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For the pressure-density Hugoniot relation for a perfect gas, the change in specific entropy is $[s]/c_v=\log P-\gamma \log D(P)$, with $c_v$ the specific heat capacity at constant volume. Its derivative isIt is positive for a compressive normal shock wave with $P>1$. For a weak shock with $P=1+\delta$, integrating the leading term givesThus the entropy production is cubic in the small pressure jump, even though the mass density and temperature changes already appear at first order.

For $N$ equal weak shocks with individual pressure ratio $1+\delta$ and total ratio $P_s=(1+\delta {)}^N$, entropy production in a perfect-gas shock givesAt fixed $P_s>1$, this tends to zero as the compression is divided into increasingly many smaller weak shocks. A single finite normal shock wave instead gives the positive value $\log P_s-\gamma \log D(P_s)$. Gradual compression can consequently approach reversible isentropic flow while an abrupt compression generates finite entropy.

For a perfect-gas normal shock wave reflected from a stationary rigid wall, apply the Rankine-Hugoniot conditions for a perfect gas using signed velocity in each shock frame. The once-shocked state is the common left state for both applications. Its two relative speeds are roots of $v^2-(\gamma +1)u_{s}v/2-\gamma p_s/{\rho }_s=0$. Their product and the pressure jump eliminate the velocity, giving the displayed relation. A strong incident shock wave gives $p_1/p_s\to (3\gamma -1)/(\gamma -1)$.

Normal shock tables tabulate upstream-to-downstream property ratios for a normal shock wave, typically as functions of upstream Mach number and heat capacity ratio. They encode the Rankine-Hugoniot conditions for a perfect gas rather than defining the discontinuity itself.

A shock wave forms when a compressive flow steepens or when gas must be decelerated or turned faster than upstream acoustic waves can communicate the change. Astrophysical examples include a supernova remnant sweeping up surrounding gas, a fast stellar wind colliding with ambient material, and supersonic accretion onto an obstacle or onto a dense gas layer. A large Mach number by itself is not sufficient: a uniform supersonic stream need not contain a shock wave.

Choose the shock frame, and denote upstream and downstream normal velocities by $u_1,u_2$. Integrating the local conservation laws over a thin control volume spanning a steady normal shock wave gives equal fluxes on both sides. Mass conservation gives constant mass flux per unit area,Pressure and advected momentum together give the normal momentum flux,For a thin non-radiating layer, with negligible gravitational-potential change across it, conservation of energy gives equal fluxes $j(h+u^2/2)$. A perfect gas has specific enthalpy $h=\gamma p/[(\gamma -1)\rho ]$, soThese are the Rankine-Hugoniot conditions for a perfect gas. Dissipation inside the thin layer produces an entropy increase and selects the physical compressive branch. Efficient radiative cooling would change the energy relation and may instead lead to an isothermal shock.

Let $U_1,U_2$ denote the total flow speeds, keeping the normal components distinct from them. The original diagram measures $\beta$ from the initial wall direction to the shock wave itself. The outgoing velocity is turned by $\theta$, so it makes angle $\beta -\theta$ with the shock.

Resolve each velocity tangent and normal to the oblique shock. Tangential momentum conservation, with nonzero mass flux and no tangential surface force, makes the tangential velocity continuous:The normal flow obeys the strong normal shock wave result from part (b):Divide the second relation by the first to obtainThe approximation is controlled by the upstream normal Mach number $M\sin \beta \gg 1$, not just by $M\gg 1$. This distinction matters on the small-angle weak branch in part (d). The angle convention is also important: formulas using the angle to the shock normal instead would interchange sine and cosine.

The Rankine-Hugoniot conditions follow by integrating conservation laws across a thin moving layer, then letting its macroscopic thickness tend to zero. For example, normal momentum conservation in one dimension has the conservative formFor a shock wave moving at speed $V$, integrate over a control interval whose endpoints move with the discontinuity. The Reynolds transport theorem giveswhere $[f]=f_R-f_L$ denotes the difference of limiting states. Mass conservation similarly gives $V[\rho ]=[\rho u]$. If $v=u-V$ is the normal velocity in the shock frame, combining the two relations yieldsThe same construction applied to conservation of energy, for a non-radiating normal shock wave with nonzero mass flux, gives $[h+v^2/2]=0$, where $h$ is the specific enthalpy. An equation of state closes these relations, and the physical compression branch has increasing entropy.

Finite dynamic viscosity replaces the ideal discontinuity by a thin smooth transition. A viscous stress contributes a derivative to the momentum flux inside the layer and converts directed kinetic energy into internal energy. If its gradients vanish in the uniform far states, the same limiting Rankine-Hugoniot conditions still hold there. Balancing nonlinear transport with viscous diffusion gives a thickness of order $\nu /|u_L-u_R|$ in the viscous Burgers equation, where $\nu$ is the kinematic viscosity.

Use a steady, plane, nonradiative normal shock wave in a perfect gas, and measure the normal speeds $v_{\mathrm{u}},v_{\mathrm{d}}>0$ in the shock frame. The labels denote upstream and downstream. Conservation of mass, normal momentum and total energy givesandThe enthalpy term includes the pressure work. For very large upstream Mach number, neglect $p_{\mathrm{u}}$ compared with ${\rho }_{\mathrm{u}}{v}_{\mathrm{u}}^2$. If $r={\rho }_{\mathrm{d}}/{\rho }_{\mathrm{u}}$ is the shock compression ratio, the first two equations implySubstitution into the energy equation givesDiscard the no-shock solution $r=1$. For a nonrelativistic monatomic gas, $\gamma =5/3$, so the Strong-shock Rankine-Hugoniot conditions becomeThe factor four requires this adiabatic index and negligible radiative, magnetic and escaping-particle contributions to the flux budget. For example, the same strong, nonradiative calculation with $\gamma =4/3$ gives compression seven, not four.

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.

Use $v=|u_r|>0$ for the radial speed and distinguish the central mass $M$ from the Mach number $\mathcal{M}=v/c$. In each smooth nonzero-flow region, mass conservation, the adiabatic equation of state, and the Bernoulli equation giveHere $K>0$ is constant on a smooth isentropic flow branch, $c$ is the adiabatic sound speed, and $\mathcal{E}$ is the Bernoulli function. Either sign of $u_r$ is possible. The positive-radius nondimensionalization in the question requires $\mathcal{E}>0$; flows with $\mathcal{E}\le 0$ do not have these positive $r_s,x,\lambda$ and require another scaling.

Indeed, the supplied adiabatic process equations give $D(p{\rho }^{-3/2})/Dt=0$, so $K$ is spatially constant along a steady nonzero radial fluid flow. The adiabatic sound speed is $c^2=(\partial p/\partial \rho {)}_s$, and the specific enthalpy is $h=c^2/(\gamma -1)=2c^2$. The radial Euler momentum equation, $vdv/dr=-{\rho }^{-1}dp/dr-GM/r^2$, integrates to $\mathcal{E}$ because $dh=dp/\rho$ on this isentropic flow branch.

Put $D=[j(3K/2{)}^2{]}^{2/5}$. Eliminating the mass density from the continuity equation givesWith $y={\mathcal{M}}^{2/5}$, the Bernoulli equation becomesTherefore the physical constants of the spherical polytropic flow with adiabatic exponent three halves are

For $t>0$, $F^{\prime }(t)=4(t^5-1)/t^2$, so $F$ decreases from infinity to its unique minimum $F(1)=5$, then increases to infinity. This completely determines the algebraic curves. For $0<\lambda <1$, there are two disconnected branches at every $x>0$, one subsonic flow with $y<1$ and one supersonic flow with $y>1$. For $\lambda =1$, the line $y=x$ and a decreasing branch meet at $(1,1)$. Their slopes are $+1$ and $-1$, since the leading expansion is $F(1+\eta )=5+10{\eta }^2+O({\eta }^3)$. For $\lambda >1$, solutions exist only where $F(x)\ge 5\lambda$: if $x_{-}<1<x_{+}$ solve $F(x_{\pm })=5\lambda$, there is a forbidden interval $x_{-}<x<x_{+}$. At either endpoint the two branches turn at $y=1$ and have an infinite slope.

A smooth sonic point needs $F^{\prime }(x)=F^{\prime }(y)=0$; hence a regular sonic transition requires $\lambda =1$ and $r=r_s$. The critical speed of a polytropic flow is then $c_s^2=GM/(2r_s)=2\mathcal{E}$, and the allowed mass flux is $j=(2\mathcal{E}{)}^{5/2}{r}_s^2/(3K/2{)}^2$. Here the subscript on $c_s$ denotes evaluation at the sonic point, rather than the isothermal sound speed convention used in the next solution.

