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View Full Version : Transforming to Eddington-Finkelstein Coordinates for Schwarzschild Geometry in GR


Astrofiend
Oct21-09, 09:59 PM
1. The problem statement, all variables and given/known data

I'm having problems seeing how the transformation to Eddington-Finkelstein in the Schwarzschild geometry works. Any help would be great!

2. Relevant equations

So we have the Schwarzschild Geometry given by:


ds^2 = -(1-2M/r)dt^2 + (1-2M/r)^-^1 dr^2 + r^2(d\theta^2+sin^2\theta d\phi^2)


and the Edd-Fink transformation assigns


t = v - r -2Mlog\mid r/2M-1 \mid


The textbook says it is straight-forward to simply sub this into the S.G line element to get the transformed geometry, but I can't seem to get it.

3. The attempt at a solution

OK, so differentiating the expression for the new t coordinates above, I get:


dt = dv - dr -2M. \frac{d}{dr} [log(r-2m)-log(2m)]


dt = dv - dr -2M. \left(\frac{1}{r-2m}\right)


dt = dv - dr - \left(\frac{2m}{r}-1\right)



...but the book says the answer is


dt = -\frac{dr}{1-\frac{2M}{r}} + dv


>where am I going wrong here? Given this last expression, it is fairly easy to sub it into the SG line element to get the transformed coordinates. Problem is, I can't seem to get that far! Any help much appreciated...

gabbagabbahey
Oct21-09, 10:09 PM
1. The problem statement, all variables and given/known data
OK, so differentiating the expression for the new t coordinates above, I get:


dt = dv - dr -2M. \frac{d}{dr} [log(r-2m)-log(2m)]



Ermm... shouldn't this be

dt = dv - dr -2M. \frac{d}{dr} [log(r-2m)-log(2m)]dr

?:wink:

Also, 2m\left(\frac{1}{r-2m}\right)
\neq \left(\frac{2m}{r}-1\right)

Astrofiend
Oct21-09, 11:16 PM
Damn it! Still not seeing it...

gabbagabbahey
Oct21-09, 11:20 PM
dt=dv-dr-2m\left(\frac{1}{r-2m}\right)dr=dv-\left[1+2m\left(\frac{1}{r-2m}\right)\right]dr

Just simplify....it's basic algebra from here.