Black Hole via charge compression

[michael879]

Here is my basic question:
Lets say we have a massless charged shell with charge q and radius R. Is there some R where a black hole is formed?

I haven't work this out with the EFE, but the calculations I've done are relativistic (GEM equations instead of EFE, for simplicity). I found that the energy contained within some radius r is:
$$M(r) = \dfrac{q^2}{2}\left(\dfrac{1}{R}-\dfrac{1}{r}\right)\Theta(r-R)$$
where $\Theta$ is the Heaviside step function. The outer event horizon of this system if it is a black hole (using GR here, not fully consistent) would be:
$$r_+ = M(r_+)+\sqrt{M^2(r_+)-q^2}$$
This can be solved to get:
$$r_+ = \dfrac{q^2}{2R}\left(1+\sqrt{1-\dfrac{8R^2}{q^2}}\right)$$
Now, if R < r₊ the shell should be considered 'collapsed' into a black hole. This condition is satisfied when $R \leq \frac{|q|}{\sqrt{8}}$.

I was a little sloppy here, and I'm pretty sure I have some factor issues because I find that when $R=\frac{|q|}{\sqrt{8}}$ $M(r_+) = \frac{3\sqrt{2}}{4}|q|$ which is > |q|, but $r_+=r_-$ so the BH should be extremal. Regardless, I think the qualitative result holds and I don't see why using GR would do anything but change the exact values. A more qualitative argument is simply that you can generate an infinite amount of mass (diverging at 0) by compressing the shell, but the charge never changes. So at some point the mass should become large enough to create a black hole

It makes sense to me that if you have a large amount of EM energy in a very small volume, it would generate a black hole. What makes NO sense to me though is what is going on inside of the black hole. The mass contained around the charged shell is ALWAYS 0 so once we stop compressing it, it should just explode! So if an event horizon is formed in the process of compressing the shell, what keeps it inside?

 

[Matterwave]

You might want to take a look at: http://en.wikipedia.org/wiki/Reissner–Nordström_metric

A charged black hole metric is possible, but physically, astrophysical objects don't tend to be charged to a significant degree (mostly charge neutrality holds). An interesting aspect of this metric is the possibility of a naked singularity (which would violate the cosmic censorship hypothesis), but it is thought that formations of such naked singularities is perhaps physically or practically impossible.

 

[michael879]

thanks, but that wasn't my question at all! I already know plenty about general relativity and black holes, my question is about the formation of a very specific charged black hole. If you start off with a massless, charged object can it be compressed into a black hole? And if it can, what keeps it within the bounds of the event horizon when there is no inward force on the object itself??

*note: the object starts off massless but obtains mass from the work you put into compressing it. I'm not talking about naked singularities here

 

[PeterDonis]

Lets say we have a massless charged shell with charge q and radius R.

First of all, you can't have a massless charged shell. If the shell is charged, then there is a nonzero electromagnetic field present, and the energy associated with that field means there is nonzero mass present. (More precisely, there can be zero EM field inside the shell, but there can't be zero EM field outside the shell.)

Basically, what you have (in the simplest case where there is zero EM field inside the shell) is an exterior Reissner-Nordstrom geometry matched across the shell to an interior Minkowski geometry. This is similar to the uncharged case where you have an exterior Schwarzschild geometry matched across a thin spherical shell to an interior Minkowski geometry. The only difference is the presence of the EM field as well as spacetime curvature, which means you have to have the right charge distribution on the shell to match up with the exterior EM field by Maxwell's Equations.

Is there some R where a black hole is formed?

Yes, by the same logic as for the uncharged case. If you take an uncharged spherical shell and compress it enough, it will form a Schwarzschild black hole. The only wrinkle here is that, as you note, the mass of the hole will be larger than the original mass of the shell before you started compressing it. That's because the compression process does work on the shell, increasing its total energy and therefore its mass. (Note, though, that an observer very far away would see *no* change in the total mass of the system, because the energy that is put into the compression process has to come from somewhere, and from far enough away that energy source's mass gets counted into the total system's mass. So from the faraway observer's viewpoint, all that is going on is an internal transfer of energy from one part of the system to another.)

You might also ask, what happens to the shell once the black hole is formed? The answer is, it collapses into the singularity at the center of the black hole and is destroyed. (At least, that's what happens classically; let's not open the quantum gravity/black hole information paradox can of worms here. ) Actually, the collapse will start *before* the hole is formed, because there is no stable equilibrium possible for an object with a radius less than 9/8 of the Schwarzschild radius corresponding to its mass. So once we've compressed the shell to that point, it will start imploding under its own weight, speeding up the process of forming the black hole.

Similarly, if you take a charged spherical shell and compress it enough, it will form a Reissner-Nordstrom black hole. (I don't know if the shell will start collapsing under its own weight at some radius larger than the horizon radius for its mass, as happens with the uncharged shell--see above. I suspect it will, but I've never seen a theorem for the R-N case similar to the one that's been proven for the Schwarzschild case.) What happens to the shell after the hole is formed is somewhat more complicated in this case (see below); but the bottom line remains the same, it can't get back out once it has fallen through the outer horizon of the R-N black hole.

The mass contained around the charged shell is ALWAYS 0

No, it isn't. The EM field energy is always there. See above.

once we stop compressing it, it should just explode!

Not once it's inside the horizon. It can't; it would have to move faster than light.

So if an event horizon is formed in the process of compressing the shell, what keeps it inside?

The geometry of spacetime inside the horizon.

Consider the uncharged case first--that one is actually closer to what you were imagining since there actually *is* zero mass outside the shell. Suppose the shell has just collapsed through the horizon, and we stop compressing it. Internal forces within the shell can certainly cause it to "explode" relative to local free-falling observers, yes. But that won't stop the shell from falling into the singularity, because inside the horizon, *all* timelike worldlines end in the singularity, even ones with arbitrarily large proper acceleration radially outward, or outward radial velocity arbitrarily close to the speed of light (relative to local free-falling observers). So there's no way for the shell to explode back outside the horizon once it's inside.

In the charged (R-N) case, the shell won't necessarily hit the singularity because the R-N singularity is timelike, not spacelike. (This brings up the whole question of how physically reasonable the R-N interior is to begin with, particularly the part inside the inner horizon. But we don't really need to delve into that here.) But the shell still can't get back outside the outer horizon once it falls through, for the same reason as in the Schwarzschild case: it would have to move faster than light. Once you're inside the outer horizon, all timelike paths lead to the inner horizon; none of them lead back out into the original exterior region.

 

[pervect]

Here is my basic question:
Lets say we have a massless charged shell with charge q and radius R. Is there some R where a black hole is formed?

There is some chance you'll form a naked singularity http://en.wikipedia.org/wiki/Naked_singularity rather than a black hole.

There is a certain minimum mass that a black hole of charge Q must have - see "extremal black hole" http://en.wikipedia.org/wiki/Extremal_black_hole

However, it's rather problematic to have a truly massless charge in the first pace, it would have to be traveling at "c". As I recall, there are some serious difficulties with "massless charge", but I don't recall the details other than it had to do with representation theory.

