# Can a particle in free fall into a black hole escape?

A friend posed this question to me, and I can't seem to figure it out.

If a particle falls into a black hole from a great distance, when it reaches the singularity (assuming it doesn't collide with anything) should have the energy required to escape. However, escape requires a velocity in excess of the speed of light, how can that be?

Intuitively to me, it seems that the particle should gain mass as it approaches the singularity, and then lose that mass on it's way out and escape. But that only raises more questions, like why can't any particle pass though a small (volume) blackhole's event horizon?

What material I have found seems to say that conservation of energy in GR isn't like Newtonian physics... bizarre as that seems to me. Any reading or multimedia material suggestions that will help me understand are welcome.

• As a general rule regarding time, gravity, or virtually anything regarding anything on the other side of the event horizon of a black hole: assuming your intuition is wrong is the first step. – Cort Ammon Dec 17 '16 at 1:03
• Very fair. But I still don't understand what happens and why the particle can't escape. – SillyInventor Dec 17 '16 at 1:23
• The spacetime curvature within the horizon is so great that the 'forward' direction in time is towards the singularity; moving away from the singularity is as impossible as moving backwards in time. – Alfred Centauri Dec 17 '16 at 1:38
• @AlfredCentauri Should be an answer. – Rob Jeffries Dec 17 '16 at 11:32
• I think I am starting to understand. To clarify, a time keeper on the event horizon will see the object as slowing such that it will never escape -- despite having the momentum to do so? – SillyInventor Dec 17 '16 at 22:02

But I still don't understand what happens and why the particle can't escape

The short answer is that the spacetime curvature within the horizon is so great that the 'forward' direction in time is towards the singularity; moving away from the singularity is as impossible as moving backwards in time.

The line element for the Schwarzschild black hole:

$$ds^2=-\left(1-\frac{2M}{r}\right)dt^2 + \frac{1}{1-\frac{2M}{r}}dr^2 + r^2d\Omega^2$$

Notice that, outside the horizon where the radial coordinate $r$ is greater than the Schwarzschild radius $2M$, a displacement in the time coordinate $dt$ makes a negative contribution to the interval $ds^2$ as opposed to a displacement in the spatial radial coordinate $dr$ which makes a positive contribution.

It is this sign difference that distinguishes a temporal displacement from a spatial displacement.

But look, inside the horizon where the radial coordinate is less than $2M$, the signs reverse and now a displacement $dr$ makes a negative contribution to the interval while a displacement $dt$ makes a positive contribution. That is, $dr$ is a timelike displacement when $r \lt 2M$.

Put another way, within the horizon, the 'arrow of time' points toward the singularity; the singularity at $r=0$ is in the future of and the 'end of time' for all world lines (even photon world lines) that cross the horizon.

I'll close with a quote from page 290 of Schutz's A first course in general relativity:

Everything inside $r=2M$ is trapped and, moreover, doomed to encounter the singularity at $r=0$, since $r=0$ is in the future of every timelike and null world line inside $r=2M$.

If a particle falls into a black hole from a great distance, when it reaches the singularity (assuming it doesn't collide with anything) should have the energy required to escape.

Wrong. In general a particle falling on a gravitational well, the earth for example, from far away can have a very large velocity, but whether it will collide, be captured as a moon, or escape by scattering off the gravitational field depends on the particular angles and velocities etc, and equations have to be solved.

However, escape requires a velocity in excess of the speed of light, how can that be?

This statement does not apply to an incoming projectile on the black hole, see the previous paragraph. It applies to hypothetical particles already within the black hole's horizon, where General Relativity has to be used for the mathematics,

Intuitively to me, it seems that the particle should gain mass as it approaches the singularity,

Relativistic mass is not a helpful concept , it is better to think of increasing in-falling momentum, supposing the newtonian gravitational well solution gives an in-falling trajectory

and then lose that mass on it's way out and escape.

Only trajectories that are not on a collision course , hyperbolic or parabolic, behave like that , do not fall in and scatter off back to infinity, losing the energy gained while falling into the gravitational well.

But that only raises more questions, like why can't any particle pass though a small (volume) blackhole's event horizon?

Black holes are a mathematical proposal from the theory of General Relativity and have been detected in astrophysical observations. It needs the mathematics of GR to understand what is happening, which cannot be handwaved away. Once caught below the horizon the particle cannot escape is what the solutions say.

In everyday Newtonian gravitation, a chemical reaction can be used to escape from a potential well by providing enough energy, that is how we have satellites . In GR the mathematical solutions are such that no amount of energy will allow escape from inside the horizon. This hyperphysics link gives a thread to follow the mathematics of the case for black holes..

• Perhaps I wasn't clear in my question. Some clarifications: -In this case, we're to imagine a nearly straight path by a very small (volume) singularity. -As it passes through the event horizon the velocity to escape back out is greater than the speed of light. -Yes, that is the question, why can't the momentum gained allow scattering within the event horizon of a black hole for incoming particles? -I know that's what the solutions say, but I don't understand why or how. Any direction toward material that can explain this would be helpful. – SillyInventor Dec 17 '16 at 21:57
• the answer you chose explains why. It is because the solutions from general realtivity constrain ALL paths to go to the singularity. For a newtonian singularity this is not so, a scattering might change the trajectory to a hyperbolic one. This cannot happen in classical GR. When quantum mechanics is introduced in the broth there is "evaporation" see this hawking.org.uk/into-a-black-hole.html – anna v Dec 18 '16 at 5:07