- #1
gitano
- 11
- 0
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
[tex] J_{\phi} = I\sin \theta' \delta (\cos \theta') \frac{\delta (r'-a)}{a} [/tex]
What I don't understand is where the [tex] \sin \theta' [/tex] 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
[tex] \vec{J}d^{3}x = Id\vec{l} [/tex] and plugging this into the equation for vector potential, but I want to understand how Jackson derives his version of the current density.
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
[tex] J_{\phi} = I\sin \theta' \delta (\cos \theta') \frac{\delta (r'-a)}{a} [/tex]
What I don't understand is where the [tex] \sin \theta' [/tex] 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
[tex] \vec{J}d^{3}x = Id\vec{l} [/tex] and plugging this into the equation for vector potential, but I want to understand how Jackson derives his version of the current density.