Relativity: Equivalence Energy-Metric

1. Vectorial expression of the geodesic equation

The Geodesic equation is :

dv^k/dτ + Γ^k_ijvⁱv^j = 0

In terms of coordinates, it is:

d²x^α/dτ² + Γ^α_μν (dx^μ/dτ)(dx^ν/dτ) = 0

Or

d²x^α/dτ² = - Γ^α_μν (dx^μ/dτ)(dx^ν/dτ)

We will take the following approximation and generalize after.

The motion of the particle is slow (v << c), and
The gravitational field is week and stationary (static).

The motion of the particle is slow implies neglecting the spatial part terms dx¹/dτ, dx²/dτ, dx³/dτ in comparison with the term dx⁰/dτ = c dt/dτ. Then the geodesic equation becomes:

d²x^α/dτ² = - Γ^α₀₀ c² (dt/dτ)²

The field is week implies: ∂g_μν/∂t = 0.
Therefore the affine connection, that is the Christoffel symbol:
Γ^k_ij = (1/2) g^kl [dg_il/∂x^j + dg_jl/∂xⁱ - dg_ij/∂x^l]
becomes:

Γ^α₀₀ = (1/2) g^αβ [dg_0β/∂x⁰ + dg_0β/∂x⁰ - dg₀₀/∂x^β]
= (1/2) g^αβ [2 dg_0β/∂x⁰ - dg₀₀/∂x^β]

We have:
2 ∂g_0β/∂x⁰ = 0 (field is week). It remains:

Γ^α₀₀ = (-1/2) g^αβ ∂g₀₀/∂x^β

Γ^α₀₀ = (-1/2) g^αβ ∂g₀₀/∂x^β

For a week field we have:
g_μν = η_μν + h_μν + O(h²)
and

∂g_μν/∂t = c ∂g_μν/∂x⁰ = 0

Where

h_μν is the first order deviation in h from the flat Minkowski space (space time) that corresponds to the absence of gravitational field: |h_μν| < 1.

Therefore:

Γ^α₀₀ = (-1/2) η^αβ ∂h₀₀/∂x^β = (-1/2) δ^α_β ∂h₀₀/∂x^β

And only the spatial parts of remains. We have the Kronecker in the Minkowski space-time: δ^α_β = - 1 for spatial parts.

So
Γ^α₀₀ = (1/2) dh₀₀/∂x^β

Γ^α₀₀ = (1/2) dh₀₀/∂x^β

The geodesic equation, with c = 1, then becomes:
d²x^α/dτ² = (- 1/2) (dt/dτ)² dh₀₀/∂x^β

d²x^α/dτ² = (- 1/2) (dt/dτ)² ∂h₀₀/∂x^β

Vectorially, for the spacial parts:
∂ d²x^α/dτ² = (- 1/2) (dt/dτ)² ∇_βh₀₀

For the approximations considered:
The motion of the particle is slow , and
the gravitational field is week and stationary, we have
dt/dτ = 1,

so
The geodesic equation, with c = 1, then becomes:

d²x^α/dt² = (- 1/2) ∇_βh₀₀

Vectorial Geodesic equation

d²x/dt² = (- 1/2) ∇_βh₀₀

2. Newtonian limit

Recall the Newtonian (classical) acceleration field g is:

g = g(r) = F/m = (GMm/r²)/m = GM/r² = - ∂Φ/∂r,
where Φ = - GM/r is the Newtonian gravitational potential.

With del or gradient operator, it can be written:

g = - ∇Φ

Its gradient has the expression:

∇. g = - 4 πGρ

And the Poisson's equation for gravity is:

∇²Φ = 4 πGρ

Now, we want the vectorial Geodesic equation:
d²x/dt² = (- 1/2) ∇_βh₀₀
coincides with the newton equation:
g = - ∇Φ

That is:
(- 1/2) ∇_βh₀₀ = - ∇Φ
or
h₀₀ = 2Φ + constant

We have the following convention : Far from any masses, where Φ = 0, and the metric is g_μν = η_μν , we have h₀₀ = 0. Then constant = 0 Hence, from the equation:
g_μν = η_μν + h_μν + O(h²)

We have in the weak-field limit:
g₀₀ = 1 + h₀₀ = 1 + 2Φ

g₀₀ = 1 + 2Φ = 1 + 2(Φ/c²)

The Poisson's equation for gravity :

∇²Φ = 4 πGρ
becomes:
∇²Φ = (1/2)∇²g₀₀ = 4 πGρ

But ρ is the T⁰⁰ component of the tensor Energy-Momentum T^μν. So
∇²Φ = 4 πGT⁰⁰
or
(1/2)∇²g₀₀ = 4 πGT⁰⁰

That is, for a particle moving slowly in a weak gravitational field:
∇²g₀₀ = 8 πGT⁰⁰

Particle moving slowly in a weak gravitational field:

∇²g₀₀ = 8 πGT⁰⁰

Now, we make a tensor:
g₀₀∇²g₀₀ = 8 πG g₀₀T⁰⁰ = 8 πG T₀₀
And denote g₀₀∇²g₀₀ by G₀₀

Therefore:

G₀₀ = 8 πG T₀₀

We generalize and write:
G_μν = 8 πG T_μν

Equivalence Energy-Metric:

G_μν = g_μν∇²g_μν = 8 πGT_μν = (8 πG/c⁴)T_μν

Recall:

G_μν = 8 πGT_μν