# Div and curl in other coordinate systems

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1. Feb 23, 2016

### Jezza

My question is mostly about notation. I know the general definitions for divergence and curl, which can be derived from the divergence and Stokes' theorems respectively, are:

$$\mathrm{div } \vec{E} \bigg| _P = \lim_{\Delta V \to 0} \frac{1}{\Delta V} \iint_{S} \vec{E} \cdot \mathrm{d} \vec{S}$$

Where $S$ is the closed surface bounding $V$, and $V$ is a volume whose limit contains the point $P$.

$$\left(\mathrm{curl } \vec{E}\right) \cdot \vec{n} \bigg|_P= \lim_{\Delta S \to 0} \frac{1}{\Delta S} \iint_{\ell} \vec{E} \cdot \mathrm{d} \vec{\ell}$$.

Where $\vec n$ is the unit normal to the closed contour $\ell$ which bounds the surface $S$ whose limit surrounds the point $P$.

I also know that in cartesian coordinates, when we define the del operator $\vec \nabla$, these definitions are equivalent to: (Including $\mathrm{grad} \phi$ where $\phi$ is a scalar field.)
\begin{align} &&\vec \nabla &= \left(\frac{\partial}{\partial x}, \frac{\partial}{\partial y}, \frac{\partial}{\partial z}\right) \nonumber \\ \nonumber \\ \mathrm{grad} \phi &=& \vec \nabla \phi &= \left(\frac{\partial \phi}{\partial x}, \frac{\partial \phi}{\partial y}, \frac{\partial \phi}{\partial z}\right) \nonumber \\ \nonumber \\ \mathrm{div } \vec{E} &=& \vec\nabla \cdot \vec E &= \frac{\partial E_x}{\partial x} + \frac{\partial E_y}{\partial y} + \frac{\partial E_z}{\partial z} \nonumber \\ \nonumber \\ \mathrm{curl } \vec{E} &=& \vec\nabla \times \vec E &= \left(\frac{\partial E_z}{\partial y} - \frac{\partial E_y}{\partial z}, \frac{\partial E_x}{\partial z} - \frac{\partial E_z}{\partial x}, \frac{\partial E_y}{\partial x} - \frac{\partial E_x}{\partial y}\right) \nonumber \end{align}
In this case, the right hand sides can be reached from the left hand sides simply by applying vector methods, which (at least as I was led to believe) is the beauty of the $\vec \nabla$ notation. I'm also perfectly happy and understand that the definition of $\vec \nabla$ changes with coordinate system. For example, in spherical polar coordinates $(r, \theta, \varphi)$:
$$\vec \nabla = \left(\frac{\partial}{\partial r}, \frac{1}{r} \frac{\partial}{\partial \theta}, \frac{1}{r \sin\theta} \frac{\partial}{\partial \varphi}\right)$$
The problem as I see it comes when you try and work out div, grad and curl in other coordinate systems. grad can be reached using standard vector methods, but the divergence in spherical polars is:

$$\mathrm{div } \vec{E} = \frac{1}{r^2} \frac{\partial}{\partial r}\left(E_r r^2\right) + \frac{1}{r \sin \theta} \frac{\partial}{\partial \theta}\left(E_\theta \sin \theta \right) + \frac{1}{r \sin \theta}\frac{\partial E_\varphi}{\partial \varphi}$$

Which clearly cannot be reached simply using vector methods. What I'm trying to get at is that despite $\vec \nabla \cdot \vec E \neq \mathrm{div} \vec E$ (in traditional vector notation), we still write it! It's a similar story for curl.

It's a similar situation with cylindrical coordinates, so I assume the ability to blindly apply vector methods when calculating div and curl in cartesian coordinates is a special case. I suppose my question is why do we use this notation for general coordinate systems? How can we call del on its own an operator when actually there's more going on for the div and curl operations? Am I just missing something about how the del operator works?

Thanks for any help!

2. Feb 23, 2016

### phyzguy

I think the point that you are missing is the following:
In rectangular coordinates, $\vec{E} = E_x \vec{e_x} + E_y \vec{e_y} + E_z \vec{e_z}$. When you take $\frac{\partial}{\partial x}(E_x \vec{e_x})$, you just get $\frac{\partial E_x}{\partial x}$ because $\vec{e_x}$ is constant. However, in curvilinear coordinate systems, like spherical coordinates, you have $\vec{E} = E_r \vec{e_r} + E_\theta \vec{e_\theta} + E_\phi \vec{e_\phi}$. When you take $\frac{\partial}{\partial r}(E_r \vec{e_r})$, you get one term which is $\frac{\partial E_r}{\partial r}$, but you get a second term that includes $E_r \frac{\partial \vec e_r}{\partial r}$, because $\vec e_r$ changes from point to point. Does this help?

3. Feb 24, 2016

### Jezza

It does help, thank you :)