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## Main Question or Discussion Point

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:

[tex]M(r) = \dfrac{q^2}{2}\left(\dfrac{1}{R}-\dfrac{1}{r}\right)\Theta(r-R)[/tex]

where [itex]\Theta[/itex] 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:

[tex]r_+ = M(r_+)+\sqrt{M^2(r_+)-q^2}[/tex]

This can be solved to get:

[tex]r_+ = \dfrac{q^2}{2R}\left(1+\sqrt{1-\dfrac{8R^2}{q^2}}\right)[/tex]

Now, if R < r

I was a little sloppy here, and I'm pretty sure I have some factor issues because I find that when [itex]R=\frac{|q|}{\sqrt{8}}[/itex] [itex]M(r_+) = \frac{3\sqrt{2}}{4}|q|[/itex] which is > |q|, but [itex]r_+=r_-[/itex] 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?

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:

[tex]M(r) = \dfrac{q^2}{2}\left(\dfrac{1}{R}-\dfrac{1}{r}\right)\Theta(r-R)[/tex]

where [itex]\Theta[/itex] 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:

[tex]r_+ = M(r_+)+\sqrt{M^2(r_+)-q^2}[/tex]

This can be solved to get:

[tex]r_+ = \dfrac{q^2}{2R}\left(1+\sqrt{1-\dfrac{8R^2}{q^2}}\right)[/tex]

Now, if R < r

_{+}the shell should be considered 'collapsed' into a black hole. This condition is satisfied when [itex]R \leq \frac{|q|}{\sqrt{8}}[/itex].I was a little sloppy here, and I'm pretty sure I have some factor issues because I find that when [itex]R=\frac{|q|}{\sqrt{8}}[/itex] [itex]M(r_+) = \frac{3\sqrt{2}}{4}|q|[/itex] which is > |q|, but [itex]r_+=r_-[/itex] 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?