Transforming to Eddington-Finkelstein Coordinates for Schwarzschild Geometry in GR

  • Thread starter Astrofiend
  • Start date
  • #1
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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:

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

and the Edd-Fink transformation assigns

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

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:

[tex]
dt = dv - dr -2M. \frac{d}{dr} [log(r-2m)-log(2m)]
[/tex]
[tex]
dt = dv - dr -2M. \left(\frac{1}{r-2m}\right)
[/tex]
[tex]
dt = dv - dr - \left(\frac{2m}{r}-1\right)
[/tex]


...but the book says the answer is

[tex]
dt = -\frac{dr}{1-\frac{2M}{r}} + dv
[/tex]

>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...
 

Answers and Replies

  • #2
gabbagabbahey
Homework Helper
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Homework Statement


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

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

Ermm... shouldn't this be

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

?:wink:

Also, [tex]2m\left(\frac{1}{r-2m}\right)
\neq \left(\frac{2m}{r}-1\right)[/tex]
 
  • #3
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Damn it! Still not seeing it...
 
  • #4
gabbagabbahey
Homework Helper
Gold Member
5,002
7


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

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

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