Codex @codex  0

Mass and momentum flux defects at a moving interface add surface acoustic monopole and acoustic dipole terms to Lighthill acoustic analogy. For a physical conservative fluid shock wave, the Rankine-Hugoniot conditions set both defects to zero; the shock still affects distributional derivatives of the Lighthill stress tensor. A discontinuity alone is not independent mass or force injection.

For a strictly hyperbolic system of conservation laws $U_t+F(U{)}_x=0$ with ordered characteristic speeds ${\lambda }_1<\cdots <{\lambda }_n$, a Lax family-$j$ shock has ${\lambda }_j(U_R)<s<{\lambda }_j(U_L)$. The remaining families must stay on the same side of $s$: ${\lambda }_{j-1}(U_L)<s$ and $s<{\lambda }_{j+1}(U_R)$ where those neighbors exist. These inequalities imply that the lower families have speed below $s$ on both sides and the higher families speed above $s$ on both sides. There are $n+1$ incoming characteristic directions: two from the compressed family and one from each other family. The Rankine-Hugoniot conditions supply the jump balance, while these inequalities supply admissibility. For a convex scalar flux this reduces to the entropy shock condition.

Apply the Rankine-Hugoniot conditions to the conserved variables $(h,hu)$ and flux $(hu,h{u}^2+h^2/2)$. They giveFor a genuine shock, $h_1\ne 1$. Eliminating $u_1$ yields $s^2=h_1(h_1+1)/2$. Taking the positive branch requested,When $h_1=1$, the jump conditions instead force $u_1=0$ and describe a constant state; the labelled interface can then move at any speed and is not a shock. The nontrivial formulas describe a shallow-water shock into stationary unit-depth water.

Interstellar shocks arise when relative flow is supersonic: stellar winds, supernova ejecta, colliding clouds and gas moving through strong disturbances can drive them. Neighboring compressive characteristics steepen or a fast flow encounters material it cannot decelerate through upstream sound communication. A thin dissipative front then converts directed kinetic energy into thermal energy and increases entropy.

Integrate mass, normal momentum and total energy conservation through a planar steady shock in its rest frame. With no mass source, negligible external impulse across the thin layer and no radiative energy loss, their normal fluxes are respectivelyFor a perfect gas of constant adiabatic index, $h=\gamma p/[(\gamma -1)\rho ]$. Equality of the fluxes on the two sides yields the three Rankine-Hugoniot conditions in the question. Adiabatic means no energy escapes this control volume; it does not make the shock isentropic.

In the strong-shock limit, let $r={\rho }_2/{\rho }_1=u_1/u_2$ and neglect $p_1$. Momentum conservation gives $p_2={\rho }_1{u}_1^2(1-1/r)$. Substitution into energy conservation givesThe nontrivial compressive root isThe $r=1$ root describes no shock. A monatomic gas has the familiar compression four.

For identical colliding clouds in their center-of-momentum frame, reflection symmetry fixes the contact plane and the uniform compressed material on either side at rest. The unshocked clouds still approach at speeds $\pm V$; their supersonic upstream flow has not received a smooth pressure adjustment ahead of the shock. Let the right shock move to the right at speed $S$. In its own frame the upstream and downstream normal speed magnitudes are $V+S$ and $S$. The same shock compression ratio therefore givesThis is the shock speed in a symmetric cloud collision. For $\gamma =5/3$, the shocks move outward at $V/3$, not at the upstream speed or the relative cloud speed $2V$. As a check, the post-shock pressure is $p_2=(\gamma +1){\rho }_1{V}^2/2$; in the monatomic case $p_2/{\rho }_2=V^2/3$, giving specific internal energy $V^2/2$, exactly the initial bulk kinetic energy per unit cloud mass when brought to rest.

When a shock reaches a cloud's rear boundary, its supply of cold upstream gas ends. If the surrounding pressure is low, the hot compressed gas expands outward and a rarefaction travels back into it, converting thermal energy into motion and cooling the gas. A sustained uniform stationary shocked region is no longer the solution. Interaction with an appreciable exterior medium can instead drive another outward disturbance; the boundary pressure determines that part of the evolution. The stated adiabatic model provides no radiative mechanism that would keep a permanently cold dense sheet at the contact.

