Deriving the transformation law for the Christoffel symbols

Question

I am a first year undergraduate teaching myself General Relativity from the book by Bernard Schutz. In one of the problems he asks to derive the transformation law for the Christoffel symbols from the definition:

\begin{equation} \Gamma^{\mu}_{\alpha\beta} \vec{e}_{\mu} = \frac{\partial \vec{e}_{\alpha}}{\partial x^\beta}.\tag{1} \end{equation} After a lot algebra and using the transformation laws for the 'known' quantities, I arrived at: \begin{equation} \Gamma^{\mu '}_{\alpha '\beta '} = \frac{\partial x^{\mu'}}{\partial x^{\gamma}}\frac{\partial^2x^\gamma}{\partial x^{\beta '}\partial x^{\alpha '}}+\frac{\partial x^{\mu'}}{\partial x^\gamma}\frac{\partial x^{\sigma}}{\partial x^{\alpha '}}\frac{\partial x^p}{\partial x^{\beta '}}\Gamma^{\gamma}_{\sigma p}.\tag{2} \end{equation} Now Wikipedia says the order of partial derivatives is swapped in the first term. But the expression that I am getting through first principles is the one above. However Schutz mentions towards the end of the chapter that for a non-coordinate basis the definition of the Christoffel symbol changes and hence the one that was used to derive the above is only valid for coordinate basis in which the partial derivatives can be swapped. So by that assumption, since the basis for this derivation is coordinate, I can swap the order of the partial derivatives above and obtain the expression: \begin{equation} \Gamma^{\mu '}_{\alpha '\beta '} = \frac{\partial x^{\mu'}}{\partial x^{\gamma}}\frac{\partial^2x^\gamma}{\partial x^{\alpha '}\partial x^{\beta '}}+\frac{\partial x^{\mu'}}{\partial x^\gamma}\frac{\partial x^{\sigma}}{\partial x^{\alpha '}}\frac{\partial x^p}{\partial x^{\beta '}}\Gamma^{\gamma}_{\sigma p}.\tag{3} \end{equation} which is the one in standard texts and Wikipedia. Am I correct in reasoning so? I apologize if I sound naive as it is my first time learning the material.

Answer 1

Yes, in physics all functions are so well behaved that you can swap partial derivatives: $$ \frac{\partial^2 x^\gamma}{\partial x^{\beta'}\partial x^{\alpha'}}= \frac{\partial^2 x^\gamma}{\partial x^{\alpha'}\partial x^{\beta'}}\,. $$

Answer 2

According to the Schwarz's theorem, any function on $n$-dimensional real space that has continuous second derivatives in every variable also has symmetric second partial derivatives: $$f\in C^2(\mathbb{R}^n) \Rightarrow \frac{\partial^2 f}{\partial x^\mu \partial x^\nu} = \frac{\partial^2 f}{\partial x^\nu \partial x^\mu}$$ where $x^\mu$ are coordinates on $\mathbb{R}^n$ here. The coordinate transform $x'^\mu(x^\kappa)$ (and its inverse) can be viewed as such a function $f$ component by component. That is, if we assume that the coordinate transforms are $C^2$ we can switch the derivatives. Generally, it is customary in physics to assume that any function appearing in your expressions is smooth enough so that any partial derivatives in your formulas can be swapped. (There are exceptions, and those are handled case by case.)