Past exam of the short options paper shared by Prelims and Parts A and B of the physics course of the University of Oxford / 2012 / S26 / 1

For a spherical star in hydrostatic equilibrium, $dP/dr=-Gm(r)\rho /r^2$. Multiply by $4\pi r^{3}dr$ and integrate to its radius $R$:Integration by parts gives $4\pi R^3{P}_s-3\int PdV=\Omega$. Neglecting the external surface pressure $P_s$, an ideal gas with constant heat capacity ratio $\gamma$ has thermal energy density $u=P/(\gamma -1)$. ThereforeThis is the hydrostatic virial theorem. A nonzero confining pressure retains the right-hand side $4\pi R^3{P}_s$; varying composition, radiation contributions or a spatially varying effective exponent require the integral relation, rather than pulling one $\gamma$ outside it.

For a fully ionized, nonrelativistic monatomic ideal gas, $\gamma =5/3$ and the virial theorem becomesSince $U$ increases with its particle-weighted mean temperature, the equilibrated self-gravitating object has negative heat capacity: adding total energy allows expansion, makes $\Omega$ less negative, and reduces $U$. Conversely, radiating energy permits contraction and raises the internal temperature. This refers to a sequence of gravitational equilibria, not to the positive local heat capacity of a parcel at fixed volume.

In an ordinary nondegenerate nuclear-burning star, a rise in nuclear heating tends to expand and cool the burning region, lowering its temperature-sensitive reaction rate. Reduced heating leads to contraction and reheating. This expansion feedback is the nuclear thermostat; it can fail in degenerate matter where pressure responds weakly to temperature. During contraction of a pre-main-sequence star, gravitational binding supplies the radiated luminosity and heats the interior. For a fixed-structure scaling, half the released gravitational energy increases $U$ and half escapes as radiation, until hydrogen burning can support a long-lived main sequence equilibrium.

Write $\Omega =-aG{M}_{\odot }^2/R_{\odot }$ with a structure factor $a$ of order unity. Then the thermal radiation timescale isFor a uniform sphere $a=3/5$, this is about $9.4\times 10^6$ years. The conventional Kelvin-Helmholtz cooling time $G{M}_{\odot }^2/(R_{\odot }{L}_{\odot })$ is about $3.1\times 10^7$ years; a centrally concentrated solar model changes the coefficient. Both estimates are far shorter than the nuclear lifetime.

Ionization and dissociation can store compression energy without the full temperature rise of a monatomic gas, lowering the effective adiabatic response in partially ionized outer layers. Radiation pressure tends towards an adiabatic exponent $4/3$. These effects, changing composition and degeneracy make the effective exponent position dependent. The local dynamical exponent is ${\Gamma }_1=(\partial \log P/\partial \log \rho {)}_s$; it equals $C_P/C_V$ in the simple fixed-composition ideal-gas model, but need not do so for an ionizing mixture.

If an envelope has an effective exponent at or below $4/3$, its pressure response is soft and sufficiently extensive such regions can promote radial pulsation, envelope instability or mass loss. In the ideal homologous global model, $\gamma =4/3$ is marginal and $\gamma <4/3$ is unstable to contraction or expansion. A low value confined to a thin outer layer does not by itself prove collapse of the entire star: global stability depends on the pressure-weighted structure. Its location and mass extent matter.

Displace a small parcel upward, with negligible heat exchange, rapid pressure balance, and the same composition as its surroundings. Define the environmental gradient $\nabla =d\log T/d\log P$. An adiabatic ideal gas parcel obeys $P{\rho }^{-\gamma }=\mathrm{constant}$ and hence $T\propto P^{(\gamma -1)/\gamma }$, giving ${\nabla }_{\mathrm{ad}}=(\gamma -1)/\gamma$.

The upward displacement has $\delta \log P<0$. If $\nabla <{\nabla }_{\mathrm{ad}}$, the parcel's temperature changes more negatively than the environment's. At their common pressure it is cooler, denser and driven back down. Therefore the Schwarzschild criterion for stability isThe reversed inequality produces buoyant growth, and equality is neutrally stable in this approximation. A composition gradient requires the Ledoux criterion; pressure equilibrium alone does not ensure equal mean molecular weight.

In the Sun's cool outer layers, partial ionization lowers the adiabatic temperature gradient, while large opacity makes radiative transport inefficient. The required radiative gradientcan exceed ${\nabla }_{\mathrm{ad}}$, so convection carries part of the outward luminosity. The Sun therefore has an outer convective envelope, above its primarily radiative interior. Here $a$ is the radiation constant, $\kappa$ opacity per unit mass and $m_r,L_r$ the enclosed mass and luminosity. Near the visible surface, radiative losses are again important and the flow need not be perfectly adiabatic.

Near a red giant's compact core, intense nuclear burning in a small region can produce a large $L_r/m_r$ and a steep required radiative gradient. Where this exceeds the adiabatic gradient, convection transports the released energy and mixes fuel. Core helium burning is a clear example: the strongly temperature-sensitive Triple-alpha process commonly produces a convective burning core; shell flashes can also make convective regions.^[ref]

The qualification matters: an inert degenerate helium core and a steady hydrogen-burning shell on the first-ascent red giant branch are not universally convective; conduction or radiation may dominate there. Red giants also commonly have deep convective envelopes. The physical condition is an excessive required radiative gradient, not proximity to the centre alone.