# Christoffel symbols derivation

• PWiz
In summary: 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} ##.In summary, the Christoffel symbols are necessary when working in a curved coordinate system, where the basis vectors vary from point to point. By operating on a basis vector and expanding the
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:
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.

Geofleur said:
Your first equation should be ## \partial_\mu e_\nu = \Gamma^{\beta}_{\nu \mu} e_\beta ##.
Whoops, forgot the basis on the right! Thanks.
Geofleur said:
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.
Can you elaborate on this please?

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} ##.

Chestermiller
Geofleur said:
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} ##.
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.

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.

PWiz

## 1. What are Christoffel symbols and what are they used for?

Christoffel symbols are a set of mathematical objects used in the field of differential geometry. They are used to represent the curvature and connection properties of a manifold, which allows for the study of geometric phenomena such as gravity and electromagnetism.

## 2. How are Christoffel symbols derived?

The derivation of Christoffel symbols involves the use of a coordinate system and the calculation of partial derivatives. The symbols are calculated using the metric tensor, which describes the distance between points in a manifold.

## 3. What is the significance of the Christoffel symbol notation?

The notation for Christoffel symbols, which uses Greek letters and superscripts, is a compact and efficient way to represent the complex mathematical relationships involved in differential geometry. It allows for easier manipulation and analysis of these relationships.

## 4. Can Christoffel symbols be calculated for any type of manifold?

Yes, Christoffel symbols can be calculated for any type of manifold with a defined metric tensor. This includes both curved and flat manifolds, as well as higher-dimensional spaces.

## 5. What are some applications of Christoffel symbols in physics?

Christoffel symbols are commonly used in physics, particularly in the study of General Relativity. They are used to calculate the geodesic equation, which describes the path of a particle under the influence of gravity. They are also used in the calculation of the Einstein field equations, which describe the curvature of spacetime due to the presence of mass and energy.

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