{c}

Write $\mathbf x$ for a <comoving coordinate>, $\mathbf u$ for the <peculiar velocity>, $H=\dot a/a$ for the <Hubble parameter>, and $\delta=\delta\rho/\bar\rho$ for the <density contrast>. For a homogeneous self-gravitating fluid, the linearized <continuity equation>, <Euler equations for an inviscid fluid>, and <cosmological Poisson equation> are
$$
\dot\delta=-\frac1a\nabla\cdot\mathbf u,
\qquad
\dot{\mathbf u}+H\mathbf u=-\frac{c_s^2}{a}\nabla\delta-\frac1a\nabla\phi,
\qquad
\nabla^2\phi=4\pi G a^2\bar\rho\delta.
$$
Here $c_s$ is the physical <adiabatic sound speed>, gradients refer to <comoving coordinates>, and the homogeneous background force has already been subtracted. Differentiating the first equation and eliminating $\mathbf u$ and $\phi$ gives
$$
\ddot\delta+2H\dot\delta-\frac{c_s^2}{a^2}\nabla^2\delta-4\pi G\bar\rho\delta=0.
$$
A <Fourier mode> with comoving <wavenumber> $k$ therefore obeys
$$
\ddot\delta_k+2H\dot\delta_k+
\left(\frac{c_s^2k^2}{a^2}-4\pi G\bar\rho\right)\delta_k=0.
$$
The <Jeans wavenumber> occurs where the coefficient in parentheses vanishes. Since a comoving <wavelength> is $2\pi/k$, the result is
$$
\boxed{k_J=\frac{a\sqrt{4\pi G\bar\rho}}{c_s},\qquad
\lambda_J^{\rm com}=\frac{c_s}{a}\sqrt{\frac{\pi}{G\bar\rho}}.}
$$
Below this <comoving Jeans length>, pressure restores a displaced fluid element: there are <acoustic waves> rather than growing <Jeans instability>. In a static background they oscillate; expansion changes their amplitude and frequency, and dissipation can damp them. Pressure support alone does not erase them. In the pre-recombination <photon-baryon fluid>, <Silk damping> supplies a separate erasure mechanism.

For collisionless matter, replace the <sound speed> by a characteristic <velocity dispersion>. For an isotropic <Maxwell-Boltzmann velocity distribution>, defining $\sigma^2=\langle u_x^2\rangle=\langle|\mathbf u|^2\rangle/3$, the static collisionless marginal-stability calculation gives
$$
\boxed{\lambda_{J,\rm collisionless}^{\rm com}
=\frac{\sigma}{a}\sqrt{\frac{\pi}{G\bar\rho}}
=\frac{v_{\rm rms}}a\sqrt{\frac{\pi}{3G\bar\rho}}.}
$$
The coefficient depends on the distribution and the convention for <velocity dispersion>. Small-scale suppression is now <collisionless free streaming> and <phase mixing>, rather than collisional <acoustic waves>. An instantaneous <collisionless Jeans length> must be distinguished from the accumulated comoving distance $\int v\,dt/a$ traveled since decoupling.

For the requested thermal histories, an <Einstein-de Sitter universe> describes the matter-era limit: it cannot literally also have a radiation-dominated epoch. Interpret the question as a spatially flat universe with negligible <cosmological constant>, passing from <radiation domination>, $a\propto t^{1/2}$, to <matter domination>, $a\propto t^{2/3}$. For the usual schematic dynamical Jeans scale use $\bar\rho_{\rm tot}$ in the estimate, so
$$
\lambda_{\rm dyn}^{\rm com}\propto
\begin{cases}
c_s a,&t\ll t_{\rm eq},\quad\bar\rho_{\rm tot}\propto a^{-4},\\
c_s a^{1/2},&t\gg t_{\rm eq},\quad\bar\rho_{\rm tot}\propto a^{-3}.
\end{cases}
$$
Near the <particle horizon> a relativistic perturbation calculation replaces the Newtonian derivation; the scaling estimate still identifies the relevant sound-crossing or streaming scale.

