= Solution
In the inertial frame the steady convective acceleration is $-R\Omega^2\mathbf e_R$. For a <barotropic fluid>, $\nabla p/\rho=\nabla h$, where $h$ is the <barotropic enthalpy function>. The equilibrium momentum equation therefore gives
$$
\nabla\left(h+\Phi-\frac12\Omega^2R^2\right)=0,
\qquad\boxed{H=h+\Phi-\frac12\Omega^2R^2=\text{constant}}
$$
on each connected fluid region. Uniform rotation is essential: it makes the <centrifugal acceleration> derivable from the <centrifugal potential> $-\Omega^2R^2/2$.
The <Cowling approximation> sets the gravitational-potential perturbation to zero while retaining background gravity. The barotropic <pressure> force linearizes as $\delta(\nabla p/\rho)=\nabla\delta h$, with $\delta h=p'/\rho=c_s^2\rho'/\rho$. Denote its mode amplitude by $W$. The advective time derivative on a scalar mode is $\partial_t+\Omega\partial_\phi$, so the stated positive-frequency exponential produces $i\sigma$ with $\sigma=\omega+m\Omega$. Linearizing the radial centrifugal term produces $-2\Omega v_\phi$, while linearizing the azimuthal convective term produces $2\Omega v_R$. Thus
$$
i\sigma v_R-2\Omega v_\phi=-W_R,\qquad
i\sigma v_\phi+2\Omega v_R=-imW/R,\qquad
i\sigma v_z=-W_z.
$$
These are the pressure-gradient and <Coriolis acceleration> terms in the corotating-frame form. Independently, linearizing <continuity equation> gives
$$
\boxed{i\sigma\frac{\rho W}{c_s^2}
=-\frac1R\partial_R(R\rho v_R)-\frac{im\rho v_\phi}R-\partial_z(\rho v_z).}
$$
The equilibrium <mass density> is axisymmetric, so no additional azimuthal background-density derivative appears.
For $\sigma\ne0$ and $D=4\Omega^2-\sigma^2\ne0$, invert the horizontal momentum system:
$$
v_R=-\frac{i}{D}\left(\sigma W_R+\frac{2m\Omega W}R\right),\qquad
v_\phi=\frac1D\left(2\Omega W_R+\frac{m\sigma W}R\right),\qquad
v_z=\frac i\sigma W_z.
$$
Inserting these into the <continuity equation>, dividing by $i\sigma$ and multiplying by $D$ gives
$$
D\frac{\rho W}{c_s^2}
=\frac1R\partial_R(R\rho W_R)-\frac{m^2\rho W}{R^2}
-\frac D{\sigma^2}\partial_z(\rho W_z)
+\frac{2m\Omega}{R\sigma}\rho_RW.
$$
The two terms proportional to $\rho W_R$ cancel; this cancellation leaves precisely the background-density derivative in the last term. Since $-D/\sigma^2=1-4\Omega^2/\sigma^2$, this is exactly
$$
\boxed{D\frac{\rho W}{c_s^2}
=\frac1R\partial_R(R\rho W_R)-\frac{m^2\rho W}{R^2}
+\left(1-\frac{4\Omega^2}{\sigma^2}\right)\partial_z(\rho W_z)
+\frac{2m\Omega}{R\sigma}\rho_RW.}
$$
The singular frequency cases require the original <velocity> equations rather than this inversion.
In the low-frequency <anelastic approximation for a rotating barotropic star>, neglect the left side. Spherical background <mass density> satisfies $\rho_R/R=\rho_z/z$, understood by its smooth limiting form on the equatorial plane. For $W=zR^m$,
$$
\frac1R\partial_R(R\rho W_R)-\frac{m^2\rho W}{R^2}
=mzR^{m-1}\rho_R,\qquad
\partial_z(\rho W_z)=R^m\rho_z.
$$
The remaining equation reduces to
$$
R^m\rho_z\left[m+1+\frac{2m\Omega}\sigma-\frac{4\Omega^2}{\sigma^2}\right]=0.
$$
Writing $q=2\Omega/\sigma$, the bracket factors as $(m+1-q)(q+1)$. The regular desired branch is therefore
$$
\boxed{\sigma=\frac{2\Omega}{m+1},\qquad W=zR^m.}
$$
The other algebraic factor corresponds to $\sigma=-2\Omega$, where the eliminated horizontal system is singular; it is not classified by this pressure-equation inversion. A constant-density background makes the bulk factor identically zero but does not invalidate the displayed solution.
For the regular branch, an explicit <velocity> check is particularly informative. Up to a common mode normalization,
$$
(v_R,v_\phi,v_z)=\frac{m+1}{2\Omega}\bigl(-izR^{m-1},\ zR^{m-1},\ iR^m\bigr).
$$
It satisfies $Rv_R+zv_z=0$ and $\nabla\cdot\mathbf v=0$. Thus $\nabla\cdot(\rho\mathbf v)=0$ for every spherical <mass density> profile, directly verifying the anelastic continuity condition. The motion is tangential to spherical shells and is the <sectoral inertial mode of a slowly rotating barotropic star>. For single-valued azimuthal modes, $m$ is a positive integer. With the source's exponential convention,
$$
\omega=\sigma-m\Omega=-\frac{(m-1)(m+2)}{m+1}\Omega.
$$
For $m\ge2$ the pattern is retrograde relative to the star but prograde in the inertial frame. This is a leading slow-rotation anelastic mode, not an exact solution of the compressible equations whose left-hand term was discarded.
Back to article page
