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Re: Charged mass exact solutions to Einstein's field equations
Posted:
Feb 11, 2013 1:23 PM
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On Feb 10, 5:36 pm, David Waite wrote:
> Implications of new exact solutions of Einstein's field
> equations I've found and a particular coordinate expression
> of the charged black hole solution are discussed at the video link
> https://www.youtube.com/watch?v=M1WcSCsCFck
The Reissner-Nordstrom metric has the following form:
** ds^2 = c^2 T (1 ? K / r + L / r^2) dt^2
- dr^2 / (1 ? K / r + L / r^2) ? r^2 dO^2
Where
** T, K, L = Constants
** dO^2 = cos^2(Latitude) dLongitude^2 + dLatitude^2
Well, it does not satisfy the vacuum field equations unless (L = 0)
which becomes the Schwarzschild metric. Thus, the Reissner-Nordstrom
metric is wrong. The closest metric to Reissner-Nordstrom metric is
the one below.
** ds^2 = c^2 T (1 ? K / R) dt^2
- dr^2 (dR/dr)^2 / (1 ? K / R) ? R^2 dO^2
Where
** R = r / (1 ? L/K / r)
Torturously listening to these monotonous rambling at 1:30 into the
boring presentation, you showed the following metric as a
transformation for the good old Schwarzschild metric.
** ds^2 = c^2 dt^2 / (1 + K / r)^2
? (1 + K / r)^2 dr^2 ? (1 + K / r)^2 r^2 dO^2
The above metric is wrong in which it does not satisfy the null Ricci
tensor. The solution you are looking for is:
** ds^2 = c^2 T (1 ? K / R) dt^2
- dr^2 (dR/dr)^2 / (1 ? K / R) ? R^2 dO^2
Or
** ds^2 = c^2 dt^2 / (1 + K / r)
? (1 + K /r) dr^2 ? (1 + K / r)^2 r^2 dO^2
Where
** R = r + K
The metric above and the Schwarzschild metric are both valid solutions
to the null Ricci tensor. The Schwarzschild metric manifests black
holes in the infinite future, but the metric above does not. <shrug>
The Reisnner-Nordstrom metric has the electric static force obeying
the inverse cubed law instead of inverse squared law. The self-styled
physicists have a better chance of fudged Coulomb?s law with the
following modification to the Schwarzschild metric.
** ds^2 = c^2 (1 ? K / r + L / r) dt^2
- dr^2 / (1 ? K / r + L / r) ? r^2 dO^2
Where
** L = Constant that would make up Coulomb?s law
Waite is becoming an embarrassment to the Einstein Dingleberries, and
GR is becoming a bigger embarrassment to self-styled physicists.
<shrug>
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