Before <cosmological recombination>, <adiabatic initial conditions> keep the entropy per <baryon> fixed in the tightly coupled <photon-baryon fluid>. With $R_b=3\bar\rho_b/(4\bar\rho_\gamma)\propto a$, its <photon-baryon sound speed> is
$$
c_s^2=\frac{c^2}{3(1+R_b)}.
$$
For example, this follows from $\delta\rho_\gamma/\bar\rho_\gamma=(4/3)\delta\rho_b/\bar\rho_b$ and $\delta P=c^2\delta\rho_\gamma/3$. In the radiation-dominated limit $R_b\ll1$, $c_s\simeq c/\sqrt3$; in the strongly baryon-loaded matter-era limit $R_b\gg1$, $c_s\propto a^{-1/2}$. Thus the <baryon Jeans length across recombination> has the asymptotic history
$$
\boxed{\lambda_{J,b}^{\rm com}\propto
\begin{cases}
a\propto t^{1/2},&t\ll t_{\rm eq},\\
\text{approximately constant},&t_{\rm eq}\ll t<t_{\rm rec},\\
a^{-1/2}\propto t^{-1/3},&t>t_{\rm rec}\text{ with adiabatic gas cooling}.
\end{cases}}
$$
The transition to the plateau is smooth: retaining $1+R_b$ gives $\lambda_{J,b}^{\rm com}\propto a^{1/2}/\sqrt{1+R_b}$ in <matter domination>. If baryon loading is still small, this intermediate segment instead rises as $a^{1/2}$ until loading becomes important.

At <photon decoupling>, radiation ceases to provide pressure support to the <baryons>, so the <sound speed> drops from the coupled-fluid value to that of a nonrelativistic <ideal gas>, $c_{s,b}^2=\gamma k_BT_b/(\mu m_p)$, with $\gamma=5/3$. The drop in <comoving Jeans length> at fixed density is the same ratio of <sound speeds>, typically several orders of magnitude. Subsequently an <isentropic process> gives $T_b\propto\bar\rho_b^{\gamma-1}\propto a^{-2}$, so $c_{s,b}\propto a^{-1}$ and $\lambda_{J,b}^{\rm com}\propto a^{-1/2}$. Residual energy exchange through <Compton scattering> can initially keep $T_b\propto a^{-1}$, producing a short approximately flat gas segment before the decreasing segment. The idealized sketch assumes immediate adiabatic cooling:

``
log(comoving baryon Jeans scale)
 ^
 | __________________
 | / |
 | / | loss of photon pressure
 | / |
 | / +---\
 | / \
 +-------------------|-----------------|----------> log(time)
 t_eq t_rec
 slope 1/2 slope ~0 slope -1/3
``

For the <cold-dark-matter kinetic Jeans scale>, take the stated toy history to mean that collisions maintain a particle <temperature> equal to the photon <temperature> until <kinetic decoupling>. This is an assumption about thermal coupling; a negligible collision rate after decoupling makes the subsequent treatment collisionless. Assume the ordering $t_{\rm NR}<t_{\rm dec}<t_{\rm eq}<t_{\rm rec}$. While relativistic, the characteristic particle speed is constant; while nonrelativistic but thermally coupled, $\sigma\propto\sqrt T\propto a^{-1/2}$; after <kinetic decoupling>, particle momentum redshifts as $a^{-1}$ and $\sigma\propto a^{-1}$. Combining these with the background dynamical density gives
$$
\boxed{\lambda_{\rm dyn,CDM}^{\rm com}\propto
\begin{cases}
a\propto t^{1/2},&t<t_{\rm NR},\\
a^{1/2}\propto t^{1/4},&t_{\rm NR}<t<t_{\rm dec},\\
\text{constant},&t_{\rm dec}<t<t_{\rm eq},\\
a^{-1/2}\propto t^{-1/3},&t>t_{\rm eq}.
\end{cases}}
$$
There is no sudden change at <cosmological recombination>: <cold dark matter> has already decoupled. Its small initial <velocity dispersion> makes its suppression scale much smaller than the pre-recombination <baryon> scale.