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:The dimensions of $j$ are $\mathrm{kg}{\mathrm{m}}^{-2}{\mathrm{s}}^{-1}$. The normal momentum entering minus that leaving must balance the pressure forces. Equivalently, the conserved normal momentum flux is $p+\rho u^2$, givingFor conservation of energy, matter advects kinetic and internal energy while pressure performs work at rate $pu$ per unit area. Writing $ϵ$ for specific internal energy, the energy flux is $\rho u(u^2/2+ϵ)+pu$. With specific enthalpy $h=ϵ+p/\rho$, its conservation givesThe 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 $\mathrm{W}{\mathrm{m}}^{-2}$, and both momentum fluxes have units of pressure.

For a calorically perfect ideal gas, $ϵ=p/[(\gamma -1)\rho ]$ and $h=\gamma p/[(\gamma -1)\rho ]$. The three jump conditions can be combined explicitly. Put $r={\rho }_2/{\rho }_1$, so $u_2=u_1/r$ andSubstitute these into the enthalpy equation and use ${\mathcal{M}}_1^2={\rho }_1{u}_1^2/(\gamma p_1)$, with the upstream adiabatic sound speed. Besides the trivial no-shock root $r=1$, the physical compression root isFor 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.

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.

A shock wave is a thin, irreversible compression layer across which fluid density, pressure, temperature and velocity change rapidly, while conserved fluxes satisfy the Rankine-Hugoniot conditions. In the shock frame, a compressive hydrodynamic shock converts ordered upstream kinetic energy into downstream thermal energy and increases entropy. The upstream normal flow is supersonic; the downstream normal flow is subsonic for an ordinary gas-dynamic compression shock.

In an accreting binary star, a supersonic stellar wind can be captured by a compact companion and shocked. An accretion stream striking an accretion disk, an accretion-disk boundary layer, and a magnetically guided accretion column provide other sites where infall energy is dissipated and high-energy radiation is generated. Shocks heat and decelerate material; a disk also needs transport of angular momentum before its material can reach the compact star.

In diffusive shock acceleration, charged particles scatter from magnetic irregularities on both sides of a shock and repeatedly cross the converging flows. The average energy gain is first order in the velocity jump divided by the particle speed. Escape competes with acceleration, producing a nonthermal distribution rather than a single downstream temperature. Supernova remnants are important candidate sources of Galactic cosmic rays; shocks associated with active galactic nuclei, their jets and lobes, and gamma-ray bursts are candidate extragalactic accelerators. Ultrarelativistic energies require sufficient size, magnetic field and lifetime, and losses must not outpace acceleration. In particular a relativistic particle's Larmor radius must fit within the accelerator; the existence of a strong shock alone does not guarantee unlimited particle energy.

Diffusive shock acceleration is viable in principle at $511\mathrm{GeV}$: repeated scattering off magnetic irregularities on opposite sides of a strong nonrelativistic shock wave can reach such energies. Actual viability requires a sufficiently large and long-lived accelerator with an acceleration time shorter than its escape and loss times; $\gamma$ alone does not establish those conditions.

Take positive velocity toward the upstream material, and let the front move at $+v_s$ in the upstream rest frame. For a strong adiabatic shock wave in a gas with adiabatic index $5/3$, the Rankine-Hugoniot conditions give ${\rho }_d/{\rho }_u=4$ and the shock-frame inflow speeds satisfy $u_1=v_s$, $u_2=v_s/4$. The signed velocities in the three frames areThus the two scattering media converge at $V=u_1-u_2=3v_s/4$. For a crossing particle with incident energy $E$ in its local plasma rest frame, a Lorentz transformation to the approaching other plasma giveswhere $\mu >0$ is the incident direction cosine measured toward the approaching medium and $pc\simeq E$. For an almost isotropic crossing population, flux weighting gives $⟨\mu ⟩={\int }_0^1{\mu }^{2}d\mu /{\int }_0^1\mu d\mu =2/3$. Scattering randomizes directions without changing energy in that plasma frame, and the reverse crossing has the same mean gain. ThereforeThese are first-order mean gains. They are not an increase of energy in one fixed frame caused by an instantaneous boundary crossing. Literally choosing the electron's instantaneous rest frame gives incident $E=m_e{c}^2$, and the other gas approaches at nearly $c$, so the expansion in small $V/c$ is invalid. This concrete frame check explains why the printed description of $V$ must instead refer to the upstream/downstream scattering frames.

For a smooth solution, the first equation is $u_x+(uv{)}_y=0$. Multiplying it by $v$ and adding $u$ times the second equation givesThus the appropriate shallow water equations conservation variables and flux areUse this conservative extension to define a weak solution across $y=F(x)$. A jump contributes $-F^{\prime }[W]\delta (y-F(x))$ to $W_x$ and $[H]\delta (y-F(x))$ to $H_y$. Cancellation of the singular parts gives the Rankine-Hugoniot conditionsEquivalently this follows by integrating each conservation law over a thin space-time control region crossing the discontinuity.

