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Existence of scalar potential

  1. Sep 29, 2011 #1
    Hi I know it's easy to prove that if a vectorfield is the gradien of a potential, [tex] \vec F = \nabla V[/tex], then [tex]\nabla \times F = 0.[/tex] But how about the converse relation? Can I prove that if [tex]\nabla \times F = 0,[/tex] then there exist a salar potential such that [tex] \vec F = \nabla V?[/tex]

    I get as far as proving the existence of a potential function

    [tex]V(\vec r) = \int_{\vec r_0}^{\vec r} \vec F \cdot d\vec r,[/tex]
    but how do I now establish that

    [tex] F = \nabla V?[/tex]

    What is the gradient of a line integral?


    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Sep 29, 2011 #2

    LCKurtz

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    To answer your last question first, that line integral evaluates to a scalar function of x,y, and z (it's the integral of a dot product). Now, if you have

    [tex]V(\vec r) = \int_{\vec r_0}^{\vec r} \vec F \cdot d\vec r,[/tex]

    you automatically have independence of path. Let me rewrite that equation:

    [tex]V(x,y,z) = \int_{P_0(x_0,y_0,z_0)}^{P(x,y,z)}\langle S(x,y,z),T(x,y,z),U(x,y,z)\rangle \cdot\langle dx,dy,dz\rangle[/tex]

    You want to show, for example, that Vx(x,y,z) = S(x,y,z)

    Look at:
    [tex]\frac {V(x+h,y,z)-V(x,y,z)}{h}=\frac 1 h\left(\int_{(x_0,y_0,z_0)}^{(x+h,y,z)}
    \vec F\cdot d\vec r - \int_{(x_0,y_0,z_0)}^{(x,y,z)}
    \vec F\cdot d\vec r\right)[/tex]

    Now that first integral can be taken along any path. So let's use the path from
    (x0,y0,z0) to (x,y,z) to (x+h,y,z) for the first integral. The first section of the path will cancel with the second integral leaving

    [tex]\frac {V(x+h,y,z)-V(x,y,z)}{h}=\frac 1 h\left(\int_{(x,y,z)}^{(x+h,y,z)}
    S(x,y,z)\,dx\right) [/tex]

    since dy and dz are 0 on this path segment. Now apply the mean value theorem for integrals on the x variable:

    [tex]\frac 1 h\left(\int_{(x,y,z)}^{(x+h,y,z)}
    S(x,y,z)\,dx\right) = \frac{hS(c,y,z)}{h}[/tex]
    for some c between x and x+h. Now if you let h→0 you get Vx=S(x,y,z).

    You do the other two partials similarly.
     
    Last edited: Sep 29, 2011
  4. Sep 30, 2011 #3
    Thank you! That was a very clear explanation.

    So one may then conclude that for a irrotational field

    [tex]\vec F = \nabla \int_{\vec r_0}^{\vec r} \vec{F} \cdot d \vec r ?[/tex]
     
  5. Sep 30, 2011 #4

    LCKurtz

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    Yes, as long as all relevant hypotheses are satisfied. I don't have a reference handy, but I think appropriate continuity of partials on an open connected set was mentioned if I remember correctly.
     
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