I have difficulty understanding the following vector calculus example. Text can be found here. It is the 5th Q&A -- starting with equation (31.1035).It concerns finding the vector potential of a current loop.

I get the argument up to $$\vec A=\frac{\mu I}{4\pi r^3}\oint \vec r'\cdot \vec r \,\,\,d\vec l'$$. But I don't understand why $$\oint \vec r'\cdot \vec r \,\,\,d\vec l'=\left(\int d\vec a\right)\times \vec r$$

[ADDED]: Definitions of the terms:

This describes a current loop, with a current $I$ flowing in it.

$\vec r$ is the position vector of an arbitrary point in space. $\vec r'$ is a position vector of a point on the loop. $d\vec l'$ is an infinitesimal line element along the loop.

$\vec A(\vec r)$ is the vector potential, given by $$\frac{\mu I}{4\pi}\oint \frac{d\vec l'}{|\vec r'-\vec r|}$$ The above expression is obtained by Taylor expanding this definition.

The first thing I tried is the vector triple product identity $$(\vec r'\cdot \vec r )\,\,\,d\vec l'=\vec r\times(d\vec l'\times \vec r')+(\vec r\cdot d\vec l')\vec r'$$ The first term looks like it's doing the right sort of thing, but I don't see why the second term vanishes (or fits in) ...Maybe I have missed out something quite obvious?

  • $\begingroup$ Thank you, @Muphrid and @Mike! (This is the OP speaking). Your answers are very helpful! :) Sorry I had to post this as an "answer" -- for some reason I can't seem to "comment"...? $\endgroup$
    – user24119
    Commented May 7, 2013 at 7:54

2 Answers 2


This is a very good question; it's not at all obvious. And you had a good idea trying the vector triple product, but we can probably do something easier, building on stuff you've seen already.

Any time you see an integral over one type of region (line, surface, or volume) on one side of an equation, and an integral over a different type of region on the other, you know it involves some form of Stokes' theorem -- such as the basic (Kelvin-)Stokes' theorem, or Green's theorem, or the divergence theorem, or the gradient theorem. So it might be helpful to remember each of those theorems, and think about which one might be useful. In this case, you might see a lot of similarity to the (Kelvin-)Stokes' theorem: \begin{equation} \oint \vec{F} \cdot d\vec{l}' = \int (\vec{\nabla} \times \vec{F}) \cdot d\vec{a}~. \end{equation} But this is a scalar, and your equation above is a vector on either side, so you need to look at components of your equation.

Let's look at the $x$ component of your equation, by taking the dot product with $\hat{x}$. Since $\hat{x}$ is constant, I can just take it inside the integral. On the left-hand side of your equation, I get an integral of $(\vec{r}' \cdot \vec{r})\hat{x} \cdot d\vec{l}'$. Now we're getting somewhere. Comparing this to the Kelvin-Stokes equation, we see that defining $\vec{F} = (\vec{r}' \cdot \vec{r})\hat{x}$ is a sensible thing to do, and we'll have to take the curl of this. Now, I can also write $\vec{r}' \cdot \vec{r} = x'\, r_x + y'\, r_y + z'\, r_z$, so the curl is \begin{equation} \vec{\nabla} \times \vec{F} = r_z\, \hat{y} - r_y\, \hat{z}~. \end{equation} Another little calculation using this expression shows you that $(\vec{\nabla} \times \vec{F}) \cdot d\vec{a}$ is exactly the $x$ component of $(\int d\vec{a}) \times \vec{r}$.

Doing this again for the $y$ and $z$ components gives you your result.


Mike has captured the gist of things, but you can be a bit more direct. The following is a known identity: $\nabla(r \cdot s) = s$ for any vector $s$ that does not depend on $r$.

Stokes' theorem tells us that, for any scalar field $\phi$,

$$\oint \phi(r') \, d\ell' = \int da' \times \nabla' \phi(r')$$

Take $\phi(r') = r' \cdot r$ to get

$$\oint (r' \cdot r) \, d\ell' = \int da' \times r$$

well, $r$ is constant with respect to $r'$--I think it poor that they chose not to leave it $da$ and not $da'$, because that would've reinforced the point--, so you can evaluate the area integral independently before doing the cross product.

  • $\begingroup$ +1 from me for elegance. Though I'm not sure that form of Stokes' theorem appears in typical intro E&M courses. :) $\endgroup$
    – Mike
    Commented May 7, 2013 at 14:40
  • 1
    $\begingroup$ @Mike Yeah,I had to derive it from scratch because I couldn't find it anywhere, just to check the sign of the cross product. $\endgroup$
    – Muphrid
    Commented May 7, 2013 at 14:48

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