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In the rest frame of a strong planar shock wave, a perfect gas has ${\rho }_2/{\rho }_1=(\gamma +1)/(\gamma -1)$, $u_2/u_1=(\gamma -1)/(\gamma +1)$ and $p_2=2{\rho }_1{u}_1^2/(\gamma +1)$. Its downstream specific internal energy is ${ϵ}_2=p_2/[(\gamma -1){\rho }_2]=2u_1^2/(\gamma +1{)}^2$. Multiplying by mass flux $j={\rho }_1{u}_1$ gives the internal-energy flux in the title. The lost bulk kinetic energy flux instead equals $j(u_1^2-u_2^2)/2=\gamma F_{\mathrm{int}}$: this includes downstream pressure work and is the increase of specific enthalpy flux in the cold-upstream limit. Neither quantity is a temperature change per unit time of a fixed target mass without further information about that target.

For a monatomic gas obeying the ideal gas law, the specific internal energy is $e=3P/(2\rho )$. With no stellar nuclear fusion heating, the first law of thermodynamics makes the effective stellar energy-generation rate $ϵ=-{\partial }_{t}e-P{\partial }_t(1/\rho )$, with the derivatives taken at fixed enclosed mass. Differentiating gives $ϵ=-3{\partial }_{t}P/(2\rho )+5P{\partial }_t\rho /(2{\rho }^2)$. In a fixed-mass stellar homology Kelvin-Helmholtz contraction, $\rho \propto R^{-3}$ and $P\propto R^{-4}$ at each mass label. Hence $ϵ=-3P\dot{R}/(2\rho R)>0$ when $\dot{R}<0$. This includes internal energy storage as well as pressure work; using the gravitational potential-energy decrease alone would overestimate the outgoing luminosity.

For a fixed amount of substance and composition of classical ideal gas, internal energy depends only on temperature: $U=U(T,n)$, with no independent volume dependence. This does not require constant heat capacity at constant volume. Between equilibrium endpoints,Here $C_V$ is total heat capacity at constant volume, $C_{V,\mathrm{m}}$ is molar heat capacity, and $U/m$ for sample mass $m$ is specific internal energy, whose change is ${\int }_{T_1}^{T_2}{c}_V(T)dT$. These endpoint relations apply even when the intervening process is not at fixed volume; they do not say that transferred heat always equals $\Delta U$. With constant $C_V$, the safe statement is $U(T)=U(T_{\ast })+C_V(T-T_{\ast })$. Writing $U=C_{V}T$ additionally chooses the reference energy; it does not prove that a physical gas remains classical and ideal down to zero temperature. ^[ref]

A thermodynamic proof outline uses the first law of thermodynamics in equilibrium form, $dU=TdS-pdV$, and the Maxwell relation $(\partial S/\partial V{)}_T=(\partial p/\partial T{)}_V$:The last step uses the ideal gas law and the molar gas constant $R$, with fixed composition and amount of substance. Microscopically, separated ideal gas particles store no intermolecular interaction energy; their translational kinetic energy and any internal molecular energy are determined by temperature. Negligible intermolecular interactions still allow a molecule to have internal vibrational potential energy.

An original analytical model makes the distinction between an ideal gas and constant heat capacity concrete. Assume classical translation, two fully active rotational terms, and one independent quantum harmonic oscillator per diatomic molecule, with fixed angular frequency $\omega$. Define the vibrational temperature scale $\Theta =\hslash \omega /k_B$, using the reduced Planck constant $\hslash =h/(2\pi )$ and Boltzmann constant $k_B$. Summing its thermal occupations givesThe constant $U_{\ast }$ includes the chosen reference and zero-point energy, whose temperature derivative vanishes. When $T\ll \Theta$ the vibrational contribution to heat capacity is exponentially small; when $T\gg \Theta$ it tends to $R$ per mole. This model assumes rotations remain active throughout the interval and excludes dissociation, electronic excitation and real gas corrections. Its low-$T/\Theta$ edge is not a model of rotational freezing or of an actual substance at absolute zero.^[ref]

For this stipulated model, heating from $T_1=0.2\Theta$ to $T_2=\Theta$ gives $\Delta U/(nR\Theta )\simeq 2.57519$. Holding the frozen-vibration value $C_{V,\mathrm{m}}=5R/2$ throughout would give $2$, underestimating the required internal energy increase by about $22.3\%$ of the model's full value. The integral measures all thermal storage activated along the interval. At fixed volume with only heat input, this is also the supplied heat; at fixed pressure, expansion work must be added through Mayer's relation.

