# Christoffel symbols derivation

1. Nov 10, 2015

### PWiz

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?

Last edited: Nov 10, 2015
2. Nov 10, 2015

### Geofleur

Your first equation should be $\partial_\mu e_\nu = \Gamma^{\beta}_{\nu \mu} e_\beta$. Otherwise, you have a vector on the left hand side and only components on the other. A free index on only one side of an equation always signals that something is amiss. Also, the main idea is that we are operating on a basis vector, $e_\nu$, and then expanding the result in terms of the same basis. This idea is very powerful and gets used in lots of other areas of physics.

3. Nov 10, 2015

### PWiz

Whoops, forgot the basis on the right! Thanks.
Can you elaborate on this please?

4. Nov 10, 2015

### Geofleur

What makes the Christoffel symbols necessary is that the basis vectors themselves are functions of position. That's why in Cartesian coordinates, with their standard basis, the issue doesn't arise - $\mathbf{i}$, $\mathbf{j}$, and $\mathbf{k}$ are the same everywhere. If we set $\mathbf{i} = \mathbf{e}_1$, $\mathbf{j} = \mathbf{e}_2$, and $\mathbf{k} = \mathbf{e}_3$, we can write $\frac{\partial \mathbf{e}_i}{\partial x^j} = 0$. That way, for an arbitrary vector field $\mathbf{v}(x,y,z)$ we can take its derivative as $\frac{\partial \mathbf{v}}{\partial x^j} = \frac{\partial}{\partial x^j}\left(v^i \mathbf{e}_i\right) = \left(\frac{\partial v^i}{\partial x^j}\right) \mathbf{e}_i$.

In contrast, suppose we wanted to use spherical coordinates, with their basis $\mathbf{e}_r$, $\mathbf{e}_{\phi}$, and $\mathbf{e}_{\theta}$. Since $\mathbf{e}_r$ varies from point to point, taking the derivative of it will yield, not zero, but another vector, call it $\mathbf{w}$, so that $\frac{\partial }{\partial x^j}\mathbf{e}_r = \mathbf{w}$. We can express $\mathbf{w}$ in terms of the other spherical basis vectors, $\frac{\partial }{\partial x^j}\mathbf{e}_r = \mathbf{w} = w^i \mathbf{e}_i = w^r \mathbf{e}_r + w^{\phi} \mathbf{e}_\phi + w^{\theta} \mathbf{e}_{\theta }$. But notice, the values of the $w^j$ will depend on which coordinate we are taking the derivative with respect to as well as which basis vector we are taking the derivative of, so we should really label them with subscripts $j$ and $r$, as in $\frac{\partial }{\partial x^j}\mathbf{e}_r = \mathbf{w} = w^i_{rj} \mathbf{e}_i$. But we've ended up with none other than the Christoffel symbols corresponding to the $\mathbf{e}_r$ basis vector! That is to say, $w^i_{rj} = \Gamma^{i}_{rj}$.

5. Nov 10, 2015

### PWiz

Hmmm, makes sense! But can we just go ahead and "add" indices to expressions? I don't expect it to be a legal mathematical operation in the general case.

6. Nov 10, 2015

### Geofleur

You can always add indices as a notational convenience, you just may not end up with a vector or, more generally, a tensor. The Christoffel symbols themselves, for example, are not components of a tensor.