Solving the Spherically Symmetric Einstein Equation

[Tomas Vencl]

TL;DR

When suppose only spherically symmetric distribution of mass-energy, how it simpify the Einstein equation ?

Can be Einstein equation rewrited into some simpler form, when suppose only spherically symmetric (but not necessarily stationary) distribution of mass-energy ?
If yes, is there some source to learn more about it ?
Thank you.
edit: by simpler form I mean something with rather expressed derivatives, than more compact form (if this makes sense)

 

[robphy]

These may be helpful:
the worksheets on
http://pages.pomona.edu/~tmoore/grw/resources.html

You might also wish to look at the General Relativity notes
(starting at say Ch 24...
Ch 25 talks about Schwarzschild, where you can see the role of the static condition)
http://home.uchicago.edu/~geroch/Course Notes

 

[PeterDonis]

Can be Einstein equation rewrited into some simpler form, when suppose only spherically symmetric (but not necessarily stationary) distribution of mass-energy ?

Yes. For the spherically symmetric case, there are only two unknown functions of the coordinates in the metric (a general metric with no symmetry has ten unknown functions of the coordinates, one for each of the independent metric components). The most common way of treating this case is to choose coordinates such that the metric is diagonal and the areal radius $r$ (given by $r = \sqrt{A / 4 \pi}$, where $A$ is the radius of the 2-sphere containing the event in spacetime whose coordinates one is evaluating) is one of the coordinates. The general form of the metric with these assumptions, along with the significant components of the Einstein tensor, is given in this Insights article:

https://www.physicsforums.com/insights/short-proof-birkhoffs-theorem/

The article only considers the actual solution of the EFE for the vacuum case; but one can easily generalize what is done there to the non-vacuum case by simply putting the appropriate stress-energy tensor components (as functions of $t$ and $r$) on the RHS of the equations.

 

[Tomas Vencl]

Thank you both, I will look at the links.