# Do partial derivatives of different coordinate systems commute?

Consider an arbitrary set of coordinates $$x^\mu$$ and another set of coordinates $$y^{\mu}$$, which is a (lorentzian) transformation from $$x^\mu$$ given by $$y^\mu = f(x^\mu)$$.

So I want to know whether $$\frac{\partial}{\partial x^\alpha}\frac{\partial}{\partial y^\beta} = \frac{\partial}{\partial y^\beta}\frac{\partial}{\partial x^\alpha}$$ holds true or false?

• Relevant: Clairuts theorem – Buraian May 8 at 11:59

## 2 Answers

It depends on the transformation at hand, but in general the answer is no. It boils down to whether $$x^\alpha$$ can change while $$y^\beta$$ is kept constant.

Denote by $$J_{\nu}^\mu(x) = \frac{\partial y^\mu}{\partial x^\nu}$$ the Jacobian of the change of coordinates.

By the chain rule

$$\frac{\partial}{\partial x^\alpha} = J_{\alpha}^\nu(y)\frac{\partial}{\partial y^\nu}$$

It is not difficult to see that it makes a difference whether this expression is acted upon by $$\partial_{y^\beta}$$ from the left or the right. The former generates an additional term

$$\frac{\partial J_{\alpha}^\nu(y)}{\partial y^\beta} \frac{\partial}{\partial y^\nu}$$ which needn't vanish.

No they don't commute in general. At least not with the usual understanding that $$\frac{\partial}{\partial x^2}$$ is the partial derivative with $$x^1$$, $$x^3$$ etc held fixed.

Here is a counterexample: Let $$x= r \cos \theta$$, $$y= r \sin \theta$$ be cartesian and polar coordinates. Then
$$\frac{\partial x}{\partial y}=0\Rightarrow \frac{ \partial}{\partial r} \frac{\partial x}{\partial y}=0.$$ but $$\frac{\partial x}{\partial r}\equiv \left(\frac{\partial x}{\partial r}\right)_\theta= \cos \theta$$ is the derivative with $$\theta$$ being held fixed. Now $$\frac {\partial} {\partial y} \frac{\partial x}{\partial r}= \frac {\partial} {\partial y}\cos \theta =\frac {\partial} {\partial y}\frac{x}{\sqrt{x^2+y^2}}\ne 0.$$ so $$\left(\frac {\partial} {\partial y}\left( \frac{\partial x}{\partial r}\right)_\theta\right)_x \ne \left(\frac {\partial} {\partial r}\left( \frac{\partial x}{\partial y}\right)_x\right)_\theta,$$ where I have made it explicit what is being held fixed for each derivative.

It may be that they commute if the transformation is linear, but even then I have doubts. Why don't you try and see if it it's OK in this restricted case?

• Yes, indeed, in the case of linear transformation, it is commuting. But I guess It is not a Lorentzian transformation in my case. Thanks anyways – Vikash Kotteeswaran May 8 at 15:48