- #36
michael879
- 698
- 7
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
[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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