Black Hole via charge compression

[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.

 

[michael879]

Right of course, there is a non-zero EM field within the shell. However, this can be made arbitrarily small by shrinking the shell. So without shrinking the shell to 0, you can still end up with negligible contributions to the effective mass AT the shell surface due to the EM field. So you would get something like:
<br /> M_{shell} = M_∞-\dfrac{Q^2}{2R} - M_{EM}(R)<br />
where M_{EM}(R) &lt;&lt; M_∞ can be accomplished for any situation by shrinking the shell right? In fact, for most situations this would be the case, since the EM field outside a thin charged shell would have much more energy than the EM field within the shell.

The rest mass of the matter content would be M_{shell}, and should still remain >= 0 regardless of the details of the shell's internal interactions. Of course it is free to change as matter is turned into EM energy (or some other form of energy), but the matter content can't go below 0!

So my argument above should still hold. If M_{shell}\geq 0 always, that means M_∞-M_{EM}(R)\geq\dfrac{Q^2}{2R} and the minimum radius the shell can be collapsed to can be solved analytically from that inequality. Assuming M_{EM}(R) &lt;&lt; M_∞ we find the approximate solution:
<br /> R_{min} = \dfrac{Q^2}{2M_∞}<br />

Working this out without any approximations (shell has thickness a) gives:
<br /> R_{min}-a = \dfrac{Q^2}{2M_∞}<br />

I may have done something wrong, but the only assumption I made was that the mass of the matter content of the system had to remain non-negative. Once the mass of the matter hits 0, there is no method to obtain more energy out of it so the shell must stop shrinking! So while the RN metric is a solution to the EFE, it seems that the only way to CREATE an ideal one would be to compress an infinitely massive shell. If the mass is infinite you get R_{min}=a, so as long as it is possible to compress the thickness of the shell you can reach a singularity at R=0