Second order partial derivatives vanish?

[George Keeling]

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
\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?

 

[stevendaryl]

The equality of the first two parts follows because partial derivatives commute.

This principle needs to be used with caution. For a fixed coordinate system, if $x$ and $y$ are two different coordinates, then it is definitely the case that $\frac{\partial}{\partial x} \frac{\partial}{\partial y} f(x,y) = \frac{\partial}{\partial y} \frac{\partial}{\partial x} f(x,y)$

However, if $x$ and $y$ are from DIFFERENT coordinate systems, then that isn't necessarily true. Here's a trivial example. Consider one dimension, so there is only one coordinate, $x$. We can switch to another coordinate system $x' = \sqrt{x}$. Consider the function $cos(x)$

Take the derivative with respect to $x$. You get $-sin(x)$.
Re-express it in terms of $x'$. You get $-sin((x')^2)$.
Take the derivative with respect to $x'$. You get $-2 x' cos((x')^2) = -2 \sqrt{x} cos(x)$.

Now do it in the opposite order.
Re-express $cos(x)$ in terms of $x'$. You get $cos((x')^2)$.
Take the derivative with respect to $x'$. You get $-2x' sin((x')^2)$.
Re-express it in terms of $x$. You get $-2\sqrt{x} sin(x)$.
Take the derivative with respect to $x$. You get $-\frac{1}{\sqrt{x}} sin(x) - 2\sqrt{x} cos(x)$.

Those aren't the same.

 

[George Keeling]

This principle needs to be used with caution.

Oh dear. I'll have to think about cautions tomorrow.

 

[George Keeling]

This principle needs to be used with caution.

Thanks for taking the trouble to work the example. It teaches me two things:
1) Apply that caution
2) Try an example when I get another 'to good to be true result'
Following rule (1) we believe$$
\frac{{\partial }^2f}{\partial x^{\rho }\partial x^{{\nu }}}=\frac{{\partial }^2f}{\partial x^{{\nu }}\partial x^{\rho }}
$$Following rule (2) test$$
f\left(x,y\right)=x^2{\mathrm{sin} y\ }
$$then $$
\frac{\partial }{\partial x}\left(\frac{\partial }{\partial y}f\right)=\frac{\partial }{\partial x}\left(x^2{\mathrm{cos} y\ }\right)=2x{\mathrm{cos} y\ }
$$and$$
\frac{\partial }{\partial y}\left(\frac{\partial }{\partial x}f\right)=\frac{\partial }{\partial y}\left(2x{\mathrm{sin} y\ }\right)=2x{\mathrm{cos} y\ }
$$That works!
My question was really when $f=x^\mu$ which is a coordinate. So I should test $f=y$ and $f=z$. The second order derivatives both rapidly vanish. I mow need to revise my rule to:
The second order derivative of a coordinate by two other coordinates in the same reference frame vanishes. Not so impressive.

I will now now have to rework my problem
Once again stevendaryl, thanks for the help you have given with with this and other problems.