The differential operator $d$ can be written as $$d=\partial_i dx^i$$ where $\partial_i\equiv\frac{\partial}{\partial x_i}$. If I have $d$ expressed in cartesian coordinate and I want to obtain the operator in terms of a new set of coordinate, polar for example ($x=r\cos\theta$ $y=r\sin\theta$), I was thinking to rewrite $$d=\partial_x dx+\partial_y dy$$ using the fact that $$\partial_x=\partial_xr\partial_r+\partial_x\theta\partial_\theta,\quad \partial_y=\partial_yr\partial_r+\partial_y\theta\partial_\theta$$ and that $$ dx=\partial_rxdr+\partial_\theta xd\theta,\quad dy=\partial_rydr+\partial_\theta yd\theta. $$ The substitution gives \begin{align} d&=A\partial_r dr+B\partial_\theta d\theta+C\partial_rd\theta+D\partial_\theta dr\\ &=(\partial_xr\partial_rx+\partial_yr\partial_ry)\partial_rdr+(\partial_x\theta\partial_\theta x+\partial_y\theta\partial\theta y)\partial_\theta d\theta+(\partial_xr\partial_\theta x+\partial_yr\partial_\theta y)\partial_rd\theta+(\partial_x\theta\partial_rx+ \partial_y\theta\partial_ry)\partial_\theta dr \end{align} and while I found correctly ($A=1$, $C=D=0$), my $B$ is different from the correct one $B=1\neq\frac{1}{r}$. I am using the following Jacobian \begin{bmatrix}\partial_rx=\cos\theta&\partial_\theta x=-r\sin\theta\\\partial_r y=\sin\theta&\partial_\theta y=r\cos\theta\end{bmatrix} and its inverse \begin{bmatrix}\partial_xr=\cos\theta&\partial_yr=\sin\theta\\\partial_x \theta=\frac{-\sin\theta}{r}&\partial_y\theta=\frac{\cos\theta}{r}\end{bmatrix}Where is my flaw?

  • $\begingroup$ Could you please write explicitly what you got and what you expected? $\endgroup$ – md2perpe Sep 7 '18 at 14:56
  • $\begingroup$ I've got B=1 while I expected B=1/r. $\endgroup$ – yngabl Sep 8 '18 at 8:35
  • $\begingroup$ $B=1$ is correct. I think that you are confusing $d$ with $\nabla$. $\endgroup$ – md2perpe Sep 8 '18 at 9:31
  • $\begingroup$ Probably you wanted to show that: $$\nabla \phi = \vec{e}_x \, \partial_x\phi + \vec{e}_y \, \partial_y\phi = \vec{e}_r \, \partial_r + \vec{e}_\theta \, \frac{1}{r} \partial_\theta$$ $\endgroup$ – md2perpe Sep 8 '18 at 10:13

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