### The Schwarzschild Solution ↑

As might be expected, the calculation of the Schwarzschild metric is a little brutal. This included in accordance with mathematical and scientific rigour. If one wants to properly grasp the meaning of the equations of general relativity, one should follow through a calculation, even if one does not want to check every minus sign oneself!In Schwarzschild Coordinates, (

*t*,

*r*, θ, φ), with an origin at the centre of an isolated spherically symmetric, non-rotating, gravitating body, such as a planet or a star, ignoring the gravity of other stellar objects, the metric is

*k*=

*e*

^{λ}and κ =

*e*

^{μ}which will simplify calculation,

*r = x*

^{1}, the non-vanishing partial derivatives of the metric are

*R*

_{00}= 0 as required by Einstein’s equation,

Also

*R*

_{11}= 0 as required by Einstein’s equation,

We also have

*R*

_{22}= 0,

The other angular component,

*R*

_{33}, contains the same information as

*R*

_{22}and does not require calculation. Using the equations for

*R*

_{00}and

*R*

_{11}

μ = −λ and κ =

So, the equation for *k*^{−1}.*R*

_{22}is

*C*is a constant of integration. Thus,

*C*= −2

*GM*. Thus the metric outside of an isolated, spherically symmetric, non-rotating, gravitating body of mass

*M*is

*Schwarzschild metric*. Return

### The Newtonian Orbit

In a Newtonian potential we have*r*.

*u*=

*r*

^{−1}.

*u*. Let

*p*=

*h*

^{2}⁄ μ. Then,

*e*and

*f*are constants of integration. Thus,

*r*a minimum for φ = 0, the orbit is recognised as an ellipse with eccentricity

*e*and semi-latus rectum

*p*,