If you consider instead a hypothetical black hole formed from electrons (rather than the problematical massless charge), I believe there are some theorems that show that you will form a black hole via compression rather than a naked singularity. The resulting black hole will have mass, because in order to get an electrons close enough togethter, they will need kinetic energy to overcome the repulsive forces that would tend to drive them apart. This kinetic energy will add to the mass (aka energy-at-infinity) of the assembly and will contribute to the "mass of the black hole" after it is formed.

This is from memory, perhaps someone else can provide / lookup a reference regarding the naked singularity / black hole issue - if it is of interest , the question has mutated a bit from the question asked by the OP, due to whether or not they really wanted the problematical "massless charge", or whether it was just an approximation gone bad.

 

[michael879]

Thanks for the responses guys, I need to go through the relevant info more closely but I want to clarify a few things first.

1) Sorry when I said massless I meant for R→infinity. i.e. the shell material has no mass, and the only energy is outside from the E&M field. And yes, I realize this is completely impossible since we know there are no massless charged particles. However, an electrons charge is sooo much larger than its mass, that you could create an analogous situation with the same qualitative features
2) The gravitational force on the shell is ALWAYS 0 because it has no intrinsic mass. All of the energy is spherically symmetric and outside of the shell.
3) I've taken general relativity, I'm a 5th year grad student working on ATLAS, and I've been obsessed with black holes and naked singularities for years. I'm not claiming to be an expert, but I've read all of the wikipedia pages and other papers on the subject many times. I'm not trying to be rude, just pointing out that I don't need all that extra info you guys are providing

 

[michael879]

Basically, what you have (in the simplest case where there is zero EM field inside the shell) is an exterior Reissner-Nordstrom geometry matched across the shell to an interior Minkowski geometry. This is similar to the uncharged case where you have an exterior Schwarzschild geometry matched across a thin spherical shell to an interior Minkowski geometry. The only difference is the presence of the EM field as well as spacetime curvature, which means you have to have the right charge distribution on the shell to match up with the exterior EM field by Maxwell's Equations.

Yes, that is where I got the formula r₊. You are right though, this problem is A LOT simpler than I made it out to be. It is described perfectly by the RN metric outside the shell, and completely flat inside the shell. So the only thing wrong in my calculations (besides whatever mistake I made to get those weird factors) is that I used the flat-space Maxwell's equations to calculate the energy outside the shell. I would expect a full relativistic treatment would change the final mass of the black hole somewhat.

Similarly, if you take a charged spherical shell and compress it enough, it will form a Reissner-Nordstrom black hole. (I don't know if the shell will start collapsing under its own weight at some radius larger than the horizon radius for its mass, as happens with the uncharged shell--see above. I suspect it will, but I've never seen a theorem for the R-N case similar to the one that's been proven for the Schwarzschild case.) What happens to the shell after the hole is formed is somewhat more complicated in this case (see below); but the bottom line remains the same, it can't get back out once it has fallen through the outer horizon of the R-N black hole.

No, it isn't. The EM field energy is always there. See above.

Not once it's inside the horizon. It can't; it would have to move faster than light.

The geometry of spacetime inside the horizon.

THIS is where my confusion is. All of the shell's "mass" is contained in the field OUTSIDE the shell. So while the shell clearly has inertial mass, it will not experience any gravitational force because all of the energy is outside of it! So I understand that a black hole must form at some R simply because M→infinity as R→0 and q stays constant. What I don't understand is what keeps the shell together after the event horizon has formed around it? The only force on it is repulsive! Is it that space-time is so incredibly deformed that an observer on the shell would see it explode but never cross the internal event horizon? Or would it explode into the second 'universe' in the RN geometry? Or is there something special about GR that causes a compressive force on something with no gravitational mass??

pervect:
I don't think a naked singularity can be formed in this situation. A singularity only arises when R=0, and at that point M=infinity so clearly M>Q.

And yes, the massless charge thing is an idealization but its theoretically allowed in GR, which is what my question is really about. However, I believe (like you and I both said above) that a shell of electrons would basically be the same thing as a massless shell and the form of the issue isn't changed at all

 

[PeterDonis]

All of the shell's "mass" is contained in the field OUTSIDE the shell.

In the limit where we assume the actual mass of the shell substance itself is negligible, yes. You can't have the shell substance itself have zero mass, because it contains charge density, and no known substance has charge density but zero mass. But I agree that we can ignore that complication for this discussion.

However, there's another complication we can't ignore: the shell's stress-energy tensor must have nonzero stress terms in it, even if we idealize its energy density to be zero. See below.

So while the shell clearly has inertial mass, it will not experience any gravitational force

Yes, it will; it is "supporting the weight" of the field outside it, as well as its own substance.

Mathematically, you can see this by considering what the metric looks like inside the shell. It will go from Reissner-Nordstrom at the shell's outer surface, to Minkowski at the shell's inner surface; but that means that, anywhere inside the shell, the metric is still curved, so the worldline of a given piece of the shell, which is static, will have nonzero proper acceleration. That means it is "feeling a force".

because all of the energy is outside of it!

Not quite. The shell will be under stress, so its stress-energy tensor will be nonzero even if the energy density part is (idealized to be) zero. This stress will also act as a source of gravity, so there will be some effective "gravitational force" due to it.

What I don't understand is what keeps the shell together after the event horizon has formed around it?

Nothing does, if we assume that the only thing holding it together before the collapse was whatever was compressing it. In that case, the shell will indeed "explode" as soon as whatever was compressing it stops compressing it. It's just that the "explosion" will not result in anything escaping outside the (outer) horizon.

Is it that space-time is so incredibly deformed that an observer on the shell would see it explode but never cross the internal event horizon?

It would indeed cross the *inner* horizon (if we assume that the entire geometry of the spacetime is indeed the maximally extended R-N geometry, which probably isn't physically reasonable, but we're leaving that out). But it won't cross the *outer* horizon, because that would require going faster than light.

Or would it explode into the second 'universe' in the RN geometry?

It could, possibly, after passing through the region inside the inner horizon but avoiding the timelike singularity. Then it would exit out the second "inner horizon", and then out the second "outer horizon" (both of these being past horizons opening out into the second "universe").

 

[michael879]

Ok now we're getting somewhere! So there are 2 issues here:

1) Stress-energy: hmm you're right, I forgot about the stress contribution to the stress-energy tensor. However, would this actually cause a compressive force on the shell?? As the shell's thickness goes to 0 I just don't see how there can be any kind of attractive force with the spherical symmetry of the stress-energy tensor that is 0 everywhere within..

2) Explosion: Well in this scenario there is no singularity anywhere, because the RN geometry becomes flat at the surface of the shell. So let's say we put enough energy into the shell that it will compress into a black hole without any further interference (which would complicate things). I'm not certain of this, but I would expect that when the event horizon first forms it is degenerate (i.e. maximal BH). So the shell would be inside the time-like inner region of the RH geometry, still getting smaller. At some point, assuming there isn't a significant attractive force, the shell will stop and begin to expand. This is where things get confusing.

