Answer to Question #275799 in Calculus for Hafsa Tahsin

Question

[SADT3] For scalar functions u and v, show that

B=( nabla u)*( nabla v)

is solenoidal and that

A= 1 2 (u nabla v-v nabla u)

is a vector potential for B, i.e. B= nabla* A

Answer 1

$B=( \nabla u)\times( \nabla v)$

$A= \frac{1}{2} (u \nabla v-v \nabla u)$

$B= \nabla\times A$

for solenoidal vector field:

$\nabla \cdot B=0$

$\nabla u=\frac{\partial u}{\partial x}i+\frac{\partial u}{\partial y}j+\frac{\partial u}{\partial z}k$

$\nabla v=\frac{\partial v}{\partial x}i+\frac{\partial v}{\partial y}j+\frac{\partial v}{\partial z}k$

$B=( \nabla u)\times( \nabla v)=\begin{vmatrix} i & j&k \\ \frac{\partial u}{\partial x}&\frac{\partial u}{\partial y}&\frac{\partial u}{\partial z}\\ \frac{\partial v}{\partial x}&\frac{\partial v}{\partial y}&\frac{\partial v}{\partial z}\\ \end{vmatrix}=$

$=(\frac{\partial u}{\partial y}\frac{\partial v}{\partial z}-\frac{\partial v}{\partial y}\frac{\partial u}{\partial z})i-(\frac{\partial u}{\partial x}\frac{\partial v}{\partial z}-\frac{\partial u}{\partial z}\frac{\partial v}{\partial x})j+(\frac{\partial u}{\partial x}\frac{\partial v}{\partial y}-\frac{\partial u}{\partial y}\frac{\partial v}{\partial x})k=0$

so, $\nabla \cdot B=0$

$A= \frac{1}{2} (u \nabla v-v \nabla u)=\frac{1}{2} (u\frac{\partial v}{\partial x}i+u\frac{\partial v}{\partial y}j+u\frac{\partial v}{\partial z}k-v\frac{\partial u}{\partial x}i-v\frac{\partial u}{\partial y}j-v\frac{\partial u}{\partial z}k)$

$\nabla\times A=[\frac{\partial}{\partial y}(u\frac{\partial v}{\partial z}-v\frac{\partial u}{\partial z})-\frac{\partial}{\partial z}(u\frac{\partial v}{\partial y}-v\frac{\partial u}{\partial y})]i+[\frac{\partial}{\partial z}(u\frac{\partial v}{\partial z}-v\frac{\partial u}{\partial z})-\frac{\partial}{\partial x}(u\frac{\partial v}{\partial z}-v\frac{\partial u}{\partial z})]j+$

$+[\frac{\partial}{\partial x}(u\frac{\partial v}{\partial y}-v\frac{\partial u}{\partial y})-\frac{\partial}{\partial y}(u\frac{\partial v}{\partial x}-v\frac{\partial u}{\partial x})]k=B$