# Hard integral

## 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 $$\vec{r_{12}}$$

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!

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gabbagabbahey
Homework Helper
Gold Member
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 $\oint _{C2}\frac{d\textbf{l}_2\cdot\mathbf{\hat{r}}_{12}}{r^2_{12}}$? Thank you very much.

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

Plus that $$\vec{dl_2}=\vec{dr_2}$$, so the above integral equals 0. Is it correct?

gabbagabbahey
Homework Helper
Gold Member
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]? 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 Homework Helper Gold Member 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 [itex]\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?

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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?