# Describing a circular current loop as delta functions

It would be really nice to see how Jackson got eqn 5.33 on his example problem for finding the vector potential of a circular current loop

$$J_{\phi}=I\sin\theta'\delta(\cos\theta')\frac{\delta(r'-a)}{a}$$

for describing this geometry: I've found a few 'explanations' googling, but I'm still confused on the origin of the $\sin\theta$

My best guess

$$\int \vec{J}\cdot\hat{n} \; d^3x=\int \vec{I}\cdot d\vec{l} \\ \nabla_\phi \left ( \int \vec{J}\cdot\hat{n} \; d^3x=\int \vec{I}\cdot d\vec{l} \right) \\ \frac{1}{\sin\theta} \frac{\partial}{\partial\phi} \int \vec{J}\cdot\hat{n} \; r^2 dr d\cos\theta d\phi = \frac{1}{\sin\theta} \frac{\partial}{\partial\phi} \int I a d\phi \\ \frac{1}{\sin\theta} \int J_\phi r^2 dr d\cos\theta = \frac{1}{\sin\theta} I a \\ \frac{1}{\sin\theta} \int A \delta(\cos\theta) \delta(r-a) r^2 dr d\cos\theta =\frac{1}{\sin\theta} I a \\ Aa^2=Ia \\ A=\frac{I}{a}\\ \therefore J_\phi = I \delta(\cos\theta) \frac{\delta(r-a)}{a}$$

I know the $\sin\theta$ doesn't really matter.. because

$$\int \sin^2\theta \delta(\theta-\pi/2)d\theta=1$$

but I'd still like to know where it comes from.

EDIT
The explanation that "it doesn't matter, so lets include it" bothers me. There must be some place in the derivation he picks it up. What possible reason would there be for adding it in at the end?

The change of variables [ref] doesn't work because

$$\delta(\cos\theta)=\frac{\delta(\theta-\pi/2)}{\sin\frac{\pi}{2}}$$ so you still end up with $$\int \sin^2\theta \delta(\theta-\pi/2)d\theta$$

I also don't think that it comes out of the Jacobian. Because he later goes on to integrate over the solid angle

$$\int J r^2 d\Omega dr = \int J r^2 \sin\theta d\theta d\phi$$ which would, again, give something of the form $\sin^2\theta$

EDIT
Jackson's next step (Eqn. 5.35)

$$A_\phi(r,\theta)=\frac{\mu_0I}{4\pi a}\int r'^2dr'd\Omega' \frac{\sin\theta'\cos\phi'\delta(\cos \theta')\delta(r'-a)}{\left | \mathbf{x}-\mathbf{x'} \right |} \\ \; \\ \text{with } d\Omega' = d\cos\theta' d\phi = \sin\theta' d\theta' d\phi$$

I would agree that a $\sin \theta$ belongs in the formula if he only integrated over $d\theta$ not $\sin\theta d\theta$

See, you are required to find volume current density $J_\phi$. Though its name is volume current density, you know it is the current flowing per unit surface area. Now the subscript $\phi$ in $J_\phi$ denotes it is flowing in the $\hat{\phi}$ direction. Now in the spherical polar co-ordinate the infinitesimal length elements along the direction $\hat{r},\hat{\theta} \rm\ and\ \hat{\phi}$ are dr, $rd\theta$ and $r\sin\theta d\phi$ respectively. So $dr\times rd\theta$ is the area you are interested in.

