Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: Proving div(F X G) = G·curl(F) - F·curl(G)

  1. Sep 21, 2008 #1
    1. The problem statement, all variables and given/known data

    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)

    3. 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)
  2. jcsd
  3. Sep 21, 2008 #2


    User Avatar
    Science Advisor

    You seem to have added rather than subtracted G·curl(F) and F·curl(G)!
  4. Sep 21, 2008 #3
    That's what it appears like, but I checked that as well. for instance, lets 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!
  5. Sep 22, 2008 #4
  6. Sep 22, 2008 #5


    User Avatar
    Science Advisor
    Homework Helper

    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: Sep 22, 2008
  7. Sep 25, 2008 #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 quesiton now, is how do I make them equal? I am thinking of product rule. On the right track?
  8. Sep 25, 2008 #7


    User Avatar
    Science Advisor
    Homework Helper

    Yes, product rule! Product rule!
  9. Sep 25, 2008 #8
    Done, thank you!
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook