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

 Messages: [ Previous | Next ]
 Koobee Wublee Posts: 1,417 Registered: 2/21/06
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 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>

>
> No the RN metric sin't wrong. Its not supposed to be vacuum.
> The electromagnetic field carries stress-energy 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 stress-energy 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 Ricci-tensor.

So, why bother? <shrug>

> It merely has a null Ricci-scalar.

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>

Date Subject Author
2/11/13 Koobee Wublee
2/12/13 Koobee Wublee
2/13/13 Brian Q. Hutchings