- #1

PWiz

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- 116

I've attempted to derive an expression for the Christoffel symbols (of the 2nd kind) solely in terms of the covariant and contravariant forms of the metric by only using the definition of the Christoffel symbols. I would like to know if my approach is correct or not.

The Christoffel symbols are defined as ##∂_{μ} e_ν = Γ^{β} _{νμ}##.(I'm using the natural basis here, not the normalized ones.)

EDIT: I forgot to put the basis on the right. It should read ##∂_{μ} e_ν = Γ^{β} _{νμ} e_β## .

The basis on a point P are defined as the tangents to the coordinate curves at that point. So ##e_ν = \frac{∂P}{∂x^{v}}##, so we get ##\frac{∂^2 P}{∂x^{μ} x^{ν}} = Γ^{β} _{νμ}## .

Since I'm doing this calculation for spacetime, I know that I'm dealing with a (pseudo) Riemannian manifold, which implies that all points on it can be continuously parametrized and that the manifold is ##C^∞##(except at a singularity). These are all the necessary conditions for Clairaut's theorem, so I can exchange the partial derivatives: ##\frac{∂^2 P}{∂x^{μ} x^{ν}} = \frac{∂^2 P}{∂x^{ν} x^{μ}}=∂_ν e_μ##.

But this means that the Christoffel symbols are symmetric in their lower indices: ##Γ^{β} _{νμ} = Γ^{β} _{μν}##.

Okay, so let's calculate some derivatives of the metric tensor now. ##∂_σ g_{μν} = ∂_σ ( e_μ⋅e_ν) = Γ^{β} _{μσ} g_{βν} + Γ^{α} _{νσ} g_{αμ}##, but ##α## is a dummy index, so I can just relabel it back to ##β##.

If I go ahead and permute the lower indices, I get ##∂_μ g_{vσ} = Γ^{β} _{νμ} g_{βσ} + Γ^{β} _{σμ} g_{βν}## and ##∂_ν g_{σμ} = Γ^{β} _{σν} g_{βμ} + Γ^{β} _{μν} g_{βσ}## .

I can exploit the symmetry of the lower indices of the Christoffel symbols and add ##∂_μ g_{vσ}## to ##∂_σ g_{μν}## to get ##2 Γ^{β} _{σμ} g_{βν} + Γ^{β} _{νσ} g_{βμ} + Γ^{β} _{νμ} g_{βσ}## , and then subtract ##∂_ν g_{σμ}## from this to get ##\frac{1}{2} (∂_μ g_{vσ}+∂_σ g_{μν}-∂_ν g_{σμ}) = Γ^{β} _{σμ} g_{βν}##.

To remove the metric on the RHS, I can just multiply both sides by ##g^{γν}## , so that ##\frac{1}{2} g^{γν} (∂_μ g_{vσ}+∂_σ g_{μν}-∂_ν g_{σμ}) = Γ^{β} _{σμ} g_{βν} g^{γν} = Γ^{β} _{σμ} \delta ^{γ} _{β} = Γ^{γ} _{σμ}##,

which is what is required (I think).

Any mess-ups?

The Christoffel symbols are defined as ##∂_{μ} e_ν = Γ^{β} _{νμ}##.(I'm using the natural basis here, not the normalized ones.)

EDIT: I forgot to put the basis on the right. It should read ##∂_{μ} e_ν = Γ^{β} _{νμ} e_β## .

The basis on a point P are defined as the tangents to the coordinate curves at that point. So ##e_ν = \frac{∂P}{∂x^{v}}##, so we get ##\frac{∂^2 P}{∂x^{μ} x^{ν}} = Γ^{β} _{νμ}## .

Since I'm doing this calculation for spacetime, I know that I'm dealing with a (pseudo) Riemannian manifold, which implies that all points on it can be continuously parametrized and that the manifold is ##C^∞##(except at a singularity). These are all the necessary conditions for Clairaut's theorem, so I can exchange the partial derivatives: ##\frac{∂^2 P}{∂x^{μ} x^{ν}} = \frac{∂^2 P}{∂x^{ν} x^{μ}}=∂_ν e_μ##.

But this means that the Christoffel symbols are symmetric in their lower indices: ##Γ^{β} _{νμ} = Γ^{β} _{μν}##.

Okay, so let's calculate some derivatives of the metric tensor now. ##∂_σ g_{μν} = ∂_σ ( e_μ⋅e_ν) = Γ^{β} _{μσ} g_{βν} + Γ^{α} _{νσ} g_{αμ}##, but ##α## is a dummy index, so I can just relabel it back to ##β##.

If I go ahead and permute the lower indices, I get ##∂_μ g_{vσ} = Γ^{β} _{νμ} g_{βσ} + Γ^{β} _{σμ} g_{βν}## and ##∂_ν g_{σμ} = Γ^{β} _{σν} g_{βμ} + Γ^{β} _{μν} g_{βσ}## .

I can exploit the symmetry of the lower indices of the Christoffel symbols and add ##∂_μ g_{vσ}## to ##∂_σ g_{μν}## to get ##2 Γ^{β} _{σμ} g_{βν} + Γ^{β} _{νσ} g_{βμ} + Γ^{β} _{νμ} g_{βσ}## , and then subtract ##∂_ν g_{σμ}## from this to get ##\frac{1}{2} (∂_μ g_{vσ}+∂_σ g_{μν}-∂_ν g_{σμ}) = Γ^{β} _{σμ} g_{βν}##.

To remove the metric on the RHS, I can just multiply both sides by ##g^{γν}## , so that ##\frac{1}{2} g^{γν} (∂_μ g_{vσ}+∂_σ g_{μν}-∂_ν g_{σμ}) = Γ^{β} _{σμ} g_{βν} g^{γν} = Γ^{β} _{σμ} \delta ^{γ} _{β} = Γ^{γ} _{σμ}##,

which is what is required (I think).

Any mess-ups?

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