Nothing can cross the first inner horizon (time points inward between the two horizons and nothing can travel back in time), but the second two horizons CAN be crossed. However, its not clear to me how the shell would ever get to the second space-time, since it needs to pass through r=0 to get to the r < 0 region. So either I'm missing something about the dual space-time, or the shell will just expand approaching the inner horizon boundary but never crossing it. But now there is another issue, because the mass of the black hole is a function of the shell's radius! So as the shell expands the mass should decrease right? But if the mass decreases the inner horizon moves out, allowing the shell to expand further. Eventually the black hole will become extremal again and once the shell expands further the event horizon should dissappear all together!

Clearly I've made a mistake somewhere here, because GR doesn't allow the evaporation of black holes, but I can't see where...

 

[michael879]

Although, now that I think about it no matter what is happening inside of a BH its mass will always stay the same! So the second the shell becomes a black hole it should stop gaining mass, meaning that the black hole should just be frozen in the extremal case with the shell right up against the event horizon...

*edit* although, I've neglected the kinetic energy of the shell here. You guys were right, the massless idealization really doesn't work. How can you apply a force on a massless charge?? So assuming infinitesimal mass, the shell would need enough kinetic energy to compress it to some radius R. So once the event horizon forms the shell can continue to shrink, and the kinetic energy will just be turned into EM energy. The BH will have mass contributions from both the EM field and the kinetic energy of the shell. Once the shell stops, this EM energy should turn back into kinetic energy and the shell would expand. Then what? As long as the charge of the shell is larger than it's rest mass the electromagnetic pressure will be larger than the gravitational attraction

 

[PeterDonis]

However, would this actually cause a compressive force on the shell??

Yes. For this scenario, stress basically "has weight", just like energy density does.

As the shell's thickness goes to 0 I just don't see how there can be any kind of attractive force with the spherical symmetry of the stress-energy tensor that is 0 everywhere within.

For a shell with finite thickness, the metric changes smoothly from Reissner-Nordstrom to Minkowski through the shell, as I said before. As the shell's thickness goes to zero, there will be a step discontinuity in the metric at the shell; it will be something like a "domain wall" separating the two regions, with nonzero stress-energy density required on the shell to account for the discontinuous change in the metric. So the stress-energy is still there even for a "zero thickness" shell. (Of course "zero thickness" is really an idealization, but hopefully you see what I mean.)

in this scenario there is no singularity anywhere, because the RN geometry becomes flat at the surface of the shell.

Not once the shell has collapsed inside the outer horizon. The spacetime inside the outer horizon is not static; the shell will inevitably continue to collapse, and the Minkowski region inside it will ultimately disappear.

At least, that's my understanding; but I admit I have not seen much discussion of the charged analog of the Oppenheimer-Snyder solution for the collapse of an uncharged object to a Schwarzschild black hole. The key difference between the two cases is that the Schwarzschild singularity at r = 0 is spacelike, whereas the R-N singularity at r = 0 is timelike. But I'm still not sure how an inner Minkowski region could persist "inside the shell" once the shell has collapsed past the inner horizon.

However, all of these considerations are really separate from the issues you're raising anyway; none of the issues you're raising will make any difference as far as I can see. See below.

So let's say we put enough energy into the shell that it will compress into a black hole without any further interference (which would complicate things). I'm not certain of this, but I would expect that when the event horizon first forms it is degenerate (i.e. maximal BH).

If we start out from a state where $M \le Q$, i.e., the mass as measured "at infinity" outside the shell is less than the total charge, then it would seem that as you add mass through compression, you would eventually reach a state where $M = Q$. But I don't think it's guaranteed that the radius of the shell, at that same instant, will be exactly equal to the appropriate horizon radius. If those things just happened to be at the same instant, then yes, the hole would be degenerate, with the degenerate horizon forming at $r = M$ just as $M = Q$ was satisfied. But if $M = Q$ became true when $r > M$ still held, $M$ would have to increase further before the horizon was formed, so the hole would not be degenerate at that point.

However, I don't think this really makes a difference anyway, because either way the singularity at $r = 0$ is timelike and there are timelike worldlines that escape it and reach a second exterior universe. See below.

So the shell would be inside the time-like inner region of the RH geometry, still getting smaller. At some point, assuming there isn't a significant attractive force, the shell will stop and begin to expand.

Yes, once it's inside the inner horizon, it can expand, since there are timelike worldlines with increasing $r$ there. But this is still true even if the hole is not degenerate (as are the additional points I make below), so I don't think that really makes much difference, as I noted just above.

Nothing can cross the first inner horizon (time points inward between the two horizons and nothing can travel back in time), but the second two horizons CAN be crossed. However, its not clear to me how the shell would ever get to the second space-time, since it needs to pass through r=0 to get to the r < 0 region.

The "second spacetime" (by which I assume you mean the second exterior "universe" region) does not have $r < 0$. It has $r > r_{+}$, i.e., it is outside a second "outer horizon" (with $r = r_{+}$), which is in turn outside a second "inner horizon" (with $r = r_{-}$). It is reachable from inside the original inner horizon because these second horizons are "past horizons", i.e., timelike worldlines emerge from them, rather than falling in. So the shell could indeed "explode" into the second universe by this route.

In fact there is *no* region with $r < 0$ in the Reissner-Nordstrom geometry. You may be confusing it with the interior of the Kerr geometry, which does have an $r < 0$ region (inside the "ring singularity" at $r = 0$).

But now there is another issue, because the mass of the black hole is a function of the shell's radius!

Not once you stop compressing it. The only reason the mass varied with radius originally was that the radius was decreasing as the shell was compressed and energy was added to it through the compression process. Once you stop compressing the shell, the mass is constant. (This should be obvious from Birkhoff's Theorem--for which there is an analog in the charged R-N case. The shell's collapse and "explosion" are spherically symmetric, so once you stop adding energy to it via compression its externally measured mass must be constant.)

 

[michael879]

Ah you're right I am confusing it with the Kerr-Newman metric... I have to be honest I have no idea how the dual universe works in the RN case, or how one would get to it.

Sorry, I'm pretty tired right now so I did make a number of mistakes in that post. The shell would be smaller than the radius of the black hole when it forms, and the black hole probably wouldn't be extremal (if the mass is purely due to the EM energy density it actually would be extremal, but the added mass contributions would probably change the situation). Also, you're right that the mass of the shell would never change (that was a dumb one). If you pump the energy into it at "infinity" then it will always have the same mass, equal to the energy put into it. The EM field will grow as the shell collapses and slows down, and at some point the event horizon will form leaving a RN BH. I'm still not clear on what happens after the BH forms though.. I know this is ENTIRELY theoretical as most people don't expect GR to be valid within a black hole, but I'd still like to know what GR predicts

As for the singularity, the only way one would form is if the shell continued to collapse rather than explode. The singularity is at r=0 in the RN geometry, but r=0 is Minkowski for R > 0.

