At marginal stability write $s=i\omega$ with real $\omega$. The imaginary part of the pressure factor is
$$
\operatorname{Im}\frac{1+i\gamma\tau\omega}
{1+i\tau\omega}
=\frac{(\gamma-1)\tau\omega}{1+\tau^2\omega^2}.
$$
For $\gamma>1$, $\tau>0$, and nonzero $k$, the dispersion relation can have zero imaginary part only when $\omega=0$. Thus every marginal mode has
$$
\boxed{\operatorname{Im}s=0,\qquad s=0}.
$$
The marginal equation is consequently
$$
c_0^2k^2-2\pi G\Sigma_0|k|+\Omega^2=0.
$$
Defining the isothermal Toomre parameter $Q_0=c_0\Omega/(\pi G\Sigma_0)$, its two roots are
$$
\boxed{|k|_\pm=\frac{\pi G\Sigma_0}{c_0^2}
\left(1\pm\sqrt{1-Q_0^2}\right)}.
$$
They exist when $Q_0\leq1$, and the interval between them is unstable when $Q_0<1$. Finite thermal relaxation therefore leaves the onset criterion
$$
\boxed{Q_0<1}
$$
equal to the isothermal criterion, regardless of $\gamma$ and the nonzero value of $\tau$.

Solved by gpt-5.6-sol high.

Solution

Past exam of the mathematics course of the University of Cambridge / 2026 / iii / Paper 321 / 3 / b / iii / Solution

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2026/iii/paper-321/3/b/iii/solution