Wednesday, February 24, 2016

general relativity - Understanding spherically symmetric metric


In these lecture notes the static isotropic metric is treated as follows (p. 71):


Take a spherically symmetric, bounded, static distribution of matter, then we will have a spherically symmetric metric which is asymptotically the Minkowski metric. It has the form (in spherical coordinates): $$ds^2=B(r)c^2dt^2-A(r)dr^2-C(r)r^2(d\theta^2+\sin^2\theta d\phi^2)$$ And then it goes on eliminating $C$ and expanding $A$ and $B$ in powers of $\frac{1}{r}$. No explanations are given on why we can assume that form for the metric. Could someone explain why, please?


Personally, I would rather assume the form (in cartesian coordinates): $$ds^2=f(r)dt^2-g(r)(dx^2+dy^2+dz^2)$$ which would certainly give a spherically symmetric metric, and then change to spherical coordinates, obtaining something looking like: $$ds^2=f(r)dt^2-g(r)(dr^2+r^2d\theta^2+r^2\sin^2\theta d\phi^2)$$ which looks substantially different from the above. Is this approach wrong? Why?


By the way, don't be afraid of getting technical. I have a pretty good mathematical basis on the subject (a course of one year on differential geometry).




No comments:

Post a Comment

classical mechanics - Moment of a force about a given axis (Torque) - Scalar or vectorial?

I am studying Statics and saw that: The moment of a force about a given axis (or Torque) is defined by the equation: $M_X = (\vec r \times \...