Electrodynamics in general relativity

On Feb 13, 11:48 am, David Waite wrote:

Heres a video I did showing how the electromagnetic field
tensor obeys general relativity's version of Maxwells equations
for the case of my extreme static charge exact solution to
Einstein's field equations.

http://www.youtube.com/watch?v=M0HU26j9tLI

Have you really derived the field equations in their useful form in
differential equations? It is very certain you have not. If you do,
why don’t you post the null Einstein tensor in polar coordinate
(diagonal metric)? shrug

All the solutions you have pulled out of your ass do not satisfy the
Einstein field equations. You are just shooting in the dark with no
fvcking idea of what you are doing. shrug

On Feb 13, 11:33 pm, David Waite wrote:
On Wednesday, February 13, 2013, Koobee Wublee wrote:

Have you really derived the field equations in their
useful form in differential equations? It is very certain
you have not. If you do, why don’t you post the null
Einstein tensor in polar coordinate (diagonal metric)? shrug

All the solutions you have pulled out of your ass do not
satisfy the Einstein field equations. You are just shooting
in the dark with no fvcking idea of what you are doing. shrug

ok:

grtw();

GRTensorII Version 1.79 (R6)

6 February 2001

Developed by Peter Musgrave, Denis Pollney and Kayll Lake

Copyright 1994-2001 by the authors.

Latest version available from:http://grtensor.phy.queensu.ca/

Who has validated this software? shrug

Eric Gisse the college dropout possessed software that gives the
following as the solutions to the null Einstein tensor (or Ricci
tensor).

** ds^2 = c^2 dt^2 / (1 + K / r) – (1 + K / r) dr^2
– r^2 (1 + K / r)^2 dO^2

Where

** U = G M / c^2 / r

The spacetime geometry satisfies the general form below where [R(r) =
r].

** ds^2 = c^2 dt^2 / (1 + K / R) – (1 + K / R) (dR/dr)^2 dr^2
– R^2 (1 + K / R)^2 dO^2

The above spacetime geometry satisfies the null Einstein or Ricci
tensor with diagonal metric in polar coordinate system. For example,
when [R(r) = r - K], the result is the Schwarzschild metric. What
should R be to get to your metric? None. Thus, you are wrong. You
have relied on others to give you solutions which you have no way of
verifying. shrug

If you want, Koobee Wublee can give you the null Einstein tensor with
diagonal metric in polar coordinate system where you can test out your
metric. shrug

shrugging verbosely,
he types the bipolar opposite of "expository verbiage;"
alas, po'Two-cow.

If you want, Koobee Wublee can give you the null Einstein tensor with
diagonal metric in polar coordinate system where you can test out your
metric.

"Koobee Wublee" wrote in message
...

On Feb 13, 11:33 pm, David Waite wrote:
On Wednesday, February 13, 2013, Koobee Wublee wrote:

Have you really derived the field equations in their
useful form in differential equations? It is very certain
you have not. If you do, why don’t you post the null
Einstein tensor in polar coordinate (diagonal metric)? shrug

All the solutions you have pulled out of your ass do not
satisfy the Einstein field equations. You are just shooting
in the dark with no fvcking idea of what you are doing. shrug

ok:

grtw();

GRTensorII Version 1.79 (R6)

6 February 2001

Developed by Peter Musgrave, Denis Pollney and Kayll Lake

Copyright 1994-2001 by the authors.

Latest version available from:http://grtensor.phy.queensu.ca/

Who has validated this software? shrug

Eric Gisse the college dropout possessed software that gives the
following as the solutions to the null Einstein tensor (or Ricci
tensor).

** ds^2 = c^2 dt^2 / (1 + K / r) – (1 + K / r) dr^2
– r^2 (1 + K / r)^2 dO^2

Where

** U = G M / c^2 / r
===========================================
Kinky Wobbly possesses hardware that gives the following as
the solution to the luminiferous ectoplasm (or Kinky aether).

## y = mx+c
Where
## Q = R/sqrt(k^4)


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