(What is) Schwarzschild Radius and Potential Energy

[Narroo]

I was thinking a bit earlier:

What if you created a particle at infinity (ignore the normal particle rules for simplicity), and allowed it to fall towards a massive body? Ignoring other effects such as the other forces, would the particle ever gain enough energy to surpass the energy required to create the particle?

Intuitively, we know the answer to be no. But I felt like doing a 5 second back-of-the-envelope calculation anyways. I did the following:

Say the gravitational potential difference from infinity to some distance $r$ is $U=-\frac{GM}{r}$. Where $M$ is the large mass and $m$ is the mass of the particle. So, by falling from infinity to $r$ the particle would gain energy $ΔE=\frac{GMm}{r}$.

The rest energy of the particle is $E_{rest}=mc^{2}$. I set the equations equal in order to see what (classical) radius I'd need in order to gain my energy back from creating the particle. The result is:

$r=\frac{GM}{c^{2}}$

Well, I recognized this as (half) the Schwarzschild Radius, the radius of a black hole! $r_{s}=\frac{2GM}{c^{2}}$

So now I'm curious! What's this mean? Why would I get half the Schwarzschild Radius? Is it simply telling me it's impossible, or is there a more interesting reason? Does anyone know?

 

[PeterDonis]

Intuitively, we know the answer to be no.

Really? How so? Remember that we're talking about relativity here, not Newtonian physics (if we were talking about Newtonian physics, the whole idea of mass-energy equivalence wouldn't make any sense in the first place, so the answer would be "undefined" or "invalid question" or something like that).

in order to see what (classical) radius I'd need in order to gain my energy back from creating the particle.

If by "classical" you mean "Newtonian", then the calculation you're trying to do is meaningless; see above.

As far as the correct relativistic answer: if a particle free-falls into a black hole, its energy (more precisely, its energy at infinity, which is the correct concept here) is constant; you can't "recover" any of it. In order to "recover" any energy from the particle, you need to either slowly lower it towards the hole, or stop its fall before it reaches the horizon and extract its kinetic energy. In an idealized case where you can do either of these processes with 100% efficiency, the energy you can recover by lowering the particle from infinity to a radius $r$ is

$$
E = m c^2 \left( 1 - \sqrt{1 - \frac{2 G M}{c^2 r}} \right)
$$

In other words, the closer to the horizon you lower the particle, the more of its rest energy you can recover; in the limit as the particle goes to the horizon, the energy recovered goes to the particle's entire rest energy. Note, however, that you can't actually lower the particle all the way to the horizon; you have to stop short of it, even if it's only by an infinitesimal amount.

 

[pervect]

Newtonian back-of-the envelope calculations aren't going to tell you anything very interesting.

The general relativistic calculations tell you the following:

You can see some of the equations online at Orbits in strongly curved space-time, they are taken from MTW's textbook "Gravitation" if you want a textbook source.

The effective gravitational potential with no angular momentum (L=0) is given by

$\sqrt{1 -\frac{2M}{r}}$ in geometric units (G=c=1), $\sqrt{1 -\frac{2GM}{r\,c^2}}$ in standard units.

The equation of motion for $r(\tau)$ , $\tau$ being proper time, for an infalling particle which we shall consider to be of unit mass, in geometric units (G=c=1) is:

$\left( \frac{dr}{d\tau} \right)^2 + 1 -\frac{2M}{r} = E^2$, where E is a constant, the energy-at-infinity of our unit mass particle. If the particle is freefalling from infinity, we can take E=1

So the energy-at-infinity of a particle is a constant of motion for a free-falling particle.

The expressions can be put in various forms, I did one conversion in https://www.physicsforums.com/showpost.php?p=4386571&postcount=11 where I used isotropic coordinates. These may not be familiar, rather than explain I'll "eyeball" the results into standard Schwarzschild coordinates.

The result will be (assuming again no angular momentum)

$$E = \frac {\sqrt{g_{00}}} { \sqrt{1 - \left( {v_{r}}/{c_r} \right)^2 } }$$

In geometric units with G=c=1, $g_{00} = 1 -\frac{2M}{r}$

Our formula for E can be interpreted as saying the energy-at-infinity of a unit mass particle is multipled by a number less than 1 to account for the loss of gravitational potential energy, and multiplied by something that looks like the relativistic equation for gamma to include it's kinetic energy. Here $v_r / c_r$ is the ratio of the coordinate radial velocity of the particle to the coordinate radial velocity of light. It would be more natural to use the actual velocity of the infalling particle relative to a local stationary observer, but I don't have that expression worked out. While the coordinate velocity used in this expression doesn't have much physical significance, the ratio of the coordinate velocity of the particle to the coordinate velocity of light has the usual physical significance.

Note that because $g_{00}$ goes to zero at the event horizon, the energy-at-infinity becomes the product of one factor going to zero by another factor going to infinity. This ill-behavior is a product of the ill-behavior of the Schwarzschild coordinates, because E turns out to be constant at all other values of r excpet the horizon, the ill-behavior is more mathematical than physical.