It’s a simple question. What is the gravitational acceleration at the event horizon?
I don’t think there are any known black holes without X-ray producing accretion disks, so the question is academic. Of course, since most black holes are discovered because of their X-ray emissions, it is possible those without accretion disks are effectively invisible, so we may never know.
Not only are there no closed orbits inside of a black hole’s event horizon, there aren’t even any for some distance outside. For a non-rotating hole, you can’t be in a closed orbit any closer than three times the radius of the event horizon (though you can still swoop in and swoop back out on a hyperbolic-like orbit). This distance is called the Innermost Stable Circular Orbit, or ISCO.
And there’s no reason a black hole must necessarily have an accretion disk. Remember, a black hole’s gravity is the same as that of any other object of the same mass. All of the black holes we know of have accretion disks, but that’s just a selection effect: The accretion disk is how we detect the black hole to begin with. But one would certainly expect there to be plenty without.
The Tao’s Revenge, it’s not correct to say that larger black holes take longer to evaporate because of their mass to surface area ratio. In fact, larger black holes have more surface area per mass, since a hole’s radius is proportional to its mass. Rather, the biggest cause of the difference in lifetimes is the temperature: Radiation is proportional to temperature to the fourth power, and a black hole’s temperature is inversely proportional to its mass.
I hope you’re not expecting the answer to be c, since speed and acceleration have different units. The gravitational acceleration at the event horizon varies with the mass of the hole. It’s inversely proportional to the mass, so a sufficiently-large hole could have a gravitational acceleration that is very small indeed.
Ignorance fought, thank you!
First of all, there are black holes within which there are stable (or at least kind-of stable) orbits, for instance certain kinds of charged, rotating black holes. (However, in the classic, static Schwarzschild black hole case, it’s indeed true that once you cross the horizon, there’s no avoiding the central singularity, anymore than there is avoiding Monday morning.)
And for a large enough black hole, it’s certainly possible to cross the event horizon pretty much unscathed. To an external observer, your time will indeed appear to grind to a halt, but to you, time flows always at a rate of one second per second.
As for the gravitational acceleration at the event horizon, that’s actually not such a simple question; the wiki article has some discussion. For a Schwarzschild black hole, the surface gravity goes as 1/4*M, and thus, gets small in the limit of large mass.
EDIT: Wow, there’s like 15 replies that weren’t there when I started typing…
Why would I be expecting c? I am expecting “infinite”. But let’s say it’s not infinite.
Using Newtonian equations, the gravitational acceleration of a black hole that has 16B solar masses is it around 100G at the event horizon. So - if I am at the event horizon, and I accelerate at 101G - do I escape?
To you, time flows at one second per second, but if you travel infinitesimally close to c, the time, subjectively, for you to reach the singularity from the event horizon is infinitesimally close to zero. Thus, to say you cross “pretty much unscathed” makes no sense if you are (subjectively, to you) destroyed immediately upon crossing.
But you aren’t. As I said, there are at least metastable orbits for certain kinds of black holes, and even for ordinary BHs, infall time is proportional to the mass, so it’s just not the case that I’m destroyed immediately upon crossing.
Also, it’s not the case that my speed at the horizon is infinitesimally close to c – it’s that the escape velocity from any point beyond the horizon is greater than c. This is different: if, for instance, I were lowered by some winch mechanism to the event horizon (or to a fraction of a milimeter above it, if you prefer), and then released, my speed will be close to zero upon crossing; I could not, however, escape from a point beyond the horizon unless my speed exceeded c.
It would not be 0, since the proper acceleration at the event horizon is infinite. AFAIU, Wald gives the equation as a=1/sqrt(r-2M) and r=2M at the event horizon.
If, at the event horizon acceleration is infinite then tidal stresses would be infinite.
Since nothing can survive infinite tidal forces, nothing can survive the event horizon. Since it is believed the tidal forces of a rather large black hole are actually not that bad, then it must not be infinite according to our understanding.
If you’re at or beyond the event horizon of a black hole, and you accelerate, that’ll just result in you getting to your ultimate destination quicker. Unfortunately, your ultimate destination is the singularity and certain death. You can’t just “accelerate outwards”, because “outwards” is a timelike direction.
How is it algebraically possible for this equation to produce an infinite value for ‘a’, unless I’m missing something?
A question. From the point of view of particle for a tiny black hole or spaceship for a big one what kind of time spans from their POV are we talking about between crossing the event horizon and getting to (or too close) to the singularity? Assuming we are taking a path to maximize the time it takes.
Do black holes have a “photon shell” of captured photons right at the event horizon? Would crossing it mean getting zapped by the mother of all laser beams?
Yes, but it’s at 1.5 times the schwarzschild radius. You wouldn’t get zapped (at least I don’t think it would be appreciably bigger than the normal zappage due to being in the accretion disk) since the orbit is unstable – any photons that go into it will eventually be perturbed and either get thrown out or fall in.
Division by 0.
So you’re suggesting that all the solutions of this equation result in division by zero? :dubious: Try again.
Edited to add: division by zero is simply undefined, which I understand is distinctly different from infinity.
As r approaches 2M, a approaches infinity.
Then by your logic, even allowing for the very novel definition of infinity as division by zero, any value of ‘r’ less than 2M results in a finite acceleration, doesn’t it?