Now looking at your figure I am interested in rewriting the current I.

$$I=I\int \delta(r-a) dr= I\int \delta(r-a) dr\ \delta\big(\cos\theta-cos(\pi/2)\big) d(cos\theta)$$ $$=I\int \delta(r-a) dr\ \delta\big(\cos\theta-0\big) d(cos\theta)$$ $$=I\int \delta(r-a) dr\ \delta\big(\cos\theta\big) d(cos\theta)$$ $$=I\int \delta(r-a) dr\ \delta\big(\cos\theta\big) \sin\theta d\theta$$ $$=I\int \delta(r-a)\delta\big(\cos\theta\big) \sin\theta dr\ d\theta$$ $$=I\int \frac{ \delta(r-a)\delta(\cos\theta) \sin\theta}{r} {dr\ r d\theta}$$

So you see your $$J_\phi = I\frac{ \delta(r-a)\delta(\cos\theta) \sin\theta}{r}$$

• Thank you for the reply. Yes, a $\sin\theta$ shows from Jacobian $\sin\theta d\theta$ But, jackson integrates over the full solid angle $d\Omega =\sin\theta d\theta d\phi$ ... which would give a $\sin^2\theta$ I'll edit the question to show his next step
– Ben
Jul 31, 2014 at 16:58
• I would agree that a $\sin \theta$ belongs in the formula if he only integrated over $d\theta$ not $\sin\theta d\theta$
– Ben
Jul 31, 2014 at 17:19
• I think you couldn't follow my answer properly. Take time to read it properly. If it doesn't answer your question completely, point out exactly where you have the problem in my answer. Aug 1, 2014 at 5:04
• Yeah you're right, I wasn't following. I was thinking the $\sin \theta$ was included when the azimuthal angle was held constant (i.e. I thought $dS_\phi=r\sin\theta dr d\theta$ ). I see what you were saying now. thank you
– Ben
Aug 1, 2014 at 5:43
• So when I did my line integral $\int I \cdot dl$ I should should have evaluated it as $\int I a\sin\theta d\phi$ right?
– Ben
Aug 1, 2014 at 6:26

If you note that $$\delta(\cos\theta)=\frac{\delta(\theta-\pi/2)}{\sin\theta}$$ Then you can see that the sine terms actually cancel out.

• I just want to note that, as I mention in my post, the sine term is completely superfluous. If it weren't, however, there would be no mathematical justification in including it. Your formula should be sin(root), which is 1, not sin(theta). Jul 30, 2014 at 13:52
• I don't think this is correct because the $\sin\theta$ should be $\sin\pi/2$. See edit
– Ben
Jul 30, 2014 at 16:26

EDITED ANSWER: The delta distribution $\delta(x)$ is not unique. It is invariant under transformations of the form $\delta(x) \to f(x)\delta(x)$ where $f(0) = 1$. This is because it is really a distribution and not a function. It is mathematically improper to talk about $\delta(x)$ instead of $\int \delta(x)dx$. Derivations of the term you're interested in will not be unique either. You can show that

$$I = I\int \delta(r-a)dr = I\int \delta(r-a)\delta(\theta - \pi/2)drd\theta = I\int\frac{\delta(r-a)}{r}\delta(\theta - \pi/2)rdrd\theta.$$

From this expression, it is apparent that we can write the current as

$$I = \int J_{\phi}rdrd\theta \implies J_\phi = I \frac{\delta(r-a)}{r}\delta(\theta - \pi/2).$$

This result leaves out the $sin(\theta)$ and replaces $a$ with $r$. It really makes no difference because distributions are mathematically represented by $(T, F)$ where $F$ is a set of continuous functions and $T$ is the map from $F \to \mathbb{R}$. While it is common for physicists to represent distributions as just the mapping (i.e., $T = (T,F)$), this is somewhat false and leads to the non-unique representation of quantities like $\delta(x)$.

