Einstein field equation

In classical mechanics, gravitational potential is

Φ=GMr. \Phi = -\frac{GM}{r}.

And it is sourced by the mass density ρm\rho_m following

2Φ=4πGρm. \nabla^2 \Phi = 4\pi G \rho_m.

We want to find a relativistic equivalent. There are several non-relativistic issues with the equation that we must address.

  1. First, 2\nabla^2 is a spatial derivative in a particular reference

frame. We want a covariant representation. Perhaps we can use the relativistic wave operator \square?

  1. Second, we have shown that gravity acts on energy, not rest mass, so

we need some covariant representation of energy. The stress-energy tensor TμνT^{\mu\nu} comes to mind.

The left hand side of the classical equation 2Φ\nabla^2 \Phi can be written as (Φ)-\nabla \cdot (-\nabla \Phi), where Φ-\nabla \Phi is the gravitational field. This makes it clear that the left side represents the divergence of the gravitational field, i.e. gravitational tides.

Doing some derivations I don’t know how to do yet, we reach the Einstein field equation

Gμν=8πGc4Tμν. G^{\mu\nu} = \frac{8\pi G}{c^4} T^{\mu\nu}.

Where the Einstein curvature tensor GμνG^{\mu\nu} represents the curvature of spacetime.

In Cartesian coordinates, the curvature tensor has units of length2\text{length}^{-2}. The numerical factor relating energy and curvature is

8πGc4=2.08×1043meter2J/meter3. \frac{8\pi G}{c^4} = 2.08 \times 10^{-43} \frac{\text{meter}^{-2}}{\text J / \text{meter}^3}.

It takes an enormous amount of energy to create even a small curvature in spacetime.