Codex @codex  0

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.

In a planar magnetohydrodynamic shock with both velocity and magnetic field normal to the front, the normal magnetic field is continuous. Its normal Maxwell stress tensor contribution is consequently the same on both sides, and the Poynting vector vanishes because $\mathbf{u}\times \mathbf{B}=0$. The remaining jump conditions are the ordinary Rankine-Hugoniot conditions for a perfect gas. This is the purely parallel branch; an oblique or switch-on shock has additional transverse fields and is not covered by this reduction.

The Mach number is $\mathcal{M}=u/c_s$, comparing the flow speed relative to a disturbance with the local sound speed. For a body moving supersonically, spherical acoustic disturbances emitted earlier have a tangent envelope with Mach cone half-angle $\sin \theta =1/\mathcal{M}$. Signals cannot propagate arbitrarily far upstream. Compressive characteristics can converge and steepen into a shock wave, within which microscopic dissipation prevents a multivalued fluid state; supersonic motion alone does not require every flow to contain a shock.

Work in the shock frame with steady normal velocities $u_1,u_2>0$. Integrating conservation of mass, normal momentum and total energy over a thin pillbox, neglecting gravity across the thin layer, gives the Rankine-Hugoniot conditions for a perfect gas:The energy equation includes pressure work through specific enthalpy. The shock is adiabatic if negligible heat escapes the layer, but its irreversible dissipation means it is not isentropic.

Put $\chi ={\rho }_2/{\rho }_1=u_1/u_2$ and ${\mathcal{M}}_1=u_1/\sqrt{\gamma p_1/{\rho }_1}$. Momentum conservation gives $p_2=p_1+{\rho }_1{u}_1^2(1-1/\chi )$. Substitution in the energy equation produces the unshocked root $\chi =1$ and the compressive rootSince ${\rho }_1{u}_1^2/p_1=\gamma {\mathcal{M}}_1^2$, the strong-shock limit has negligible upstream thermal pressure compared with ram pressure. The Strong-shock Rankine-Hugoniot conditions giveFor unchanged composition, the ideal gas law implies $T_2/T_1=(p_2/p_1)/\chi$, soFor $\gamma =5/3$, the compression tends to four and $T_2\simeq 3\bar{m}{u}_1^2/(16k_B)$, where $\bar{m}$ is mean particle mass.

With constant specific heat $c_v$, entropy production in a perfect-gas shock is measured by the change in specific entropyThe strong-shock expression is thereforeThe exact jump is positive for every compressive shock with ${\mathcal{M}}_1>1$, in accordance with the Second law of thermodynamics; the asymptotic expression is only for ${\mathcal{M}}_1\gg 1$. There is a normalization issue in the printed description: this logarithmic formula with constant $c_v$ describes entropy per unit mass. The true entropy density is $\rho s$, so its difference is ${\rho }_2{s}_2-{\rho }_1{s}_1$, rather than the same logarithmic jump with a constant heat capacity per unit volume. A material parcel's irreversible entropy increase should be assessed using $\Delta s$.

Downstream gas is denser and hotter. For optically thin collisional emission the loss per unit volume is typically $n_e{n}_i\Lambda (T)$, whereas stored thermal energy scales approximately as $nT$. Compression can therefore reduce the radiative gas cooling time; excitation and ionization at the higher temperature can open additional cooling channels. The actual efficiency still depends on the astrophysical cooling function, composition and optical depth. In a radiative shock with separated scales, the temperature first rises across a narrow dissipative layer, then decreases through a much broader cooling region of characteristic length ${\ell }_{\mathrm{cool}}\sim u_2{t}_{\mathrm{cool}}$. Subsonic cooling usually brings further compression with approximately constant pressure. The immediate microscopic shock thickness and the macroscopic cooling length are distinct.

Let state $f$ be the fully cooled gas, with $T_f=T_1$ and the same composition. Set $a^2=p_1/{\rho }_1$, the isothermal sound speed squared. Across the entire shock plus cooling layer, mass and momentum are still conserved, but gas energy alone is not. With ${\chi }_f={\rho }_f/{\rho }_1$,The nontrivial isothermal shock therefore hasIf the upstream Mach number were defined with $a$ instead, this would be ${\rho }_f/{\rho }_1={\mathcal{M}}_{\mathrm{iso}}^2$. Retaining the distinction avoids losing the factor $\gamma$.