Joule's free-expansion experiment tests the independence from volume in a different way: an insulated rigid vessel allows gas into an evacuated compartment, so no external work is delivered and ideally no heat is exchanged. The first law of thermodynamics gives $\Delta U=0$; positive $C_V$ then predicts equal initial and final equilibrium temperatures for the ideal gas. The apparatus, thermal-capacity controls and limitations of this inference are described in Joule's free-expansion experiment and temperature sensitivity of a free-expansion experiment. This endpoint test differs from the reversible adiabatic expansion of a loaded piston, which delivers work and lowers internal energy.

An ice bath can absorb heat while its temperature barely changes: the heat converts ice into water. The latent heat of fusion quantifies this melting budget. For a pure substance at its equilibrium melting temperature $T_m(p)$ and specified pressure $p$, define it here per unit mass:Here $h_l$ and $h_s$ are the specific enthalpies of the coexisting bulk liquid and solid, and $H$ is enthalpy. For ordinary melting, $L>0$, in $\mathrm{J}\mathrm{k}{\mathrm{g}}^{-1}$; an amount $m$ has a total fusion enthalpy $mL$ in joules. A molar fusion enthalpy instead has units $\mathrm{J}\mathrm{mo}{\mathrm{l}}^{-1}$ and equals $ML$, where $M=m/n$ is mass divided by amount of substance in $\mathrm{kg}\mathrm{mo}{\mathrm{l}}^{-1}$. Always check the normalization before inserting a tabulated value into an energy balance.^{[ref] [ref]}

For a closed system converting completely from solid to liquid at fixed pressure, with only pressure-volume work and negligible changes in bulk kinetic energy and potential energy, the first law of thermodynamics gives $Q_p=\Delta H=mL$. Both endpoints are at $T_m(p)$: warming initially colder ice or subsequently warming the resulting water requires additional sensible heat. Since $H=U+pV$, the corresponding specific internal energy jump is $\Delta u=L-p\Delta v$, where $v=V/m$; $L$ includes the work associated with the change in volume. Heat is transferred energy, whereas enthalpy is a state function. ^[ref]

Let $x=m_l/m$ be the liquid phase fraction by mass, rather than by volume. Neglecting interfacial contributions to enthalpy, equilibrium phase coexistence at fixed pressure givesIn the model above, reaching $T_m$ takes $2.10\mathrm{kJ}$ of sensible heat; the plateau then absorbs $33.40\mathrm{kJ}$, ending at a cumulative $35.50\mathrm{kJ}$. At $18.80\mathrm{kJ}$, half the mass is liquid and the temperature is still $T_m$. The final warming takes another $4.18\mathrm{kJ}$. The rounded $L$ and $c_l$ are inspired by the introductory ice-melting calorimetry values in Case Western's demonstration; constant properties and exact equilibrium throughout are stipulated model assumptions. ^[ref]

Within one thermodynamic phase, specific heat capacity at constant pressure is $c_p=(\partial h/\partial T{)}_p$, in $\mathrm{J}\mathrm{k}{\mathrm{g}}^{-1}{\mathrm{K}}^{-1}$; a sample's heat capacity at constant pressure is $C_p=m{c}_p$, in $\mathrm{J}{\mathrm{K}}^{-1}$. The plateau is a finite enthalpy change at zero temperature change through varying $x$, so neither a finite single-phase $c_p$ nor $Q=m{c}_p\Delta T$ represents it. The heating curve needs separate sensible heat intervals and a fusion contribution. ^[ref]

The plateau assumes a pure substance, a stable bulk solid-liquid phase coexistence pair, fixed pressure, and sufficiently slow, uniform heating to remain near thermodynamic equilibrium. The total mass and composition must remain fixed, with negligible evaporation and no other phase transition. A fast heater can produce temperature gradients, so a local thermometer need not read the equilibrium $T_m$. Neither $T_m$ nor $L$ is universal across pressures or compositions: common-pressure melting-point slope describes the pressure dependence of $T_m$, while mixtures can melt over a temperature range as their coexisting compositions change. Their liquidus must be consulted instead of assuming a pure-substance plateau. ^[ref]

This makes ice useful as a temperature buffer: under these conditions it can absorb $m_{s}L$ before the remaining solid mass $m_s$ disappears, after which the water warms. If a stipulated constant net input power $P$ goes entirely into melting, that buffer lasts $m_{s}L/P$. Reversing the equilibrium conversion releases the same fusion enthalpy during freezing. The beautiful thermodynamic connection is $L=T_m(s_l-s_s)$, with $s$ the specific entropy and $T_m$ absolute temperature: equal Gibbs free energies at phase coexistence allow different enthalpies and entropies. The nearly stationary thermometer hides a finite, measurable change of state. ^[ref]

A monatomic gas consists of individual atoms or ions rather than molecules. In the classical dilute ideal-gas regime, its three translational degrees of freedom give $U=3N{k}_{B}T/2$ by equipartition theorem. Thus $C_V=3N{k}_B/2$, while $C_P-C_V=N{k}_B$ follows by differentiating the ideal-gas enthalpy $H=U+PV=5N{k}_{B}T/2$. Consequently the adiabatic index is $\gamma =C_P/C_V=5/3$ and the specific internal energy is $e=3P/(2\rho )$. The same translational result applies to a fully ionized mixture when the particle numbers are fixed. This excludes additional excitation, ionization and radiation contributions to the heat capacity.

