Curl and Product Rule

Question

I know the famous identity

$\nabla \times (\vec{A} \times \vec{B}) = A(\nabla \cdot \vec{B}) -B(\nabla \cdot \vec{A}) + (B\cdot \nabla) \vec{A}- (A\cdot \nabla) \vec{B}$

My question: Can I distribute the curl like a normal differential operator as follows:

$\nabla \times (\vec{A} \times \vec{B}) =(\nabla \times \vec{A})\times B + (\nabla \times \vec{B})\times A$

With the del operator acting only on A in $(\nabla \times \vec{A})\times \vec{B}$ and only acting on B in the second term?

Or are there any "product rules" for the curl of a cross product other than the well-known one I gave in the beginning?

EDIT:

I posted the question because in the solution to my homework we are given the following equation which I don't understand:

$\nabla \times (\vec{e}_r \times \phi \vec{A}) = (\nabla \phi)\times (\vec{e}_r \times \vec{A})$ with the radial unit vector $\vec{e}_r$ and $\nabla \times \vec{A}=0$.

So does anyone have an idea where this comes from, then?

Answer 1

No. Choose some simple parallel vectors like $$\vec A = \vec B = \langle x, x, x\rangle$$ Then $\vec A \times \vec B = \vec 0$, but

$$\nabla \times \vec A = \langle 0, -1, 1\rangle$$ so that $$(\nabla \times \vec A) \times \vec B = \langle -2x, x, x\rangle$$

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I'm not aware of any similar rules that involve only curl, although there is a product rule for multiplying by a scalar function, e.g. $\nabla (f \vec A)$. Some identities for second derivatives are given here as well.

Answer 2

There is a general fact about derivatives that seems to not be as widely known as it should: the derivative of an expression is the sum of the derivatives of its subexpression. Applying this fact requires precisely keeping track of what is being differentiated, so I will use the following convention: $\nabla$ acts as it usually does, but $\dot\nabla$ only differentiates other dotted variables. The dot notation is powerful since all vectors identities are valid even with $\dot\nabla$ so long as we properly keep track of differentiation like this and so long as $\dot\nabla$ appears in a linear slot.

As a good example of all this in action, let's consider your last equation. As best as I can tell, this just isn't true; what is true though is $$ \nabla\cdot(e_r\times(\phi A)) = (\nabla\phi)\cdot(e_r\times A) $$ so perhaps it was a typo. More generally, $$ \nabla\cdot(A\times(\phi B)) = (\nabla\phi)\cdot(A\times B) \tag{$*$} $$ for any $A, B$ with $\nabla\times A = 0$ and $\nabla\times B = 0$. To see why, start with $$ \nabla\cdot(A\times(\phi B)) = \dot\nabla\cdot(\dot A\times(\dot\phi\dot B)) = \dot\nabla\cdot(\dot A\times(\phi B)) + \dot\nabla\cdot(A\times(\dot\phi B)) + \dot\nabla\cdot(A\times(\phi\dot B)). $$ Since $\phi$ is a scalar quantity, the second term is $$ \dot\nabla\cdot(A\times(\dot\phi B)) = (\nabla\phi)\cdot(A\times B), $$ so we need to show that the two other terms sum to $0$. In fact, they are both individually $0$; consider that the triple product is stable under cyclic permutations: $$ a\cdot(b\times c) = c\cdot(a\times b) = b\cdot(c\times a). $$ Thus $$ \dot\nabla\cdot(\dot A\times(\phi B)) = \phi B\cdot(\dot\nabla\times\dot A) = 0, $$$$ \dot\nabla\cdot(A\times(\phi\dot B)) = \phi A\cdot(\dot B\times\dot\nabla) = 0, $$ proving ($*$).

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This is also how we can prove your first equation. Using the identity $$ a\times(b\times c) = (a\cdot c)b - (a\cdot b)c $$ we see $$\begin{aligned} \nabla\times(A\times B) &= \dot\nabla\times(\dot A\times B) + \dot\nabla\times(A\times\dot B) \\ &= (\dot\nabla\cdot B)\dot A - (\dot\nabla\cdot\dot A)B + (\dot\nabla\cdot\dot B)A - (\dot\nabla\cdot A)\dot B \\ &= (B\cdot\nabla)A - (\nabla\cdot A)B + (\nabla\cdot B)A - (A\cdot\nabla)B. \end{aligned}$$

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Your second equation is false simply because $$ a\times(b\times c) \not= (a\times b)\times c. $$ However, we do have the Jacobi identity $$ a\times(b\times c) + c\times(a\times b) + b\times(c\times a) = 0, $$ so applying this the best we can say is $$\begin{aligned} \nabla\times(A\times B) &= \dot\nabla(\dot A\times B) + \dot\nabla(A\times\dot B) \\ &= -B\times(\dot\nabla\times\dot A) - \dot A\times(B\times\dot\nabla) - \dot B\times(\dot\nabla\times A) - A\times(\dot B\times\dot\nabla) \\ &= (\nabla\times A)\times B + A\times(\nabla\times B) + (B\times\nabla)\times A- (A\times\nabla)\times B \end{aligned}$$ so the degree to which your identity fails is measured by the term $$ (B\times\nabla)\times A- (A\times\nabla)\times B. $$