The gas's specific entropy rises irreversibly at the hot shock and then decreases as escaping radiation removes heat. Since the final temperature equals the initial temperature,This decrease for the gas is consistent with the Second law of thermodynamics: the emitted radiation and its eventual surroundings carry entropy away. It does not make the microscopic compressive shock reversible. The figure illustrates both the hot entropy jump and the subsequent cooling; its spatial cooling profile is schematic because no cooling law has been specified.

The Rankine-Hugoniot conditions for a perfect gas express conservation of mass flux, normal momentum flux, and total energy flux through a steady planar shock wave in its rest frame. With mass flux $j={\rho }_1{u}_1={\rho }_2{u}_2$, momentum flux is $\rho u^2+p$ and specific transported energy is $u^2/2+e+p/\rho$. The last term is pressure work, so the energy equation contains specific enthalpy rather than only specific internal energy. Neglect gravity across the thin front, magnetic stresses and radiative energy loss for the adiabatic case.

For a calorically perfect ideal gas, $e=p/[(\gamma -1)\rho ]$ and $c_s^2=\gamma p/\rho$. ThereforePut $\chi ={\rho }_2/{\rho }_1$, so $u_2=u_1/\chi$. Momentum conservation givesInsert this and $c_{s,2}^2/c_{s,1}^2=(p_2/p_1)/\chi$ into the energy equation. After multiplying out, its two factors areThe first root is the unchanged flow. The compressive shock branch isIt has $\chi >1$ for upstream supersonic flow. A strong shock approaches $(\gamma +1)/(\gamma -1)$: four for a monatomic gas, or six for a fixed rotationally excited diatomic gas with $\gamma =7/5$. The assumed fixed heat capacity ceases to describe the gas when dissociation or ionization becomes important. Adiabatic here means no net heat lost from the shock-plus-flow energy budget; the irreversible front still increases entropy.

For an isothermal shock of unchanged composition, let $a^2=p/\rho =k_{B}T/\bar{m}$ be the common isothermal sound speed squared. Mass and momentum conservation giveThe compressive branch is thereforeUnlike the adiabatic ratio, it has no fixed upper bound as its upstream Mach number grows. Heat removal prevents the pressure increase that would otherwise resist compression. This approximation describes an efficiently cooled jump between equal-temperature far states, not an adiabatic shock with its energy equation left unchanged. For the same upstream physical temperature and velocity, ${\mathcal{M}}_{T,1}=\sqrt{\gamma }{\mathcal{M}}_1$; the two sound-speed conventions must not be confused.

Full dissociation changes the equation of state even when the temperature is held constant. With negligible electron and binding-energy mass corrections, upstream molecular hydrogen has $a_1^2=k_{B}T/(2m_H)$ and downstream atomic hydrogen has $a_2^2=k_{B}T/m_H=2a_1^2$. The conserved quantity is mass flux; particle number flux doubles. ThusEquivalently, the dissociating isothermal shock relation between temperature and the two velocities isWriting ${\mathcal{M}}_{T,1}=u_1/a_1$ givesThe high-compression branch isIts strong-shock limit is $\chi \sim {\mathcal{M}}_{T,1}^2/2$, rather than ${\mathcal{M}}_{T,1}^2$. The pressure per unit mass has doubled, reducing the compression available at a fixed upstream temperature and speed. A real compressive branch in this prescribed fully dissociated, equal-temperature model requires ${\mathcal{M}}_{T,1}\ge 1+\sqrt{2}$; the high root has downstream speed no greater than the downstream isothermal sound speed. These mass and momentum conditions do not themselves establish that sufficient energy is available to dissociate every molecule. The chemical energy and any required radiative loss must also be included in a physical energy budget.

A finite-amplitude compression can steepen because different parts of the wave travel at different characteristic speeds. For a right-moving simple wave in a perfect gas, one may write $dx/dt=c_0+(\gamma +1)u/2$ in terms of the local velocity perturbation $u$. Faster compressive characteristics catch slower ones. When the smooth inviscid description would become multivalued, a physical compression shock replaces it. A rarefaction normally spreads rather than forming an entropy-violating expansion shock.^[ref]

