Thursday, 14 March 2019

Second order partial derivatives vanish?

At the end of a long proof I came across something in tensor calculus that seems too good to be true. And if something seems too good to be true .....

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
\frac{{\partial }^2x^{\mu }}{\partial x^{\rho }\partial x^{{\mu }^{'}}}=\frac{{\partial }^2x^{\mu }}{\partial x^{{\mu }^{'}}\partial x^{\rho }}=0 & \phantom {10000}(1) \\
\end{align}The equality of the first two parts follows because partial derivatives commute. We have
\frac{{\partial }^2x^{\mu }}{\partial x^{{\mu }^{'}}\partial x^{\rho }}=\frac{\partial }{\partial x^{{\mu }^{'}}}\left(\frac{\partial x^{\mu }}{\partial x^{\rho }}\right) & \phantom {10000}(2) \\
\frac{\partial x^{\mu }}{\partial x^{\rho }}={\delta }^{\mu }_{\rho } & \phantom {10000}(3) \\
\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 (2) is the derivative of a constant which always vanishes. QED.

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

On PF at
The good stevendaryl pointed out my error and gave a little example that proved$$
\frac{{\partial }^2x^{\mu }}{\partial x^{\rho }\partial x^{{\mu }^{'}}}≠\frac{{\partial }^2x^{\mu }}{\partial x^{{\mu }^{'}}\partial x^{\rho }}
$$Partial derivatives only commute if they are in the the same coordinate system, so this IS true$$
\frac{{\partial }^2x^{\mu }}{\partial x^{\rho }\partial x^{{\nu }}}=\frac{{\partial }^2x^{\mu }}{\partial x^{{\nu }}\partial x^{\rho }}$$Say we had$$
f\left(x,y\right)=x^2{\mathrm{sin} y\ }
\frac{\partial }{\partial x}\left(\frac{\partial }{\partial y}f\right)=\frac{\partial }{\partial x}\left(x^2{\mathrm{cos} y\ }\right)=2x{\mathrm{cos} y\ }
\frac{\partial }{\partial y}\left(\frac{\partial }{\partial x}f\right)=\frac{\partial }{\partial y}\left(2x{\mathrm{sin} y\ }\right)=2x{\mathrm{cos} y\ }$$Say we had
$$x^{\mu }=z$$then $$\frac{\partial }{\partial x}\left(\frac{\partial }{\partial y}z\right)=0$$and
$$\frac{\partial }{\partial y}\left(\frac{\partial }{\partial x}z\right)=0$$Say we had$$x^{\mu }=y
\frac{\partial }{\partial x}\left(\frac{\partial }{\partial y}y\right)=\frac{\partial }{\partial x}\left(1\right)=0$$and$$
\frac{\partial }{\partial y}\left(\frac{\partial }{\partial x}y\right)=\frac{\partial }{\partial y}\left(\frac{\partial }{\partial x}y\right)=0$$

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