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Integrating each conservation law across a vanishingly thin interval around the stationary shock wave leaves equal conservation law fluxes on its two sides. HenceThese are the Rankine-Hugoniot conditions for a perfect gas. Signed velocities may both be negative when the material travels from positive to negative $z$; no sign change is needed in the conservation law fluxes.

Put $r={\rho }_2/{\rho }_1=u_1/u_2$, $P=p_2/p_1$ and $m=M_1^2={\rho }_1{u}_1^2/(\gamma p_1)$. Momentum conservation givesDivide the energy condition by the nonzero mass flux. Equality of kinetic energy plus specific enthalpy givesSubstituting $P$ and multiplying by $2(\gamma -1)r^2/\gamma$ yieldsThe factor $r-1$ is the continuous, no-shock solution. On the nontrivial normal shock wave branch,An admissible compressive gas shock wave has $M_1>1$, so $r>1$ and $P>1$. The algebraic jump equations alone also allow a reversed expansive discontinuity; the entropy production in a perfect-gas shock excludes that branch. At $M_1=1$ the nontrivial branch joins the continuous solution.

Let $D=u_1>0$ denote the front speed in the stationary upstream frame. In the shock frame, the upstream and downstream velocities are $-D$ and $-D/r$. Taking $M_1\to \infty$ in the Rankine-Hugoniot conditions for a perfect gas givesThe downstream laboratory velocity is $D-D/r=2D/(\gamma +1)$, which distinguishes the gas speed from the front speed.

For the planar blast-wave energy scaling of a self-similar blast wave, integration of the total energy density over the shocked interval $0<z<Z$ gives the energy per unit area on this side:The similarity solution makes $I$ time-independent; a finite positive explosion energy requires $0<I<\infty$. Conservation of energy therefore gives $\dot{Z}=(E_{+}/({\rho }_{1}I){)}^{1/2}{Z}^{-1/2}$ for the expanding front. Integrating from $Z(0)=0$,If $E$ denotes the one-sided energy, $C=(9/(4I){)}^{1/3}$. If the released energy $E$ feeds two symmetric fronts, $E_{+}=E/2$ and $C=(9/(8I){)}^{1/3}$. In either convention the requested scaling is $Z=C(E/{\rho }_1{)}^{1/3}{t}^{2/3}$. The constant depends on the similarity profiles and the energy convention; energy conservation determines the exponent without solving those profiles. The Strong-shock Rankine-Hugoniot conditions additionally fix $f(1)=(\gamma +1)/(\gamma -1)$ and $g(1)=h(1)=2/(\gamma +1)$.

Work in the shock frame, assume a calorically perfect gas with fixed specific-heat ratio $\gamma >1$, and neglect magnetic stresses and body forces across the thin shock. The internal energy per unit mass is $e=p/[(\gamma -1)\rho ]$, and the specific enthalpy is $w=\gamma p/[(\gamma -1)\rho ]$. Integrating the local conservation laws across a thin stationary control volume makes the mass flux $\rho u$, momentum flux $p+\rho u^2$, and energy flux $u(\rho u^2/2+\gamma p/(\gamma -1))$ continuous. Hence the Rankine-Hugoniot conditions for a perfect gas areMultiplication of the last equality by $m$ gives continuity of the stated energy flux. These express that mass, momentum and energy cannot accumulate in an infinitesimally thin steady shock, although entropy production occurs on the physical compression branch. The same constant in the polytropic equation of state $p=K{\rho }^{\gamma }$ must therefore not be imposed on both sides. For flow from $z>0$ to $z<0$, both signed velocities are negative; the flux equations and ratios remain valid.

In the Strong-shock Rankine-Hugoniot conditions, $p_1/({\rho }_1{u}_1^2)=1/(\gamma M_1^2)$ tends to zero. Put $r={\rho }_2/{\rho }_1$. The first two flux balances give $u_2=u_1/r$ and $p_2={\rho }_1{u}_1^2(1-1/r)$ to leading order. The energy balance becomesDiscarding the unchanged-state root $r=1$ gives the compressive shock compression ratio and the required limits:

Work in the shock frame. The undisturbed upstream gas enters the stationary discontinuity at speedwhile the downstream gas, which moves with the piston in the laboratory frame, leaves atThe Rankine-Hugoniot conditions for a perfect gas areandEliminating the velocities between these equations gives the perfect-gas density jump

The mass and momentum equations also implysoSubstitution of the density ratio into mass conservation, followed by this momentum relation, givesUsing the adiabatic sound speed $c_1^2=\gamma p_1/{\rho }_1$ therefore yieldsFinally, eliminating $U$ from $V=\beta p_1/({\rho }_{1}U)$ gives the piston-driven normal shock relationFor a compressive shock $\beta >0$, the first boxed equation has $U>c_1$: the shock propagates supersonically into the undisturbed gas, approaching the sound speed in the weak shock limit $\beta \to 0$.

In the shock frame, integrate mass conservation, momentum conservation, and total-energy conservation across a thin interval containing the stationary normal shock wave. No mass, momentum, or energy is stored in the vanishingly thin interval, so each flux has the same value on both sides. For a perfect gas, the specific enthalpy is $h=\gamma p/[(\gamma -1)\rho ]$. The resulting Rankine-Hugoniot conditions for a perfect gas areThe momentum flux includes both transported momentum and pressure force. The energy flux includes kinetic energy and enthalpy, the latter accounting for internal energy and the work needed to push gas through the interval. Entropy production can occur inside the normal shock wave even though total energy is conserved.

Integrating the three conservation laws across a stationary normal shock wave gives the Rankine-Hugoniot conditions for a perfect gasEliminate $u_{x2}$ with the conserved mass flux and write $v_s^2=\gamma p/\rho$ and $M_x=u_x/v_s$. Solving the remaining two algebraic equations givesFor a compressive shock, $M_{x1}>1$, the density and pressure increase, and the downstream normal flow is subsonic in the shock frame.

In the shock frame, conservation of mass, normal momentum, and energy givesThese are the Rankine-Hugoniot relations.

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.

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 $(\gamma +1)/(\gamma -1)$.