 

[PeterDonis]

I have to be honest I have no idea how the dual universe works in the RN case, or how one would get to it.

There are good Penrose-style spacetime diagrams on Andrew Hamilton's site:

http://casa.colorado.edu/~ajsh/rn.html

http://casa.colorado.edu/~ajsh/rne.html

The diagrams in question are at the bottom of each page; the first is for the $M > Q$ case and the second is for the $M = Q$ case. Basically, to reach the "dual universe", you just keep going up the diagram.

(Note that this is actually pretty much the same as the Kerr metric geometry, or at least the "equatorial plane" of that geometry. The only real difference with the latter is presence of the $r < 0$ region, which appears as a little triangle on the "other side" of the $r = 0$ singularity. It's there because the locus $r = 0$ in the Kerr geometry is a ring, not a point, so you can go "through the ring" to the $r < 0$ region on the other side.)

I'm still not clear on what happens after the BH forms though.. I know this is ENTIRELY theoretical as most people don't expect GR to be valid within a black hole, but I'd still like to know what GR predicts

As I said, I haven't seen an actual solution for this case (collapsing shell matched up to exterior R-N geometry), so everything I've said is based on educated guesses from the analogous Oppenheimer-Snyder model of collapse to a Schwarzschild black hole.

As for the singularity, the only way one would form is if the shell continued to collapse rather than explode.

This is one point I'm not sure about. In the Oppenheimer-Snyder model, once the collapsing object (whether it's a shell or anything else) gets inside the horizon radius, it *has* to collapse to $r = 0$ (so if it's a shell with Minkowski geometry inside, that interior geometry has to disappear eventually); but the singularity at $r = 0$ is spacelike for that case, whereas it's timelike for the R-N case. So yes, it appears to be possible, just from looking at the nature of the singularity if it were to form, that the shell collapse could stop at some $r > 0$ and it could expand again. But I'm not sure how consistent such a scenario would be, because I'm not sure if the "dual universe" could be there if the singularity were not there, and yet if the "dual universe" were not there I'm not sure what could even be inside the inner horizon.

Another wrinkle here, which is mentioned on the Andrew Hamilton pages, is that in the standard R-N geometry (i.e., the maximally extended R-N spacetime, with no shells or anything else to gum up the works), gravity is repulsive inside the inner horizon. That is, timelike geodesics are "repelled" by the singularity at $r = 0$, so something that free-falls through the inner horizon will be "pushed" back out the second inner horizon (and toward the "dual universe"). So it might be that, once the shell collapses inside the inner horizon, the singularity has to form anyway--more precisely, the scenario "shell stops collapsing and expands again" implicitly *includes* the singularity being there, because the electromagnetic repulsion that causes the shell to stop collapsing and start expanding *is* the singularity, in some sense.

Again, all this is really hand-waving because I have not seen an actual mathematical solution describing this scenario.

 

[michael879]

hmm, ok let's look at it this way then:

We put some energy M into the "massless" charged shell Q so it collapses to some radius R which is within the inner horizon of the black hole created. Treat this system (with the shell at rest) as stationary for the moment, and let's imagine how a test charge acts.

Send a test charge q into the black hole with enough energy so that it will stop at the surface of the shell. What would happen next? This is a fairly simple problem, since the charge is outside the shell its just in the RN geometry with same-sign charge as the "singularity". Would this charge exit the black hole into the second universe?

 

[PeterDonis]

Would this charge exit the black hole into the second universe?

I can answer this question for the R-N geometry without any massless shell present: the answer is yes, it would. Once the charge gets inside the inner horizon, if it does not hit the singularity, it has nowhere else to go but the second universe (by exiting the second inner horizon and then the second outer horizon).

My guess is that the answer is the same if the massless shell is present inside the inner horizon, because, as I said in my last post, I don't think that scenario is really any different than the R-N scenario without a shell present. If both horizons are there, and whatever charge is present is "localized" inside the inner horizon, then that's basically the R-N geometry. The only real difference might be that the shell has a nonzero surface area, so the geometry in the shell case would be a "truncated" R-N geometry where the "singularity" (the 2-sphere where the charge is located) is at $r > 0$ instead of $r = 0$ (and possibly with a small Minkowski region inside the shell, but I haven't tried to draw a Penrose diagram of that to see if it actually makes sense).

 

[michael879]

Ok you did answer my question (so thanks , but I think you're confused on one point. In this scenario, assuming the sphere is not changing size at all, the metric is perfectly determined at all points, and there is no singularity. Outside the sphere there is the R-N metric, which has a singularity at 0. However the inside of the sphere is flat, so there is no singularity anywhere!

Of course, I don't think situation can possibly be static except maybe with finely tuned parameters. So either the sphere will collapse or expand, and that's what I'm trying to figure out. You said the charge would have no other choice but to exit through the event horizon. But it could also keep falling to r=0 couldn't it? Clearly once it crosses the shell things get more complicated, but I'm really just interested in what direction the charge will go if it is at rest directly outside the shell (which is possible because there is assumed to be within the time-like region of RN)

 

[PeterDonis]

In this scenario, assuming the sphere is not changing size at all, the metric is perfectly determined at all points, and there is no singularity. Outside the sphere there is the R-N metric, which has a singularity at 0. However the inside of the sphere is flat, so there is no singularity anywhere!

Yes, I wasn't clear enough about that, but this is basically what I was referring to when I mentioned a "truncated" R-N geometry.

Of course, I don't think situation can possibly be static except maybe with finely tuned parameters.

I don't think it can be static even *with* finely tuned parameters; I think the only possible static geometry with charge present is the complete R-N geometry, with the singularity at the center. See below.

You said the charge would have no other choice but to exit through the event horizon. But it could also keep falling to r=0 couldn't it?

More precisely, I said that if the charge does not fall into the singularity, it has no choice but to exit through the (second) horizon. It seems like in principle it could hit the singularity at $r = 0$ instead; however, again, I haven't actually done the math, so it's possible that this can't happen--that the combination of EM repulsion (singularity charge <-> infalling charge) and repulsive gravity inside the inner horizon will always force an incoming charge to turn around and head back out.

I'm really just interested in what direction the charge will go if it is at rest directly outside the shell (which is possible because there is assumed to be within the time-like region of RN)

If the charge is at rest directly outside the shell, it will obviously be repelled and move outward. The only possible way it could fall further in is if it had sufficient inward velocity when it was directly outside the shell.

Similar reasoning applies to the shell itself: if at some instant the shell is at rest with respect to the singularity (i.e., its $r$ coordinate is momentarily constant), then it will be repelled and will start moving outward. So I don't think a static geometry is possible with a shell present.

