Is the Divergence of the Cross Product of Gradients Zero?

[Payne0511]

Homework Statement

div(grad f x grad g)=0. I need to prove this somehow.

Homework Equations

The Attempt at a Solution

I don't really know where to even start this at >.< any help is greatly appreciated.

 

[Dick]

Write grad(f) and grad(g) out in components and take their cross product. Now take div of that. Use that mixed partial derivatives are equal e.g. d^2(f)/(dxdy)=d^2(f)/(dydx). That will get the job done.

 

[VKint]

Personally, I like using index notation for this type of problem; it's a good deal more compact and, once you get used to it, very transparent. Instead of writing out all the components longhand, write $$\nabla f \times \nabla g = \tilde{\varepsilon}^{ijk} f_{,i} g_{,j} \hat{\mathbf{e}}_{k}$$ , where $$\tilde{\varepsilon}$$ is the Levi-Civita symbol (not being careful about index placement here, since we're in flat space). Then you have
$$\begin{align*} \nabla \cdot (\nabla f \times \nabla g) &= \tilde{\varepsilon}^{ijk} (f_{,i} g_{,j})_{,k}\\ &= \tilde{\varepsilon}^{ijk} f_{,i,k} g_{,j} + \tilde{\varepsilon}^{ijk} f_{,i} g_{,j,k} \textrm{.} \end{align*}$$
Both summands in the last equation vanish. Indeed, let $$P = \tilde{\varepsilon}^{ijk} f_{,i,k} g_{,j}$$ . Then, since partials commute, we have $$P = \tilde{\varepsilon}^{ijk} f_{,k,i} g_{,j}$$ ; by a trivial relabeling of dummy indices, this gives $$P = \tilde{\varepsilon}^{kji} f_{,i,k} g_{,j} = -\tilde{\varepsilon}^{ijk} f_{,i,k} g_{,j} = -P$$ (since $$\tilde{\varepsilon}$$ is completely antisymmetric), so $$P = 0$$. The logic for the other summand is identical.

 

[gabbagabbahey]

Alternatively, if you've already proven the following identities in class, you can simply combine them with $\textbf{A}=\mathbf{\nabla}f$ and $\textbf{B}=\mathbf{\nabla}g$ and the result becomes apparent.

Identities:

$$\mathbf{\nabla}\cdot(\textbf{A}\times\textbf{B})=\textbf{B}\cdot(\mathbf{\nabla}\times\textbf{A})-\textbf{B}\cdot(\mathbf{\nabla}\times\textbf{B})$$

$$\mathbf{\nabla}\times(\mthbf{\nabla}f)=0$$