Divergence of a cross product

1. Oct 2, 2015

1. The problem statement, all variables and given/known data
The problem is given in the following photo:

Actually I did the first proof but I couldn't get the second relation. (Divergence of E cross H).

2. Relevant equations
They are all given in the photo. (a) (b) and (c).

3. The attempt at a solution
What I tried is to interchange divergence and cross products as it was given in (a). But I couldn't figure out how I am going to get 2 terms at the end. I also tried to apply the relation in (c), but it does not have any cross product, and I think there is no way to use equation in (b). So how can I prove the equation given at the end by using (a) (b) or (c) without decomposing into components or using Einsteins notation.

2. Oct 2, 2015

Fredrik

Staff Emeritus
The product rule, as it appears in (c), is a vector equation. Its ith component is $\partial_i (fg)=(\partial_i f)g+f\partial_ig$. If you use the definition of the cross product to rewrite the cross products in the problem, you will encounter expressions of the form $\partial_i (fg)$.

Edit: In this problem, you don't even have to use the definition, since (c) also tells you that if f and g are vector-valued functions, you're allowed to use that $\partial_i (f\cdot g)=(\partial_i f)\cdot g+f\cdot\partial_ig$ and $\partial_i (f\times g)=(\partial_i f)\times g+f\times\partial_i g$.

3. Oct 2, 2015

That is right. I didn't think using that for cross product. After that I can use (a) to prove the given relation.

It seems this was a little bit dummy question.

Thank you very much!

4. Oct 2, 2015

Fredrik

Staff Emeritus
Looking at the problem again, I see that the final sentence tells you NOT to use the definition of the cross product to rewrite it in terms of components. But you can still use the comment I added when I edited my previous post.