 

[michael879]

Ok good point, if something manages to slow the charge down to rest then obviously it will continue to push it out of the black hole. So I guess my last question would be:
In a complete RN geometry where M > Q, what happens to a charge q if it falls into the black hole? Clearly there are only two options: it hits the singularity or it emerges in the 2nd universe. But I'm curious which of these is the case, and if it is dependent on the initial conditions at all (i.e. for some charge and initial energy it will hit the singularity, for others it will escape)

 

[PeterDonis]

In a complete RN geometry where M > Q, what happens to a charge q if it falls into the black hole? Clearly there are only two options: it hits the singularity or it emerges in the 2nd universe. But I'm curious which of these is the case, and if it is dependent on the initial conditions at all (i.e. for some charge and initial energy it will hit the singularity, for others it will escape)

Looking at the Penrose diagram of the R-N geometry for M > Q, there are clearly timelike worldlines that hit the singularity, and others that pass on through the inner region and reach the second universe. So in principle both are possible, and which one happens in a specific case will depend on the details.

The only question in my mind is whether an infalling object with the same sign of charge as the hole (so that its motion will not be geodesic--it will experience proper acceleration in the outward radial direction) can ever overcome the combined EM and gravitational repulsion in the inner region to reach the singularity. (I guess it always could if you attached a sufficiently powerful rocket to it that provided inward thrust.)

 

[michael879]

O actually I do have 1 more question related to this, but kindof off topic:

Lets say you put just enough energy into this shell to NOT make it a black hole. It will come to rest right outside where the event horizon would be, but Q > M. The metric outside the shell would be described by the RN geometry of a naked singularity correct? Is this a problem? I'm a little rusty on this, but problems only occur with singularities when it is possible to reach them and return to your original location right? So in this case, since there is no singularity there wouldn't be any CTCs or anything

*edit* it would come to rest well within where the event horizon would have formed. Pushing it a tiny bit further would cause an event horizon to form further away

 

[PeterDonis]

Lets say you put just enough energy into this shell to NOT make it a black hole. It will come to rest right outside where the event horizon would be, but Q > M.

I'm not sure a static equilibrium is possible at an arbitrary radius. In the Schwarzschild case the limit is 9/8 of the horizon radius; inside that point no static equilibrium is possible for a shell, planet, or other such object that does not expend energy to maintain altitude. (A rocket can still "hover" closer than that, but it must expend energy to do so.) As I think I mentioned earlier in this thread, I have not seen an analogous theorem for the R-N case, but I would expect there to be one. I think I would expect such a limit to be present even in the Q > M case, where there is no horizon, but again, I haven't actually looked at the math.

The metric outside the shell would be described by the RN geometry of a naked singularity correct?

It would be described by R-N geometry with Q > M down to whatever radial coordinate the shell occupies. The metric inside the shell is Minkowski, so there's no singularity anywhere. "The RN geometry of a naked singularity" is not a good description of this.

*edit* it would come to rest well within where the event horizon would have formed.

Would have formed on what assumptions?

Pushing it a tiny bit further would cause an event horizon to form further away

I'm not sure this is consistent. First, the event horizon doesn't form at a nonzero radius; if there is an EH present in the spacetime but it is not eternal, it forms at r = 0 and moves outward. (For example, consider the Oppenheimer-Snyder scenario of a spherically symmetric uncharged object collapsing to a Schwarzschild black hole; the EH forms at r = 0 and moves outward, just reaching r = 2M at the point where the outer surface of the collapsing matter is falling through that radius.) So in this case, if the shell is collapsing in such a way that there will be a horizon present, I would expect the horizon to form at r = 0 and move outward to meet the shell as it reaches the horizon radius. (More precisely, this would be the outer horizon radius in the charged case.)

Second, if you're imagining taking a shell with Q > M and increasing M by compressing it (adding energy), remember that M is a global parameter that includes all the energy present in the spacetime, including the stored energy that you are drawing on for the compression process. So if there's enough stored energy to increase what you are thinking of as "M" so that it is larger than Q, then the actual M of the spacetime must already be larger than Q, and the full geometry must already be the Q < M R-N geometry with a horizon present (and the shell must already be inside the horizon and therefore falling inward, not static).

 

[michael879]

Sorry for gravedigging this post, but I came back to this problem recently and actually worked out the entire problem in detail. What I found is a little strange though, since it suggests that we could "create" and enter an alternate universe without ever getting near the quantum scale (which is the usual argument why the weird effects of GR can be ignored)..

The problem:
You have a charged shell whose radius R(t) is the only free parameter of the system. Outside of the shell is the Reissner–Nordström metric and electric field. Inside the shell space-time is completely flat. Requiring that these two solutions match at the shell constrains the mass to be $$M=\dfrac{Q^4}{2R}$$
The shell will form an event horizon when the outer horizon of the Reissner–Nordström metric is greater than R. This gives the inequality:
$$R \leq M + \sqrt{M^2-Q^4} = \dfrac{Q^4}{2R} \left(1 + \sqrt{1-\dfrac{4R}{Q^4}}\right)$$
which has the solution
$$R \leq \dfrac{1}{2}Q^2$$
If you imagine compressing the shell until this point, you get an extremal black hole with radius
$$r_{H}=M=Q^2$$
I also worked out the Lagrangian for this system in terms of R (which turns out to be exactly the same for both flat and curved space) to get the dynamics of this sytem. If you assume the shell has mass m at infinity, the equation of motion are:
$$\dfrac{m\dot{R}}{(1-\dot{R}^2)^{-3/2}} = \dfrac{Q^2}{2R^2}$$
which ends up being very trivial, but it does show that there is an outward pressure for all R even after the event horizon has formed. (as a side note, if you take m to be 0 as I did in the OP the lagrangian becomes nonsensical and the EOM becomes $\frac{Q^2}{2R^2}=0$ which can only be satisfied for 0 charge or infinite radius... Is massless charge not even possible theoretically??)

So if you compress a charged sphere to this extremal radius, an event horizon will form and the shell will be quickly expelled from the white hole in the second space-time. What bothers me about this is that if Q is large enough, this entire experiment could be carried out at macroscopic scales. The only good argument I've seen to dismiss strange effects of GR such as naked singularities, wormholes, and alternate space-times is that quantum gravity is expected to remove them from the theory. However I don't see how quantum mechanics could possibly play a role in this scenario?

 

[PeterDonis]

Requiring that these two solutions match at the shell constrains the mass to be $$M=\dfrac{Q^4}{2R}$$

But this means $R$ can't vary with time, because that would require $M$ to vary with time, and you ruled that out by hypothesis, because you said that $R$ was the only free parameter in the system. Also, even if we leave that aside, $M$ can't vary just because we say $R$ varies; $M$ can only vary if extra stress-energy comes into the system or leaves the system, which you've ruled out by hypothesis.

So what you have here is a way of specifying a *family* of models where each model has a fixed $R$, not a single model where $R$ can vary. In other words, you have specified a family of static, unchanging systems, not a single system which can change with time.

(Also, the formula you give doesn't look right, because the units don't work out; $Q^2 / 2 R$ would give the right units, since $Q$, $M$, and $R$ all have the same units.)

