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I need to show that the del operator in 2D polar coordinates is $\nabla=e_r\partial_r+\frac{1}{r}e_r+\frac{1}{r}e_{\phi}\partial_{\phi}$. I try the following approach:

$\nabla=\partial_xe_x+\partial_ye_y=\bigg(\frac{\partial r}{\partial x}\partial_r+\frac{\partial \phi}{\partial x}\partial_{\phi}\bigg)\bigg(\cos(\phi)e_r-\sin(\phi)e_{\phi}\bigg)+\bigg(\frac{\partial r}{\partial y}\partial_r+\frac{\partial \phi}{\partial y}\partial_{\phi}\bigg)\bigg(\sin(\phi)e_r+\cos(\phi)e_{\phi}\bigg) $

and then use $r=\sqrt{x^2+y^2}$, $y=\arctan(y/x)$, $\partial_{\phi}e_r=e_{\phi}$ and $\partial_{\phi}e_{\phi}=-e_r$.

The problem is, I end up with $\nabla=e_r\partial_r+\frac{1}{r}e_{\phi}\partial_{\phi}$, missing a term compared to the expression above. Any ideas?

Best, Jorgen

EDIT: spelling

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    $\begingroup$ I think a good idea would be to find $\frac{\partial (x,y)}{\partial (r,\theta)}$. $\endgroup$ – Theorem Jul 19 '12 at 18:29
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You have to remember not to actually apply the partials unless you've fed a function into $\nabla$. Here's what I mean: In rectangular coordinates, you feed a function $f$ to $\nabla$ by doing $\nabla f = (\partial_xf)e_x + (\partial_yf)e_y$. If the partials had already been applied to $e_x$ and $e_y$, then $\nabla$ would just be identically zero, and so would $\nabla f$. In your case, you can't apply the partials to $e_r$ and $e_\phi$ --- doing so gets rid of the partials, leaving $\nabla$ as a simple vector function, which is not what we want.

To make sure that I don't accidentally fall into this trap when working with $\nabla$, partials, and other differential operators, I tend to follow David Griffiths' advice from his intro quantum book: Always apply the operator to a dummy function until you're done working with it. For you're case, this means, instead of determining $\nabla$ itself in polar coordinates, you determine $\nabla f$ in polar coordinates, where $f$ is a "dummy function". Formally, these two approaches are identical, but the latter helps make sure we don't make mistakes.

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As you know, the operator $\nabla$ is defined by $$\nabla\equiv i\partial_x+j\partial_y+k\partial_z$$ It is a vector operator so, it needs to have $\partial_{.}$ terms. Your calculation above is right in 2D polar system and I think something is missing in your original expression where you wrote $\frac{1}{r}e_r$. In fact you don't need to have such this term when you want to calculate del operator in 2D polar coordinates.

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  • $\begingroup$ Nice job, Babak! + $\endgroup$ – amWhy Mar 12 '13 at 1:49

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