Divergence of Cross Product Relation

In summary, the problem involves proving a relation using the product rule and the given equations (a), (b), and (c). The product rule can be used in the form of ##\partial_i (f\cdot g)=(\partial_i f)\cdot g+f\cdot\partial_i g## and ##\partial_i (f\times g)=(\partial_i f)\times g+f\times\partial_i g##. This can be used in conjunction with equation (a) to prove the given relation without decomposing into components or using Einstein's notation.
  • #1
advphys
17
0

Homework Statement


The problem is given in the following photo:
problem.png

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

Homework Equations


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

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.
 
Physics news on Phys.org
  • #2
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
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
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.
 

1. What is the formula for calculating the divergence of a cross product?

The formula for calculating the divergence of a cross product is div(a x b) = b · curl(a) - a · curl(b).

2. How is the divergence of a cross product related to the vector calculus identities?

The divergence of a cross product is related to the vector calculus identities through the curl-curl identity, which states that the curl of a curl is equal to the divergence of the gradient minus the Laplacian operator.

3. What is the physical significance of the divergence of a cross product?

The physical significance of the divergence of a cross product is that it represents the rate of change of rotational forces in a vector field. It can also be interpreted as the net flow of the curl of a vector field through a small surface surrounding a point.

4. How do you interpret a negative or positive value for the divergence of a cross product?

A negative value for the divergence of a cross product indicates that the vector field is circulating in a clockwise direction, while a positive value indicates a counterclockwise direction. This can also be interpreted as a net flow into or out of a point in the vector field.

5. Can the divergence of a cross product be used in applications outside of mathematics and physics?

Yes, the divergence of a cross product has applications in areas such as fluid mechanics, electromagnetism, and even computer graphics and animation. It is a useful tool for understanding the behavior of vector fields in various physical and virtual systems.

Similar threads

  • Calculus and Beyond Homework Help
Replies
5
Views
975
  • Calculus and Beyond Homework Help
Replies
13
Views
2K
  • Introductory Physics Homework Help
Replies
5
Views
770
  • Precalculus Mathematics Homework Help
Replies
5
Views
579
  • Calculus and Beyond Homework Help
Replies
6
Views
1K
  • Calculus and Beyond Homework Help
Replies
2
Views
1K
Replies
14
Views
1K
  • Calculus and Beyond Homework Help
Replies
7
Views
1K
Replies
8
Views
841
  • Calculus and Beyond Homework Help
Replies
7
Views
3K
Back
Top