The shell will form an event horizon when the outer horizon of the Reissner–Nordström metric is greater than R. This gives the inequality:
$$R \leq M + \sqrt{M^2-Q^4} = \dfrac{Q^4}{2R} \left(1 + \sqrt{1-\dfrac{4R}{Q^4}}\right)$$
which has the solution
$$R \leq \dfrac{1}{2}Q^2$$

Ok so far (except for the unit issue I brought up above). But note that all this tells you is which of the models in your family of static models will have an event horizon outside the shell. Which, since the shell can't be static if it's inside an event horizon, actually tells you which of the models in your family of models are internally inconsistent, since $R$ can't change with time but it would have to for the shell to be inside a horizon.

If you imagine compressing the shell until this point

But you can't, because of what I said above: your model can't describe a single system that changes with time. This is in addition to the issue I raised before, that you need to include the energy used to compress the shell in the total energy of the spacetime, and you haven't.

As far as I can tell, this invalidates the rest of your post.

 

[michael879]

But this means $R$ can't vary with time, because that would require $M$ to vary with time, and you ruled that out by hypothesis, because you said that $R$ was the only free parameter in the system. Also, even if we leave that aside, $M$ can't vary just because we say $R$ varies; $M$ can only vary if extra stress-energy comes into the system or leaves the system, which you've ruled out by hypothesis.

So what you have here is a way of specifying a *family* of models where each model has a fixed $R$, not a single model where $R$ can vary. In other words, you have specified a family of static, unchanging systems, not a single system which can change with time.

You're right, for M to be time-dependent the system can't be isolated. I think I mixed up the massless and massive models here a bit, but I think the conclusions still hold. You could just pump energy into the shell and that would cause R to go down and M to go up with time.

(Also, the formula you give doesn't look right, because the units don't work out; $Q^2 / 2 R$ would give the right units, since $Q$, $M$, and $R$ all have the same units.)

if you're taking Q to be $Q=\frac{Ge^2}{4\pi\epsilon_0c^4}$ then yea, I quess Q is in the same units as M. I'm using "e" for the charge though, so it actually has units of sqrt(length)

Ok so far (except for the unit issue I brought up above). But note that all this tells you is which of the models in your family of static models will have an event horizon outside the shell. Which, since the shell can't be static if it's inside an event horizon, actually tells you which of the models in your family of models are internally inconsistent, since $R$ can't change with time but it would have to for the shell to be inside a horizon.



But you can't, because of what I said above: your model can't describe a single system that changes with time. This is in addition to the issue I raised before, that you need to include the energy used to compress the shell in the total energy of the spacetime, and you haven't.

As far as I can tell, this invalidates the rest of your post.

Why exactly can't R change with time? I don't see why compressing the shell until it forms an event horizon would be impossible, and I'm not really sure where I went wrong in the math (granted, I did explain the problem incorrectly but the math should still hold if you are pumping energy into the shell)

 

[PeterDonis]

You could just pump energy into the shell and that would cause R to go down and M to go up with time.

No, it won't, because $M$ is the mass of the entire spacetime, as seen at infinity. If energy is pumped into the shell, then $M$ at infinity already contains that energy. See further comments below.

if you're taking Q to be $Q=\frac{Ge^2}{4\pi\epsilon_0c^4}$

That's $Q^2$, not $Q$. Check the dimensions of all the constants; you'll see that $G / \epsilon_0 c^4$ has dimensions of meters squared per Coulomb squared, so you have to take its square root to convert charge to length.

Why exactly can't R change with time?

My original answer was, because that would require $M$ to change with time, and it can't, because of how it's defined. However, on re-reading I realized that your formula for $M$ in terms of $R$ is wrong, and correcting it also changes how I would answer this question.

Let me walk through how I would derive the correct formula. First, by hypothesis, we have a spacetime where the only stress-energy present is a charged spherical shell and its electric field. (We'll consider how we might model adding other stress-energy below.) So, as you say, we have a region of R-N geometry extending in from infinity to some finite radius $R$ where the shell is located; then we have (in the idealized model we're using) an instant transition from R-N geometry outside the shell to flat Minkowski geometry inside the shell. (A real shell would have finite thickness, of course, but we're idealizing it as zero thickness to keep things simple.)

Consider the R-N geometry region. We can define a function $m(r)$ which gives the "mass inside radius $r$", as follows:

$$
m(r) = M - \frac{Q^2}{2r}
$$

In other words, at a finite radius $r$, the mass inside that radius is just the mass at infinity, $M$, minus the mass associated with the electric field stress-energy that is outside radius $r$.

What happens at the shell radius, $R$? The "mass inside radius $r$" must transition from the value given above for the R-N geometry at radius $R$, to zero. I.e., the mass of the shell itself must be

$$
m_{shell} = m(R) = M - \frac{Q^2}{2R}
$$

But now we can see why $R$ can't change with time, at least in this simple model: that would require $m_{shell}$ to change with time (because $M$ and $Q$ are constant) and how is it going to do that if there is no other stress-energy present?

We can also see that the equation you wrote down, $M = Q^2 / 2R$, is the point at which $m_{shell} = 0$; if we flip this equation around it gives us the lower bound of the radius the shell can have in order to have a positive mass: for $m_{shell} > 0$, we must have

$$
R > \frac{Q^2}{2M}
$$

For any R-N geometry with a horizon, i.e., with $Q \le M$, the minimum $R$ for the shell is inside the inner horizon, which means I was wrong in my previous post to say that the shell could not be inside the horizon: I think it *is* possible to have a static shell inside the inner horizon (but not between the outer and inner horizons), as well as outside the outer horizon. For an R-N geometry without a horizon, i.e., with $Q > M$, AFAIK the spacetime is static down to $r = 0$, so the shell could be at any $R$ that meets the above condition.

Now for the question of how we could model $R$ changing with time. To do that, as the above analysis shows, we must add some other stress-energy besides the shell and the electric field energy. And, because changing $R$ requires changing $m_{shell}$, this added stress-energy must be traveling to or from the shell, i.e., it must be traveling to or from radius $R$. For example, we might think about a spherical shell of stress-energy emitted by the charged shell, decreasing $m_{shell}$ and $R$. (Note, btw, that this is backwards from what you were imagining: if the charged shell contracts, its mass goes down, not up. So you can't create a horizon in the inner R-N region by this method if one didn't already exist.) So at some finite radius $r > R$, the "mass inside radius $r$" will be time-dependent, i.e., we will have a function $m(r, t)$ instead of just $m(r)$. But at infinity, we will still have $M$, the total mass of the spacetime, including the mass of the stress-energy emitted by the charged shell.

What would this more complicated geometry look like? Inside the second shell (the spherical shell of stress-energy emitted by the charged shell--we will assume this second shell is electrically neutral), we will have an R-N geometry outside the charged shell, and Minkowski geometry inside it, as before; but it is true that the $M$ associated with this R-N geometry will not include the mass of the second shell (see below). Outside the second shell, I think what we will have is an R-N geometry with $M$ being the total mass of the spacetime, including the mass of the second shell; this is what will be seen at infinity.

