Past exam of the mathematics course of the University of Cambridge / 2014 / iii / Paper 54 / 4

For cylindrical rotation $u_{\varphi }=R\Omega (R)$, the equilibrium Euler momentum equation is $-R{\Omega }^2{\mathbf{e}}_R=-\nabla p/\rho -\nabla \Phi$. HenceThe integral is the centrifugal potential of cylindrical rotation. On a regular barotropic fluid branch define the specific enthalpy $h(\rho )$ by $dh=dp/\rho$. ThenFor a regular equation of state with $dp/d\rho >0$, $h$ is invertible, so $\rho$ and $p=p(\rho )$ depend only on $\Psi$ within a connected equilibrium region. In particular $dp/d\Psi =-\rho$. This local conclusion uses an invertible barotropic branch; distinct disconnected fluid regions can have different integration constants.

The linearized azimuthal Euler momentum equation isFor the time dependence $e^{-i\omega t}$ and $u_R^{\prime }=-i\omega {\xi }_R$, it givesThis expresses conservation of the displaced element's specific angular momentum. It applies directly to nonzero-frequency modes, with the zero-frequency limit taken in the displacement formulation.

The radial advective acceleration supplies $-2\Omega v_{\varphi }$, while the pressure force perturbation is $\delta \rho \nabla p/{\rho }^2-\nabla \delta p/\rho$. Eliminate $v_{\varphi }$ and retain the vertical equation. Under the Cowling approximation, $\delta \Phi =0$, soThe continuity equation gives $\delta \rho =-\nabla \cdot (\rho \mathbf{\xi })$. The Lagrangian adiabatic relation $\Delta p/p=\gamma \Delta \rho /\rho$, with $\Delta \rho =-\rho \nabla \cdot \mathbf{\xi }$, givesHere $\kappa$ is the radial epicyclic frequency. These formulas define the axisymmetric adiabatic displacement operator.

Let $a_{\xi }=\mathbf{\xi }\cdot \nabla \Psi$, $d_{\xi }=\nabla \cdot \mathbf{\xi }$ and ${\rho }_{\Psi }=d\rho /d\Psi$. Since $\nabla p=-\rho \nabla \Psi$, the pressure and mass density perturbations are $\delta p_{\xi }=\rho a_{\xi }-\gamma p{d}_{\xi }$ and $\delta {\rho }_{\xi }=-\rho d_{\xi }-{\rho }_{\Psi }{a}_{\xi }$. Integrating the pressure-gradient term by parts in the mass density-weighted inner product givesThe boundary term vanishes for regular admissible displacements because $p,\rho$ vanish there and $\delta p=\rho a-\gamma pd$. Completing the pressure square giveswhereThis effective-potential stratification coefficient is real. All coefficients of the bilinear form are real, so $⟨\eta ,\mathcal{F}\xi ⟩=⟨\mathcal{F}\eta ,\xi ⟩$: the operator is symmetric on the stated boundary domain, giving the usual self-adjoint realization of the stellar normal-mode problem.

Set $\eta =\xi$ and use $\mathcal{F}\xi =-{\omega }^2\xi$. The Cowling energy principle for a rotating barotropic star isThe Rayleigh quotient $Q[\xi ]/\int \rho |\xi {|}^2$ tests stability. If $Q\ge 0$ for every admissible displacement, no mode has ${\omega }^2<0$, so there is no exponentially growing mode. If an admissible trial displacement has $Q<0$, the Rayleigh-Ritz variational principle puts negative spectrum below zero; in the usual discrete stellar mode problem this gives a mode with $\omega =i\sigma$, $\sigma >0$. Equality allows neutral modes, rather than establishing strictly positive frequencies. A locally negative coefficient alone is not a complete instability proof: a trial function must also control the pressure and other positive terms.

The pressure-square term in the Cowling energy principle for a rotating barotropic star is nonnegative, and ${\mathcal{N}}^2>0$ makes its stratification term nonnegative. Write $j=R^2\Omega$ for specific angular momentum. ThenFor a regular star reaching the rotation axis, $j(0)=0$. The given $dj/dR>0$ therefore makes $j>0$ for $R>0$, and consequently ${\kappa }^2>0$. Equivalently, with the usual nonnegative angular-velocity convention the sign follows directly. All three energy terms are nonnegative, soThis is stability within the Cowling approximation used throughout. The orientation-independent rotational condition is $d(j^2)/dR\ge 0$, the Rayleigh discriminant criterion. If a fluid region excludes the axis and negative angular velocity is allowed, $dj/dR>0$ alone needs the additional sign of $j$; the squared-angular-momentum criterion avoids that ambiguity.