= Solution
The <Mach number> is $\mathcal M=u/c_s$, comparing the flow speed relative to a disturbance with the local <sound speed>. For a body moving supersonically, spherical acoustic disturbances emitted earlier have a tangent envelope with <Mach cone> half-angle $\sin\theta=1/\mathcal M$. Signals cannot propagate arbitrarily far upstream. Compressive characteristics can converge and steepen into a <shock wave>, within which microscopic dissipation prevents a multivalued fluid state; supersonic motion alone does not require every flow to contain a shock.
Work in the <shock frame> with steady normal velocities $u_1,u_2>0$. Integrating <conservation of mass>, normal <momentum> and total <energy> over a thin pillbox, neglecting gravity across the thin layer, gives the <Rankine-Hugoniot conditions for a perfect gas>:
$$
j=\rho_1u_1=\rho_2u_2,\qquad
p_1+\rho_1u_1^2=p_2+\rho_2u_2^2,
$$
$$
j\left(h_1+\frac{u_1^2}{2}\right)
=j\left(h_2+\frac{u_2^2}{2}\right),\qquad
h=\frac{\gamma p}{(\gamma-1)\rho}.
$$
The <energy> equation includes <pressure> work through <specific enthalpy>. The shock is adiabatic if negligible <heat> escapes the layer, but its irreversible dissipation means it is not isentropic.
Put $\chi=\rho_2/\rho_1=u_1/u_2$ and $\mathcal M_1=u_1/\sqrt{\gamma p_1/\rho_1}$. <Momentum> conservation gives $p_2=p_1+\rho_1u_1^2(1-1/\chi)$. Substitution in the <energy> equation produces the unshocked root $\chi=1$ and the compressive root
$$
\chi=\frac{(\gamma+1)\mathcal M_1^2}{(\gamma-1)\mathcal M_1^2+2},\qquad
\frac{p_2}{p_1}=\frac{2\gamma\mathcal M_1^2-(\gamma-1)}{\gamma+1}.
$$
Since $\rho_1u_1^2/p_1=\gamma\mathcal M_1^2$, the strong-shock limit has negligible upstream thermal <pressure> compared with ram <pressure>. The <Strong-shock Rankine-Hugoniot conditions> give
$$
\boxed{\frac{\rho_2}{\rho_1}=\frac{u_1}{u_2}\longrightarrow\frac{\gamma+1}{\gamma-1},\qquad
\frac{p_2}{p_1}\simeq\frac{2\gamma}{\gamma+1}\mathcal M_1^2.}
$$
For unchanged composition, the <ideal gas law> implies $T_2/T_1=(p_2/p_1)/\chi$, so
$$
\boxed{\frac{T_2}{T_1}\simeq\frac{2\gamma(\gamma-1)}{(\gamma+1)^2}\mathcal M_1^2.}
$$
For $\gamma=5/3$, the compression tends to four and $T_2\simeq3\bar m u_1^2/(16k_B)$, where $\bar m$ is mean particle <mass>.
With constant specific heat $c_v$, <entropy production in a perfect-gas shock> is measured by the change in <specific entropy>
$$
\Delta s=c_v\ln\left[\frac{p_2}{p_1}\left(\frac{\rho_1}{\rho_2}\right)^\gamma\right].
$$
The strong-shock expression is therefore
$$
\boxed{\Delta s\simeq c_v\ln\left[\frac{2\gamma\mathcal M_1^2}{\gamma+1}
\left(\frac{\gamma-1}{\gamma+1}\right)^\gamma\right].}
$$
The exact jump is positive for every compressive shock with $\mathcal M_1>1$, in accordance with the <Second law of thermodynamics>; the asymptotic expression is only for $\mathcal M_1\gg1$. There is a normalization issue in the printed description: this logarithmic formula with constant $c_v$ describes <entropy> per unit <mass>. The true <entropy density> is $\rho s$, so its difference is $\rho_2s_2-\rho_1s_1$, rather than the same logarithmic jump with a constant <heat capacity> per unit <volume>. A material parcel's irreversible <entropy> increase should be assessed using $\Delta s$.
Downstream gas is denser and hotter. For optically thin collisional emission the loss per unit <volume> is typically $n_en_i\Lambda(T)$, whereas stored thermal <energy> scales approximately as $nT$. Compression can therefore reduce the <radiative gas cooling time>; excitation and ionization at the higher <temperature> can open additional cooling channels. The actual efficiency still depends on the <astrophysical cooling function>, composition and <optical depth>. In a radiative shock with separated scales, the <temperature> first rises across a narrow dissipative layer, then decreases through a much broader cooling region of characteristic length $\ell_{\rm cool}\sim u_2t_{\rm cool}$. Subsonic cooling usually brings further compression with approximately constant <pressure>. The immediate microscopic shock thickness and the macroscopic cooling length are distinct.
Let state $f$ be the fully cooled gas, with $T_f=T_1$ and the same composition. Set $a^2=p_1/\rho_1$, the <isothermal sound speed> squared. Across the entire shock plus cooling layer, <mass> and <momentum> are still conserved, but gas <energy> alone is not. With $\chi_f=\rho_f/\rho_1$,
$$
a^2+u_1^2=a^2\chi_f+\frac{u_1^2}{\chi_f},\qquad
(\chi_f-1)(a^2\chi_f-u_1^2)=0.
$$
The nontrivial <isothermal shock> therefore has
$$
\boxed{\rho_f=\rho_1\frac{u_1^2}{a^2}
=\gamma\mathcal M_1^2\rho_1.}
$$
If the upstream Mach number were defined with $a$ instead, this would be $\rho_f/\rho_1=\mathcal M_{\rm iso}^2$. Retaining the distinction avoids losing the factor $\gamma$.
The gas's <specific entropy> rises irreversibly at the hot shock and then decreases as escaping radiation removes heat. Since the final <temperature> equals the initial <temperature>,
$$
\boxed{s_f-s_1=-(\gamma-1)c_v\ln\chi_f<0.}
$$
This decrease for the gas is consistent with the <Second law of thermodynamics>: the emitted radiation and its eventual surroundings carry <entropy> away. It does not make the microscopic compressive shock reversible. The figure illustrates both the hot <entropy> jump and the subsequent cooling; its spatial cooling profile is schematic because no cooling law has been specified.
\Image[/past-exam-of-the-astrophysical-fluid-dynamics-course-of-the-university-of-cambridge/2014/paper-1-radiative-shock.png]
{title=Temperature and specific entropy across an adiabatic shock followed by radiative cooling}
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