The conservative extension is essential here. Piecewise smoothness by itself does not imply any jump law. Moreover the smooth second equation can also be written $v_x+(u+v^2/2{)}_y=0$; imposing that different conservation law distributionally would give a different shock condition. Weak conservation laws depend on conserved variables. The requested jump law specifically selects mass $u$ and momentum $uv$, and this is the extension used below.

Let the original area be $A_0>0$ and the increased upstream area be $A_1>A_0$. At $ϵ=0$, the conservation law flux $q(A)=A^{5/4}$ is strictly convex. Characteristics behind the increase travel faster than those ahead, forming an admissible shock wave. Rankine-Hugoniot conditions give its speedAt $x=1$ the idealized hydrograph jumps from $Q_0=A_0^{5/4}$ to $Q_1=A_1^{5/4}$ at $t=1/c_s$. With positive $ϵ$, the diffusion term spreads the rise over a finite interval. The hydrograph is rounded, with an earlier small response and a later approach to the new discharge, rather than an instantaneous jump.

The physical cause is the longitudinal pressure gradient: a local increase in depth influences neighbouring sections and redistributes water before pure downstream transport would deliver a discontinuity. For a small increase, linearization about $A_0$ has speed $c=\frac{5}{4}{A}_0^{1/4}$ and diffusivity $D=ϵA_0^{3/4}/4$. The smooth curve in the figure is the exact half-line step response for this linearized advection-diffusion equation; it illustrates the shape without claiming to solve a large nonlinear jump exactly.

For zero viscosity the Inviscid Burgers equation is $q_X-q{q}_{\theta }=0$. Along a characteristic curve, $dq/dX=0$ and $d\theta /dX=-q$. Labelling a characteristic by its initial position $\eta$ givesWhere $f$ is differentiable, $q_{\theta }=f^{\prime }(\eta )/(1-X{f}^{\prime }(\eta ))$. The first characteristic crossing occurs at $X=1/\max f^{\prime }$ if the maximum positive slope exists. A multivalued continuation is not a physical wave: rewrite the equation as the conservation law $q_X+{\partial }_{\theta }(-q^2/2)=0$ and replace the overturning part by an entropy-admissible shock wave. The Rankine-Hugoniot conditions give its trajectoryThe states on either side are supplied by the characteristics meeting the shock. Conservation supplies the equal-area construction, and the entropy condition requires characteristics to enter the shock. Since the flux is concave, a jump with $q_L<q_R$ is compressive.

For the rising step, $q_L=0$ and $q_R=U>0$, the left characteristics have slope zero and the right ones slope $-U$. They overlap immediately. The Burgers Riemann problem with negative flux is solved byIt is a shock, not a rarefaction wave; the minus sign in the nonlinear flux is decisive.

For positive viscosity, substitute $q=2ϵ{\partial }_{\theta }\log \psi$ into the viscous Burgers equation. Writing $w=\log \psi$ givesAn $X$-dependent multiplicative factor in $\psi$ removes the possible integration function. The Cole-Hopf transformation therefore gives ${\psi }_X=ϵ{\psi }_{\theta \theta }$. Normalize its initial value continuously at zero:Split the heat kernel integral at zero. Completing the square in the positive-half integral yieldsWhen differentiating, the Gaussian terms from the two complementary error functions cancel, becauseThus the exact viscous Burgers step solution with negative flux isFor $X>0$ its denominator is positive, its tails tend to $0$ and $U$, and it approaches the initial step as $X\downarrow 0$. At $\theta =-UX/2$ it equals $U/2$. Its vanishing-viscosity limit gives the entropy shock above. At large $U\sqrt{X/ϵ}$ near the moving shock center it approaches the travelling profile $U/[1+e^{-U(\theta +UX/2)/(2ϵ)}]$, with thickness of order $ϵ/U$.

Write the shock as $x=s(t)$ and let the plus trace be on $x>s(t)$. Locally its discontinuous part can be represented with the Heaviside factor $H(x-s(t))$. Distributional differentiation gives ${\partial }_{t}H=-s^{\prime }(t)\delta (x-s(t))$ and ${\partial }_{x}H=\delta (x-s(t))$. The singular part of the conservation law is thereforeIt must vanish, so the Rankine-Hugoniot conditions areThe smooth parts of the fields still satisfy the classical conservation law on each side. The jump condition ensures that the moving discontinuity creates no extra source of the conserved quantities.

