Proving div(F X G) = G·curl(F) - F·curl(G)

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In summary, the homework statement says that if F = F1i + F2j + F3k and G = G1i + G2j + G3k are differentiable vector functions of (x,y,z), then div(F X G) = G·curl(F) - F·curl(G)
  • #1
t_n_p
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Homework Statement



If F = F1i + F2j + F3k and G = G1i + G2j + G3k are differentiable vector functions of (x,y,z) prove that div(F X G) = G·curl(F) - F·curl(G)

The Attempt at a Solution



If computed both sides of the equation, but they are not the same. My left hand side is
div(F X G) = (∂/∂x)(F2G3-G2F3) + (∂/∂y)(G1F3-F1G3) + (∂/∂z)(F1G2-F2G1)

My right hand side is
G·curl(F) - F·curl(G) = (∂/∂x)(2F2G3-2G2F3) + (∂/∂y)(2G1F3-2F1G3) + (∂/∂z)(2F1G2-2F2G1)

This weird two has appeared out of nowhere! I've checked over many, many times and still can't spot any arithmatic mistakes. I'll break it down below and hopefully somebody may be able to point out a mistake.

G·curl(F) = G1F3(∂/∂y) - G1F2(∂/∂z) - G2F3(∂/∂x) + G2F1(∂/∂z) + G3F2(∂/∂x) - G3F1(∂/∂y)

F·curl(G) = F1G3(∂/∂y) - F1G2(∂/∂z) - F2G3(∂/∂x) + F2G1(∂/∂z) + F3G2(∂/∂x) - G1F3(∂/∂y)
 
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  • #2
You seem to have added rather than subtracted G·curl(F) and F·curl(G)!
 
  • #3
That's what it appears like, but I checked that as well. for instance, let's just group the ∂/∂x terms.

From G·curl(F) we have:
∂/∂x (G3F2 - G2F3)

and from F·curl(G) we have:
∂/∂x (F3G2- F2G3)

Then G·curl(F) - F·curl(G) = [∂/∂x (G3F2 - G2F3)] - [∂/∂x (F3G2- F2G3)]
= ∂/∂x (2F2G3 - 2F3G2)

as shown above!
 
  • #4
bump!
 
  • #5
From G.curl(F) I get G3*d/dx(F2)-G2*d/dx(F3). That's different from your result. Why? G.curl(F) shouldn't have ANY derivatives of G, right?
 
Last edited:
  • #6
ok, I've taken that onboard. I am still struggling though.

My left hand side remains the same,
div(F X G) = (∂/∂x)(F2G3-G2F3) + (∂/∂y)(G1F3-F1G3) + (∂/∂z)(F1G2-F2G1)

Now taking into account the help from Dick, my RHS G·curl(F) - F·curl(G) now becomes
G1(∂/∂y)F3 - G1(∂/∂z)F2 - G2(∂/∂x)F3 + G2(∂/∂z)F1 + G3(∂/∂x)F2 - G3(∂/∂y)F1 - [F1(∂/∂y)G3 - F1(∂/∂z)G2 - F2(∂/∂x)G3 + F2(∂/∂z)G1 + F3(∂/∂x)G2 - G1(∂/∂y)F3]

My question now, is how do I make them equal? I am thinking of product rule. On the right track?
 
  • #7
Yes, product rule! Product rule!
 
  • #8
Done, thank you!
 

1. What is the meaning of "div(F X G)"?

The term "div(F X G)" refers to the divergence of the cross product of the vector fields F and G. It is a mathematical operation that measures the rate at which the vector field is flowing out of a small closed surface around a point.

2. How do you prove that div(F X G) = G·curl(F) - F·curl(G)?

To prove this equation, you can use the vector calculus identities and the definition of divergence and curl. First, expand the cross product F X G using its components. Then, use the product rule for divergence and the vector triple product identity to simplify the equation. Finally, use the definition of curl to express it in terms of the components of F and G, and rearrange the terms to match the given equation.

3. What is the significance of "div(F X G) = G·curl(F) - F·curl(G)" in vector calculus?

This equation is known as the Product Rule for Divergence and plays an important role in vector calculus. It is used to simplify the calculation of divergence in more complex vector fields and is often used in fluid dynamics and electromagnetism.

4. Can you provide an example of a vector field that satisfies "div(F X G) = G·curl(F) - F·curl(G)"?

Yes, a simple example would be the vector field F = 2xyi + 3xj + 4zk and G = x^2yi + 2xyj + 3xzk. Using the definition of divergence and curl, it can be shown that div(F X G) = G·curl(F) - F·curl(G) = 6x + 2y + 2z, which satisfies the equation.

5. How is "div(F X G) = G·curl(F) - F·curl(G)" related to Stokes' Theorem?

Stokes' Theorem is a fundamental theorem in vector calculus that relates the surface integral of a vector field to the line integral of its curl. By using the Product Rule for Divergence, the equation "div(F X G) = G·curl(F) - F·curl(G)" can be derived, which is a crucial step in proving Stokes' Theorem.

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