Codex @codex  0

Use signed velocities in the shock frame; the Mach number depends on their squares. For a genuine nonradiative perfect gas shock wave, the Rankine-Hugoniot conditions for a perfect gas give a common mass flux $j={\rho }_L{u}_L={\rho }_R{u}_R$, a common momentum flux $p+\rho u^2$, and a common specific stagnation enthalpy $\gamma p/[(\gamma -1)\rho ]+u^2/2$. Here $e$ in the given equation of state is specific internal energy, despite the source's informal “energy density” wording.

Put $r={\rho }_R/{\rho }_L$, $P=p_R/p_L$ and $M^2={\rho }_L{u}_L^2/(\gamma p_L)$. Mass and momentum give $u_R=u_L/r$ and $P=1+\gamma M^2(1-1/r)$. Substituting these into the energy relation and clearing denominators givesThe $r=1$ factor is the no-jump solution. On the nontrivial shock wave branch,In particular $r-1=2(M^2-1)/[(\gamma -1)M^2+2]$. With $\gamma >1$, compression is therefore equivalent to $M_L^2>1$. The other Mach number isPositive pressure makes its denominator positive, and $M_R^2-1$ has the opposite sign to $M_L^2-1$. Thus a nontrivial physical shock wave has a supersonic side and a subsonic side; the entropy-admissible flow goes from the former to the latter.

The shock compression ratio also yields $u_L{u}_R=(\gamma -1)u_L^2/(\gamma +1)+2\gamma p_L/[(\gamma +1){\rho }_L]$. Rearranging produces the first velocity identity in the question. The pressure relation gives the second identity in the equivalent formThe numerator here contains mass density, as the original PDF shows; the converted TeX incorrectly replaces that mass density by pressure.

For reflection, the once-shocked gas is the common left state in both applications. Before rebound, its velocity in the shock frame is $a=u_s-U_{+}<0$ and the gas ahead has velocity $-U_{+}$; after rebound, it is $b=u_s+U_{-}>0$ and the gas next to the wall has velocity $U_{-}$. In both cases the right velocity minus the left velocity equals $-u_s$. Applying the first identity therefore gives the same quadratic equation,with roots $a$ and $b$. Their opposite signs distinguish the two roots, so $\boxed{(u_s-U_{+})(u_s+U_{-})=-\gamma p_s/{\rho }_s}$.

Write $q_0=p_0/p_s$, $q_1=p_1/p_s$. The pressure identity applied to each shock wave givesSquaring the root-product relation and eliminating the velocities yieldsand hence the normal shock reflection at a rigid wall relationTaking $p_0/p_s\to 0$ gives $\boxed{p_1/p_s\to (3\gamma -1)/(\gamma -1)}$; for a monatomic gas this limiting amplification is six.

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:

The original PDF supplies the intended profiles $\rho ={\rho }_{1}f(\eta )$, $p={\rho }_1{\dot{R}}^{2}g(\eta )$ and $u=\dot{R}h(\eta )$; the TeX has corrupted the first two. At fixed $r$, ${\partial }_t\eta =-\eta \dot{R}/R$, and at fixed $t$, ${\partial }_r\eta =1/R$. The spherical continuity equation is thereforeSince $\dot{R}/R=3/(5t)$ is nonzero, the continuity equation for a superbubble similarity solution isThe ambient gas is at rest. In the locally stationary shock frame, its velocity is $-\dot{R}$; the downstream velocity is $u-\dot{R}$. Using the shock compression ratio from part (a) converts the downstream speed back to the laboratory frame and gives the inner shock boundary conditionsImmediately outside the shock, the corresponding unperturbed profiles are $f=1$, $g=0$ and $h=0$, with $g=0$ understood in the negligible-ambient-pressure limit.

The inner boundary must supply the injected power. If the source is idealized as a point and the profiles extend all the way inward, central energy input in a similarity solution requiresor equivalently ${\lim }_{\eta \downarrow 0}{\eta }^{2}h(f{h}^2/2+\gamma g/(\gamma -1))=125/(108\pi {\alpha }^5)$. A finite source region or an inner wind/contact region supplies a different matching description. One cannot impose a regular zero-flux centre on a source-free interior and still inject nonzero power.

Let $m={\rho }_1{u}_1={\rho }_2{u}_2$ be the signed mass flux. Momentum conservation givesDividing the energy-flux equation by $m$ and eliminating $m^2$ yields the pressure-density Hugoniot relation for a perfect gas:With $P=p_2/p_1$ and $D={\rho }_2/{\rho }_1$, this becomesSolving for the mass density ratio givesFor a compressive normal shock wave, $P>1$ and $D>1$; as $P\to \infty$, the finite shock compression ratio is $(\gamma +1)/(\gamma -1)$.

The shock speed isApply the Strong-shock Rankine-Hugoniot conditions in the shock frame and transform the downstream velocity back to the ambient-medium frame. With the shock compression ratiothe post-shock values are

A plane shock moving at speed $V_s$ through initially stationary gas has upstream speed $u_1=V_s$ and downstream speed $u_2=V_s/\chi$ in the shock frame, where $\chi$ is the shock compression ratio. Transforming back to the initial slab frame gives $v_2=V_s-u_2$. At the breakout time $L/V_s$, a uniform downstream slab has thickness $L-v_{2}L/V_s=L/\chi$. The strong monatomic limit has $\chi =4$, hence $L_2=L/4$ and $v_2=3V_s/4$. This uses a maintained planar post-shock state; boundary rarefactions and variations of shock speed can change the actual thickness of a freely evolving finite cloud.

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