Proving: Closed Curve Integral in 3D Space

[hikaru1221]

Homework Statement

Giving 2 closed curves in 3-dimension space C1 and C2, prove that:\oint _{C1} \oint _{C2}\frac{(\vec{dl_2}.\hat{r_{12}})\vec{dl_1}}{r^2_{12}}=0

Where:

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

Homework Equations

No idea.

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!

 

[gabbagabbahey]

Hint: What is \mathbf{\nabla}_2\left(\frac{1}{r_{12}}\right) (the subscript of the del operator indicates that the gradient should be taken with respect to \textbf{r}_2, the position vector of the second curve)? What does that make \oint _{C2}\frac{d\textbf{l}_2\cdot\mathbf{\hat{r}}_{12}}{r^2_{12}}?

 

[hikaru1221]

Thank you very much.

After some steps, I got this one:
\oint _{C2}\frac{\vec{dr_2}.\hat{r_{12}}}{r^2_{12}} = - \oint _{C2}d(\frac{1}{r_{12}}) =0Plus that \vec{dl_2}=\vec{dr_2}, so the above integral equals 0. Is it correct?

 

[gabbagabbahey]

I'm not sure what you mean by d\left(\frac{1}{r_{12}}\right)...what did you get for \mathbf{\nabla}_2\left(\frac{1}{r_{12}}\right)[/tex]?

 

[hikaru1221]

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 \vec{grad}(\frac{1}{r_{12}}), because it's a vector. So to relate it to the integral, I calculate d (\frac{1}{r_{12}}) by setting \vec{r_1}=const. Not sure if I use the right symbol, but that's what I understand.

 

[gabbagabbahey]

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?).

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

I'm not sure why calculate \vec{grad}(\frac{1}{r_{12}}), because it's a vector.

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

So to relate it to the integral, I calculate d (\frac{1}{r_{12}}) by setting \vec{r_1}=const.

I'm still not exactly sure what you mean by d \left(\frac{1}{r_{12}}right)...how do you calculate the differential of a multivariate function?

 

[hikaru1221]

The problem is I haven't learned vector calculus yet I just do it in a physical sense.
I see. So I have to apply the gradient theorem.

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}

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}

Since \hat{r_2}\vec{dr_2}=dr_2 and \vec{dr_2}=\frac{\partial{\vec{r_2}}}{\partial{r_2}}dr_2 we have:

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

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