The partial derivative of a tensor of rank $n$, $T_{...i}$, with respect to $x^j$ can be expressed using the transformation rule:

\begin{equation} \frac{\partial}{\partial x^j}T'_{...i}=\frac{\partial}{\partial x^j}\sum_{...k}...\frac{\partial x^k}{\partial x'^i}T_{...k} \end{equation}

Since the derivative is linear:

\begin{equation} \frac{\partial}{\partial x^j}T'_{...i}=\sum\frac{\partial}{\partial x^j}(...\frac{\partial x^k}{\partial x'^i}T_{...k}) \end{equation}

If I'm correct, applying the product rule and knowing that $\frac{\partial x^i}{\partial x^j}=\delta^i_j$ yields:

\begin{equation} \frac{\partial}{\partial x^j}T'_{...i}=\sum_{...k}...\frac{\partial x^k}{\partial x'^i}\frac{\partial T_{...k}}{\partial x^j} \end{equation}

I'm having trouble interpreting the result. I believe it's not even correct. What's the $\frac{\partial T_{...k}}{\partial x^j}$ factor?

  • $\begingroup$ You mean the covariant derivative. $\endgroup$
    – J.G.
    Jul 25 at 6:28

I think the way to proceed is:

$\frac{\partial}{\partial x^j} T'_{i}=\frac{\partial}{\partial x^j} \left(\frac{\partial x^s}{\partial x'^i}T_{s}\right)=\frac{\partial}{\partial x^j} \left(\frac{\partial x^s}{\partial x'^i}\right)T_{s}+\frac{\partial x^s}{\partial x'^i}\frac{\partial T_{s}}{\partial x^j}=\frac{\partial x'^q}{\partial x^j} \,\frac{\partial^2 x^s}{\partial x'^i \partial x'^q}\,T_{s}+\frac{\partial x^s}{\partial x'^i}\frac{\partial T_{s}}{\partial x^j}$

I don't think you get cancellation there. Casing point. Polar and Cartesian coordinates in 2d:

$\frac{\partial}{\partial \phi}\,\frac{\partial r}{\partial x}=\frac{\partial}{\partial \phi}\,\cos\phi=-\sin\phi\neq0$


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