Christoffel symbols: symmetry

Reference: Moore, Thomas A., A General Relativity Workbook, University Science Books (2013) – Chapter 17; Box 17.3.

The Christoffel symbols are defined in terms of the basis vectors in a given coordinate system as:

\displaystyle  \boxed{\frac{\partial\mathbf{e}_{i}}{\partial x^{j}}=\Gamma_{\; ij}^{k}\mathbf{e}_{k}} \ \ \ \ \ (1)

Remember that the basis vectors {\mathbf{e}_{i}} are defined so that

\displaystyle   ds^{2} \displaystyle  = \displaystyle  d\mathbf{s}\cdot d\mathbf{s}\ \ \ \ \ (2)
\displaystyle  \displaystyle  = \displaystyle  \left(dx^{i}\mathbf{e}_{i}\right)\cdot\left(dx^{j}\mathbf{e}_{j}\right)\ \ \ \ \ (3)
\displaystyle  \displaystyle  = \displaystyle  \mathbf{e}_{i}\cdot\mathbf{e}_{j}dx^{i}dx^{j}\ \ \ \ \ (4)
\displaystyle  \displaystyle  \equiv \displaystyle  g_{ij}dx^{i}dx^{j} \ \ \ \ \ (5)

In a locally flat frame using rectangular spatial coordinates, the basis vectors {\mathbf{e}_{i}} are all constants, so from 1, all the Christoffel symbols must be zero: {\Gamma_{\; ij}^{k}=0}.

Now let’s look at the second covariant derivative of a scalar field {\Phi}:

\displaystyle   \nabla_{i}\nabla_{j}\Phi \displaystyle  = \displaystyle  \nabla_{i}\left(\partial_{j}\Phi\right)\ \ \ \ \ (6)
\displaystyle  \displaystyle  = \displaystyle  \partial_{i}\partial_{j}\Phi-\Gamma_{ij}^{k}\partial_{k}\Phi \ \ \ \ \ (7)

where in 6 we used rule 1 for the covariant derivate: the covariant derivative of a scalar is the same as the ordinary derivative.

In the locally flat frame, this equation reduces to

\displaystyle  \nabla_{i}\nabla_{j}\Phi=\partial_{i}\partial_{j}\Phi \ \ \ \ \ (8)

Since the covariant derivative is a tensor, this is a tensor equation, and since ordinary partial derivatives commute, this equation is the same if we swap the indices {i} and {j}. Tensor equations must have the same form in all coordinate systems, so this implies that 7 must also be invariant if we swap {i} and {j}. This means that the Christoffel symbols are symmetric under exchange of their two lower indices:

\displaystyle  \boxed{\Gamma_{ij}^{k}=\Gamma_{ji}^{k}} \ \ \ \ \ (9)

At first glance, this seems wrong, since from the definition 1 this symmetry implies that

\displaystyle  \frac{\partial\mathbf{e}_{i}}{\partial x^{j}}=\frac{\partial\mathbf{e}_{j}}{\partial x^{i}} \ \ \ \ \ (10)

In 2-D polar coordinates, if we take the usual unit vectors {\hat{\mathbf{r}}} and {\hat{\boldsymbol{\theta}}} then both these vectors are constants as we change {r} and both of them change when we change {\theta}, so it’s certainly not true that {\partial\hat{\mathbf{r}}/\partial\theta=\partial\hat{\boldsymbol{\theta}}/\partial r}, for example. However, remember that the basis vectors we’re using are not the usual unit vectors; rather they are defined so that condition 4 is true. In polar coordinates, we have

\displaystyle  ds^{2}=dr^{2}+r^{2}d\theta^{2} \ \ \ \ \ (11)

so

\displaystyle   \mathbf{e}_{r} \displaystyle  = \displaystyle  \hat{\mathbf{r}}=\cos\theta\hat{\mathbf{x}}+\sin\theta\hat{\mathbf{y}}\ \ \ \ \ (12)
\displaystyle  \mathbf{e}_{\theta} \displaystyle  = \displaystyle  r\hat{\boldsymbol{\theta}}=-r\sin\theta\hat{\mathbf{x}}+r\cos\theta\hat{\mathbf{y}} \ \ \ \ \ (13)

For the derivatives, we have

\displaystyle   \frac{\partial\mathbf{e}_{r}}{\partial\theta} \displaystyle  = \displaystyle  -\sin\theta\hat{\mathbf{x}}+\cos\theta\hat{\mathbf{y}}=\hat{\boldsymbol{\theta}}\ \ \ \ \ (14)
\displaystyle  \frac{\partial\mathbf{e}_{\theta}}{\partial r} \displaystyle  = \displaystyle  -\sin\theta\hat{\mathbf{x}}+\cos\theta\hat{\mathbf{y}}=\hat{\boldsymbol{\theta}} \ \ \ \ \ (15)

Thus the condition 10 is actually satisfied here.

3 thoughts on “Christoffel symbols: symmetry

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