Solving the Transformation to Eddington-Finkelstein in Schwarzschild Geometry

[Astrofiend]

Homework Statement

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

Homework 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.

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]

Homework Statement

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

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

?

Also, $$2m\left(\frac{1}{r-2m}\right)<br /> \neq \left(\frac{2m}{r}-1\right)$$

 

[Astrofiend]

Damn it! Still not seeing it...

 

[gabbagabbahey]

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

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