Past exam of the astrophysics course of the University of Oxford / 2022 / 2

There are two different physical interpretations. The supplied logarithmic integral signals free fall in a fixed singular isothermal sphere, with a test particle moving through a frozen background. Write $C={\rho }_0{r}_0^2$. By the shell theorem, the enclosed mass is $M(r)=4\pi Cr$ and the radial equation of motion isThe second equality follows by multiplying by $\dot{r}$ and using the initial rest condition. The free fall trajectory is given implicitly byThus the intended fixed-profile answer isA collapsing material shell gives a different answer. If the entire cloud undergoes pressureless gravitational collapse from rest, the edge follows the fixed enclosed mass $M_0=4\pi {\rho }_0{r}_0^3$, not the instantaneous fixed-profile value $4\pi Cr$. With no shell crossing, $\ddot{r}=-G{M}_0/r^2$, soThis is a concrete counterexample to treating the fixed-profile result as the exact collapse time of the whole nebula. The initial mean density inside $r_0$ is $3{\rho }_0$, consistent with the usual free-fall time of a uniform sphere expressed using that mean density. The logarithmic integral is valid for the explicitly frozen background approximation.

Assume an isolated, bound, nonrotating Newtonian gas cloud, negligible surface pressure, negligible magnetic and radiation stresses, no persistent bulk flow, and a stationary or time-averaged virialized state. The stellar virial theorem then reduces towhere $\Omega <0$ is the total gravitational potential energy. Define the volume-averaged pressure $\overline{P}=V^{-1}\int PdV$ and average gravitational energy density $u_g=\Omega /V$. HenceEquivalently, for a nonrelativistic monatomic gas, the thermal internal energy $U=(3/2)\int PdV$ obeys $2U+\Omega =0$. This second derivation must not count the thermal motions again as a separate bulk kinetic energy. If the external pressure is $P_s$, the correct result is $\overline{P}-P_s=-u_g/3$; non-negligible bulk motion adds $2T_{\mathrm{bulk}}$ to the virial theorem. The initial sharp-edged $r^{-2}$ profile need not survive virialization; maintaining that profile hydrostatically to its edge would generally require a surface term.

For a family of spherical profiles with a fixed dimensionless shape, write the gravitational potential energy as $\Omega =-\alpha G{M}^2/R$, where $\alpha >0$ depends on that shape. The virial theorem in b and the sphere volume $V=4\pi R^3/3$ giveFor a uniform density, $\alpha =3/5$; formally integrating $-{\int }_0^{R}GM(r)dM(r)/r$ for $M(r)=Mr/R$ gives $\alpha =1$. Neither the virial theorem nor the initial profile guarantees a specific final shape, so the proportionality requires the stated homologous-profile assumption. Surface pressure must be handled consistently with b before using either coefficient as an equilibrium model.

For a regular spherical star, let ${\rho }_c=\rho (0)$ and $P_c=P(0)$. The stellar mass conservation equation and stellar hydrostatic equation areAssuming the smoothness needed for the requested Taylor series, write $M=a_0+a_{1}r+a_2{r}^2+a_3{r}^3+\cdots$. There is no point mass at the centre, so $a_0=0$, and matching the stellar mass conservation equation gives $a_1=a_2=0$, $a_3=4\pi {\rho }_c/3$. Substitution in the stellar hydrostatic equation gives $P^{\prime }(r)=-(4\pi G/3){\rho }_c^{2}r+O(r^3)$, henceThe remainder orders use a smooth spherically symmetric scalar density, for which $\rho (r)={\rho }_c+O(r^2)$. Merely finite continuous density gives the same leading coefficients with $o(r^3)$ and $o(r^2)$ remainders. In particular $P^{\prime }(0)=0$, despite the apparent $r^{-2}$ in the stellar hydrostatic equation.

For a gas-pressure-supported stellar core of fixed mass and composition, neglect boundary terms for the illustrative whole-core estimate. The virial theorem gives $E=U+\Omega =-U$, and the monatomic gas gives $U=3M_c{k}_B\overline{T}/(2\mu m_u)$. Thus the effective heat capacity along the equilibrium sequence isThis negative heat capacity of a virialized gravitational system means that net energy loss makes a self-gravitating core contract and heat. It is different from the positive local material specific heat capacity. On the main sequence, hydrogen burning supplies most of the radiated energy, so rapid contraction is arrested. Hydrogen depletion increases the mean molecular weight; slow structural readjustment usually raises the central temperature rather than making it fall.

After core hydrogen exhaustion, a massive stellar core contracts and heats more rapidly until a later nuclear fuel ignites. There is no general reversal to cooling simply upon leaving the main sequence. If radiation pressure becomes dominant, the internal energy becomes $U=U_{\mathrm{gas}}+U_{\mathrm{rad}}$ with $3\int PdV=2U_{\mathrm{gas}}+U_{\mathrm{rad}}$. The virial theorem therefore gives $E=-U_{\mathrm{gas}}$; the radiation-dominated limit is marginally bound and its effective adiabatic index approaches $4/3$.

Quasistatic contraction can still maintain virial equilibrium. It fails dynamically only when sufficient support is lost, for example if pair creation reduces the effective adiabatic index below $4/3$, or when an iron core undergoes electron capture and photodisintegration. A fixed-composition gas-dominated contracting core with $T\propto 1/R$ is already a counterexample to any assertion that leaving the main sequence alone reverses the heating or destroys virial equilibrium.