While revisiting the transformation laws - under a diffeomorphism - of the partial derivative of a vector field I got confused by the following result. So, for a vector field $A$ the usual transformation law is given by,

$$ A^{\alpha'}_{,\beta'}=\partial_{\beta'} A^{\alpha'}= \frac{\partial x^ {\alpha'}}{\partial x^{\alpha}} \frac{\partial x^{\beta}}{\partial x^{\beta'}} A^\alpha_{,\beta} + \frac{\partial^2 x^{\alpha'}}{\partial x^{\alpha}\partial x^{\beta}} \frac{\partial x^{\beta}}{\partial x^{\beta'}} A^\alpha~. $$

However, for the last term on the RHS, using the properties of the partial derivatives:

\begin{equation} \begin{aligned} \frac{\partial^2 x^{\alpha'}}{\partial x^{\alpha}\partial x^{\beta}} \frac {\partial x^{\beta}}{\partial x^{\beta'}} A^\alpha & = \frac{\partial}{\partial x^\beta}\left(\frac{\partial x^{\alpha'}}{\partial x^{\alpha}}\right) \frac{\partial x^ {\beta}}{\partial x^{\beta'}} A^\alpha \\ & =\frac{\partial}{\partial x^{\beta'}}\left(\frac{\partial x^{\alpha'}}{\partial x^{\alpha}}\right) A^\alpha \\ & = \frac{\partial}{\partial x^\alpha}\left(\frac{\partial x^{\alpha'}}{\partial x^{\beta'}}\right) A^\alpha \\ & = \frac{\partial \delta^{\alpha'}_{\beta'}}{\partial x^\alpha} A^\alpha =0~, \end{aligned} \end{equation}

where $\delta^{\alpha'}_{\beta'}$ is the Kronecker delta and in the last equality it was used the fact that its derivative is zero.

Now, I know that this is wrong but I don't know where, could somebody help me find the error?


1 Answer 1


Hint: Derivatives $\frac{\partial}{\partial x^{\alpha}}$ wrt. the unprimed coordinate system and derivatives $$\frac{\partial}{\partial x^{\prime\beta}}~=~\frac{\partial x^{\gamma}}{\partial x^{\prime \beta}}\frac{\partial}{\partial x^{\gamma}} $$ wrt. the primed coordinate system do not necessarily commute $$\left[\frac{\partial}{\partial x^{\alpha}},\frac{\partial}{\partial x^{\prime\beta}}\right]~=~\left[\frac{\partial}{\partial x^{\alpha}},\frac{\partial x^{\gamma}}{\partial x^{\prime \beta}}\right]\frac{\partial}{\partial x^{\gamma}},$$ since the Jacobian matrix $\frac{\partial x^{\gamma}}{\partial x^{\prime \beta}}$ may be non-constant.

  • 1
    $\begingroup$ I think your completely correct but rather laconic answer should be a bit more verbose for clarification. $\endgroup$
    – hyportnex
    Jan 12, 2019 at 12:26
  • 1
    $\begingroup$ I updated the answer. $\endgroup$
    – Qmechanic
    Jan 12, 2019 at 12:43

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.