But now for an interesting question: suppose we start off with a single charged shell, the original simple geometry we considered above, with $Q < M$, and we let the shell emit a second shell of stress-energy to decrease its mass, according to the more complicated model we just discussed. Can this process achieve $Q > M$ for the "inner" R-N region (the one inside the second shell that gets emitted)? I originally thought the answer to this was "no", but now I'm not sure.

 

[michael879]

No, it won't, because $M$ is the mass of the entire spacetime, as seen at infinity. If energy is pumped into the shell, then $M$ at infinity already contains that energy. See further comments below.



That's $Q^2$, not $Q$. Check the dimensions of all the constants; you'll see that $G / \epsilon_0 c^4$ has dimensions of meters squared per Coulomb squared, so you have to take its square root to convert charge to length.

Oops you're right, that's me copying the RN metric incorrectly. Just replace every instance of $Q^2$ with |Q| in the post above

 

[michael879]

Now for the question of how we could model $R$ changing with time. To do that, as the above analysis shows, we must add some other stress-energy besides the shell and the electric field energy. And, because changing $R$ requires changing $m_{shell}$, this added stress-energy must be traveling to or from the shell, i.e., it must be traveling to or from radius $R$. For example, we might think about a spherical shell of stress-energy emitted by the charged shell, decreasing $m_{shell}$ and $R$. (Note, btw, that this is backwards from what you were imagining: if the charged shell contracts, its mass goes down, not up. So you can't create a horizon in the inner R-N region by this method if one didn't already exist.) So at some finite radius $r > R$, the "mass inside radius $r$" will be time-dependent, i.e., we will have a function $m(r, t)$ instead of just $m(r)$. But at infinity, we will still have $M$, the total mass of the spacetime, including the mass of the stress-energy emitted by the charged shell.

What would this more complicated geometry look like? Inside the second shell (the spherical shell of stress-energy emitted by the charged shell--we will assume this second shell is electrically neutral), we will have an R-N geometry outside the charged shell, and Minkowski geometry inside it, as before; but it is true that the $M$ associated with this R-N geometry will not include the mass of the second shell (see below). Outside the second shell, I think what we will have is an R-N geometry with $M$ being the total mass of the spacetime, including the mass of the second shell; this is what will be seen at infinity.

But now for an interesting question: suppose we start off with a single charged shell, the original simple geometry we considered above, with $Q < M$, and we let the shell emit a second shell of stress-energy to decrease its mass, according to the more complicated model we just discussed. Can this process achieve $Q > M$ for the "inner" R-N region (the one inside the second shell that gets emitted)? I originally thought the answer to this was "no", but now I'm not sure.

Yea you're right, I was neglecting the source of the mass increase which is insane when you put it like that :P One thing I think you're off about is the mass m(r), which isn't actually the shell's mass AFAIK. I believe m(r) is the effective mass experienced by a test particle at r which has an acceleration Gm/r^2. When you match the RN geometry outside the shell with the flat geometry inside the shell you find that m(R)=0 always, which doesn't really make sense if m_shell = m(R).

I need to think about this more, because you're right: this needs to be done with additional spherical shells either incoming or outgoing with respect to the original charged shell.

 

[PeterDonis]

One thing I think you're off about is the mass m(r), which isn't actually the shell's mass AFAIK.

The function $m(r)$ in a static, spherically symmetric spacetime is, as I said, the "mass inside radius $r$". That's true of *any* static, spherically symmetric spacetime. I wasn't claiming that $m(r)$ must give the mass of the shell just because; I was showing that we can obtain the mass of the shell by reasoning using $m(r)$, as follows:

Just outside the shell, the function $m(r)$ must have the value $m(R) = M - Q^2 / 2R$, because that's just applying the functional form of $m(r)$ that's true in any R-N spacetime region, including the region outside the shell, to the value $r = R$. But just *inside* the shell, there is no "mass inside radius $r$" at all; spacetime inside the shell is Minkowski, so we must have $m(r) = 0$ for $0 < r < R$. That means the mass of the shell itself must be equal to the "mass inside radius $r$" at $r = R$; i.e., $m_{shell} = m(R) = M - Q^2 / 2R$.

I believe m(r) is the effective mass experienced by a test particle at r which has an acceleration Gm/r^2.

Yes, that's a valid physical interpretation of $m(r)$ in a general static, spherically symmetric spacetime (except that there should be a redshift factor in the denominator of the acceleration).

When you match the RN geometry outside the shell with the flat geometry inside the shell you find that m(R)=0 always

No, you don't. You find that $m(R) = 0$ just inside the shell, but $m(R) = M - Q^2 / 2R$ just outside the shell. Remember that we're idealizing the shell as having zero thickness, so there is a jump discontinuity at $r = R$; functions like $m(r)$ don't have a single value there, they behave like a step function. (If we were squeamish about jump discontinuities and step functions, which I'm not, we could be more pedantic and define everything in terms of limits taken as $r \rightarrow R$ from above and below.)

A more realistic model would have a shell with some finite thickness $\delta R$, so the outer surface would be at $r = R$ and the inner surface would be at $r = R - \delta R$. Then we would have $m(R) = M - Q^2 / 2R$ and $m(R - \delta R) = 0$, with a smooth transition between the two values over the range $R - \delta R < r < R$, and we would have R-N geometry for $R \le r < \infty$ and Minkowski geometry for $0 < r \le R - \delta R$, and a more complicated geometry including the effect of the shell's stress-energy for $R - \delta R < r < R$.

 

[PeterDonis]

Yes, that's a valid physical interpretation of $m(r)$ in a general static, spherically symmetric spacetime (except that there should be a redshift factor in the denominator of the acceleration).

Actually, on consideration, I'm not sure this is exactly right. Computing the proper acceleration of a "hovering" observer in R-N spacetime gives:

$$
a = \sqrt{g_{ab} a^a a^b} = \sqrt{g_{rr}} \left( u^a \nabla_a u \right)^r = \sqrt{g_{rr}} u^t \Gamma^r{}_{tt} u^t
$$

Substituting $u^t = 1 / \sqrt{g_{tt}}$ and $\Gamma^r_{tt} = (1/2) g^{rr} \partial_r g_{tt}$, and using $g_{tt} = 1 / g_{rr}$, gives

$$
a = \frac{1}{2} \sqrt{g_{rr}} \partial_r g_{tt} = \left( \frac{M}{r^2} - \frac{Q^2}{r^3} \right) \frac{1}{\sqrt{1 - 2M / r + Q^2 / r^2}}
$$

But computing $m(r) / r^2$ and adding in the redshift factor gives:

$$
a' = \frac{m(r)}{r^2 \sqrt{1 - 2m(r) / r}} = \left( \frac{M}{r^2} - \frac{Q^2}{2 r^3} \right) \frac{1}{\sqrt{1 - 2M / r + Q^2 / r^2}}
$$

These are not the same. I think what's going on here is that the proper acceleration is not just caused by mass; it is caused by "gravity", where other components of stress-energy besides the 0-0 component (the "mass") can contribute to "gravity". If we write things in terms of stress-energy tensor components, and assume the SET is diagonal, then the "source of gravity" is not just $T_{00}$, it's $T_{00} + T_{11} + T_{22} + T_{33}$.

