:49