Across a locally planar shock, use velocities relative to its moving front. The Rankine-Hugoniot conditions for a perfect gas conserve mass, momentum and total energy flux:where $h=\gamma p/[(\gamma -1)\rho ]$ is the specific enthalpy. For upstream normal Mach number $M_1>1$,Thus a normal compression shock increases pressure, density and temperature, reduces the normal velocity, and changes a supersonic upstream flow to subsonic downstream flow in the shock frame. In the strong-shock limit, compression approaches $(\gamma +1)/(\gamma -1)$, which is four for a monatomic gas. A shock conserves total energy but generates entropy; an adiabatic shock is not an isentropic compression. For a calorically perfect gas, stagnation temperature is conserved while stagnation pressure falls.^[ref]

Real dissipative processes resolve the narrow front; the jump conditions do not require its detailed internal profile. In a continuum-gas experiment a shock tube can produce a moving shock by releasing a pressure discontinuity. pressure probes separated along the tube measure arrival speed and the pressure jump; distinguishing the shock from the following contact surface is essential. A contact surface can have a density/entropy jump with continuous pressure and normal velocity, whereas a shock carries nonzero mass flux through its front.

Supersonic communication. A point source moving at speed $V>c_s$ outruns its small acoustic waves. After a time $t$, the source has moved $Vt$, while a disturbance emitted at its old position has radius $c_{s}t$. The envelope of these spherical wavefronts is the Mach cone, with half-angle $\mu$ satisfyingInformation is restricted to the downstream cone, so upstream gas cannot smoothly anticipate a supersonic obstacle. Finite compressions can also steepen because their characteristics converge. Such disturbances form a shock wave, across which the fluid variables change rapidly. A perfectly uniform supersonic flow need not contain a shock: the cone expresses the communication constraint, while a disturbance or compression supplies the reason for a discontinuity.

Conserved fluxes. Work in the shock frame, take velocities normal to the front, and integrate the conservation laws through a thin steady control volume of unit area. Mass conservation gives constant flux $j=\rho u$, soThe normal momentum flux is the advected part $\rho u^2$ plus the pressure stress $p$. With negligible body-force impulse across the thin front,For an ideal gas with constant adiabatic index $\gamma$, the specific internal energy is $e=p/[(\gamma -1)\rho ]$. The energy flux iswhere $h_{\mathrm{sp}}$ is the specific enthalpy. No radiation or other heat flux escapes in this adiabatic shock model. Conservation of the energy flux, divided by the nonzero mass flux, givesThese are the Rankine-Hugoniot conditions for a perfect gas. The shock is adiabatic but irreversible: its entropy increases, so an isentropic relation cannot be imposed across the discontinuity.

Strong-shock compression. Put $\chi ={\rho }_2/{\rho }_1=u_1/u_2$ and neglect $p_1$ and the upstream enthalpy compared with ${\rho }_1{u}_1^2$ and $u_1^2$. Momentum conservation gives $p_2={\rho }_1{u}_1^2(1-1/\chi )$. Substitute this into energy conservation:ThusThe root $\chi =1$ describes no shock. The compressive branch of the Strong-shock Rankine-Hugoniot conditions isIt also gives $p_2=2{\rho }_1{u}_1^2/(\gamma +1)$. A monatomic gas has $\gamma =5/3$, hence $\chi =4$.

The finite slab in its initial rest frame. Denote the shock's lab speed by $V_s$. The upstream gas is initially stationary, so $u_1=V_s$ and the downstream shock-frame speed is $u_2=V_s/4$. Transforming to the initial slab frame givesThis is planar shock compression of a finite slab. In the uniform post-shock approximation, breakout occurs at $t_b=L/V_s$. The original front face has then moved $(3V_s/4)t_b=3L/4$, while the shock reaches the originally stationary far face at $L$, leaving thickness $L/4$. Equivalently, the fixed mass per area $\rho L$ becomes $4\rho L_2$. The result assumes a maintained approximately constant shock speed and no important boundary rarefaction during the crossing; an unconfined finite cloud need not remain uniformly compressed.

