How to do the curl of the product of a function and a vector field

Question

I have the following property which says:

$∇ × (f G) = ∇f × G + f (∇ × G)$

Where $f$ is a differentiable function in an open set $S$ on $\mathbb {R}^3$ , and $G$ is a $C^1$ class vector field on $S$

I can't wrap my mind around where should I start for proving it, since I don't know how to do, from the left side, the curl of a product of a given function and a vector field (the $f G$ part)

From the right side, I don't know what $∇f × G $ means , while I guess that $f (∇ × G)$ is the function of the curl of G

Any hints or even solutions are welcome. I'm a bit green on this...

Solved:

$∇f×G$=$(\frac{∂f}{∂x},\frac{∂f}{∂y},\frac{∂f}{∂z}) × (G_1,G_2, G_3)=(\frac{∂f}{∂y}G3-\frac{∂f}{∂z}G2,\frac{∂f}{∂z}G1-\frac{∂f}{∂x}G3,\frac{∂f}{∂x}G2-\frac{∂f}{∂y}G1)$ ,

$f(∇×G)=f((\frac{∂G3}{∂y}-\frac{∂G2}{∂z},\frac{∂G1}{∂z}-\frac{∂G3}{∂x},\frac{∂G2}{∂x}-\frac{∂G1}{∂y})$ , (which to be expanded, ending up with the equivalent of the above with $f$ and $G_x$ "switched")

and on the left side:

$∇×(fG)=(\frac{∂fG3}{∂y}-\frac{∂fG2}{∂z},\frac{∂fG1}{∂z}-\frac{∂fG3}{∂x},\frac{∂fG2}{∂x}-\frac{∂fG1}{∂y})$, (which is to be expanded with the product rule)

Answer 1

From the left, the product $fG$ is still a vector field, so if you know how to calculate curls, go right ahead, since $f$ just scales the components of $G.$

On the right, $\nabla f×G$ is the cross between the gradient of $f$ (a vector by definition), and $G,$ also a vector, both three-dimensional, so the product is defined; also, $f(\nabla × G)$ is just $f,$ a scalar field, times the curl of $G,$ a vector. This is also defined. So you have two vectors on the right summing to the vector on the left.

As for proving, just go component wise; it might be easier working from right to left.

Finally, note that this can be remembered easily by the analogous Leibniz rule in single-variable calculus for differentiating the product of two functions.

Answer 2

Why not just write it out? Let $G=(G_1, G_2, G_3)$. Then \begin{align} \nabla \times (fG) &= (\partial_1, \partial_2, \partial_3) \times (fG_1, fG_2, fG_3)\\ &= (\partial_2(fG_3) - \partial_3(fG_2), \partial_3(fG_1)-\partial_1(fG_3), \partial_1(fG_2)-\partial_2(fG_1))\\ &= \ldots \end{align}

Answer 3

$$∇ × (f G) =\sum{\hat{i} \times \frac{\partial}{\partial x}(f G)} =\sum{\hat{i}\times ( \frac{\partial f}{\partial x} G + f \frac{\partial}{\partial x}G)}= (\sum{\hat{i}}\frac{\partial f}{\partial x}) \times G + f (\sum{\hat{i}}\frac{\partial G}{\partial x}) = ∇f × G + f (∇ × G)$$

Answer 4

The left-hand side is a vector with $i$th component $\epsilon_{ijk}\partial_j(fG_k)$, with implicit summation over the repeated indices $j,\,k$. By the product rule, this splits into $\epsilon_{ijk}(\partial_jf)G_k+f\epsilon_{ijk}\partial_jG_k$. The first of these terms is the $i$th component of the vector $(\nabla f)\times G$; the second term is the $i$th component of the vector $f(\nabla\times G)$, where the parentheses indicate multiplication of the vector $\nabla\times G$ by the scalar $f$. Thus $\nabla\times fG=\nabla f\times G+f\nabla\times G$ as required.

Answer 5

This can be done much faster with differential forms! The first thing to notice is that for a scalar field $f$ and a vector field $\vec{F}$ there exists corresponding 0 form and one form field respectively. In $R^3$, we can write:

$$ (f \vec{F}) \to f \left( P dx + Q dy + Rdz\right)= f \alpha$$

Now, taking the curl of the product of scalar field and vector field corresponds to taking the exterior derivative of the form field on the right, hence:

$$ d \left[ (f \alpha) \right] = df \wedge \alpha + (-1)^0 f \wedge d \alpha $$

Now note that following correspondances: $df \to \nabla F$ and $ d \alpha \to \nabla \times \vec{F}$, wedge product of two one forms the cross product, and finally, the wedge product of a 0 form and the exterior derivative of a oneform corresponds to multiplying the curl of the vector field corresponding to that form by the scalar function

$$ (f \vec{F}) \to d( f \vec{F}) =df \wedge \alpha + f \wedge d \alpha \to \nabla f \times \vec{F} + f \nabla \times \vec{F} $$

Hence,

$$ \nabla (f \vec{F}) = \nabla f \times \vec{F} + f \nabla \times \vec{F}$$ Done!

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