On Tue, 15 Mar 2005 17:07:53 GMT,
(Bilge) wrote:
John C. Polasek:
On Mon, 14 Mar 2005 23:10:31 GMT,
(Bilge) wrote:
snip
Since you are apparently too dense to either work out the result
yourself or search google or whatever, I've worked it for you and
included the pressure. Start with the oppenheimer-volkov equation,
dP/dr = -(P + \rho)(m + 4\pi r^3 P)/[r(r - 2m)]
where P is pressure and \rho is the (uniform) density. With
m = (4\pi/3)\rho r^3, you have:
dP/dr = -(P + \rho)(4\pi\rho r^3/3 + 4\pi r^3 P)/[r(r - (8\pi/3)\rho r^2)]
= -4\pi r (P + \rho)(\rho + 3P)/(3 - 8\pi\rho r^2)
Integrate from P(r) = P_0 to P(r=R) = 0 to get,
(\rho + P_0)/(\rho + 3P_0) = sqrt(1 - (8\pi\rho r^2)/3)
Solve for P_0. Then use the relation,
(\rho + P) d\phi/dr = -dP/dr
and the continuity with the exterior metric to obtain,
\exp(\phi) = [3 sqrt(1 - k) - sqrt(1 - k)]/2
k = 2 Gm/rc^2 (after reinserting the constants G and c^2)
Then, the red shift is given by, z = \exp(-\phi) - 1.
And so, finally you are in a position to respond directly with numbers
to settle the theme of this thread namely,
"Re: Why isn't Earth's surface clock slower then one at Earth's
Centre?"
Congratulations on the algebra that produced the admirably terse
expression for z.
But you stopped short. Now it should be simple (for you) to evaluate
(and authenticate) your function z to produce numeric values of
redshift at earth's surface and at the earth's center (no shell game
this time) as seen from infinity. That's the aim of this piece.
Mr. Dual Space
If you have something to say, write an equation.
If you have nothing to say, write an essay
.
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