Consider the following:

In the process of emission of Hawking radiation, a pair of particles is created near the event horizon such that, while one particle falls through the event horizon and approaches the center of a black hole, the other escapes.

However, the time reversed process, the one where the pair is created far apart, approximates near the horizon and then just vanishes, cannot happen. This is because the particle inside the black hole would need to travel faster than c to do so.

Is there something wrong with the above reasoning?
Furthermore, does the converse of Noether's theorem implies a time-symmetry since the energy-momentum is conserved($T^{\mu\nu}_{\phantom\u\ ;\mu} = 0$)?

Long story short:

  • Is General Relativity time reversible?
  • Does the converse of Noether's theorem implies a time symmetry?
  • What is happening with the black hole and the pair of particles?
  • $\begingroup$ conservation of energy requires that the particle/antiparticle air are co-located when they are created? Black hole evaporation due to hawking radiation is due to the particles inability to recombine as same post creation scenario continues. $\endgroup$
    – JMLCarter
    Mar 30, 2017 at 1:02
  • $\begingroup$ never-the-less a particle captured by an event horizon can not escape it even if its momentum were theoretically reversed. $\endgroup$
    – JMLCarter
    Mar 30, 2017 at 1:04
  • 1
    $\begingroup$ @JMLCarter the particle would escape if everything is the universe were reversed (not just its own momentum). The black hole would become a white hole. $\endgroup$
    – user126422
    Mar 30, 2017 at 3:58
  • $\begingroup$ @HughMungus Shouldn't the emission process itself be time symmetric? Wouldn't reversing everything in the universe be a reversal of a larger process than the emission itself? $\endgroup$
    – user140561
    Apr 2, 2017 at 16:28
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    $\begingroup$ Is this not a duplicate of physics.stackexchange.com/q/39383 ? $\endgroup$
    – Rococo
    Apr 3, 2017 at 16:27

2 Answers 2


The answer to this question actually depends on the understanding of different singularities. The singularity of $R = 2m$ is a co-ordinate singularity and it's not a real singularity. Thus the observation of this phenomenon is different for different observers.

For Observer at Infinity: For this observer one particle(the outgoing one) is reaching towards him and the other particle(going to black hole) is going away. Now because of the gravitational field of black hole the 2nd particle will be redshifted as it gets closer and closer to the event horizon. It will take infinite time for the particle to reach R = 2m surface in his co-ordinates. Thus if this observer time reverses, there is no ambiguity in seeing this particle coming back and annihilating with it's anti-particle through pair annihilation as the particle never crossed the event horizon.

The Observer in Particle's frame: In particle's frame the R = 2m surface is not a singularity. There is nothing special about it and it can happily pass through it and can also cross it when time reversed. The only singularity it can hit is at R = 0 which is a real singularity. But if the particle reaches there then Einstein's equation(along with every other physics) ceases to be true. Then $T^{\mu\nu}_{\phantom\u\ ;\mu} = 0$ no longer holds.

  • $\begingroup$ I found your insight about observers very smart(+1). For the observer at infinity, what happens in the case where one particle is created inside the horizon($R_{particle1} < 2M$) while the other is created outside it ($R_{particle2} > 2M$)? (Also, the pair is created sufficiently far apart such that the uncertainty in position won't be a problem) $\endgroup$
    – user140561
    Apr 6, 2017 at 18:26
  • $\begingroup$ Also, for the observer in the particle's frame, how can the particle go from $R < 2M$ to $R > 2M$ if it needs to travel faster than c to do so? (Is this comment question frased badly when it comes to observers?) $\endgroup$
    – user140561
    Apr 6, 2017 at 18:26
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    $\begingroup$ To answer your first question, at the time of pair production the particles must be created at a single space-time point. Thus both their co-ordinate must be same. Otherwise it leads to non-locality of the theory. So the $R_{particle1,2}$ can be either both > 0 or <0. I hope this solves this problem. $\endgroup$
    – Ari
    Apr 7, 2017 at 4:21
  • $\begingroup$ For second question, the particle does not require to travel at velocity $>c$ to go from $R <2M to R > 2M$. It's the observer at infinity for whom this is true. The particle can come out at finite time in it's frame but that will take infinite time in the frame of asymptotic observer. $\endgroup$
    – Ari
    Apr 7, 2017 at 4:24
  • $\begingroup$ Thanks for the great answer. I tried calculating the (proper) time interval from the outgoing orbit equation, and it is indeed the same time for an infalling particle. By the way, do you know some resource containing an elaboration of this matter on the escape from a black hole for different observers? $\endgroup$
    – user140561
    Apr 7, 2017 at 21:01

We can ask the same question without mixing in Hawking radiation and a pair of particles. If a particle falls into the black hole, it can't come out. And time reversibility won't get it out of there either. In fact, it is not a time reversible process. The reason for this is explained in A. Zee's book “Einstein Gravity in a Nutshell” (pp. 416--417). I'll just quote it:

A common confusion about plunging into a black hole

Confusio speaks up: “I have learned that the fundamental laws in classical physics (and also quantum physics) are time reversal invariant, that is, they are unchanged upon $t\rightarrow-t$. I read that if we take a movie depicting a microscopic process and run it backward, the reversed process must also be allowed by the laws of physics. So why can’t I run the film of the observer radially plunging into a black hole and watch him come flying out?”

Well, well, that Confusio is more astute than we think. Indeed, the Lagrangian $$L=\left[\left(1-\frac{r_S}{r}\right)\left(\frac{dt}{d\tau}\right)^2-\left(1-\frac{r_S}{r}\right)^{-1}\left(\frac{dr}{d\tau}\right)^2-r^2\left(\frac{d\theta}{d\tau}\right)^2-r^2\sin^2\theta\left(\frac{d\varphi}{d\tau}\right)^2\right]^{\frac{1}{2}}$$ governing the motion of a particle in Schwarzschild spacetime is manifestly invariant under $t\rightarrow-t$. So where is the catch in the standard arguments about time reversal invariance?

The catch, as I have already mentioned, is that the coordinate time $t$ increases to $+\infty$ as $r\rightarrow r^+_S$ and then decreases from $+\infty$ after the observer crosses the horizon. Indeed, as is evident from the Lagrangian just displayed, $t$ and $r$ exchange roles for $r<r_S$. The letter “$t$” no longer denotes time! Much more on this in the next chapter.

The standard arguments about time reversal invariance work perfectly well as long as $r>r_S$. Thus, if we could somehow install a trampoline at $r^+_S$ just outside the black hole, the observer in radial plunge could bounce back out to $r=\infty$, retracing his trajectory.


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