The simple transonic branch is $y=x$. Direct substitution givesThus the radial velocity has constant magnitude, while the mass density and pressure scale as $r^{-2}$ and $r^{-3}$. For an outward fluid flow, this is a subsonic flow becoming a supersonic flow as its adiabatic sound speed decreases.

On the other transonic branch, $y$ decreases as $x$ increases. At large $r$, the terms $4/y$ and $x^4$ dominate, giving $y\sim 4x^{-4}$. At small $r$, the terms $y^4$ and $4/x$ dominate, giving $y\sim 4^{1/4}{x}^{-1/4}$. The resulting asymptotic expansions areFor inward fluid flow, these are the nearly static reservoir at infinity and the inner free fall region of Bondi accretion. The specific enthalpy is smaller than $GM/r$ in the inner limit, which explains free fall. Reversing the radial velocity gives a decelerating outflow with the same profiles. The formal small-$r$ limit describes the point-mass model: if the spherical gravitating body has a finite surface, the exterior solution stops there.

A stationary, nonradiative normal shock wave conserves the mass flux and the sum $v^2/2+h$, while the Newtonian gravitational potential is continuous across its thin layer. Hence $\mathcal{E}$ and $r_s$ are unchanged. The entropy production in a perfect-gas shock increases $K=p/{\rho }^{3/2}$, soMore explicitly, the Rankine-Hugoniot condition for $\gamma =3/2$ gives, with upstream $m={\mathcal{M}}_1^2>1$,Thus the shock wave jumps vertically, at the same $x$, from the supersonic flow to a subsonic flow on a larger-$\lambda$ curve. The figure illustrates an outward flow on $y=x$ with a shock wave at $x=1.4$; the downstream subsonic flow has $\lambda \simeq 1.224$ and continues to larger $x$. The shock wave radius is an additional boundary-data choice, rather than being determined by the smooth equations alone.

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.

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

For a perfect gas with constant specific heat capacity $c_v$, the first law of thermodynamics givesUsing the pressure-density Hugoniot relation for a perfect gas, the entropy production in a perfect-gas shock is thereforeThis is positive for $P>1$, as required by the Second law of thermodynamics for the physical compressive normal shock wave.

For $N$ equal weak shocks, $(1+\delta {)}^N=P_s$, henceAdding the individual increases in specific entropy gives the entropy production in successive weak shocksBy contrast, a single normal shock wave with the same total pressure ratio givesThus a single strong shock produces more entropy than the chain of sufficiently small weak shocks. At fixed $P_s$, the chain approaches zero entropy production as $\delta \to 0$, while the single-shock result stays positive. The gradual compression approaches reversible isentropic flow.

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.

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 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.

Let the bed be $z=0$ and the free surface $z=h(x,t)$. In a long, shallow layer, vertical acceleration is smaller than gravity, so the vertical Euler equation gives hydrostatic pressure $p=p_{\mathrm{atm}}+\rho g(h-z)$. Incompressibility makes the vertical speed smaller than the horizontal speed by the aspect ratio; irrotationality then makes the leading horizontal velocity depth-independent. The horizontal Euler equation becomesIntegrating $u_x+w_z=0$ across the depth, using bed impermeability $w(0)=0$ and the surface condition $w(h)=h_t+u{h}_x$, yieldsCombining these gives the conservative momentum equation $(hu{)}_t+(h{u}^2+g{h}^2/2{)}_x=0$. Integral mass and momentum balance across a stationary thin jump therefore requireThese conservation laws still apply across a jump where the smooth hydrostatic/irrotational derivation does not resolve the internal turbulent structure.

Put $A=h_{-}$, $B=h_{+}$ and $q=A{u}_{-}=B{u}_{+}>0$, orienting the flow from minus to plus. For a nontrivial jump the momentum condition gives $q^2=gAB(A+B)/2$. Its change in mechanical-energy flux isA passive jump cannot create mechanical energy, so the change is nonpositive. Since $q>0$, $\boxed{h_{+}>h_{-}}$ for a genuine jump; equality describes no jump. The upstream flow is supercritical and the downstream flow subcritical, since $F{r}_{-}^2=B(A+B)/(2A^2)>1$ and $F{r}_{+}^2=A(A+B)/(2B^2)<1$.

The gas-dynamic analogue is a compressive normal shock: density and pressure increase and flow changes from supersonic to subsonic. Entropy increases; directed mechanical energy is irreversibly converted into internal energy. Total gas energy, including internal energy, is conserved across an adiabatic shock. This distinguishes that total-energy balance from the reduced shallow-water mechanical-energy loss.

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