- #1

Jezza

- 37

- 0

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

Where [itex]S[/itex] is the closed surface bounding [itex]V[/itex], and [itex]V[/itex] is a volume whose limit contains the point [itex]P[/itex].

[tex]\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}[/tex].

Where [itex]\vec n[/itex] is the unit normal to the closed contour [itex]\ell[/itex] which bounds the surface [itex]S[/itex] whose limit surrounds the point [itex]P[/itex].I also know that in cartesian coordinates, when we define the del operator [itex]\vec \nabla[/itex], these definitions are equivalent to: (Including [itex] \mathrm{grad} \phi[/itex] where [itex]\phi[/itex] is a scalar field.)

[tex]

\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}

[/tex]

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 [itex]\vec \nabla[/itex] notation. I'm also perfectly happy and understand that the definition of [itex]\vec \nabla[/itex] changes with coordinate system. For example, in spherical polar coordinates [itex](r, \theta, \varphi)[/itex]:

[tex]

\vec \nabla = \left(\frac{\partial}{\partial r}, \frac{1}{r} \frac{\partial}{\partial \theta}, \frac{1}{r \sin\theta} \frac{\partial}{\partial \varphi}\right)

[/tex]

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:

[tex]

\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}

[/tex]

Which clearly cannot be reached simply using vector methods. What I'm trying to get at is that despite [itex]\vec \nabla \cdot \vec E \neq \mathrm{div} \vec E[/itex] (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!