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To give context, suppose I have an uncharged and uniformly magnetized sphere with its magnetization in the $z$ direction with magnitude $M_0$. For this object, the volume bound current density $\mathbf{J}_b$ is obviously $0$, but it has a nonvanishing surface bound current density equal to$$\mathbf{K}_b=M_0\:\mathbf{e}_z \times (\sin\theta\cos\phi\:\mathbf{e}_x + \sin\theta\sin\phi\:\mathbf{e}_y + \cos \theta \:\mathbf{e}_z)=M_0\sin\theta\:\mathbf{e}_\phi.$$

This object is equivalent to a charged spherical shell rotating at constant angular velocity $\omega$ in the $z$ direction such that $$\sigma \omega R=M_0$$ because such charge distribution produces a surface current density equal to $\mathbf{K}_b$ of the magnetized sphere, and thus they produce the same magnetic fields. Here are my questions:

1) If I stand above the rotating sphere on the $z$ axis and rotate at the same frequency $\omega$, then my intuition says that I measure a zero $\mathbf{B}$ field, since all charges residing in the sphere are at rest in my view. Is this reasoning correct? What is, explicitly, the transformation of $\mathbf{B}$ (or better yet, $F_{\mu\nu}$) between an inertial and a rotating frame?

2) Because the stationary uniformly magnetized sphere is essentially equivalent to the spinning charged spherical shell, I concur that I should also measure a zero $\mathbf{B}$ field if my angular velocity is $\omega\mathbf{e}_z $ standing above the magnetized sphere on the $z$ axis. I would go as far as to conclude that the effective magnetization is $\mathbf{M}=\mathbf{0}$ in my rotating frame. How does magnetization transform, if at all?

Thank you in advance for any help!

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  • $\begingroup$ It might be worth mentioning that the rotating frame is non-inertial, and hence Maxwell's equations will not have the same form. Furthermore, it is not possible to synchronize clocks in a rotating reference frame, which might lead to difficulties in defining the electric and magnetic fields from the field tensor $\endgroup$ Mar 29, 2023 at 20:34
  • $\begingroup$ If your sphere is globally uncharged, you’ll need to have a stationary oppositely charged sphere to cancel the charge of the spinning sphere. “Non relativistically” (which is not possible with Maxwell’s theory) you would therefore expect the same configuration just with opposite velocity in the rotating frame. $\endgroup$
    – LPZ
    Mar 29, 2023 at 21:10
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    $\begingroup$ I am not sure, however, I guess that the non-inertial observer in the second case would detect a non-zero magnetic field. $\endgroup$ Mar 29, 2023 at 21:52

2 Answers 2

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This object is equivalent to a charged spherical shell rotating at constant angular velocity

No, such an object is not equivalent to a charged spherical shell. Both objects would have the same current density, but they have dramatically different charge densities.

To approach your questions 1) and 2) the best path is to use the magnetization-polarization tensor (in units where $c=1$: $$M^{\mu \nu }=\left( \begin{array}{cccc} 0 & P_x & P_y & P_z \\ -P_x & 0 & -M_z & M_y \\ -P_y & M_z & 0 & -M_x \\ -P_z & -M_y & M_x & 0 \\ \end{array} \right)$$ This is a standard tensor, so we can transform it to the rotating frame using the standard tensor transformation rules: $$M^{\mu \nu }=\left( \begin{array}{cccc} 0 & \cos (T \omega ) P_x+\sin (T \omega ) P_y & \cos (T \omega ) P_y-\sin (T \omega ) P_x & P_z \\ -\cos (T \omega ) P_x-\sin (T \omega ) P_y & 0 & -M_z+\omega \sin (T \omega ) \left(X P_y-Y P_x\right)+\omega \cos (T \omega ) \left(X P_x+Y P_y\right) & -\sin (T \omega ) M_x+\cos (T \omega ) M_y+Y \omega P_z \\ \sin (T \omega ) P_x-\cos (T \omega ) P_y & M_z+\omega \sin (T \omega ) \left(Y P_x-X P_y\right)-\omega \cos (T \omega ) \left(X P_x+Y P_y\right) & 0 & -\cos (T \omega ) M_x-\sin (T \omega ) M_y-X \omega P_z \\ -P_z & \sin (T \omega ) M_x-\cos (T \omega ) M_y-Y \omega P_z & \cos (T \omega ) M_x+\sin (T \omega ) M_y+X \omega P_z & 0 \\ \end{array} \right)$$ where the $(T,X,Y,Z)$ are the coordinates in the rotating reference frame. This is a fairly complicated expression, however for your specific scenario all of the magnetization and polarization terms vanish except for $M_z$. So the above simplifies to $$M^{\mu \nu }=\left( \begin{array}{cccc} 0 & 0 & 0 & 0 \\ 0 & 0 & -M_z & 0 \\ 0 & M_z & 0 & 0 \\ 0 & 0 & 0 & 0 \\ \end{array} \right)$$ meaning that for this particular geometry the magnetization is unchanged in the rotating reference frame. The magnetization does not vanish in the rotating reference frame.

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Note that your result for inertial frame magnetization is a vector spherical harmonic, ${\bf \Phi}_{10}$ (https://en.wikipedia.org/wiki/Table_of_spherical_harmonics).

Maxwell's equation in the conventional form are only valid in inertial frames. In a rotating frame with $|\omega r|\lt c$:

$$ {\bf \nabla}\times{\bf E} = -\dot{\bf B} $$ $$ {\bf \nabla}\cdot{\bf E} = \frac{\rho}{\epsilon_0} + {\bf\nabla}\cdot\ ({\bf v} \times {\bf B}) $$

$$ {\bf\nabla}\cdot {\bf B}=0$$ $$ {\bf\nabla}\times {\bf B}=\mu_0\epsilon_0\dot{\bf E}+\mu_0{\bf I} - {\bf\nabla}\times\big[{\bf v}\times({\bf E}-{\bf v}\times {\bf B}) \big] -{\bf v}\times({\bf\nabla}\times{\bf E}) $$

where ${\bf v} = {\bf\omega}\times{\bf r}$.

So that should simplify.

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