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 length−2\text{length}^{-2}. The numerical factor relating energy and curvature is

8πGc4=2.08×10−43meter−2J/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.