= Solution
Work in the rest frame of a thin planar <shock wave>, with normal upstream and downstream velocities $u_1,u_2>0$. Assume a steady flow, negligible gravitational change across its thickness, no radiative energy loss, and negligible viscous stress and heat flux at the two remote faces of a control volume. Dissipation within the shock is permitted and is essential to the physical discontinuity.
Integrating <mass conservation> through that volume gives equal <mass flux> per unit area:
$$
\boxed{\rho_1u_1=\rho_2u_2=j.}
$$
The dimensions of $j$ are $\mathrm{kg\,m^{-2}\,s^{-1}}$. The normal momentum entering minus that leaving must balance the <pressure> forces. Equivalently, the conserved normal <momentum flux> is $p+\rho u^2$, giving
$$
\boxed{p_1+\rho_1u_1^2=p_2+\rho_2u_2^2.}
$$
For <conservation of energy>, matter advects kinetic and internal energy while <pressure> performs work at rate $pu$ per unit area. Writing $\epsilon$ for <specific internal energy>, the <energy flux> is $\rho u(u^2/2+\epsilon)+pu$. With <specific enthalpy> $h=\epsilon+p/\rho$, its conservation gives
$$
\boxed{\frac{u_1^2}{2}+h_1=\frac{u_2^2}{2}+h_2.}
$$
The division by $j$ is valid for a genuine through-flow shock. Together these are the three <Rankine-Hugoniot conditions>. The conserved <energy flux> has units $\mathrm{W\,m^{-2}}$, and both momentum fluxes have units of <pressure>.
For a calorically perfect <ideal gas>, $\epsilon=p/[(\gamma-1)\rho]$ and $h=\gamma p/[(\gamma-1)\rho]$. The three jump conditions can be combined explicitly. Put $r=\rho_2/\rho_1$, so $u_2=u_1/r$ and
$$
p_2=p_1+\rho_1u_1^2(1-r^{-1}).
$$
Substitute these into the enthalpy equation and use $\mathcal M_1^2=\rho_1u_1^2/(\gamma p_1)$, with the upstream <adiabatic sound speed>. Besides the trivial no-shock root $r=1$, the physical compression root is
$$
\boxed{r=\frac{(\gamma+1)\mathcal M_1^2}{(\gamma-1)\mathcal M_1^2+2},\qquad
\frac{p_2}{p_1}=\frac{2\gamma\mathcal M_1^2-(\gamma-1)}{\gamma+1}.}
$$
For unchanged composition, $T_2/T_1=(p_2/p_1)/r$. A compressive shock has $\mathcal M_1>1$ and increases the <specific entropy>. Thus an adiabatic shock conserves total energy but is not an <isentropic process>.
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