Past exam of the formation of Structure in the Universe course of the University of Cambridge / 2016 / Paper 1 / 2

For the binding energy of a truncated singular isothermal sphere, build the mass distribution from concentric shells. Its enclosed mass and mass element areThe interaction of a newly added shell with the interior contributes $-GM(r)\mathrm{d}M/r$. Counting each pair once,Eliminating $k$ givesThe central density is singular, but the mass and this energy integral are finite. The result assumes the stated truncation and no additional gravitating component.

The gravitational potential of a truncated singular isothermal sphere is a separate quantity from this shell-assembly energy allocation. With zero potential at infinity, both interior and exterior shells contribute:It matches $-G{M}_0/r_0=-2k$ at the edge. The equivalent total energy is $W=\frac{1}{2}\int \rho {\Phi }_{\mathrm{grav}}\mathrm{d}V$; the half prevents double counting. By contrast, $\int \rho [-{\Phi }_{\mathrm{grav}}]\mathrm{d}V$ counts the escape work of test parcels in a fixed potential and equals $-2W$. Confusing these quantities would change a local feedback threshold.

Introduce the constant $\nu >0$ by $\psi (r,t)=\nu {\rho }_g(r,t)$, and set $q=\gamma E_{\mathrm{SN}}$, the injected energy per unit mass of newly formed stars. Before a local wind switches on, no return or matter exchange impliesIn this independent-volume supernova quenching model, the accumulated injection per volume isFor any specified positive binding energy per unit remaining gas mass $B(r)$, the local threshold isGas conversion must be included in this comparison; replacing the integrated star formation by its initial rate times $t$ is only a short-time approximation.

For a quenching front in a fixed galactic potential, take $B(r)=-{\Phi }_{\mathrm{grav}}(r)=2k[1+\ln (r_0/r)]$. This treats the gravitating profile as fixed: newly formed stars initially replace gas at the same radius, and later changes of the potential caused by expelled gas are neglected. Gas first becomes unbound at the outer edge, atThereafter the inner edge of the quenched exterior region isThere is no quenched radius inside the galaxy before this onset. The remaining star-forming region is $0<r<r_{\mathrm{quench}}$; the front moves inward. The ideal logarithmically divergent central potential delays complete removal to infinite time in this fixed-profile model. It should not be interpreted as a physical unresolved central core.

The local binding-energy allocation in galactic feedback also admits a different convention, because it calls $\varphi$ an energy of a volume element without specifying how the nonlocal self-gravitating energy is assigned to that element. If the intended prescription is instead the shell-assembly contribution from part (a),then $B_{\mathrm{shell}}=2k$ at every radius. The same time $t_{\mathrm{start}}$ applies everywhere: all shells quench simultaneously, and there is no continuously propagating radial front. This is a shell-energy toy model, not the escape potential including exterior shells.

Both results follow from explicit definitions. A convention assigning $\frac{1}{2}{\rho }_g{\Phi }_{\mathrm{grav}}$ locally replaces $B$ by $B/2$ in $Q(t)=B(r)$; it is again an energy allocation rather than test-parcel escape. Updating the potential after gas loss or choosing a fixed initial-gas binding budget instead of the remaining-gas budget would require an additional closure. The stated independent-volume assumptions do not uniquely choose between these. The displayed escape-potential result and shell prescription must therefore not be presented as interchangeable universal laws.

For the shallower initial density profile,Including outer shells, the initial gravitational potential isIts central value is finite. With the same remaining-gas escape criterion and fixed-potential approximation as above,This front is inside the galaxy only when $k^{\prime }{r}_0\le Q(t)\le 2k^{\prime }{r}_0$. ThusQuenching still moves from the outside inward, but now reaches the centre in a finite time. The contrast with the $r^{-2}$ escape model is the finite central potential, not a reversed gravitational force.

Under the alternative shell-energy quenching prescription, the conclusion is different:Here the quenched region is inside the front and grows from the centre outward, unlike the simultaneous quenching of the isothermal shell-allocation model. Its edge reaches $r_0$ at ${\nu }^{-1}\ln (1+k^{\prime }{r}_0/q)$. This change of direction is a consequence of the chosen local energy model: $GM(r)/r$ increases with radius for this profile, whereas the true binding potential including all outer shells always becomes shallower outward. The source does not identify which local allocation it intends; keeping the conventions explicit resolves the different conditional predictions without discarding the exterior-shell potential.

To test radial stability of an axisymmetric circular orbit, let $\Phi (r)$ be the time-independent gravitational potential per unit stellar mass in the midplane. Axial symmetry conserves the specific angular momentum $j=r^2\dot{\phi }$. The radial equation isA circular orbit at $r_0$ obeys $j_0^2=r_0^3{\Phi }^{\prime }(r_0)$. Choose the guiding circular orbit with the same conserved angular momentum as the perturbed star. Setting $r=r_0+x$ and expanding at fixed $j=j_0$ giveswhere the radial epicyclic frequency isHere ${\Omega }^2(r)={\Phi }^{\prime }(r)/r$ and $j_c=r^2\Omega$ refer to the family of circular orbits. They are differentiated as functions of radius only after the perturbation equation has been expanded at conserved stellar $j$.

For a radially stable circular orbit, ${\kappa }^2>0$, so to linear orderThus the orbit performs small simple harmonic radial oscillations about its guiding radius.

Symmetry alone is insufficient to guarantee stability: ${\kappa }^2<0$ gives exponentially growing displacement, and ${\kappa }^2=0$ is marginal. The intended statement assumes a stable unperturbed orbit.

For a Kepler potential $\kappa =\Omega$; for a flat circular-speed curve $\kappa =\sqrt{2}\Omega$; for solid-body rotation $\kappa =2\Omega$. Conservation of $j$ also gives $\dot{\phi }\simeq {\Omega }_0(1-2x/r_0)$, explaining the accompanying azimuthal epicycle rather than a purely radial motion in the inertial plane.