= Solution
The <azimuthal derivative of a cylindrical vector Fourier mode> must include basis rotation: the azimuthal dependence refers to the components in the rotating cylindrical basis. Since $\partial_\phi\widehat{\mathbf s}=\widehat{\boldsymbol\phi}$ and $\partial_\phi\widehat{\boldsymbol\phi}=-\widehat{\mathbf s}$, a vector amplitude $\mathbf v\propto e^{im\phi}$ obeys
$$
\partial_\phi\mathbf v=im\mathbf v+\widehat{\mathbf z}\times\mathbf v.
$$
Also $\mathbf B=(J/2)(-y,x,0)$ in <Cartesian coordinates>, so $(\mathbf v\cdot\nabla)\mathbf B=(J/2)\widehat{\mathbf z}\times\mathbf v$. Consequently
$$
\boxed{(\mathbf B\cdot\nabla)\mathbf b+(\mathbf b\cdot\nabla)\mathbf B
=J\widehat{\mathbf z}\times\mathbf b+\frac{imJ}{2}\mathbf b,}
$$
and the basis-rotation terms cancel in the <ideal magnetohydrodynamic induction equation>:
$$
\boxed{(\mathbf B\cdot\nabla)\mathbf u-(\mathbf u\cdot\nabla)\mathbf B=\frac{imJ}{2}\mathbf u.}
$$
Here the printed $J$ is the coefficient in the specified <magnetic field>: $\nabla\times\mathbf B=J\widehat{\mathbf z}$. If $J_{\rm phys}$ denotes SI <electric current density>, then $J=\mu_0J_{\rm phys}$. The field and all subsequent printed coefficients are mutually consistent with this normalization.
For a <normal mode> with $\sigma\ne0$, induction gives $\mathbf b=imJ\mathbf u/(2\sigma)$. Substitution into the momentum equation, including the <Coriolis force>, gives \b[the reduced velocity equation]
$$
\boxed{\left(\sigma+\frac{m^2J^2}{4\mu_0\rho\sigma}\right)\mathbf u
+\left(2\Omega-\frac{imJ^2}{2\mu_0\rho\sigma}\right)\widehat{\mathbf z}\times\mathbf u=-\nabla p.}
$$
Thus the coefficients are the printed $\lambda$ and $\nu$. Let $C=J^2/(\mu_0\rho)$. Inserting the allowed relation $\lambda=i\delta\nu$ and multiplying by $\sigma$ yields
$$
\sigma^2-2i\delta\Omega\sigma+\frac C4(m^2-2m\delta)=0.
$$
Hence \b[the <uniform-current rotating magnetohydrodynamic wave dispersion> is]
$$
\boxed{\sigma=i\delta\Omega\pm\sqrt{\frac C4(2m\delta-m^2)-\delta^2\Omega^2}.}
$$
The radicand is real. For $|m|\geq2$, $2m\delta-m^2<0$ because $|\delta|<1$, so both roots are purely imaginary. For $m=0$ the roots of the quadratic are also imaginary; any zero-frequency case must be checked in the original equations, since the elimination divided by $\sigma$. A positive real part requires
$$
2m\delta-m^2>0,\qquad
\boxed{\frac{J^2}{\mu_0\rho}>\frac{4\delta^2\Omega^2}{2m\delta-m^2}.}
$$
For integer $m$ this is possible only when $|m|=1$ and $m\delta>1/2$. Taking the conventional representative $m\geq0$ gives the printed exception $m=1$, with $\delta>1/2$ and sufficiently large $J$. Literally allowing negative integers also gives $m=-1$, $\delta<-1/2$; <complex conjugation> maps $(m,\delta,\sigma)$ to $(-m,-\delta,\overline\sigma)$, so these describe the conjugate real disturbance. Purely imaginary roots describe neutral oscillatory modes; a repeated root at threshold does not itself prove boundedness of arbitrary initial perturbations.
Back to article page
