# Black Hole via charge compression

1. Apr 10, 2014

### 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?

2. Apr 10, 2014

### 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.

3. Apr 10, 2014

### 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

4. Apr 10, 2014

### Staff: Mentor

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.

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.

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.

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.

5. Apr 10, 2014

### pervect

Staff Emeritus
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 travelling 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.

6. Apr 10, 2014

### 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

7. Apr 10, 2014

### michael879

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.

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

8. Apr 10, 2014

### Staff: Mentor

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.

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".

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.

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.

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.

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").

9. Apr 10, 2014

### 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...

10. Apr 10, 2014

### 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

Last edited: Apr 10, 2014
11. Apr 10, 2014

### Staff: Mentor

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

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.)

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.

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.

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.

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 $r = 0$).

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.)

12. Apr 10, 2014

### 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.

13. Apr 10, 2014

### Staff: Mentor

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

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.)

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.

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.

14. Apr 15, 2014

### michael879

hmm, ok lets 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 lets 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?

15. Apr 15, 2014

### Staff: Mentor

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).

16. Apr 15, 2014

### 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)

17. Apr 15, 2014

### Staff: Mentor

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.

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.

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.

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.

18. Apr 16, 2014

### 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)

19. Apr 16, 2014

### Staff: Mentor

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.)

20. Apr 16, 2014

### 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