Relativity: Geodesic equation

1. 4-Vector velocity

Let's consider a particle moving in the space-time coordinate frame. The worldline of the particle is its path function of its proper time as it travels through this space.

The worldline of a moving particle can be written, in its reference frame ax r(τ) = xⁱ(τ), where r(τ) = xⁱ(τ) is its position vector and τ is the proper time (the time in the moving frame).

The particle is being placed in the origin of its frame, so dx = dy = dz = 0, and ds² = dτ² (with c = 1).

In the moving frame of the particle: ds = dτ

Now, let's define a 4-Velocity as:
vⁱ = dxⁱ(τ)/dτ = dxⁱ/dτ
In the frame of the moving particle:
vⁱv_i = vⁱ g_ij v^j =
g_ij (dxⁱ/dτ) (dx^j/dτ) =
ds²/dτ² = 1.
The velocity 4-vector is constant.

In the frame of the moving particle:

vⁱ = dxⁱ(τ)/dτ = dxⁱ/dτ

ds = dτ

The velocity 4-vector vⁱ is constant.

2. Geodesic equation

In Euclidean space, straight line is the shortest distance between two points . Curves of minimum length are geodesics.

The geodesic equation is the equation of motion of a freely-falling particle.

The velocity 4-Vector vⁱ = dxⁱ/dτ in the moving frame of the particle is constant, therefore its proper time derivative is zero:

dvⁱ/dτ = d²xⁱ/dτ² = 0

Geodesic equation:

dvⁱ/dτ = d²xⁱ/dτ² = 0

We have:

v = g_i vⁱ, so
dv = (∂v/∂x^j) dx^j, then
dv/dτ = (∂v/∂x^j) (dx^j/dτ)
∂v/∂x^j = ∂(g_i vⁱ)/∂x^j =
g_i ∂vⁱ/∂x^j + vⁱ ∂g_i/∂x^j =
g_i ∂vⁱ/∂x^j + vⁱ Γ^k_ijg_k =
g_k [(g_ig_k^-1)∂vⁱ/∂x^j + vⁱ Γ^k_ij]

We have
g_ig_k^-1 = g_ig^k = δ_i^k

Therefore:
∂v/∂x^j = g_k [∂v^k/∂x^j + vⁱ Γ^k_ij]

Hence
dv/dτ = g_k [∂v^k/∂x^j + vⁱ Γ^k_ij](dx^j/dτ)
= g_k [∂v^k/∂x^j + vⁱ Γ^k_ij](dx^j/dτ)

We have:
dv^k = (∂v^k/∂x^j)dx^j
and
(dx^j/dτ) = v^j

Therefore
dv/dτ = g_k [dv^k/dτ + Γ^k_ij vⁱ v^j]

Hence the geodesic equation is :
[dv^k/dτ + Γ^k_ij vⁱ v^j] = 0

Geodesic equation:

dv^k/dτ + Γ^k_ijvⁱv^j = 0

3. Geodesic equation from Euler-Lagrage equation

The metric is ds² = g_μν dx^μ dx^ν.

The path length S is

S = ∫ ds = ∫ [g_μν dx^μ dx^ν]¹/2;

Parameterising by λ, the path length S becomes:

S = ∫ ds = ∫ [g_μν (dx^μ/dλ) (dx^ν/dλ)]^1/2 dλ

The Lagrangian is L = ds/dλ = [g_μν (dx^μ/λ) (dx^ν/λ)]^1/2

Euler-Lagrange equation to use is:

∂L/∂x^α = d(∂L/∂x'^α)/dλ

L does not depend partially on λ, so instead of using L = [g_μν (dx^μλ) (dx^ν/λ ]^1/2, we use L² = g_μν (dx^μλ) (dx^ν/λ) . It leads to the same equation.

∂L²/∂x^α = d(∂L²/∂x'^α)/dλ

The derivative gives:

2L∂L/∂x^α - d(2L∂L/∂x'^α)/d&lambda} = 0 or
2L{∂L/∂x^α - d(∂L/∂x'^α)/dλ} = 0

That is

∂L/∂x^α - d(∂L/∂x'^α)/dλ = 0

With

x'^μ = d x^μ/dλ

1.First term:

∂L/∂x^α = ∂[g_μν (dx'^μ) (dx'^ν)]/∂x^μ = dx'^μ dx'^ν ∂g_μν/∂x^α

2. Second term:

d(∂L/∂x'^α)/dλ

2.1.

∂L/∂x'^α = ∂[g_μν (dx'^μ) (dx'^ν)]/∂x'^α

We have:

∂ dx'^μ/∂ dx'^ν = δ(μν).
So d(∂L/∂x'^α)/dλ= g_αν (dx'^ν) + g_μα (dx'^μ)

2.2

d(∂L/∂x'^α)/dλ

We have

∂g_αν/dλ = (∂g_αν/∂x^β) (dx^β/dλ)

Therefore

d(∂L/∂x'^α)/dλ = (∂g_αμ/∂x^β)(dx^β/dλ) dx'^μ + (∂g_αν/x^β)(dx^β/dλ) dx'^ν + g_αν (dx'^ν/dλ) + g_μα (dx'^μ/dλ)
= (∂g_αμ/∂x^β)(x'^β) dx'^μ + (∂g_αν/x^β)(x'^β) dx'^ν + g_αν (x''^ν) + g_μα (x''^μ)

The two last terms are the same, so
d(∂L/∂x'^α)/dλ =
(∂g_αμ/∂x^β)(x'^β) dx'^μ + (∂g_αν/x^β)(x'^β) dx'^ν + 2g_αν (x''^ν)

3.Euler-Lagrange equation :

dx'^μ dx'^ν ∂g_μν/∂x^α =
(∂g_αμ/∂x^β)(x'^β) dx'^μ + (∂g_αν/x^β)(x'^β) dx'^ν + 2g_αν (x''^ν)

or

2g_αν (x''^ν) = dx'^μ dx'^ν ∂g_μν/∂x^α - (∂g_αμ/∂x^β)(x'^β) dx'^μ - (∂g_αν/x^β)(x'^β) dx'^ν

Multiplying the two hands of this equation by (1/2)g^ασ , we get by changing β in ν into the second term and β into μ in the third term:

x''^σ = (1/2)g^ασ { dx'^μ dx'^ν ∂g_μν/∂x^α - (∂g_αμ/∂x^ν)(x'^ν) dx'^μ - (∂g_αν/x^μ)(x'^μ) dx'^ν } =
(1/2)g^ασ { ∂g_μν/∂x^α - ∂g_αμ/∂x^ν - ∂g_αν/x^μ }dx'^μ dx'^ν

Or

x''^σ + (1/2)g^ασ { - ∂g_μν/∂x^α + ∂g_αμ/∂x^ν + ∂g_αν/x^μ }dx'^μ dx'^ν = 0

We have the following Christoffel symbol:

Γ^σ_μν = (1/2)g^ασ { ∂g_αμ/∂x^ν + ∂g_αν/x^μ - ∂g_μν/∂x^α }

That is

x''^σ + Γ^σ _μν x'^μ x'^ν = o

x''^σ + Γ^σ_μν x'^μ x'^ν = 0

Which is the geodesic equation.