Birkoff theorem

On Apr 25, 3:17 am, waitedavidmsphysics wrote:

Heres a video I did on an exact vacuum solution's line element
I found and the relevance of the Brikoff theorem.
http://www.youtube.com/watch?v=sfKIFyzEmMs

Dude, after all these years, you are still talking garbage. Almost 9
minutes of bull**** in fact. shrug

First of all, the third equation is wrong. Secondly, your
interpretation of invariance is fvcked up. Let’s go back to a
solution to the field equations that is static and spherically
symmetric.

** ds^2 = c^2 (1 – R / u) dt^2 – du^2 / (1 – R / u) – u^2 dO^2

Where

** dO^2 = dLongitude^2 cos^2(Latitude) + dLatitude^2
** R = Integration constant
** u = ANY FUNCTION OF r

Allow Him to emphasize that any function of u(r) within the content of
the equation above is a solution to the field equations. shrug

** When [u = r],

The geometry is the Schwarzschild metric which was derived by Hilbert.

** When [u = r (1 – R^3 / r^3)^(1/3)],

The geometry becomes Schwarzschild’s original solution --- the first
solution ever derived. Schwarzschild’ original solution does not
manifest black holes and also degenerates into Newtonian law of
gravity.

** When [u = r – R],

The geometry manifests no black holes but also explains Newtonian law
of gravity.

** When [u = r / (1 + r^2 / R / S)] where (S = another constant),

The geometry also degenerates into Newtonian law of gravity at
relative short distances (galactic scale) but antigravity at very
large distances (cosmic scale).

** When [u = R^2 / r],

The geometry is not asymptotically flat which proves Birkhoff Theorem
wrong.

Just to toss another bone for you to play with, the following is also
a solution to the field equations.

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

Where

** u = ANY FUNCTION OF r

On Apr 25, 6:05*pm, Koobee Wublee wrote:
On Apr 25, 3:17 am, waitedavidmsphysics wrote:

Heres a video I did on an exact vacuum solution's line element
I found and the relevance of the Brikoff theorem.
http://www.youtube.com/watch?v=sfKIFyzEmMs

Dude, ...
First of all, the third equation is wrong. *Secondly, your
interpretation of invariance is fvcked up. *...

Not my fault that you have so little background in relativity that you
don't know what the general principle of relativity is or what
isotropic coordinates are.

On Apr 26, 4:05*am, Koobee Wublee wrote:
On Apr 25, 3:17 am, waitedavidmsphysics wrote:

Heres a video I did on an exact vacuum solution's line element
I found and the relevance of the Brikoff theorem.
http://www.youtube.com/watch?v=sfKIFyzEmMs

Dude, after all these years, you are still talking garbage. *Almost 9
minutes of bull**** in fact. *shrug

First of all, the third equation is wrong. *Secondly, your
interpretation of invariance is fvcked up. *Let’s go back to a
solution to the field equations that is static and spherically
symmetric.

** *ds^2 = c^2 (1 – R / u) dt^2 – du^2 / (1 – R / u) – u^2 dO^2

Where

** *dO^2 = dLongitude^2 cos^2(Latitude) + dLatitude^2
** *R = Integration constant
** *u = ANY FUNCTION OF r

Allow Him to emphasize that any function of u(r) within the content of
the equation above is a solution to the field equations. *shrug

** *When [u = r],

The geometry is the Schwarzschild metric which was derived by Hilbert.

** *When [u = r (1 – R^3 / r^3)^(1/3)],

The geometry becomes Schwarzschild’s original solution --- the first
solution ever derived. *Schwarzschild’ original solution does not
manifest black holes and also degenerates into Newtonian law of
gravity.

** *When [u = r – R],

The geometry manifests no black holes but also explains Newtonian law
of gravity.

** *When [u = r / (1 + r^2 / R / S)] where (S = another constant),

The geometry also degenerates into Newtonian law of gravity at
relative short distances (galactic scale) but antigravity at very
large distances (cosmic scale).

** *When [u = R^2 / r],

The geometry is not asymptotically flat which proves Birkhoff Theorem
wrong.

Just to toss another bone for you to play with, the following is also
a solution to the field equations.

