A Second order partial derivatives vanish?

George Keeling

Gold Member
37
4
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

Staff Emeritus
Science Advisor
Insights Author
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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

Gold Member
37
4
This principle needs to be used with caution.
Oh dear. I'll have to think about cautions tomorrow.
 

George Keeling

Gold Member
37
4
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.:frown:

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

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