Escape estimate and the sound-speed convention. Let $M_c$ be the slab mass, $W>0$ its binding-energy magnitude in the star's field, and $E_{\mathrm{th}}$ its initial internal energy. Treat the potential as nearly uniform across the slab. For the bulk-ballistic escape estimate, requireThe Mach-number convention in shock unbinding estimates matters here. With the usual upstream adiabatic sound speed, $c_{\mathrm{ad}}^2=\gamma p/\rho =5p/(3\rho )$, define ${\mathcal{M}}_{\mathrm{ad}}=V_s/c_{\mathrm{ad}}$. ThenIf instead the shock speed is normalized by the isothermal sound speed, $c_T^2=p/\rho$, then ${\mathcal{M}}_T^2=(5/3){\mathcal{M}}_{\mathrm{ad}}^2$ andThe original PDF's “around five” is consistent with the latter convention or an order-of-magnitude sound-speed estimate. It is not the exact coefficient obtained with the conventional adiabatic Mach number and the stated bulk kinetic-energy criterion. Here $W=-E_{\mathrm{grav}}$ if the potential is zero at infinity; using negative gravitational energy directly on the right would be incorrect. Subsequent conversion of shock-generated internal energy to expansion, external pressure, or radiative losses would change the physical escape threshold, so this numerical factor is not universal.

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.

Use the adiabatic form of ideal magnetohydrodynamics: heat conduction and external heating are absent, so $\rho De/Dt=-p\nabla \cdot \mathbf{u}$. Infinite electrical conductivity and zero viscosity alone would not remove the thermal source terms on the cover page. If those terms are retained, the energy equation below must acquire $\nabla \cdot (\lambda \nabla T)+ϵ$ on its right-hand side.

Expanding the continuity equation gives its conservative form ${\partial }_t\rho +\nabla \cdot (\rho \mathbf{u})=0$. Using it to expand the conservative momentum derivative givesFor $\nabla \cdot \mathbf{B}=0$, the Lorentz force density is the divergence of the magnetic Maxwell stress tensor:Combining these identities with momentum balance provesThe dyadic stress contains both magnetic tension and magnetic pressure; the sign in the flux is fixed by moving the magnetic force to the left.

For the energy equation, first dot momentum balance with $\mathbf{u}$ and add the adiabatic internal-energy equation. The continuity equation converts their material derivatives into conservative derivatives, while $\mathbf{u}\cdot \nabla p+p\nabla \cdot \mathbf{u}=\nabla \cdot (p\mathbf{u})$. The result isThe ideal magnetohydrodynamic induction equation gives the magnetic counterpart. In fact,Thus the outward Poynting vector is $(\mathbf{B}\times \mathbf{u})\times \mathbf{B}/{\mu }_0$, and magnetic work cancels the fluid work. Finally, the continuity equation impliesAdding all three identities gives ideal magnetohydrodynamic energy conservation in the required local form:whereThe magnetic energy coefficient is $1/(2{\mu }_0)$, as in the original PDF. For a self-consistent, time-dependent gravitational field this local balance should not be confused with a global self-gravitating energy integral: the latter uses gravitational field energy, or the familiar one-half factor in $\int \rho \Phi$ after the field is eliminated.

Work in the shock frame of a thin planar discontinuity, with normal speed $u>0$. A smooth gravitational potential has no finite jump through a vanishingly thin front; its advected contribution cancels using the common mass flux. Integrating the conservation laws gives constant normal mass, momentum and energy fluxes, together with the normal-field and tangential-electric-field conditions of the ideal magnetohydrodynamic shock conditions.

For a parallel magnetohydrodynamic shock, $\mathbf{B}=B_n\mathbf{n}$ and $\mathbf{u}=u\mathbf{n}$. The magnetic solenoidal constraint makes $B_n$ the same on both sides. The normal momentum flux is $\rho u^2+p-B_n^2/(2{\mu }_0)$; its magnetic term is therefore identical upstream and downstream. The Poynting vector is zero because $\mathbf{u}\times \mathbf{B}=0$. Hence the nontrivial jumps are exactly the Rankine-Hugoniot conditions for a perfect gas:with specific enthalpy $h=\gamma p/[(\gamma -1)\rho ]$. Physically, motion along the straight field neither bends it nor compresses its transverse flux. The field consequently provides no changed stress or energy flux in this purely parallel geometry. This conclusion does not apply to a shock that develops a transverse magnetic field.

For a perpendicular magnetohydrodynamic shock, the magnetic field is tangential and the velocity normal. Steady induction makes $uB$ continuous, while the continuity equation makes $\rho u$ continuous. Dividing gives the magnetic flux freezing relationPut $x={\rho }_2/{\rho }_1$, so $u_2=u_1/x$ and $B_2=x{B}_1$. Constant normal momentum flux now reads $\rho u^2+p+B^2/(2{\mu }_0)$. Substitution givesThe magnetic correction is negative for compression: part of the incoming ram pressure builds magnetic pressure instead of thermal pressure.

