Vector Potential of Spinning Sphere

[Shinobii]

Homework Statement

I am just wondering, in Griffiths text, he solves this problem for a spinning shell. He states that the problem is easier if you tilt the sphere so it is spinning in the xz plane.

Homework Equations

When solving for the current density, Griffiths writes,

$$
(\omega \times r') =
\begin{vmatrix}
\hat{x} & \hat{y} & \hat{z} \\
\omega \sin(\psi') & 0 & \omega \cos(\psi') \\
r' \sin(\theta') \cos(\phi') & r' \sin(\theta') \sin(\phi') & r' \cos(\theta')
\end{vmatrix}
$$

The Attempt at a Solution

But suppose we want to try it on the z-axis such that,

$$
(\omega \times r') =
\begin{vmatrix}
\hat{x} & \hat{y} & \hat{z} \\
0 & 0 & \omega \\
r' \sin(\theta') \cos(\phi') & r' \sin(\theta') \sin(\phi') & r' \cos(\theta')
\end{vmatrix}
$$

Doing this however yields an answer of 0! Since,

$$
\int_0^{2 \pi} sin(\phi') d\phi' = \int_0^{2 \pi} cos(\phi') d\phi' = 0
$$

Does anyone have insight to if this problem is solvable using the second method? I ask this, because this is what I would do in an exam situation!

 

[Mandelbroth]

First, you might consider losing the determinant calculation. If you are at this level of mathematics, you might as well use linear algebra rather than a mnemonic, if not using the tensor definition of the cross product.

Define the cross product of two vectors, $\vec{a}$ and $\vec{b}$, as $\vec{a}\times\vec{b}=\left[\begin{matrix} 0 & -a_3 & a_2 \\ a_3 & 0 & -a_1 \\ -a_2 & a_1 & 0 \end{matrix}\right]\left[\begin{matrix} b_1 \\ b_2 \\ b_3 \end{matrix}\right]$, where $a_i$ is the i^th component of the $\mathbb{R}^3$ vector $\vec{a}$, with $b_i$ defined similarly.

Assuming that $\omega$ and r' are not scalar multiples of each other, the cross product should not reduce to the zero (probably not 0!=1 either) vector.

 

[Shinobii]

Hmm, mathematics aside (I completely disagree with you, the mnemonic is simple to understand physically, while tensor notation is required for more difficult problems (such as relativistic electrodynamics), this problem is rather simple) you will still get zero. I am not worried about my terms cancelling out, it is the integration that sets my answer to zero. . .

 

[Shinobii]

Ah, I see now that I can exploit spherical harmonics and symmetry which leaves me with the correct result! Since,

$$
\sin(\theta')\cos(\phi') = - \sqrt{\frac{8 \pi}{3}}{\rm{Re}}[Y_{1,1}(\theta', \phi')]
$$

Then by orthogonality everything should fall into place (I have not done this but I will do so right away and verify that it works).

Yep, this method works fine and dandy! Thanks again Mandelbroth, I appreciate you taking the time to reply.