Given a metric in Schwarzschild coordinates, where instead of the usual gtt=-(1-2GM/rc2) and grr=1/(1-2GM/rc2) I have the following:
gtt=-1/(1+GM/rc2)2
grr=(1+GM/rc2)2
What would be the corresponding metric in isotropic coordinates?
ds2 = giso_r(dx2+dy2+dz2) - giso_tc2dt2
And how did you go about figuring it out?
Need to 2nd order, otherwise by approximation it would be just the same as Schwarzschild isotropic coordinates.
Robert,
you can look that up in my paper on "Classroom reconstruction of the Schwarzschild metric". Essentially you have to solve the differential equation
g(r) r'(R)2=r2/R2,
where r is your current and R your new coordinate and g(r) = grr of the metric. I have taken over Charles's notation for the new radial coordinate (in the paper, it is ρ).
In your case, the solution is
R = a r exp(-GM/rc2),
where a is some arbitrary constant. (With the choice a=1, R and r will behave the same at infinity.) Let r(R) the inverse function of R(r).
Then the isotropic form of the metric will be
ds2 = -1/(1+GM/r(R)c2)2 dt2 + a-2 exp(2GM/r(R)c2) (dR2+R2(dθ2+sin θ dφ2)).
A much nicer form would result, if you took grr = 1/(1-GM/rc2)2.
Thanks guys, especially Klaus for working it all out! Charles, I looked for your site before posting this, but Google is pointing to an old defunct version, rqgravity.net.
When I got a transcendental function I thought I had done something wrong and so posted this question. Funny how one form of these close approximations is more complicated than another. I might have more questions when I look it over carefully.