Dividing two extensive quantities with the same replication factor produces an intensive quantity:Division by volume gives a density, such as mass density $\rho =m/V$ or energy density $U/V$. Division by mass gives a specific quantity, such as specific internal energy $u=U/m$ or specific volume $v=V/m$. Division by amount of substance gives a molar quantity, such as molar volume $V_m=V/n$. Their units are respectively $\mathrm{kg}{\mathrm{m}}^{-3}$, $\mathrm{J}{\mathrm{m}}^{-3}$, $\mathrm{J}\mathrm{k}{\mathrm{g}}^{-1}$, ${\mathrm{m}}^3\mathrm{k}{\mathrm{g}}^{-1}$ and ${\mathrm{m}}^3\mathrm{mo}{\mathrm{l}}^{-1}$. The denominator must be stated: “per particle”, “per mole” and “per unit mass” are different normalizations.

For a hypothetical uniform parcel with mass $2\mathrm{kg}$, volume $0.002{\mathrm{m}}^3$ and internal energy $40\mathrm{kJ}$ relative to a specified reference, $u=20\mathrm{kJ}\mathrm{k}{\mathrm{g}}^{-1}$ and $\rho =1000\mathrm{kg}{\mathrm{m}}^{-3}$. Three independent copies have $6\mathrm{kg}$, $0.006{\mathrm{m}}^3$ and $120\mathrm{kJ}$; both ratios remain unchanged. This permits comparison of materials without confusing the amount tested with the state tested.

If $X$ and $Y$ are additive but the parcels have different $r_i=X_i/Y_i$, their composite ratio is a weighted arithmetic mean:The denominator supplies the weights: specific internal energy uses mass weights, while mass density uses volume weights. This identity is accounting for the specified subsystems; predicting a new state after mixing also needs the relevant equation of state.

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.

Let $e=c_{v}T$ be the specific internal energy of a calorically perfect gas and define its specific gas constant $\mathcal{R}=R^{\ast }/\mu$, so $p=\mathcal{R}\rho T$. A reversible adiabatic change has no heat input and obeysUsing $c_p-c_v=\mathcal{R}$ and $\gamma =c_p/c_v$ givesIntegration yields $p/{\rho }^{\gamma }=K$, or $\boxed{p=K{\rho }^{\gamma }}$, with $K$ fixed on that isentrope. Equivalently $s-s_{\mathrm{ref}}=c_v\ln [(p/{\rho }^{\gamma })/K_{\mathrm{ref}}]$, so $K$ specifies entropy as well as composition. In a smooth nondissipative flow this means $D(p{\rho }^{-\gamma })/Dt=0$ along each parcel. A single common $K$ throughout the fluid additionally requires a homentropic flow. No heat transfer alone does not forbid entropy production: different sides of an adiabatic shock generally have different $K$ values. Constant composition and heat capacities are needed for the fixed exponent used here.

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.

Interpret $v$ as the normal upstream speed in the shock rest frame and $\rho$ as its upstream mass density. If the quoted jet speed is in another frame, the shock speed must first be supplied to obtain this relative velocity. Use a calorically perfect gas of mean particle mass $\bar{m}=\mu m_u$.

In the strong-shock limit, ${\mathcal{M}}_1\gg 1$. The Strong-shock Rankine-Hugoniot conditions giveThe pressure expression also follows from the supplied ratio by inserting ${\mathcal{M}}_1^2=\rho v^2/(\gamma p_1)$. The ideal gas law then gives the immediate postshock temperatureThe upstream thermal contribution is negligible at this order. This is a temperature in kelvin, rather than a temperature change per unit time.

To turn the postshock thermal energy into a rate, the processed mass per unit time is $\dot{m}=\rho vS$. Its downstream specific internal energy isThus the thermal internal energy carried into the shocked gas per unit time isPer unit area, divide this expression by $S$. Its dimensions are watts; the mean molecular weight cancels because the heat capacity per unit mass is inversely proportional to it, while $T_2$ is proportional to it.