• I completely agree that it's unnecessary. So why would he put it in? Just to make our lives miserable??? ....what am I saying, that probably is the reason.
– Ben
Jul 30, 2014 at 16:31
• +1 I get what you were saying now. I'm just thick and needed someone to break it down I guess
– Ben
Aug 1, 2014 at 5:50
• I tried to do some more research on your question. It seems that everyone has a similar problem and most other textbooks/lectures leave out the sine term. Unless you come across something especially enlightening, I would just accept that it serves no purpose. Also, it doesn't have a physical interpretation either. The natural way to think about current density in this situation is "circulating" about the wire in the xy plane. Why would the angle from the plane influence this expression if it is 0 everywhere but along the circle? Aug 1, 2014 at 5:54

the correct answer can be easily obtained when you use the definition of \begin{equation} \overrightarrow{J}=\rho \overrightarrow{v} \end{equation}

with $\rho=q\delta(\overrightarrow{x}-\overrightarrow{x'})$, you will have:

\begin{equation} \overrightarrow{J}=q \overrightarrow{v} \delta(\overrightarrow{x}-\overrightarrow{x'}) \end{equation}

a small moving charge $dq$ yields to a small current density $d\overrightarrow{J}$, so

\begin{equation} d\overrightarrow{J}=dq \overrightarrow{v} \delta(\overrightarrow{x}-\overrightarrow{x'}) \end{equation}

the charge will move along the distance $d\overrightarrow{l}$:

\begin{equation} d\overrightarrow{J}=dq {{d\overrightarrow{l}}\over dt} \delta(\overrightarrow{x}-\overrightarrow{x'}) \end{equation}

but one defines $I={dq\over dt}$

\begin{equation} d\overrightarrow{J}= I {d\overrightarrow{l}}\delta(\overrightarrow{x}-\overrightarrow{x'}) \end{equation}

\begin{equation} \overrightarrow{J}= \int I \hspace{0.2cm}{d\overrightarrow{l}}\hspace{0.2cm}\delta(\overrightarrow{x}-\overrightarrow{x'}) \end{equation}

now you must write ${d\overrightarrow{l}}$ according the problem as well as the Dirac's delta function. For example, if you have a circular current loop with radius a in a x-y plane and with its center and the origin, you can write ${d\overrightarrow{l}}=a\hspace{0.1cm}d\phi \hat{\phi}$ and using cylindric coordinates for Dirac's delta:

\begin{equation} \overrightarrow{J}= \int I \hspace{0.2cm} a\hspace{0.1cm}d\phi \hat{\phi}\hspace{0.2cm}{\delta(r-a)\over r} \delta(\phi) \delta(z) \end{equation}

\begin{equation} \overrightarrow{J}= I \hspace{0.2cm} \hspace{0.1cm} \hat{\phi}\hspace{0.2cm}\delta(r-a) \delta(z) \end{equation}

In order to represent the current, we must make use of the azimuthal symmetry to reduce the problem to one involving two coordinates $$(u,z)$$, check the figure below. Where $$u$$ represents a point in the x-y plane. This is related to the spherical coordinates by $$u=r\sin\theta$$ $$z=r\cos\theta$$ Notice that rotating $$u$$ about the z-axis would give us the original problem. The current wrt u and z is given by $$\vec{J}=i\delta(u-R)\delta(z)\hat{\phi},$$ where $$i$$ is the current flowing in the wire of radius $$R$$ and $$\hat{\phi}$$ is the unit vector corresponding tangential to the ring. Now using the transformation law for delta functions, we obtain $$\vec{J}=i\dfrac{\delta(r-R)\delta(\theta-\frac{\pi}{2})}{|J|}\hat{\phi},$$ where $$|J|$$ is the Jacobian of the transformation. This is given by $$|J|=\vert det\begin{pmatrix}\dfrac{\partial u}{\partial r} & \dfrac{\partial z}{\partial r}\\ \dfrac{\partial u}{\partial \theta} & \dfrac{\partial z}{\partial \theta}\end{pmatrix}\vert=\vert det\begin{pmatrix}\sin\theta & \cos\theta\\ r\cos\theta & -r\sin\theta\end{pmatrix}\vert=r.$$
Plugging in, we obtain $$\vec{J}=i\dfrac{\delta(r-R)\delta(\theta-\frac{\pi}{2})}{r}\hat{\phi},$$ which is the desired result.