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Monday, June 20, 2016

Einstein's Field Equations Simplified

Start with light speed (c) squared equal to velocity (v) squared:

Divide both sides of the equation by c^2:


We know that g is acceleration due to gravity, and it is equal to Gm/r^2. Remove an r from the denominator and we get a velocity squared:


Now we can make a substitution:


If we take the derivative of the sphere area, we get a line with a magnitude of 8(pi)r:

Multiply both sides by 8(pi)r:

We need to convert the mass (m) to energy (E), so we find mass in terms of energy:

Replace m with E/c^2 and divide both sides of the equation by r^3. Doing so gives us a number over an area or r^2. E is replaced by T (energy density). When T is increased, the r^2 on the left side must decrease. If we think of space as an imaginary sphere with radius (r), the smaller r gets, the greater the curvature. An arbitrary distance around the circumference covers a greater angle.

You might have noticed that when r gets smaller, the energy density grows larger, which in turn causes r to shrink even more. Here's where the field equation predicts a star collapsing into a black hole.

If we don't want to dwell on black holes, we need to multiply the left side by a coefficient (g), so when T increases, g increases. The variable g will also be a measure of curvature.

Convert g into the metric tensor and T into the Energy-stress tensor, and we have the equivalent of the field equations in simplified form:

But why settle for simple when you could have complicated? LOL! What about the Einstein tensor which contains the Ricci tensor and all those lovely Christoffel symbols? I show how to derive those here and here.

But what is the logic behind the left side of the field equations? Let's start with the gravitational energy minus the escape velocity energy. When those two values are equal, spacetime is considered flat and gravity is zero.

As you can see, we don't have to think of gravity in terms of spacetime. We could think of gravity as the difference between two energies. But Einstein insisted upon spacetime, we need to express energy in terms of space, or a number over r^2, so we multiply the left side by 8(pi)G/(C^4)(r^3):

Let's replace the energies with R's, then make tensors out of them and add a cosmological constant (since empty space still has energy). In the final step, those are replaced by Gij--the Einstein tensor.

Update: Below is the Lagrangian (L) derived from the field equations--further clarifying the logic behind the left side of the field equations.

1 comment:

  1. The way you derive that langragian seems too good. Last I checked it is L=R/2k + {matter fields}

    ReplyDelete