Current Density of Circular Current Loop (Jackson)

[gitano]

Hi,

I was reading through Jackson's Electrodynamics trying to reason through example 5.5 for the vector potential of a circular current loop of radius a centered at the origin. I pretty much understand everything except when he defines the current density as

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

What I don't understand is where the \sin \theta' term comes from. I understand that the delta functions restrict flow to a radius of 'a' and a theta value of pi/2 and that we divide by 'a' to account for the normalization of the dirac delta function in spherical coordinates. Is the sine term part of this normalization?

Also, later in the example he goes on to expand in spherical harmonics instead of using elliptic integrals. When he does this it is as if he completely ignores the sine term [eq. (5.43) 3rd Ed.] when he plugs everything in.

I know that you can arrive at the correct expression by simply using
\vec{J}d^{3}x = Id\vec{l} and plugging this into the equation for vector potential, but I want to understand how Jackson derives his version of the current density.

 

[RedX]

Hi,

I was reading through Jackson's Electrodynamics trying to reason through example 5.5 for the vector potential of a circular current loop of radius a centered at the origin. I pretty much understand everything except when he defines the current density as

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

What I don't understand is where the: \sin \theta' term comes from.

\delta (\cos \theta')=\frac{\delta(\theta'-\pi/2)}{sin \theta'}

So the sin(theta')'s cancel.

 

[Bill_K]

What do you know, I have an older edition, and the sin θ' does not appear in either place! Looks like he corrected one equation and not the other. Doesn't matter though, since δ(cos θ') sets θ' = π/2 anyway.

As far as the reasoning behind it, J_φ is a current density, to be integrated over one of the planes φ = const. In this plane you use *plane* polar coordinates, in which the area element is r dr dθ'. The corresponding delta function is (1/a)δ(r)δ(θ'). However when you go to calculate A, you're going to want to use spherical coordinates, where the volume element contains d(cos θ'). Anticipating that, Jackson changes variables ahead of time:

δ(θ') → sin θ' δ(cos θ')

 

[Meir Achuz]

I know that you can arrive at the correct expression by simply using
\vec{J}d^{3}x = Id\vec{l} and plugging this into the equation for vector potential, but I want to understand how Jackson derives his version of the current density.

As you say, it is simple and correct to use Idl.
J just wanted to add some complexity there.

 

[vanhees71]

Obviously, Jackson works in spherical coordinates (I'd choose cylinder coordinates, but perhaps it's even more clever in spherical coordinates; I've to think about that). For the circular loop around the origin with radiuis, a in the xy plane, you have only a component in \varphi-direction, and for an infinitesimally thin wire you have

\vec{j}(r,\vartheta,\varphi)=\vec{e}_{\varphi} I \frac{1}{a} \delta(\vartheta-\pi/2) \delta(r-a)

Integrating this over a little surface perpendicular to the wire, you immediately see that this is the correct expression for the magnitude since this gives the total current I.

Now, in spherical coordinates, it's sometimes more convenient to rewrite the Dirac distribution wrt. to the angle as

\delta(\vartheta-\pi/2)=\sin \vartheta \delta[\cos(\vartheta)].

Here, it's important to note the validity range of the spherical coordinates, which cover not the whole \mathbb{R}^3, but you have to take out the polar axis (in the standard coordinates the z axis of the corresponding Cartesian system):

r>0, \quad 0 < \vartheta < \pi, \quad 0 \leq \varphi<2 \pi.

 

[RedX]

Obviously, Jackson works in spherical coordinates (I'd choose cylinder coordinates, but perhaps it's even more clever in spherical coordinates; I've to think about that). For the circular loop around the origin with radiuis, a in the xy plane, you have only a component in \varphi-direction, and for an infinitesimally thin wire you have

\vec{j}(r,\vartheta,\varphi)=\vec{e}_{\varphi} I \frac{1}{a} \delta(\vartheta-\pi/2) \delta(r-a)

Integrating this over a little surface perpendicular to the wire, you immediately see that this is the correct expression for the magnitude since this gives the total current I.

Now, in spherical coordinates, it's sometimes more convenient to rewrite the Dirac distribution wrt. to the angle as

\delta(\vartheta-\pi/2)=\sin \vartheta \delta[\cos(\vartheta)].

Here, it's important to note the validity range of the spherical coordinates, which cover not the whole \mathbb{R}^3, but you have to take out the polar axis (in the standard coordinates the z axis of the corresponding Cartesian system):

r>0, \quad 0 < \vartheta < \pi, \quad 0 \leq \varphi<2 \pi.

That is some beautiful font. Can you \var every greek letter?

 

[martiniux]

I agree that he should use \vec{J}(r,\vartheta,\varphi)=I \frac{1}{a} \delta(\vartheta'-\pi/2) \delta(r'-a)\vec{e}_{\varphi}
where r',\theta',\phi'are spherical coordinates for the current

At risk of repeating what some of you guys have already posted, I'll write the procedure I'd usually follow:

*) As current is restricted to flow in ring, set
\vec{J} = J_{\varphi} \vec{e}_{\varphi} where:
J_{\varphi} = \alpha \delta(r'-a) \delta(\vartheta' -\pi/2)
*) To find alpha, integrate both sides over the whole space in spherical coordinates (and remember J = I/area):
<br /> \iiint JdV&#039; = \int I_{\varphi}dl&#039; =\int_0^{2\pi}I_{\varphi}a d\varphi&#039; =<br /> \int_0^{\infty} \int_0^{\pi} \int_0^{2\pi}\alpha \delta(r&#039;-a) \delta(\vartheta&#039; -\pi/2) r&#039;^2 sin\theta&#039; dr&#039;d\theta&#039; d\varphi&#039;<br />
*) Integrating, we have:
2\pi a I_{\varphi} = \alpha 2\pi a^2 sin(\pi/2) -&gt; \alpha = I_{\varphi} /a

*) Also as \delta(cos{\vartheta&#039;}) = \frac{\delta({\vartheta&#039; - \pi/2})}{\frac{d}{d{\vartheta&#039;}}\left(cos({\vartheta&#039;})\right) | _{\vartheta&#039; = \pi/2}} -&gt; \delta({\vartheta&#039; - \pi/2}) = \delta(cos{\vartheta&#039;})

And thus after replacing, we obtain the expression claimed above.

(Note, strictly we should obtain a negative sign in the last equation (due to the derivative of cos), but we can always redefine our direction vector. Also, I've dropped the sub index phi for the current)

 

[dish_boggett]

Also: Jackson uses the notation

dcos(θ)d\phi where dcos(θ) = sin(θ)dθ​

Normaly you'd write:

δ(cosθ-cosθ')​

in this case:

δ(cosθ-cos\pi/2)​

so the argument is 0 when θ=\pi/2 and the delta function is 1



I think this is true:
having δ(cosθ-cosθ')sinθdcosθ allows you to measure θ from the x axis. For δ(cosθ) cosθ is 0 when θ = \pi/2 and sinθ=1