-1

There are a lot of questions and answers about speed falling into a black hole, none of them answer my question. Most of them only talk about speeds outside the event horizon. The only relevant answer's last paragraph is by John Rennie:

Will an object always fall at an infinite speed in a black hole?

But this does not answer my question.

Outside the black hole, the postulates of SR assure us, that the photon always wins over the neutrino in vacuum, when measured locally.

Now inside the event horizon of the black hole, this is not so simple. There are no inertial frames of reference at all, and honestly I am not even sure whether the postulates of SR still hold. The only inertial frame would be the singularity, but that is not well defined in my understanding.

Now if we shoot a photon and a neutrino (starting together) from the event horizon to the singularity (since the singularity might not be a point in space, but rather in time, let's shoot them perpendicular to the surface of the horizon, towards the "center") we can still calculate the time for the neutrino, being massive.

Calculating the lapsed time to fall from the horizon to the singularity of an existing black hole is a standard exercise in GR, and the result is: $$ \tau \approx 6.57 \frac{M}{M_{Sun}} \mu s $$ That is, for a black hole of 10 solar masses the fall takes 65.7 microseconds!

A Hollow Black Hole

So it is possible in this example to see how fast the neutrino reaches the singularity. Now we would just have to do the same for the photon, but here we encounter some problems.

If you want to know how much time a material object like a rock takes to hit the singularity, the most natural way of defining this is to ask how much time will elapse on a clock attached to the rock. The answer is then a finite number that depends on the size of the black hole. It isn't infinite. In your example of a photon, we don't have this option. We can't duct-tape a clock to a photon, so it's not possible to define the amount of time experienced by a photon. Therefore the answer to your question is fundamentally undefined.

Photons hitting the Black Hole Singularity

So based on this, are we saying that we do not know whether a photon would win over a neutrino?

More fundamentally, we do not know whether massive or massless particles move faster inside a black hole?

Question:

  1. Race to the singularity, who wins, the photon or the neutrino?
  • Comments are not for extended discussion; this conversation has been moved to chat. – tpg2114 Dec 27 '20 at 04:12
  • 1
  • 1
    What about "We can't duct-tape a clock to a photon, so it's not possible to define the amount of time experienced by a photon. Therefore the answer to your question is fundamentally undefined." is unclear to you? You cite an answer explaining that "travel time" is ill-defined for a photon, and then you go on to ask a question that wants to compare the photon's time to the neutrino's time. I do not see what you want from an answer here that is not already provided by the very answer you cite. – ACuriousMind Dec 29 '20 at 13:33
  • @ACuriousMind photons travel approx. 8 minutes to the Earth from the Sun, as far as I understand, so what I do not understand, is why we cannot do the same with the travel from the event horizon to the singularity (but I guess you are talking about the proper time) . But anyway, I am not asking for a time specifically. All I am asking for is a proof (might be mathematical), that the photon is faster (then the neutrino) even inside the horizon, meaning, that the postulates of SR are valid inside too, but I do not see any mathematical proof of that, that is why I am asking for some. – Árpád Szendrei Dec 29 '20 at 16:45

1 Answers1

6

For the concept of a race having a winner to be meaningful, the finish line has to be timelike or lightlike. Then the events of the racers' worldlines intersecting the finish line are causally ordered, and you can identify the earliest one.

The singularity of a Schwarzschild black hole is spacelike, so there's no way to put an ordering on the final moments of worldlines that end on it. Nobody wins.

The weird theoretical vacuum solutions for charged and/or rotating holes have timelike singularities, but there are good reasons to think that in realistic holes there would be a spacelike singularity just outside the inner event horizon, and everything beyond that, including the timelike singularity, the white holes leading to other universes, and the antigravity region through the ring singularity, is unphysical. (arXiv:1907.05292 is an interesting review article about this and other aspects of black hole interiors.)

$\frac43 G M/c^3 \approx 6.57 \mu s\,M/M_\odot$ is the proper time from horizon to singularity of a massive particle that freefalls from rest at infinity into a Schwarzschild black hole. The horizon-to-singularity proper time of an arbitrarily moving massive particle can be larger or smaller, but it will always be strictly between zero and $πGM/c^3 \approx 15.5 \mu s\,M/M_\odot$ (which is the proper time of a particle that freefalls from rest just outside the event horizon). The proper time for a massless particle, though, will always be zero, so the photon wins this and any other proper-time race.

benrg
  • 26,103
  • Thank you so much! Can you please tell me if I use the radius of the black hole, r and divide it simply by the speed of light 300 000 km/s, will this give anything that makes sense? – Árpád Szendrei Dec 27 '20 at 04:27
  • @ÁrpádSzendrei The last paragraph was actually wrong and I've now fixed it. You can get the maximum time by integrating the metric distance along a radial line from $r=r_s$ to $r=0$, and then dividing by $c$ (though usually you'd use units where $c=1$). If you just take $r_s/c$ then you'll be off by a factor of $π/2$. – benrg Dec 27 '20 at 05:24
  • 1
    @ChiralAnomaly I was talking about the "mass inflation instability" which is classical. When I was writing the answer I couldn't remember the name of it. I just added a reference to an interesting recent review article. The firewall is supposedly near the outer horizon, which would break general covariance, but the inner horizon has always been suspicious because it's in the "future of future infinity", so it makes sense that something terrible would happen there if the exterior isn't a perfect vacuum. – benrg Jan 03 '21 at 06:03
  • @benrg Excellent. Thank you! (I deleted my previous comment because you answered it in your edit.) – Chiral Anomaly Jan 03 '21 at 14:21