# Do black holes exert an infinite amount of energy at the event horizon?

An interesting thought crossed my mind when reading about Hawkings' apparent horizon theory.

If we assume that event horizons do actually exist, it would seem that black holes violate basic laws of physics. My (limited) understanding is that in traditional black holes, the event horizon is the place at which nothing, not even light, can escape falling into the singularity. I also believe that I have learned that for something other than light to travel at the speed of light, an infinite amount of energy would be required.

Therefore, since a traditional event horizon is defined as the so-called "point of no return," wouldn't an infinite amount of energy be exerted at the event horizon? If light itself cannot escape, that would seem to mean that the gravitational force would be at least equal to the speed of light, and wouldn't that require infinite energy?

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No it's just curvature. For example, the Earth doesn't exert any energy to keep the Moon in orbit. The curvature is such that there isn't any path out. –  Brandon Enright Feb 6 '14 at 18:33

That's a subtle question because it depends on what you mean by energy.

Suppose we take some object a long way from the black hole and let it fall. As it falls it's speed increases (obviously) and therefore it's kinetic energy increases. However if you are sat outside the black hole watching the object you'll see something strange as the object nears the event horizon. You'll see the object start to slow as it approaches the horizon and indeed at the horizon it will come to a complete stop. This is the source of the numerous questions asking whether anything can really fall into a black hole e.g. this and this. Incidentally while searching for such questions I found the question A Conflict with Black Holes that is relevant to your question.

So from our perspective outside the black hole the kinetic energy of the body rises to a maximum then falls again as it nears the horizon.

However there is more to this. Suppose you jump on a powerful rocket and hover at some distance $d$ from the event horizon, then you measure the velocity at which the object passes you. As you approach the horizon, i.e. as $d \rightarrow 0$, the velocity the object passes you will tend to $c$ and therefore it's energy will tend to infinity. However you cannot hover at the event horizon (that would take an infinitely powerful rocket) so no observer will ever see the velocity become $c$ and the kinetic energy become infinite.

The other option you might take is to jump into the black hole alongside the object so you are falling with it. This is actually the least interesting option (until you hit the singularity and the associated spectacularly messy death of course) because you will simply see the object float, weightless, beside you and you won't measure any change in it at all.

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Look into the sky through all 4(pi) steradians. Every direction is equidistant from the same Big Bang. There is no visible direction, no path to escape this universe. There is no energy expended in enforcing this. It is a consequence of geometry.

Exactly how black holes are structured, "surface" and center, remains hotly debated. The only incontestable observation is that they externally gravitate indistinguishable from general relativity. An interior photon attempting to exit a black hole's gravitational potential would red shift without limit. It must pay back its easily calculated binding energy. One fails to see an infinity in that other than Zeno's paradox.

One might be naughty and declare all the gravitation of a black hole resides its "surface," its interior then being wholly unremarkable. There is no added gravitation internal to a thin spherical shell. That removes the naked singularity problem, and the problem of assigning large external angular momenta (Kerr black holes) given all interior mass resides at a point.

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All those words, but none answers the question. –  Chris White Feb 7 '14 at 18:56