Question regarding GR wrt potential energy

[PeterDonis]

Suppose each of the tiny spherical dust particles had a charge such that that their electrostatic force established a pressure which maintained the static position of the shell. Then at t=0, the charge vanishes.

No, this won't work because it violates charge conservation. (It also violates conservation of the stress-energy tensor because there is stress-energy associated with the charge--the spacetime outside a charged shell is Reissner-Nordstrom spacetime, not Schwarzschild spacetime--it's a different solution of the EFE, although the two have many similarities.)

the concept is well defined in Newtonian physics in terms of the conservation of energy: potential plus kinetic equals zero.

That's because in Newtonian physics, space is always static, so there is always a well-defined meaning to potential energy.

Are you saying that the care required for solving a dynamic GR relationship about energy includes a need to consider states preceding the initial conditions at some t=0?

Not necessarily (but see below). I'm saying that in GR, there is no analogue in general to the absolute, static space of Newtonian physics. Only particular scenarios have an analogue to that, so only particular scenarios have a well-defined notion of potential energy.

In your scenario, if we view the shell as a self-contained, isolated system, then the region outside the shell's maximum radius is static, so it can serve to provide a well-defined notion of potential energy. The problem with that is that we then have to explain how the shell got to its condition at time t = 0, at rest at a given radius but with zero pressure. If it had pressure before t = 0 and so was static then, the pressure can't just vanish; that would violate the covariant divergence condition on the stress-energy tensor. If it didn't have pressure before t = 0, then it couldn't have been at the same radius; it must have been expanding outward before t = 0, reached maximum radius at t = 0, and then would start collapsing after t = 0 (like a rock thrown upward in a gravitational field at less than escape velocity--it reaches maximum height, stops, then starts falling again). But in that case, how did it come to be expanding outward? And even if we answer these questions, they won't give us an answer for the shell's binding energy, because we have no "unbound" state of the shell to compare with; we can define a notion of potential energy using the static region outside the shell, but how do we know it's the "right" one?

If, OTOH, we view the shell as having been brought inward from infinity, slowly lowered, then released from rest at t = 0 to collapse freely with zero pressure, then the shell is not self-contained; we have to include in our analysis whatever it is that does the slow lowering and extracts energy from the shell while doing it. Once we've accounted for that extra whatever it is, we could then split the total energy of the system into two parts, the shell and everything else, and focus on just the energy of the shell and how it changed from infinity to when it is released at t = 0. This gives an obvious comparison between the bound and "unbound" states of the shell, and that's why I recommended this method in an earlier post. This method, I believe, also answers the question I asked at the end of the previous paragraph, by showing that the answer given by this method is the same as the answer we get using the method in the previous paragraph and adopting the notion of potential energy defined by the static region outside the self-contained shell. But I haven't confirmed that with a full calculation.

 

[PeterDonis]

I haven't confirmed that with a full calculation.

To expand on this: the calculations I did in post #52 are relevant to answering, not quite this exact question, but a closely related one: what is the binding energy of a spherical ball of dust (not a shell--no vacuum region inside)? The process described by the sequence of values $M_0$, $M$, and $M_1$ in that post can also be viewed as follows: we take a large number of dust particles at infinity and slowly lower them until they are at rest in a static, spherical configuration with radial coordinate $R$ at its surface. (We assume here that this configuration is not supported by internal pressure but by strings hanging from infinity--this is a thought experiment so we can have the strings supporting dust particles in the interior of the sphere as well as on its surface without worrying about the fact that the strings would have to pass through other dust particles. ) Then, we detach all the dust particles from their strings and let the spherical ball of dust collapse.

The dust starts out with mass $M_0$, which is the same as what was called $N m$ in the OP--it's the number $N$ of dust particles times the rest mass $m$ of a particle in isolation. (We assume that the mass of the strings is negligible, though their tension will not be, as we will see in a moment. This is unrealistic since it means the strings will violate energy conditions, but it is mathematically consistent, so again, since this is a thought experiment. we can do it.) The slow lowering process extracts enough energy to infinity to reduce the mass to $M$ (the energy extracted, for the case of uniform density of dust and weak fields, is $3/5 M/R$, as I derived in post #52). But a portion of the mass $M$ is "stored" in the tension in the strings, since that is what is holding the ball in a static configuration at radius $R$. When we release the dust particles, the tension in the strings goes away, and this can be viewed as extracting another $3/5 M/R$ of energy out to infinity, leaving behind mass $M_1$ stored in the ball of dust itself. So the total "binding energy" of the ball of dust will be the total energy that was extracted to infinity, or $6/5 M/R$, i.e., twice the Newtonian binding energy of a static sphere with uniform density. (And this figure will be constant as the ball collapses--because the spacetime outside the ball is static, the collapse process can be viewed in Newtonian terms as simply converting potential energy to kinetic energy, without changing the externally measured mass $M_1$ of the system.)

