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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?

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  • $\begingroup$ Relevant: Clairuts theorem $\endgroup$ – Buraian May 8 at 11:59
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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.

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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?

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  • $\begingroup$ 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 $\endgroup$ – Vikash Kotteeswaran May 8 at 15:48

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