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Thursday, April 20, 2017

Deriving the Schwarzschild Solution to Einstein's Field Equations

Step one: Beginning with Einstein's field equations, derive the Scharzschild radius (equation 13 below):

Next, we call on Pythagoras and a right triangle to derive a basic metric equation (equation 15 below):

Using the same right triangle we derive the Lorentz factor (equation 19 below):

Now check out equation 20:

Because of equation 20, we can make a substitution and derive equations 22 and 23:

Equations 22 and 23 allow us to make more substitutions. The result is something that resembles the Schwarzschild metric (equation 24):

Here's the actual Scharzschild metric:

We can replace equation 24's cdt' with dr (differential radius) to get the following:

It would be great if the middle term (vdt)^2 had a plus sign instead of a minus sign in front of it. With some trigonometric slight of hand we change the minus sign to a plus. The result is equation 29:

So far we've used a triangle with only two space dimensions. We are one dimension short, but we can fix that:

Each dimension in space is a hypotenuse of a right triangle with two other dimensions which can replace the hypotenuse. We make a final substitution and we get the Schwarzschild metric (equation 32). Schwarzschild used spherical coordinates. For clarity and to help you visualize this type of coordinate system, I provide the diagrams below. The first two diagrams show the front and side view of a sphere of spacetime with a mass in the center. The position in spacetime is given by the radius (r), the first angle (top diagram), and a shorter radius (rsin[first angle]) and the second angle (bottom diagram).

The variables used in the Schwarzschild metric, however, are differential--a tiny piece of the radius and each angle. The value of each space variable is indicated in red below:

If we take the limit of these variables we get a point in spacetime indicated by the red dot in the diagram below:

The objective is to figure out the spacetime curvature in that tiny (red) region of space. To solve the field equations, we need to know the metric tensor components; i.e., the g's.

We can find the value of each g component within the Scharzschild metric:

Thus the metric tensor is as follows:

With the information we have, we can derive equation 42 below.

We can solve for R44--the spacetime curvature--by plugging in the mass (m) of a star, planet or black hole; the volume (V) of the space, mass and energy within an imaginary sphere with radius (r); and radius (r).


  1. I don't really understand how to derive eq29 from eq28. Can you show the steps how you make -vdt to +vdt? If we make it to +vdt, then it is not the original ds^2 (?)