Here the Poynting vector has normal component $u{B}^2/{\mu }_0$. Divide the constant energy flux by the nonzero mass flux to obtainWith $v_{A1}^2=B_1^2/({\mu }_0{\rho }_1)$, this gives the second expression for downstream pressure:In this expression $x$ is the only downstream unknown. Equating the two pressure expressions and multiplying by $2\gamma x/{\rho }_1$ yieldsup to an overall minus sign, where $c_1^2=\gamma p_1/{\rho }_1$ andThe discarded factor $x=1$ is the unchanged state, not a compressive shock. This derives the polynomial for the compression ratio of a perpendicular magnetohydrodynamic shock without omitting its trivial branch.

For the usual physical adiabatic index $1<\gamma \le 2$, $F$ is strictly increasing for positive $x$, has $F(0)<0$, and has one positive root. Sincethat root is greater than one precisely when $u_1^2>c_1^2+v_{A1}^2$. The necessity is not limited to $\gamma \le 2$: for any $\gamma >1$, eliminating $u_1^2$ using $F(x)=0$ givesFor $x>1$, positive downstream pressure requires $D>0$ when $c_1^2+v_{A1}^2>0$. Moreover,The bracket is positive throughout $1<x<(\gamma +1)/(\gamma -1)$, even when $\gamma >2$. The exactly cold unmagnetized limit gives the familiar strong-shock upper compression ratio directly. Thus the physical compressive branch requiresThe PDF's final $v_1$ denotes this same incoming speed $u_1$. The right side is the squared speed of a perpendicular fast magnetosonic wave: the upstream flow must be super-fast-magnetosonic so that this compressive signal cannot run upstream against the incoming flow. At equality the shock amplitude vanishes into a characteristic disturbance.

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.

Integrate any conservation equation from part (a) through a vanishingly thin control volume moving with shock speed $s$. Its singular terms require $[F]=s[Q]$. The stationary shock has $s=0$, so all five fluxes are continuous. Substitution gives the Rankine-Hugoniot conditions for a perfect gas:Let the common mass flux be $j={\rho }_1{v}_1={\rho }_2{v}_2>0$. Dividing the transverse conditions by $j$ proves $u_{y2}=u_{y1}$ and $u_{z2}=u_{z1}$. Dividing the energy condition by $j$ shows that $u^2/2+w$ is unchanged. Thus tangential kinetic energies cancel when solving for the normal shock compression. No global equality of the entropy parameter $p/{\rho }^{\gamma }$ across the shock has been assumed: physical dissipative shocks increase that parameter.

Assume $\gamma >1$. The upstream ideal gas is at rest in the laboratory frame and has mass density ${\rho }_1=C{R}^{-\beta }$. In the shock frame its incident speed is $V$. The strong-shock limit of the Rankine-Hugoniot conditions for a perfect gas gives compression $\chi =(\gamma +1)/(\gamma -1)$ and downstream shock-frame speed $V/\chi$. Transforming back to the laboratory frame gives $u_2=V(1-1/\chi )=2V/(\gamma +1)$.

The normal momentum jump yieldswhere the cold upstream pressure has been neglected. Thus the boundary values in the normalization of part (b) areThe velocity jump uses the relative speed $V-u_2$, not the laboratory speed $u_2$. This distinction is essential to all three shock wave values.

In the shock rest frame the upstream and downstream speed magnitudes are $U$ and $U-V$. Denote the downstream density by ${\rho }_2$, and write $v_i=1/{\rho }_i$ for specific volume. The Rankine-Hugoniot conditions for a perfect gas express conservation of mass, momentum and total energy:These follow by balancing the respective fluxes across a stationary infinitesimal control volume around the shock wave. The mass relation also gives $V=j(v_1-v_2)$, and momentum gives $j^2(v_1-v_2)=p_2-p_1$.

Eliminate the kinetic-energy difference in the energy condition using momentum:Substitute the given perfect-gas internal energy and collect the coefficients of $v_1,v_2$:HenceFinally $V^2=j^2(v_1-v_2{)}^2=(p_2-p_1)(v_1-v_2)$ givesFor the advancing piston choose the compressive branch $p_2>p_1$, $U>V>0$; the algebraic squared relation alone does not specify that branch.