= Solution
The vanishing total <pressure> must be imposed at the displaced <free surface>, not at fixed $r=R$. The first-order <Taylor expansion> is
$$
0=p_0(R+\eta)+\pi(R+\eta,z,t)
=p_0(R)+\eta p'_0(R)+\pi(R,z,t)+O(\eta^2).
$$
Since $p_0(R)=0$ and $p'_0(R)=-2\pi G\rho^2R$, the <Lagrangian pressure perturbation> condition is
$$
\pi(R,z,t)=2\pi G\rho^2R\eta.
$$
The supplied gravitational perturbation and the combined potential, evaluated at $R$, then give
$$
\eta\Pi_0I_0(\kappa)=W(R)
=-4\pi G\rho R K_0(\kappa)I_0(\kappa)\eta+2\pi G\rho R\eta.
$$
Here $K_0$ is the <Modified Bessel function of the second kind>. Equating mode amplitudes and using part (c) yields
$$
\frac{\omega^2I_0(\kappa)}{kI'_0(\kappa)}
=4\pi G\rho R\left[\frac12-K_0(\kappa)I_0(\kappa)\right].
$$
Rearrange to obtain the <dispersion relation> for the <axisymmetric gravitational instability of an incompressible cylinder>:
$$
\boxed{-\frac{\omega^2}{4\pi G\rho}
=\kappa\frac{I'_0(\kappa)}{I_0(\kappa)}
\left[K_0(\kappa)I_0(\kappa)-\frac12\right].}
$$
Every quantity on the right is dimensionless, consistent with $G\rho$ having units $\mathrm{s^{-2}}$. The stabilizing term $1/2$ comes from displacing the background <hydrostatic pressure> profile; omitting the <Lagrangian pressure perturbation> would lose it.
Back to article page
