Black Hole via charge compression

  • Thread starter michael879
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  • #26
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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, thats me copying the RN metric incorrectly. Just replace every instance of [itex]Q^2[/itex] with |Q| in the post above
 
  • #27
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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.
 
  • #28
PeterDonis
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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##.
 
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  • #29
PeterDonis
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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##.
 
  • #30
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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 [itex]m_{matter}=m(R)[/itex], the effective mass becomes [itex]m_{eff}(r)=m_∞ - Q^2/r = m_{matter} + Q^2(R^{-1}-r^{-1})[/itex] 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 [itex]m_{matter}\rightarrow 0[/itex] 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 [itex]m_{matter}[/itex]

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!
 
  • #31
PeterDonis
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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 [itex]m_{matter}=m(R)[/itex], the effective mass becomes [itex]m_{eff}(r)=m_∞ - Q^2/r = m_{matter} + Q^2(R^{-1}-r^{-1})[/itex]
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.
 
  • #32
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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!
 
  • #33
PeterDonis
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Charge by itself (w/o EM field)
There's no such thing; you can't have charge present without having an EM field present.
 
  • #34
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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.
 
  • #35
PeterDonis
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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.
 
  • #36
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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:
[tex]
M_{shell} = M_∞-\dfrac{Q^2}{2R} - M_{EM}(R)
[/tex]
where [itex]M_{EM}(R) << M_∞[/itex] 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 [itex]M_{shell}[/itex], 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 [itex]M_{shell}\geq 0[/itex] always, that means [itex]M_∞-M_{EM}(R)\geq\dfrac{Q^2}{2R}[/itex] and the minimum radius the shell can be collapsed to can be solved analytically from that inequality. Assuming [itex]M_{EM}(R) << M_∞[/itex] we find the approximate solution:
[tex]
R_{min} = \dfrac{Q^2}{2M_∞}
[/tex]

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

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 [itex]R_{min}=a[/itex], so as long as it is possible to compress the thickness of the shell you can reach a singularity at R=0
 
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