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Across a stationary normal shock wave, the Rankine-Hugoniot conditions for a perfect gas conserveMomentum conservation and $1/\rho =u/j$ giveDirect elimination of the two pressures and densities from these three jump conditions givesPart c identifies the right side with the square of the critical speed, so the Prandtl shock relation isIt maps the unique upstream supersonic state on a given Bernoulli streamline to its downstream subsonic state.

Work in the shock frame, and let the positive upstream and downstream speeds be $w_0$ and $w_1$. The Rankine-Hugoniot conditions for a perfect gas give conservation of mass and momentum:PutMass conservation gives $w_1=w_0/r$, and momentum conservation then gives

Applying the energy integral from part (a) on the two sides givesSubstitution of (1) into (2), followed by cancellation of $p_0/{\rho }_0$, yieldsSolving this linear equation for the density ratio gives

In the frame in which the shock is stationary, the upstream gas approaches it with speed $V$, while the downstream gas recedes from it with speed $V-u_1$. The Rankine-Hugoniot conditions for a perfect gas are thereforeandThese equations express conservation of mass flux, momentum flux, and total specific enthalpy respectively.

If $B_{y1}=B_{y2}=0$, alignment also gives $u_{y1}=u_{y2}=0$. The common normal magnetic field contributes the same constant magnetic stress on both sides and carries no Poynting flux in the aligned frame. The remaining mass, normal-momentum, and energy conditions are therefore exactlywhich are the Rankine-Hugoniot conditions for a perfect gas.

Work in the locally planar shock frame, taking $v_u,v_d>0$ towards the surface. For a stationary normal shock wave with negligible radiative loss inside the discontinuity, the Rankine-Hugoniot conditions for a perfect gas express conservation of mass, conservation of momentum, and conservation of energy per unit area asHere $h$ is specific enthalpy, including the pressure work; using only internal energy would omit that work. The gravitational potential is continuous across an infinitesimally thin front. For cold upstream gas, $P_u$ and $h_u$ are negligible, while a monatomic gas has $\gamma =5/3$ and $h=5P/(2\rho )$.

The hydrodynamic equations apply to flow parallel to the guiding magnetic field when its stress contributions cancel across the front. An oblique magnetohydrodynamic shock requires magnetic stresses and Poynting flux; a radiation-mediated front or appreciable escaping radiation requires the corresponding radiative momentum and energy flux terms. These are additional physics, not terms hidden in $h$ here.

Integrate each steady one-dimensional conservation law across a thin region enclosing the shock. Mass conservation gives constant mass flux $j=\rho u$. Momentum conservation includes both advected momentum and pressure force, so $p+\rho u^2$ is constant. The total energy flux is $u(\rho e+p)=j(e+p/\rho )$, where $e=c_{v}T+u^2/2$. For a genuine shock $j\ne 0$, division by the mass flux and the equation of state $c_{v}T=p/[(\gamma -1)\rho ]$ give the conserved stagnation enthalpyThus the Rankine-Hugoniot conditions for a perfect gas areThe velocities here are measured in the stationary shock frame; a moving-shock problem must first be transformed to that frame.

Let $P=p_2/p_1$ and $D={\rho }_2/{\rho }_1$ be the pressure and mass density ratios across a normal shock wave in a perfect gas with specific-heat ratio $\gamma$. Eliminating the velocities from the Rankine-Hugoniot conditions for a perfect gas givesAs $P\to \infty$, the ratio tends to the shock compression ratio $(\gamma +1)/(\gamma -1)$.