$V(R)=Af(R\cdot B)$, where A and B are constant, prove that curl V is perpendicular to both A and B

Question

If $V(R)$ can be expressed as $V(R)=Af(R\cdot B)$, where $A$ and $B$ are constant, prove that curl $V$ is perpendicular to both $A$ and $B$.

Answer 1

If $\phi: \mathbb{R^3} \to \mathbb{R}$, $F: \mathbb{R^3} \to \mathbb{R^3}$ and we let $g(x) = \phi(x) F(x)$, then a standard result is $\nabla \times g = \nabla \phi \times F + \phi (\nabla \times F)$.

Here we have $\phi(r) = f(b \cdot r)$, so we have $\nabla \phi(r) = f'(b \cdot r) b$, and $F(r) = a$, so we have $\nabla \times F = 0$, hence we get $$\nabla V(r) = f'(b \cdot r) ( b \times a) $$

It follows from the standard properties of $\times$ that $\nabla V(r)$ is perpendicular to both $a,b$.

Answer 2

Writing out $\nabla \times V$, we obtain

$\nabla \times V = (V_{z,y} - V_{y, z})\mathbf i + (V_{x,z} - V_{z, x})\mathbf j + (V_{y,x} - V_{x, y})\mathbf k, \tag{1}$

Where the comma notation is used for partial derivatives, i.e. $V_{z, y} = \frac{\partial V_z}{\partial y}$ etc., and $\mathbf i, \mathbf j,\mathbf k$ are the usual basis vectors in $\Bbb R^3$. Let's evaluate $V_{z, y}$:

$V_{z, y} = A_z \frac{\partial f(R \cdot B)}{\partial y} = A_z B_y f'(R \cdot B), \tag{2}$

where

$\frac{\partial f(R \cdot B)}{\partial y} = f'(R \cdot B) B_y \tag{3}$

by the chain rule and the fact that $R \cdot B = xB_x + yB_y + zB_z$. Likewise

$V_{y, z} = A_y \frac{\partial f(R \cdot B)}{\partial z} = A_y B_z f'(R \cdot B), \tag{4}$

so that

$V_{z, y} - V_{y, z} = f'(R \cdot B)(A_zB_y - A_yB_z), \tag{5}$

which is exactly the $\mathbf i$-component of $B \times A$. Performing similar operations on the $\mathbf j$- and $\mathbf k$-components of $\nabla \times V$, one sees that

$\nabla \times V = f'(R \cdot B)(B \times A); \tag{6}$

this formula immediately implies that $\nabla \times V$ is perpedicular to both $A$ and $B$.

Hope this helps. Cheers,

and as always,

Fiat Lux!!!