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I have been told that the following equation is true, but I don't think it is all that obvious... could someone please explain why it is necessarily true?

Suppose $C_1$ and $C_2$ are closed paths in $\mathbb R^3$, and $\vec{r}$ is a position vector, then $$\oint_{C_1} d\vec{s}\times \left(\oint_{C_2}{d\vec{s'}\times \hat{r}\over |\vec{r}|^2}\right)=-\oint_{C_2} d\vec{s'}\times \left(\oint_{C_1}{d\vec{s}\times \hat{r}\over |\vec{r}|^2}\right)$$

My guess would be that we can somehow write the expressions on both sides as a simpler cross-product and $\oint_{C_1} d\vec{s}\times \left(\oint_{C_2}{d\vec{s'}\times \hat{r}\over |\vec{r}|^2}\right)$ say equals to $\vec{a}\times \vec{b}$ and $\oint_{C_2} d\vec{s'}\times \left(\oint_{C_1}{d\vec{s}\times \hat{r}\over |\vec{r}|^2}\right)$ equals $\vec{b}\times \vec{a}$?

Just to be clear, $C_1,C_2$ are fixed in shape and location in $\mathbb R^3$.

As @joriki has pointed out, $\vec{r}$ should be $s-s'$ (resp. $s'-s$)

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    I'm not sure I understand your notation: is $\oint_{C_2}{d\vec{s'}\times \hat{r}\over |\vec{r}|^2}$ a constant vector?2012-03-01
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    If I understand correctly, $\vec{r}$ is different between $C_1$ and $C_2$ so in total we have 4 different integrals here...2012-03-01
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    @anon: You are very right :)2012-03-01
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    @MartinArgerami: Since I am assuming that $C_1$ and $C_2 $ are fixed in space and shape, the integral mentioned should be constant.2012-03-01
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    @Baffled: Are you sure the intended meaning isn't $\vec r=\vec s-\vec s'$? That would make more sense.2012-03-01
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    I don't think this is true (if $\vec{r}$ is just the position vector). Counterexample: let $C_1$ be a circle centered around the origin, and let $C_2$ be another circle centered around a point $x\in C_1$ which is also normal to $C_1$'s tangent at $x$. Then the LHS will be zero and the RHS nonzero, simply from geometric reasoning.2012-03-01
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    @joriki: Thanks for the comment, sorry, I don't quite understand... would you mind explaining why $\vec r=\vec s-\vec s'$?2012-03-01
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    @joriki: Maybe it would help if I give some context. That is what I was told in class about action and reaction forces...2012-03-01
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    @Baffled: Yes, that's what it looked like; that's why I suggested $\vec r=\vec s-\vec s'$. This looks like the magnetic forces that two current loops exert on each other. The inner integrands are the magnetic fields generated by infinitesimal current elements, the inner integrals are the total magnetic fields generated by one of the current loops, the outer integrands are the forces acting on infinitesimal current elements of the other current, and the outer integrals are the total forces, which must be equal and opposite. This only works if $\vec r$ is the *relative* position.2012-03-01
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    @joriki: Wow you saw all that from my not-to-greatly phrased question? Hats off to you! :)2012-03-01
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    @joriki: After the writing $\vec r=\vec s-\vec s'$ how can I rearrange the integrals to see explicitly that one is minus the other one?2012-03-01
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    @Baffled: Still, if you knew it all along, why phrase the question so badly?2012-03-01
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    @Raskolnikov: I didn't realize it was relevant, my bad.2012-03-01

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This is related to the rather interesting fact that the magnetic forces exerted by two moving charges, and thus by two infinitesimal current elements, on each other are not equal and opposite and thus at first sight appear to violate Newton's third law. Electrodynamics cannot be properly understood without special relativity, and a proper treatment would have to take into account that the forces don't act instantaneously at a distance but are mediated by a field that can carry momentum.

In a stationary setting, the current elements must form a closed loop, and in this case, which these integrals describe, the force contributions do add up to equal and opposite total forces exerted by the current loops on each other. To see this, express the triple vector product in terms of scalar products using $\vec a\times(\vec b\times\vec c)=(\vec a\cdot\vec c)\vec b-(\vec a\cdot\vec b)\vec c$:

$$ \oint_{C_1} \oint_{C_2} \mathrm d\vec{s}\times{\mathrm d\vec{s'}\times (\vec s-\vec s')\over |\vec s-\vec s'|^3} =\oint_{C_1} \oint_{C_2} {(\mathrm d\vec{s}\cdot(\vec s -\vec s'))\mathrm d\vec s'-(\mathrm d\vec s\cdot\mathrm d\vec s')(\vec s-\vec s')\over |\vec s-\vec s'|^3}\;, \\ \oint_{C_1} \oint_{C_2} \mathrm d\vec{s}'\times{\mathrm d\vec{s}\times (\vec s'-\vec s)\over |\vec s'-\vec s|^3} =\oint_{C_1} \oint_{C_2} {(\mathrm d\vec{s}'\cdot(\vec s' -\vec s))\mathrm d\vec s-(\mathrm d\vec s'\cdot\mathrm d\vec s)(\vec s'-\vec s)\over |\vec s-\vec s'|^3}\;. $$

If we add these two total forces, the two second terms with forces along the line connecting the current elements cancel, and we're left with the two first terms with forces along the directions of the currents. The integrals over these terms vanish since they can be written in terms of line integrals of a gradient along a closed curve:

$$ \begin{eqnarray} \oint_{C_1} \oint_{C_2} {(\mathrm d\vec{s}\cdot(\vec s -\vec s'))\mathrm d\vec s'\over |\vec s-\vec s'|^3} &=& \oint_{C_2}\mathrm d\vec s'\oint_{C_1}\mathrm d\vec s\cdot\frac{\vec s-\vec s'}{|\vec s-\vec s'|^3} \\ &=&-\oint_{C_2}\mathrm d\vec s'\oint_{C_1}\mathrm d\vec s\cdot\vec\nabla_{\vec s}\frac{1}{|\vec s-\vec s'|} \\ &=& 0\;, \end{eqnarray} $$

and analogously for the first term in the second equation. Thus the sum of the total forces vanishes, as it should in a stationary setting.

For more on this, including a treatment of the force exerted by a closed current loop on itself and references to a relativistic treatment, see The Ampère and Biot–Savart force laws by G. Cavalleri, G. Spavieri and G. Spinelli, Eur. J. Phys. 17 (1996) 205–207.

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    @joriki: Thanks a lot, joriki! This is an absolutely wonderful answer!2012-03-01
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    @Baffled: You're welcome!2012-03-01