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Topic:
Charged mass exact solutions to Einstein's field equations
Replies:
4
Last Post:
Feb 13, 2013 10:10 PM




Re: Charged mass exact solutions to Einstein's field equations
Posted:
Feb 12, 2013 2:49 AM


On Feb 11, 7:18 pm, David Waite <waitedavid1...@yahoo.com> wrote: > On Monday, February 11, 2013, Koobee Wublee wrote:
> > The ReissnerNordstrom 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 > > ReissnerNordstrom metric is wrong. The closest metric to > > ReissnerNordstrom 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 ReisnnerNordstrom metric has the electric static force > > obeying the inverse cubed law instead of inverse squared law. > > The selfstyled 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 selfstyled > > physicists. <shrug> > > No the RN metric sin't wrong. Its not supposed to be vacuum. > The electromagnetic field carries stressenergy and so its > not intended to be a vacuum solution.
Waite, you know not what you are talking about. All energy momentum tensors take place in vacuum except cosmology. Null energy momentum tensors also cover all binary stars. <shrug>
Just what do you think the energy momentum tensor is for electromagnetism anyway? <shrug>
> Its merely an exact solution for the stressenergy tensor of > the electric field from a point charge.
A point charge, by definition, is in vacuum or in air which is close enough to vacuum for the practical purpose of this discussion. <shrug>
> No, no where did I transform the RN metric into Schwarzschild.
Koobee Wublee never said you did. <shrug>
> No the RN metric is not a solution for a null Riccitensor.
So, why bother? <shrug>
> It merely has a null Ricciscalar.
You don?t make any sense as usual. You are merely an alchemist of GR. You know not what you are doing, but you are trying this and that to see if something happens. You have not understood GR. Better study up on GR before putting more of your feet in your mouth. <shrug>



