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

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