In the case of R-N spacetime, the electric field has an energy density equivalent to $Q^2 / 2r$ (where "equivalent to" means we're converting energy density to the contribution to $m(r)$ due to that energy density), but it also has a radial stress which is minus the energy density, and tangential stresses which are equal to the energy density. So in terms of the SET components, we have $T_{11} = - T_{00}$ and $T_{22} = T_{33} = T_{00}$, so the full contribution of the electric field to the "source of gravity" is the sum of all four diagonal components, which comes out to $2 T_{00}$. (And this contribution comes in with a minus sign, because at finite $r$ we're subtracting the "source of gravity due to the electric field" that is outside $r$.) That's why the term that appears in the proper acceleration is $- Q^2 / r^3$ instead of $- Q^2 / 2 r^3$.

 

[michael879]

So I've been thinking about this some more and I realized a few things:

1) Many of my problems came from assuming the shell had 0 thickness but the density did not diverge (by assuming the metric was continuous). If instead you use just a really thin shell, you see that the metric gradually goes from RN to flat with a slope based on the mass of the shell, and in the limit of 0 thickness becomes discontinuous (unless the mass is 0)

2) I don't think a charged object CAN collapse to a singularity. I can't even count how many times people claim "charge has negative mass" because of that very effective mass function you provided. However, if you think about what is actually going on the effective mass is just the mass at infinity minus the mass of the EM field outside of your radius. So if the object of radius R has mass $m_{matter}=m(R)$, the effective mass becomes $m_{eff}(r)=m_∞ - Q^2/r = m_{matter} + Q^2(R^{-1}-r^{-1})$ and it's clear that the charge actually increases the gravitational pull.

So a collapsing charged object would continue to collapse for as long as it could convert rest mass to electromagnetic energy. However, at some point this ceases to be possible (as $m_{matter}\rightarrow 0$ at the very least) at which point the electrostatic repulsion balances out the gravitational pull. It makes no sense for the object to have negative mass, and as you pointed out the only way to force it smaller would be to add energy to the system. Adding energy would increase the total mass M without decreasing $m_{matter}$

3) Even if a singularity can not be formed in the presence of charge, and event horizon is surely possible. So one would think that my original scenario would still occur if you took a charged shell with inward kinetic energy: it would shrink past the event horizon, forming a black hole, and then bounce back into a second space-time. So my problem with this being possible at a purely macroscopic scale still holds, and I'm still a little confused about it. The "unphysical" effects of GR are almost always explained away as being forbidden by quantum gravity, but this one can not!

 

[PeterDonis]

If instead you use just a really thin shell, you see that the metric gradually goes from RN to flat with a slope based on the mass of the shell

Yes, agreed.

the effective mass is just the mass at infinity minus the mass of the EM field outside of your radius.

Yes. Note, though, that the EM field mass is $Q^2 / 2r$, so we have $m_{eff} = m_{\infty} - Q^2 / 2 r$.

if the object of radius R has mass $m_{matter}=m(R)$, the effective mass becomes $m_{eff}(r)=m_∞ - Q^2/r = m_{matter} + Q^2(R^{-1}-r^{-1})$

What you're doing is defining $m_{matter} = m_{\infty} - Q^2 / 2 R$; but you can't arbitrarily say that $m_{matter}$ only contains rest mass energy and not EM field energy. On the face of it, that looks like it can't be right, because where is the charge density? It can't be in the region outside the object of radius $R$, since by hypothesis that region has R-N geometry and therefore can only contain a static electric field, with zero charge density. The obvious thing would be to view the charge density as attached to the object of radius $R$; but that means there is also EM field energy contained inside radius $R$. So I think you need to refine your model some more if you really want to separate out rest mass energy from EM field energy.

I need to think some more about the rest of your post.

 

[michael879]

What you're doing is defining $m_{matter} = m_{\infty} - Q^2 / 2 R$; but you can't arbitrarily say that $m_{matter}$ only contains rest mass energy and not EM field energy. On the face of it, that looks like it can't be right, because where is the charge density? It can't be in the region outside the object of radius $R$, since by hypothesis that region has R-N geometry and therefore can only contain a static electric field, with zero charge density. The obvious thing would be to view the charge density as attached to the object of radius $R$; but that means there is also EM field energy contained inside radius $R$. So I think you need to refine your model some more if you really want to separate out rest mass energy from EM field energy.

I need to think some more about the rest of your post.

Ah my bad about the factor of 2. I don't understand your problem with the model though.. What is wrong what the charge density being identical to the mass density? Charge by itself (w/o EM field) has no energy, so the energy enclosed within a sphere of mass M and charge Q is just M as long as the EM field vanishes, which it does because of the spherical symmetry!

 

[PeterDonis]

Charge by itself (w/o EM field)

There's no such thing; you can't have charge present without having an EM field present.

 

[michael879]

There's no such thing; you can't have charge present without having an EM field present.

I meant mathematically "without", i.e. ignoring the EM field's contribution to stress-energy. Inside a spherical shell there is no EM field, and there can't be any gravitational effects from the outside EM field because of the spherical symmetry. So if you look at the total energy within R you should get no contributions from the EM field, and only from the mass density of the shell right? That energy doesn't need to be constant, but surely it can't fall below the rest mass of the shell which is > 0. This should be true regardless of any conversion of rest mass to some other form of energy, which would only affect the minimum rest mass of the shell.

 

[PeterDonis]

I meant mathematically "without", i.e. ignoring the EM field's contribution to stress-energy.

You can't do that in any region where charge is present, but you can make an idealization that makes this kind of separation. See below.

Inside a spherical shell there is no EM field

Which means there's no charge there either. See below.

and there can't be any gravitational effects from the outside EM field because of the spherical symmetry.

Yes, agreed.

So if you look at the total energy within R you should get no contributions from the EM field, and only from the mass density of the shell right?

Only if you are modeling the charge as a pure "surface layer" at the outer surface of the shell; i.e., at radius $R$, all the (non-charge, non-EM field) mass is still inside, but all the charge is now "outside", along with its EM field. This is OK as an idealization, but of course it's a big idealization: any real charge has to be attached to some piece of matter, so any real shell will have charge (and hence EM field) distributed within it, not just on the outer surface.

With this idealization, yes, the mass inside radius $R$ will be only uncharged rest mass, with no contribution from the EM field. So we will have

$$
M_0 = M - \frac{Q}{2R}
$$

where $M_0$ is the rest mass of the shell, $M$ is the total mass of the spacetime (as measured at infinity), and $Q$ is the charge.