How to find div$(∇f × ∇g).?$

Question

Let $f $ and $g$ be two smooth scalar valued functions. Compute div$(∇f × ∇g).$

My attempt : div$(∇f × ∇g)=$ div$( \frac{\partial}{\partial x } (f.g) i + \frac{\partial}{\partial y } (f.g)j +\frac{\partial}{\partial z } (f.g)k ) $

after that im not able to proceed further

Answer 1

Let $\boldsymbol{F} = \left(F_1, F_2, F_3\right)$ and $\boldsymbol{G} = \left(G_1, G_2, G_3\right)$ be two vector fields.

Then, their vector product is defined as

$$ \boldsymbol{F}\times \boldsymbol{G} = (F_2G_3-F_3G_2, F_3G_1-F_1G_3, F_1G_2-F_2G_1) \Rightarrow. $$

$$ \begin{aligned} \text{div}\boldsymbol{F}\times \boldsymbol{G} &= \frac{\partial}{\partial x}(F_2G_3-F_3G_2) + \frac{\partial}{\partial y}(F_3G_1-F_1G_3) + \frac{\partial}{\partial z}(F_1G_2-F_2G_1) = \\ &= G_3\frac{\partial}{\partial x}F_2 + F_2\frac{\partial}{\partial x}G_3 - G_2\frac{\partial}{\partial x}F_3 - F_3\frac{\partial}{\partial x}G_2 + \\ &+ G_1\frac{\partial}{\partial y}F_3 + F_3\frac{\partial}{\partial y}G_1 - G_3\frac{\partial}{\partial y}F_1 - F_1\frac{\partial}{\partial y}G_3 + \\ &+ G_2\frac{\partial}{\partial z}F_1 + F_1\frac{\partial}{\partial z}G_2 - G_1\frac{\partial}{\partial z}F_2 - F_2\frac{\partial}{\partial z}G_1 = \\ &= G_1\left(\frac{\partial}{\partial y}F_3-\frac{\partial}{\partial z}F_2\right)+ G_2\left(\frac{\partial}{\partial z}F_1-\frac{\partial}{\partial x}F_3\right)+ G_3\left(\frac{\partial}{\partial x}F_2-\frac{\partial}{\partial y}F_1\right)- \\ &-F_1\left(\frac{\partial}{\partial y}G_3-\frac{\partial}{\partial z}G_2\right)+ F_2\left(\frac{\partial}{\partial z}G_1-\frac{\partial}{\partial x}G_3\right)+ F_3\left(\frac{\partial}{\partial x}G_2-\frac{\partial}{\partial y}G_1\right) = \\ &= \boldsymbol{G}\cdot\text{curl}\boldsymbol{F}-\boldsymbol{F}\cdot\text{curl}\boldsymbol{G}, \end{aligned} $$

where $\text{curl}\boldsymbol{F}$ is the the curl of the vector field $\boldsymbol{F}$, and it is defined as

$$ \text{curl}\boldsymbol{F} = \left(\frac{\partial}{\partial y}F_3-\frac{\partial}{\partial z}F_2, \frac{\partial}{\partial z}F_1-\frac{\partial}{\partial x}F_3, \frac{\partial}{\partial x}F_2-\frac{\partial}{\partial y}F_1\right). $$

Now, we have

$$ \text{div}\nabla{f}\times\nabla{g} = \nabla{g}\cdot\text{curl}(\nabla{f})-\nabla{f}\cdot\text{curl}(\nabla{g}). $$

Further, for any scalar function $f$, by definition of the $\nabla$ and $\text{curl}$,

$$ \begin{aligned} \text{curl}(\nabla{f}) &= \left(\frac{\partial}{\partial y}\left(\frac{\partial}{\partial z}f\right)-\frac{\partial}{\partial z}\left(\frac{\partial}{\partial y}f\right), \frac{\partial}{\partial z}\left(\frac{\partial}{\partial x}f\right)-\frac{\partial}{\partial x}\left(\frac{\partial}{\partial z}f\right), \frac{\partial}{\partial x}\left(\frac{\partial}{\partial y}f\right)-\frac{\partial}{\partial y}\left(\frac{\partial}{\partial x}f\right)\right) = (0, 0, 0) = \boldsymbol{0}. \end{aligned} $$

So, it is a zero vector.

Finally,

$$ \text{div}\nabla{f}\times\nabla{g} = \nabla{g}\cdot\text{curl}(\nabla{f})-\nabla{f}\cdot\text{curl}(\nabla{g}) = \nabla{g}\cdot\boldsymbol{0} - \nabla{f}\cdot\boldsymbol{0} = 0. $$

Answer 2

we know $ \vec{A} \times \vec{B}=(a_y b_z-a_zb_y )\hat{i}-(a_xb_z-a_zb_x)\hat{j}+(a_xb_y-a_yb_x) \hat{k}$ applying this to the cross product of the gradients we see that: $$\nabla f \times\nabla g= (\frac{\partial f}{\partial y} \frac{\partial g}{\partial z}-\frac{\partial f}{\partial z}\frac{\partial g}{\partial y} )\hat{i}-(\frac{\partial f}{\partial x}\frac{\partial g}{\partial z}-\frac{\partial f}{\partial z}\frac{\partial g}{\partial x})\hat{j}+(\frac{\partial f}{\partial x}\frac{\partial g}{\partial y}-\frac{\partial f}{\partial y}\frac{\partial g}{\partial x}) \hat{k}$$ Then we take divergence to get $$div(\nabla f \times\nabla g)=\frac{\partial }{\partial x}(\frac{\partial f}{\partial y} \frac{\partial g}{\partial z}-\frac{\partial f}{\partial z}\frac{\partial g}{\partial y} )-\frac{\partial }{\partial y}(\frac{\partial f}{\partial x}\frac{\partial g}{\partial z}-\frac{\partial f}{\partial z}\frac{\partial g}{\partial x})+\frac{\partial }{\partial z}(\frac{\partial f}{\partial x}\frac{\partial g}{\partial y}-\frac{\partial f}{\partial y}\frac{\partial g}{\partial x}) $$ There's some simplifying you can do but I will leave it to you. Use the product rule and what you know about switching the order of differentiation with partial derivatives.

Answer 3

You might also write your expression as:

$$ \nabla f = grad(f) = \sum_{i=1}^n \frac{\partial f}{\partial x_i} e_i $$ Which for 3 dimensions $i = 1,2,3$ and $e = i,j,k$, note $e$ is your basis or in this case unit vector. Thus, you obtain a 3x1 or 1x3 vector depending on your convention.

The cross product you can find in any basic physics book, but you can make your life easier by utilizing the cross product matrix. Assume $a,~b$ are just 3 dimensional vectors $$ a \times b = a^{\times} b = \begin{bmatrix} 0 & -a_3 & a_2 \\ a_3 & 0 & -a_1 \\ -a_2 & a_1 & 0\end{bmatrix} \begin{bmatrix}b_1 \\ b_2 \\ b_3\end{bmatrix} $$

Now, substitute $a$ and $b$ for $grad(f),~grad(g)$. The vector you obtain is identical to what you would get if you did this "manually" like the teach you in high school.

Finally, $div$ is the divirgence operator which is the dot product of the gradient with some vector, $$ div(h) =\nabla \cdot h= \sum_{i=1}^n \frac{\partial h_i}{\partial x_i} $$

You know $h$ from the above steps, now you can compute $\nabla \cdot ((\nabla f)\times (\nabla g)) $

You can also do this very quickly using index notation, but that is typically not taught at your level, so don't worry about it.