There is a second common convention for shock heating power: the loss of directed bulk kinetic energy flux. The energy fluxes of a strong adiabatic shock giveThis equals the downstream enthalpy flux, and includes pressure work as well as the advected internal energy. For $\gamma =5/3$, the compression ratio is $4$, $T_2=3\mu m_u{v}^2/(16k_B)$, ${\dot{E}}_{\mathrm{int}}=9\rho S{v}^3/32$, and ${\dot{E}}_{\mathrm{kin},\mathrm{loss}}=15\rho S{v}^3/32$. Specifying the energy convention avoids conflating these rates. A rate $dT/dt$ for a fixed slab cannot be inferred without its participating mass and thermal evolution; the supplied data fix the postshock temperature and the processed energy fluxes.

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.

The local conservation laws. The first equation is the radial Euler equation: the material acceleration $Du/Dt={\partial }_{t}u+u{\partial }_{r}u$ is caused by the pressure gradient force per unit mass, $-{\partial }_{r}p/\rho$. It neglects gravity and viscosity. The second equation is mass conservation in spherical coordinates,The term $2\rho u/r$ appears because the area of a spherical surface grows as $r^2$.

To obtain the third equation, let $e=p/[(\gamma -1)\rho ]$ be the specific internal energy of an ideal gas with constant heat capacity ratio $\gamma$. The first law of thermodynamics for a smooth adiabatically moving element isDifferentiating $e$ and collecting terms givesand henceThus the entropy parameter $p/{\rho }^{\gamma }$ is conserved along each smooth pathline; for fixed composition, specific entropy differs from $c_v\ln (p/{\rho }^{\gamma })$ only by a constant. This is local isentropic flow behind the shock. A physical shock wave itself generates entropy and must be treated with jump conditions, rather than by applying this smooth equation across the discontinuity.

The front-radius scaling. In the self-similar blast wave approximation there is no scale set by ambient pressure, cooling, gravity, or the initial source radius. Write $R\propto E^a{\rho }_0^b{t}^d$. Using $[E]=\mathsf{M}{\mathsf{L}}^2{\mathsf{T}}^{-2}$ and $[{\rho }_0]=\mathsf{M}{\mathsf{L}}^{-3}$, dimensional analysis givesThus $a=1/5$, $b=-1/5$, $d=2/5$ andThe dimensionless coefficient ${\xi }_0$ depends on $\gamma$ and the resolved similarity profiles; dimensions alone do not fix it. Define the length scale $\ell (t)=(E{t}^2/{\rho }_0{)}^{1/5}$, so $\xi =r/\ell$ and the front is at the fixed coordinate $\xi ={\xi }_0$.

Kinetic energy. Substitute $\rho ={\rho }_0{f}_{\rho }(\xi )$ and $u=(r/t)f_u(\xi )$ into the volume integral. Since $r=\ell \xi$ and $dr=\ell d\xi$,But ${\rho }_0{\ell }^5/t^2=E$, so the first of the spherical blast-wave similarity energy integrals isThe conclusion requires a fixed, integrable similarity profile, as in the finite-energy blast solution. Expansion increases the swept-up mass while reducing the characteristic squared speed, leaving this product constant.

Pressure and internal energy. The derivatives of the similarity variable areWriting a prime for differentiation with respect to $\xi$, the material acceleration becomesThe radial momentum equation therefore requires a pressure gradient on the scale ${\rho }_0\ell /t^2$. Integrating over radius gives pressure similarity in a spherical blast wave:whereThe prefactor has units $\mathrm{kg}{\mathrm{m}}^{-1}{\mathrm{s}}^{-2}$, as required for pressure. Strictly, the momentum equation alone permits an additional pressure uniform in space and arbitrary in time. In the scale-free strong-blast solution, shock matching and the remaining thermodynamic equation fix that freedom with the same similarity scaling; any allowed constant of integration is included in $f_p$. An independently imposed external pressure would introduce another scale and invalidate the simple dimensional ansatz.

For constant $\gamma >1$, the internal energy is consequentlyIf $E$ is the total excess explosion energy, its normalization isThe two constant fractions need not be equal. No explicit solution for $f_p$ is needed to establish their time independence.

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.

Use $\mathcal{R}$ for the stellar gas constant, reserving $R$ for the stellar radius. The specific internal energy of a monatomic gas obeying the ideal gas law is $e=P/[(\gamma -1)\rho ]=3P/(2\rho )$. At fixed enclosed mass, the first law of thermodynamics gives the heat available to the outgoing luminosity as $ϵ=-{\partial }_{t}e-P{\partial }_t(1/\rho )$ when there is no stellar nuclear fusion heating. ThereforeThe second coefficient includes both the mass density dependence of $e$ and pressure work. This is gravitational energy generation in a homologously contracting ideal-gas star.