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Hard integral

  1. Jun 21, 2010 #1
    1. The problem statement, all variables and given/known data
    Giving 2 closed curves in 3-dimension space C1 and C2, prove that:[tex]\oint _{C1} \oint _{C2}\frac{(\vec{dl_2}.\hat{r_{12}})\vec{dl_1}}{r^2_{12}}=0[/tex]


    _ [tex]\vec{dl_1}[/tex] and [tex]\vec{dl_2}[/tex] are the vector elements of the curves C1 and C2 respectively.
    _ [tex]r_{12}[/tex] is the distance between the two above elements.
    _ [tex]\hat{r_{12}}[/tex] is the unit vector of vector [tex]\vec{r_{12}}[/tex]

    2. Relevant equations
    No idea.

    3. The attempt at a solution
    No idea too. It's out of my knowledge and ability. This is not a homework question, but on the way to solve a physics problem, I encountered this. Any idea about the way to solve would be very much appreciated. Thank you!
  2. jcsd
  3. Jun 21, 2010 #2


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    Hint: What is [itex]\mathbf{\nabla}_2\left(\frac{1}{r_{12}}\right)[/itex] (the subscript of the del operator indicates that the gradient should be taken with respect to [itex]\textbf{r}_2[/itex], the position vector of the second curve)? What does that make [itex]\oint _{C2}\frac{d\textbf{l}_2\cdot\mathbf{\hat{r}}_{12}}{r^2_{12}}[/itex]? :wink:
  4. Jun 23, 2010 #3
    Thank you very much.

    After some steps, I got this one:
    [tex]\oint _{C2}\frac{\vec{dr_2}.\hat{r_{12}}}{r^2_{12}} = - \oint _{C2}d(\frac{1}{r_{12}}) =0[/tex]

    Plus that [tex]\vec{dl_2}=\vec{dr_2}[/tex], so the above integral equals 0. Is it correct?
  5. Jun 23, 2010 #4


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    I'm not sure what you mean by [itex]d\left(\frac{1}{r_{12}}\right)[/itex]....what did you get for [itex]\mathbf{\nabla}_2\left(\frac{1}{r_{12}}\right)[/tex]?
  6. Jun 23, 2010 #5
    I guess I must use the symbol for partial differentiation instead of "d" ? (oh, what is the Latex code for that symbol by the way?). I'm not sure why calculate [tex]\vec{grad}(\frac{1}{r_{12}})[/tex], because it's a vector. So to relate it to the integral, I calculate [tex]d (\frac{1}{r_{12}})[/tex] by setting [tex]\vec{r_1}=const[/tex]. Not sure if I use the right symbol, but that's what I understand.
  7. Jun 23, 2010 #6


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    The [itex]\LaTeX[/itex] code for [itex]\partial[/itex] is"\partial", but that makes even less sense than using "d" in this context.

    Yes, it's a vector...but this is vector calculus, so what's the problem? You want to calculate [itex]\mathbf{\nabla}_2\left(\frac{1}{r_{12}}\right)[/itex] because the fundamental theorem of gradients allows you to easily calculate path integrals of the gradient of a scalar function.

    I'm still not exactly sure what you mean by [itex]d \left(\frac{1}{r_{12}}right)[/itex]....how do you calculate the differential of a multivariate function?
    Last edited: Jun 23, 2010
  8. Jun 24, 2010 #7
    The problem is I haven't learned vector calculus yet :biggrin: I just do it in a physical sense.
    I see. So I have to apply the gradient theorem.

    [tex]grad(\frac{1}{r_{12}})=\frac{\partial (\frac{1}{r_{12}})}{\partial {r_2}}\hat{r_2}=(-\frac{1}{r_{12}^2}\hat{r_{12}}\frac{ \partial {\vec{r_2}}} {\partial {r_2}})\hat{r_2}[/tex]

    [tex]grad(\frac{1}{r_{12}})\vec{dr_2}= (-\frac{1}{r_{12}^2}\hat{r_{12}}\frac{\partial{\vec{r_2}}}{\partial{r_2}})\hat{r_2}\vec{dr_2}[/tex]

    Since [tex]\hat{r_2}\vec{dr_2}=dr_2[/tex] and [tex]\vec{dr_2}=\frac{\partial{\vec{r_2}}}{\partial{r_2}}dr_2[/tex] we have:

    [tex]grad(\frac{1}{r_{12}})\vec{dr_2}=- \frac{1}{r_{12}^2}\hat{r_{12}}\vec{dr_2}[/tex]

    Do the integration over the curve C2 and we have the result in post #3. Is this correct?
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