The calculation for a thin shell of dust should be similar, with a binding energy of order $M/R$ for the weak field (Newtonian limit) case, but the coefficients might be something other than the $3/5$ and $6/5$ that appear in the case of a ball of dust. I have not calculated this case in detail.

Also, as Jonathan Scott commented earlier and as I agreed, if we look at the process described above from the viewpoint of someone orbiting the system at a large distance, they will see their orbit change as energy is extracted to infinity. If we assume that the orbit is far enough away so that all the dust and all the stored mass in the tension in the strings is inside the orbital radius, then the mass that the person in orbit will compute from his orbital parameters will change from $M_0$ to $M$ during the slow lowering process, and will further change from $M$ to $M_1$ during the string release process (as the energy stored in the string tension is extracted to infinity).

Finally, the above does not address the issue of how to analyze the completely self-contained case where no energy is extracted to infinity and we just have a ball or shell of dust that expands to a maximum radius, stops for an instant, and then collapses again; and it does not answer the question of whether the "binding energy" for such a case, if it can be defined, is the same as what I calculated above. There are other complications with the self-contained case, to do with the fact that, strictly speaking, it implies a white hole in the past as well as a black hole in the future (the dust ball or shell will collapse to a black hole in the case I analyzed above). I have not analyzed that case in detail, and I'm not sure whether it is worth doing so; IMO the above analysis gives enough of an understanding of what's going on.

 

[PeterDonis]

One other caveat to my previous post: I assumed that potential energy is well-defined because there is at least some region, extending from some finite radius out to infinity, which is static. But if energy is being extracted to infinity, that isn't actually true. To really do the analysis right, we would need to put the location where the energy is extracted to at some finite radius $R_e$ which is much larger than $R$, and look at how energy/mass moves from one radius to another within the finite limits between the two radius values. For this case, an observer much farther away than $R_e$ would measure a mass of $M_0$ from his orbital parameters, regardless of what was happening down below. An observer located just inside $R_e$ would measure a mass of zero while the dust was far away (because the dust would be at a larger radius than he is). He would measure a mass of $M_0$ right at the start of the slow lowering process (because he's just inside $R_e$, so as soon as the dust is lowered past him he sees basically its entire original mass--a negligible amount of it will have been extracted by slow lowering at that point), and a gradual reduction of mass to $M$ during the slow lowering (as more and more energy is extracted to $R_e$, outside where he is). He would then observer a fairly rapid reduction of mass from $M$ to $M_1$ as the energy stored in the string tension was extracted (because he wouldn't see the change until the energy got outside his radius, which will be some time after the dust particles are released).

Hopefully this helps to illustrates the issues and caveats with trying to define binding energy/potential energy in a non-static region of spacetime. Note that we are still relying on spherical symmetry to make things work; in a general non-static spacetime without such symmetry there might not be any workable notion of binding energy/potential energy.

 

[PeterDonis]

One more caveat:

In analyzing the ball of dust case, I assumed that my calculation of $M_1$ in post #52 would also be a valid calculation of the externally measured mass of the dust ball after it has been released to start collapsing. That amounts to assuming that the metric I used in post #52 is also valid (or at least a good enough approximation) as a metric for a ball of dust with zero pressure. On its face that assumption seems obviously false: the metric in post #52 was derived on the assumption of a static mass distribution, and the pressure is essential in making it static; you can't just take away the pressure and expect the metric to be the same.

To put it another way, the integral used to calculate $M_1$ was an attempt to isolate the contribution of the pressure to the externally observed mass of the system; but that does not mean that just subtracting the pressure contribution to a static configuration gives a result that can be interpreted as the mass of a zero pressure configuration that is not static (because it's starting to collapse). That's not to say that $M_1$ is necessarily the wrong answer either; it's just that what we've done so far is not a rigorous or even semi-rigorous demonstration that it's right (or at least a good enough approximation). It's just a plausible heuristic guess.

One reason for doing the computation of the fully isolated case (of a spherical ball or shell that expands, stops for an instant, then collapses again) would be to check to see if it does in fact give the same answer as $M_1$, or at least something close to it. If it did, that would make the heuristic guess into something that could be called a reasonable (though still not rigorous) demonstration.