Why do relativisitic partial derivatives commute?

Question

\begin{eqnarray*} F_{\mu\nu} &=& \partial_\mu (A_\nu(x)) - \partial_\nu (A_\mu(x)) \\ &\rightarrow& \partial_\mu (A_\nu(x)+ \partial_\nu(\alpha(x))) - \partial_\nu (A_\mu(x)+ \partial_\mu(\alpha(x))) \\ &=& \partial_\mu (A_\nu(x))+ \partial_\mu\partial_\nu(\alpha(x)) - \partial_\nu (A_\mu(x)) - \partial_\nu\partial_\mu(\alpha(x)) \\ &=& \partial_\mu (A_\nu(x)) - \partial_\nu (A_\mu(x)) \\ &=& F_{\mu\nu} \end{eqnarray*}

Why is $\partial_\mu\partial_\nu(\alpha) = \partial_\nu\partial_\mu(\alpha)$?

Why is the mixed partials rule from high school math,

$$\frac{\partial}{\partial x}\frac{\partial}{\partial y} f= \frac{\partial}{\partial y}\frac{\partial}{\partial x} f$$

true?

Answer 1

They are symmetric, the last term in third line shoud be $−\partial_\nu\partial_\mu\alpha$, you forgot the minus sign

Answer 2

Remember that $$\partial_\mu = {\partial \over \partial x^\mu}$$ and that $x$ corresponds to $x^1$, $y$ to $x^2$, so there is no difference here (apart from notation) from high school maths.

As for why it is true, this is Clairaut's theorem, or Schwarz's theorem. It holds provided that second partial derivatives are continuous, and it is a minor pain to prove. I confess, I do think the proof I give in my third book is simpler and more elegant, but it still seems like too much effort to transcribe it into mathjax as the theorem is so natural that it is more surprising that there are functions (without continuous second partial derivatives) for which it does not hold.

Answer 3

This is a theorem in elementary analysis, often called Clairut's theorem or the Schwarz theorem:

Let $U \subseteq \mathbb{R^n}$ be open and $f: \mathbb{R^n} \to \mathbb{R^m}$ be twice differentiable. Then $$ \frac{\partial^2 f}{\partial x \partial y} = \frac{\partial^2 f}{\partial y \partial x}.$$

A proof can be found here. Its essential component is applying the mean value inequality to a suitable choosen function. Note that the theorem is wrong if the function is only partially differentiable twice where the partial derivatives are not continuous. A counterexample can be found on the Wikipedia page as well. Note that if the function is twice differentiable and the derivatives are discontinuos it remains true.

In physics most of the time we assume that all involved functions are smooth unless stated otherwise, so no problem arises.