``
log(comoving CDM dynamical Jeans scale)
 ^
 | _______________
 | ___/ \
 | ___/ \ no jump
 | / \ at t_rec
 | / \
 +-----------|------------|--------------|----|------> log(time)
 t_NR t_dec t_eq t_rec
 slope 1/2 slope 1/4 slope 0 slope -1/3
``

The density convention matters. If one instead defines a species' self-gravitating <collisionless Jeans length> using only its own nonrelativistic density $\bar\rho_m\propto a^{-3}$, its thermally coupled segment is constant and its decoupled segment decreases as $a^{-1/2}$ even during <radiation domination>. The constant radiation-era segment in the second sketch is the usual instantaneous background dynamical streaming estimate, not an exact single-species self-gravity threshold. A complete multicomponent calculation uses the separate perturbation equations and their combined <cosmological Poisson equation>.

Solution

Past exam of the mathematics course of the University of Cambridge / 2019 / iii / Paper 346 / 1 / a / Solution

Macroscopically, an <adiabatic process> exchanges no <heat> with its surroundings: $\delta Q=0$. A reversible adiabatic process is therefore <isentropic process>[isentropic], since $dS=\delta Q_{\rm rev}/T=0$. Microscopically, the system is thermally isolated, so changes in its energy arise from work done by changing external parameters rather than random energy exchange with a <heat bath>; in a sufficiently slow change, the populations of the corresponding microscopic states are carried along without producing entropy.

For repeated operation the working substance must execute a <thermodynamic cycle>. Reversible isothermal contact at $T_2$ absorbs heat $Q_2$, a reversible adiabat changes the temperature to $T_1$, reversible isothermal contact at $T_1$ rejects heat $Q_1>0$, and a second adiabat returns to the initial state. This is the <Carnot cycle>. Reversibility and zero net entropy change give
$$
\frac{Q_2}{T_2}-\frac{Q_1}{T_1}=0,
\qquad
\frac{Q_1}{Q_2}=\frac{T_1}{T_2}.
$$
The <thermal efficiency> is the net work output divided by the absorbed heat:
$$
\eta=\frac{W}{Q_2}=\frac{Q_2-Q_1}{Q_2}.
$$
Consequently
$$
\boxed{\eta_{\max}=1-\frac{T_1}{T_2}.}
$$
The <Carnot theorem> shows that any irreversible cycle has smaller efficiency, so the two isotherms joined by two adiabats are optimal.

Past exam of the mathematics course of the University of Cambridge / 2019 / ii / Paper 1 / 35D / a / Solution

For one molecule whose translational motion separates from its internal degrees of freedom, the classical one-particle <partition function> is
$$
z_1(T,V)=\frac{V}{\lambda^3}q_{\rm int}(T),
\qquad
\lambda=\sqrt{\frac{2\pi\hbar^2}{mk_BT}},
$$
where $q_{\rm int}$ contains rotational, vibrational, and other internal states. For $N$ indistinguishable noninteracting molecules, the <classical ideal-gas partition function> is $Z_N=z_1^N/N!$. The thermodynamic pressure therefore obeys
$$
P=k_BT\left(\frac{\partial\log Z_N}{\partial V}\right)_{T,N}
=\frac{Nk_BT}{V},
$$
so
$$
\boxed{PV=Nk_BT.}
$$
The internal structure changes the energy, entropy, and heat capacity through $q_{\rm int}(T)$, but it does not change the <ideal gas> law while the internal partition function is independent of volume and the molecules have negligible interactions.

For a dilute monatomic gas $q_{\rm int}=1$. Using the <Stirling formula> in $Z_N=(V/\lambda^3)^N/N!$ and differentiating the <Helmholtz free energy> $F=-k_BT\log Z_N$ gives the <Sackur-Tetrode equation>
$$
\boxed{S=Nk_B\left[
\log\left(\frac{V}{N}\left(\frac{2\pi mk_BT}{h^2}\right)^{3/2}\right)
+\frac52\right].}
$$
With $N$ fixed, an <isentropic process> therefore satisfies $VT^{3/2}=\text{constant}$, or $TV^{2/3}=\text{constant}$. Eliminating $T$ with the ideal-gas law yields the monatomic <reversible ideal-gas adiabat>
$$
\boxed{PV^{5/3}=\text{constant}.}
$$

Past exam of the mathematics course of the University of Cambridge / 2019 / ii / Paper 2 / 35D / Solution

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An isentropic process keeps the <entropy> constant. Every reversible <adiabatic process> is isentropic, while an irreversible adiabatic process can increase entropy.

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 Incoming links: Isentropic process