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The Product Rules

The product rules are much more difficult. We have two ways of making a scalar product - $ fg$ and $ \vA\cdot\vB$ . We can make two vector products as well - $ f\vA$ and $ \vA \times \vB$ (note that we will not worry about the ``pseudo'' character of the cross product unless it matters to the point we are trying to make). There are as it turns out six different product rules!

$\displaystyle \grad(fg) = f\grad g + g \grad f $

$\displaystyle \grad(\vA \cdot \vB) = \vA \times (\curl \vB) + \vB \times (\curl
\vA) + (\vA \cdot \grad)\vB + (\vB \cdot \grad)\vA $

The first is obvious and simple, the second is difficult to prove but important to prove as we use this identity a fair bit. Note well that:

$\displaystyle (\vA \cdot \grad) = A_x \partialdiv{}{x} + A_y \partialdiv{}{y} + A_z
\partialdiv{}{z} $

We have two divergence rules:

$\displaystyle \deldot (f\vA) = f(\deldot \vA) + (\vA \cdot \grad)f $

$\displaystyle \deldot (\vA \times \vB) = \vB\cdot(\curl \vA) - \vA\cdot(\curl \vB)
$

The first is again fairly obvious. The second one can easily be proven by distributing the divergence against the cross product and looking for terms that share an undifferentiated component, then collecting those terms to form the two cross products. It can almost be interpreted as an ordinary product rule if you note that when you pull $ \grad$ ``through'' $ \vA$ you are effectively changing the order of the cross product and hence need a minus sign. The product has to be antisymmetric in the interchange of $ \vA$ and $ \vB$ , so there has to be a sign difference between the otherwise symmetric terms from distributing the derivatives.

Finally, we have two curl rules:

$\displaystyle \curl (f\vA) = f(\curl \vA) - (\vA \times \grad)f $

$\displaystyle \curl (\vA \times \vB) = (\vB\cdot \grad)\vA - (\vA\cdot\grad)\vB
+ \vA(\deldot\vB) - \vB(\deldot\vA) $

The first is again rememberable as the usual product rule but with a minus sign when we pull $ \vA$ to the other side of $ \grad$ . The second one is nasty to prove because there are so very many terms in the fully expanded curl of the cross-product that must be collected and rearranged, but is very useful. Note that in electrodynamics we will often be manipulating or solving vector partial differential equations in contexts where e.g. $ \deldot \vE = 0$ or $ \deldot \vE = 0$ , so several of these terms might be zero.


next up previous contents
Next: Second Derivatives Up: Vector Derivatives Previous: The Sum Rules   Contents
Robert G. Brown 2017-07-11