#### George Keeling

Gold Member

- 33

- 4

The something is that a second order partial derivative vanishes if one of the parts in the denominator is in the same reference frame as the numerator. That is for example

\begin{align}

\frac{{\partial }^2x^{\mu }}{\partial x^{\rho }\partial x^{{\mu }^{'}}}=\frac{{\partial }^2x^{\mu }}{\partial x^{{\mu }^{'}}\partial x^{\rho }}=0 \\

\end{align}The equality of the first two parts follows because partial derivatives commute. We have \begin{align}

\frac{{\partial }^2x^{\mu }}{\partial x^{{\mu }^{'}}\partial x^{\rho }}=\frac{\partial }{\partial x^{{\mu }^{'}}}\left(\frac{\partial x^{\mu }}{\partial x^{\rho }}\right) \\

\end{align}but\begin{align}

\frac{\partial x^{\mu }}{\partial x^{\rho }}={\delta }^{\mu }_{\rho } \\

\end{align}where ##{\delta }^{\mu }_{\rho }## is the Kronecker delta which is a constant. (3) seems very reasonable because when ##\mu \neq \rho ##, ##\partial x^{\mu }## and ##\partial x^{\rho }## are orthogonal so ##{\partial x^{\mu }}/{\partial x^{\rho }}## vanishes and when ##\mu =\rho ##, ##{\partial x^{\mu }}/{\partial x^{\rho }}=1##.

So equation (1) is the derivative of a constant which always vanishes. QED.

Have I made a very stupid mistake? Or am I stating something everybody knows?