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

Where

** *u = ANY FUNCTION OF r



Idiot

On Apr 25, 6:05*pm, Koobee Wublee wrote:
On Apr 25, 3:17 am, waitedavidmsphysics wrote:

** *ds^2 = c^2 (1 – R / u) dt^2 – du^2 / (1 – R / u) – u^2 dO^2

That IS Schwarzschild's solution. What you are calling u everyone else
calls r. When you parameterise u as a function of some other
coordinate like u(r) what you are calling r just isn't what everyone
else on the planet calls r. They are calling what you are calling u as
r.

On Apr 26, 1:29 am, Tonico wrote:
On Apr 25, Koobee Wublee wrote:

Let’s go back to a solution to the field equations that
is static and spherically symmetric.

** ds^2 = c^2 (1 – R / u) dt^2 – du^2 / (1 – R / u) – u^2 dO^2

Where

** dO^2 = dLongitude^2 cos^2(Latitude) + dLatitude^2
** R = Integration constant
** u = ANY FUNCTION OF r

Allow Him to emphasize that any function of u(r) within the
content of the equation above is a solution to the field
equations. shrug

** When [u = r],

The geometry is the Schwarzschild metric which was derived by
Hilbert.

** When [u = r (1 – R^3 / r^3)^(1/3)],

The geometry becomes Schwarzschild’s original solution --- the
first solution ever derived. Schwarzschild’ original solution
does not manifest black holes and also degenerates into Newtonian
law of gravity.

** When [u = r – R],

The geometry manifests no black holes but also explains Newtonian
law of gravity.

** When [u = r / (1 + r^2 / R / S)] where (S = another constant),

The geometry also degenerates into Newtonian law of gravity at
relative short distances (galactic scale) but antigravity at very
large distances (cosmic scale).

** When [u = R^2 / r],

The geometry is not asymptotically flat which proves Birkhoff
Theorem wrong.

Just to toss another bone for you to play with, the following
is also a solution to the field equations.

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

Where

** u = ANY FUNCTION OF r

Thanks for the demystification.

You are welcome. shrug

On Apr 26, 8:32 am, waitedavidmsphysics wrote:
On Apr 25, Koobee Wublee wrote:

Let’s go back to a solution to the field equations that
is static and spherically symmetric.

** ds^2 = c^2 (1 – R / u) dt^2 – du^2 / (1 – R / u) – u^2 dO^2

Where

** dO^2 = dLongitude^2 cos^2(Latitude) + dLatitude^2
** R = Integration constant
** u = ANY FUNCTION OF r

That IS Schwarzschild's solution. What you are calling u everyone else
calls r. When you parameterise u as a function of some other
coordinate like u(r) what you are calling r just isn't what everyone
else on the planet calls r. They are calling what you are calling u as
r.

No, you are wrong again. When you have a function f(x), in general
f(x) != x. You need to understand the basics of mathematics. shrug

Allow Him to emphasize that any function of u(r) within the
content of the equation above is a solution to the field
equations. shrug

** When [u = r],

The geometry is the Schwarzschild metric which was derived by
Hilbert.

** When [u = r (1 – R^3 / r^3)^(1/3)],

The geometry becomes Schwarzschild’s original solution --- the
first solution ever derived. Schwarzschild’ original solution
does not manifest black holes and also degenerates into Newtonian
law of gravity.

** When [u = r – R],

The geometry manifests no black holes but also explains Newtonian
law of gravity.

** When [u = r / (1 + r^2 / R / S)] where (S = another constant),

The geometry also degenerates into Newtonian law of gravity at
relative short distances (galactic scale) but antigravity at very
large distances (cosmic scale).

** When [u = R^2 / r],

The geometry is not asymptotically flat which proves Birkhoff
Theorem wrong.

Just to toss another bone for you to play with, the following
is also a solution to the field equations.

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

Where

** u = ANY FUNCTION OF r

[Waite’s self-hating comments snipped]

Einstein Dingleberries are just a bunch of retards. This episode is
another example. shaking head

On Apr 26, 9:30*am, Koobee Wublee wrote:


No, you are wrong again. *When you have a function f(x), in general
f(x) != x. *...
Damn you're clueless. x is not r. f is r. Just because you are calling
x by r does not make it that.