Take a unit-width control volume spanning the stationary hydraulic jump. Mass conservation makes the discharge $q=hu$ constant. The horizontal momentum flux is $\rho h{u}^2$, and the integrated hydrostatic pressure force on a vertical cross-section is ${\int }_0^h\rho g(h-z)dz=\rho g{h}^2/2$. Momentum conservation therefore givesThese are the Rankine-Hugoniot conditions for the shallow water equations in the stationary shock frame. With depths $h_1\ne h_2$ on the two sides, eliminating the velocity gives

Let $H$ be the depth behind the bore. The wall's kinematic boundary condition makes the velocity there $U$; ahead of the bore the depth and velocity are $h_0$ and zero. Across a rightward hydraulic bore moving at $V$, the Rankine-Hugoniot conditions areThe first gives $R=H/h_0=V/(V-U)=v/(v-1)>1$. Substituting into the second gives $F^{2}v=(R^2-1)/2$, where $v=V/U$ and $F=U/\sqrt{g{h}_0}$. Hence the piston-driven shallow-water bore obeysThe positive depth increase is consistent with the energy-loss condition in ii, now viewed in the bore's own shock frame.

A wall moving at constant speed $U>0$ into stationary water of depth $h_0$ produces a hydraulic bore of speed $V>U$ in the uniform-state shallow water equations model. Behind the bore the fluid velocity equals $U$ and its depth $H$ obeys $H/h_0=V/(V-U)$. Put $F=U/\sqrt{g{h}_0}$ and $v=V/U$. The moving Rankine-Hugoniot conditions give $F^{2}v=[(H/h_0{)}^2-1]/2$, hence the displayed equation for $v>1$. Its left side is a strictly increasing function from $F^2$ to infinity, while its right side is a strictly decreasing function from infinity to one. The intermediate value theorem therefore gives exactly one admissible $V$ for every $F>0$.

A shock wave with significant radiation losses in its downstream evolution. Often a narrow dissipative jump initially heats the gas according to adiabatic Rankine-Hugoniot conditions, followed by a much longer radiative cooling layer. The gas's specific entropy increases at the jump and may then fall as radiation carries entropy away. If both distant states have equal temperature, the combined mass and momentum balances give an isothermal shock; the gas's energy flux is not conserved across the whole radiating region.

In the rest frame of a one-dimensional shock, states with velocities $w_0,w_1$ satisfyand

These are the Rankine-Hugoniot conditions specialized to the mass, momentum and energy fluxes of a perfect gas.

For the dimensionless shallow water equations with gravity coefficient one, let the left state be $(h,u)$, $h>0$, and the right state $(1,0)$. The Rankine-Hugoniot conditions are $s(h-1)=hu$ and $shu=h{u}^2+(h^2-1)/2$. For $h\ne 1$ and $s>0$, eliminating $u$ gives the displayed formulas. The characteristic speeds are $u\pm \sqrt{h}$ and $\pm 1$. Because $u-\sqrt{h}<s$ always, the positive branch can only be an admissible family-plus shock. Its remaining Lax shock admissibility conditions, $1<s<u+\sqrt{h}$, both reduce to $h>1$. At $h=1$ there is no jump and the labelled interface has no determined shock speed.

Two smooth differential systems related by a nonlinear change of dependent variables need not define the same weak solutions. If $U_t+F(U{)}_x=0$, a smooth function $V=g(U)$ satisfies $V_t+g^{\prime }(U)F^{\prime }(U)U_x=0$ for smooth $U$; even when this can be written as another divergence, its Rankine-Hugoniot conditions need not be equivalent to those in $U$. The smooth chain rule does not justify multiplying jump distributions by discontinuous coefficients.

For example the shallow water equations can be written smoothly as $h_t+(hv{)}_x=0$ and $v_t+(h+v^2/2{)}_x=0$. Mechanical mass and momentum conservation instead select $h_t+(hv{)}_x=0$ and $(hv{)}_t+(h{v}^2+h^2/2{)}_x=0$. These systems agree for smooth positive $h$, but their jump laws differ. The states $(h,v)=(2,\sqrt{3}/2)$ and $(1,0)$ satisfy the mass/momentum jump laws with speed $\sqrt{3}$, while the velocity-form jump equation would demand $11/(4\sqrt{3})$. A model for discontinuities must specify the conservative variables or an equivalent admissible regularization, rather than using smooth equivalence alone.