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Problem with vector field proof

  1. Aug 15, 2012 #1
    Suppose that F and G are vector fields and that F-G = ▽μ for some real-valued function μ(x,y). Prove that

    ∫ F.dx = ∫ G.dx for all piecewise smooth curves C in the xy-plane

    I just need some help in getting started really. Thanks
     
  2. jcsd
  3. Aug 15, 2012 #2

    LCKurtz

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    If ##dx## is a scalar, what does ##\vec F \cdot dx## mean? Have you stated the problem correctly? Is C a closed curve?
     
  4. Aug 15, 2012 #3
    Thanks LCKurtz,

    F.dx is the line integral of the vector field and yes C is closed.
     
  5. Aug 15, 2012 #4

    LCKurtz

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    That's unusual notation. And since it matters that C is closed, don't you think it would have been good to mention that?

    Let's call ##\vec H = \vec F -\vec G##. So say ##\vec H = \langle h_1,h_2\rangle## and you are talking about$$
    \oint_C \vec H \cdot d\vec r =\oint_C h_1\, dx + h_2\, dy$$So now if you use ##\vec H(x,y) =\nabla \mu(x,y)## what happens?
     
    Last edited: Aug 15, 2012
  6. Aug 20, 2012 #5
    Hi im doing a similar problem so sooner than start a new thread...
    [itex] \oint_C \vec H \cdot d\vec r =\oint_C \frac{dμ}{dx}\, dx + \frac{dμ}{dy}\, dy [/itex]?
     
  7. Aug 20, 2012 #6

    LCKurtz

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    Yes. Now do you see how to argue that it is zero?
     
  8. Aug 20, 2012 #7
    No. in a word.
    I thought i might be able to work something out using greens theorem(as the curve is closed) but its not working out...
     
  9. Aug 20, 2012 #8

    LCKurtz

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    Green's theorem is what you want. What do you get and why don't you think it works?

    [Edit]I have to go for a couple of hours. I'll check back and see if you have it by then.
     
    Last edited: Aug 20, 2012
  10. Aug 20, 2012 #9
    [itex] \oint_C \vec H \cdot d\vec r =\oint_C \frac{dμ}{dx}\, dx + \frac{dμ}{dy}\, dy [/itex] = [itex]\int\int\frac{d^2μ}{dxdy}-\frac{d^2μ}{dxdy} dxdy = 0 [/itex] but this can only be true if [itex]\oint F.dr = \oint G.dr [/itex] qed?
     
  11. Aug 20, 2012 #10

    LCKurtz

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    Yes. It would be more clear if the left side had started with$$
    \oint_C \vec F\cdot d\vec R -\oint_C\vec G\cdot d\vec R =\oint_C (\vec F-\vec G)\cdot d\vec R=\oint_C\vec H\cdot d\vec R =\ ...$$
     
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