
Originally Posted by
37818
As it is found in this equation:
ds^2 = - (1 - 2GM/(rc^2)) dt^2 + dr^2/{1 - 2GM/(rc^2))+ r^2 d theta^2 + r^2 sin^2 theta d phi^2
How is the

arrived at?
The above is the Schwarzschild Metric, which is a vacuum solution to the Einstein Field Equations, and describes space-time in the surroundings of a static, non-rotating, uncharged spherical mass. I am not certain where you got the above form of the metric from, normally the convention is to give the metric tensor a signature of (+,-,-,-), so that the line element looks like this :
This is the conventional form as normally found in most GR textbooks. In order to derive this, one starts with a set of assumptions like so :
1. The mass is spherically symmetric
2. The mass is static
3. The mass does not have angular momentum
4. The mass does not carry electrical charge
Given these one can then make an
ansatz to the ( as yet unknown ) metric, which is really only a generalization of the normal line element in standard spherical coordinates :
reducing the problem to two unknown metric coefficients A(r) and B(r); this is possible because of the spherical symmetry of the body of mass, and it being static. Now, in vacuum ( remember we are looking at the space surrounding the body of mass, not its interior ) one has
and thus the Einstein field equations reduce to
Inserting our ansatz into this tensor equation yields a system of partial differential equations for the two unknown functions A(r) and B(r), which can be solved exactly and analytically. Doing the maths one finds
The form you have given ( where did you get that from, anyway ? ) then results from inserting the A(r) and B(r) into the relativistic Lorentz factor, though I haven't seen the Schwarzschild Metric written in that form before.