Does conservation of energy make black holes impossible?

Question

I was musing today about black holes, and came across what seems to be a contradictory fact about black holes. If any matter were to actually fall into a black hole, it seems like it would need to gain infinite energy.

This is seen by considering an object close to the event horizon of a black hole. If we consider escape velocity for a particle at the event horizon, it has an escape velocity of the speed of light. The energy required to achieve this escape velocity would therefore be infinite. As we approach the event horizon, it seems like we should be able, then, to define an arbitrarily small distance across which the particle gains an arbitrary amount of energy.

As the object approaches arbitrarily close to the event horizon, it seems like its relativistic mass would therefore grow arbitrarily high, resulting in a corresponding increase in the gravity of the black hole, which seems impossible.

If we extrapolate back to the formation of the black hole, it then seems like it should be impossible for any particles to fall into the black hole, even as it forms. Instead, mass should be distributed in a probability cloud whose density asymptotically approaches zero at the event horizon of the black hole. New particles could then enter this cloud of incredibly dense (but not black hole dense) matter, where they would be captured rather than falling past the event horizon, within which there would be a vacuum.

If this were true, there would be no black holes, only hollow spheres of radiation and matter which externally look like black holes, since their gravity would be great enough that particles entering the sphere would exit at an incredibly slow rate. Interestingly, such spheres wouldn't suffer from the apparent information paradox that black holes do.

That or I'm missing an obvious explanation for why that's all wrong, which seems more likely because I'm neither a physicist nor an astronomer. What's the critical flaw in this argument?

Answer 1

For simplicity, let's consider a Schwarzschild black hole, so that the spacetime is spherically symmetric and static. In particular, the Schwarzschild time $t$ coordinate gives a direction in which the geometry 'stays the same' (a Killing vector field), and its inner product with the orbital four-velocity $u$ is conserved: $$\epsilon = -g(u,\partial_t) = \left(1-\frac{2M}{r}\right)\frac{\mathrm{d}t}{\mathrm{d}\tau}\text{,}$$ where $\tau$ is the proper time of the orbiting particle. One can think of this as the specific (per-mass) energy, including mass-energy: a particle escaping to infinity that becomes asymptotically at rest with respect to stationary observers would have $\epsilon = 1$.

If we consider escape velocity for a particle at the event horizon, it has an escape velocity of the speed of light. The energy required to achieve this escape velocity would therefore be infinite.

The energy as measured by whom? Imagine a family of stationary observers everywhere surrounding the black hole, or at least along an infalling particle's trajectory. Those observers measure the speed and energy of the particle as it falls past them. As the particle nears the horizon, they will report speeds arbitrarily close to the speed of light and arbitrarily high energies.

But to a far-away stationary observer, those observers near the horizon are experiencing increasingly divergent gravitational time dilation, and the orbital energy of the particle stays constant. Flinging a particle of mass $m$ into the black hole will increase the mass of the black hole by $m\epsilon$.

What's the critical flaw in this argument?

All energy gravitates, so I would say the main flaw is forgetting to add the gravitational potential energy of the particle to the mass of the black hole as well. Or better put, you should be concerned with the orbital energy, not just its mass+kinetic parts ('relativistic mass').

In this situation, you have a well-defined conserved orbital energy. If you insist on measuring the Lorentz factor according to close-by stationary observers, then yes, it diverges to $+\infty$, but then you would have to admit a gravitational potential energy term that diverges to $-\infty$, because their sum must resolve to be the orbital energy.

Answer 2

Let's reformulate your argument as follows (using idealized assumptions for simplicity):

1. An object some distance away from the black hole has potential energy equal to the kinetic energy it would have at the event horizon.

2. This is the same amount of energy that would be required to pull the object from the event horizon back to its original position (and velocity).

3. We would need to accelerate the object to the escape velocity to return it to its position.

4. Since the escape velocity is greater than c, it would take an infinite amount of energy to return it to its position.

The incorrect point is 3. We can pull an object out of a gravitational field slowly by exerting a force on it the whole way. We would only need to accelerate it to the escape velocity if we wanted it to escape "on its own", i.e. without us exerting a force to keep pushing it away from the black hole.

EDIT: I'm less sure of my answer after doing some back-of-the-envelope calculations. It would be helpful for a real expert to weigh in. I'll leave my answer here anyway.