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.

 

[Nickie]

The vector potential of a spinning sphere is a complex problem and there are multiple ways to approach it. It is true that the problem becomes simpler if you tilt the sphere so that it is spinning in the xz plane, as Griffiths suggests. This is because when the sphere is tilted, the angular velocity vector is no longer aligned with the z-axis, and therefore the cross product in the current density expression becomes non-zero.

However, attempting to solve the problem using the z-axis as the axis of rotation can also be done, but as you have observed, it yields an answer of 0. This is because the integral of sine and cosine over a full period is indeed 0. In this case, the solution is not physically meaningful as it implies that there is no magnetic field generated by the spinning sphere.

In an exam situation, it is important to carefully consider the given problem and choose the appropriate approach. In this case, it would be more appropriate to follow Griffiths' suggestion and tilt the sphere to simplify the problem. However, it is always good practice to understand the reasoning behind the suggested approach and to explore alternative methods of solving the problem.