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The energy-conserving self-similar blast wave requires a stage in which the original stellar radius, ejecta mass, ambient pressure and cooling have ceased to supply important scales. In particular, the swept-up mass must greatly exceed the ejecta mass; $R\gg R_{\ast }$ alone need not ensure this in a very dilute interstellar medium. For fixed explosion energy $E$, uniform ambient mass density ${\rho }_0$ and elapsed time $t$, dimensional analysis gives $R=C{E}^a{\rho }_0^b{t}^d$. Matching mass, length and time gives $a=1/5$, $b=-1/5$, $d=2/5$. Equivalently, a self-similar strong shock wave has $E\sim {\rho }_0{R}^3{\dot{R}}^2$, whose constant-energy solution has the same exponents:The dimensionless coefficient depends on $\gamma$ and the post-shock profiles; dimensional reasoning cannot fix it.

For completeness, obtain the Strong-shock Rankine-Hugoniot conditions in the shock frame. Let $v_1=V-u$ denote downstream speed relative to the outward-moving front. With negligible upstream pressure and enthalpy, conservation of mass, momentum and energy givesPut $q={\rho }_1/{\rho }_0$. Substituting $v_1=V/q$ into the last two equations and selecting the compressive, nontrivial branch yields $q=(\gamma +1)/(\gamma -1)$. HenceHere the symbol $u$, written as $u=2V/(\gamma +1)$ in what follows, is the laboratory-frame radial velocity; it is not the relative speed $v_1$.

Use the thin-shell approximation for a spherical blast wave. Set $W=4\pi R^3/3$ and $M_s={\rho }_{0}W$. Equating the swept-up mass to $4\pi R^{2}d{\rho }_1$ gives $d/R=(\gamma -1)/[3(\gamma +1)]$. The growing shell's momentum is $M_{s}u$. Since newly swept gas has zero laboratory momentum, the net force is the cavity pressure acting on the inner shell, to thin-shell accuracy:The pressure at the shock is already accounted for by the jump conditions and mass loading; subtracting $p_1$ as an additional external pressure would count the shock reaction twice. With $u=2V/(\gamma +1)$, the last equation becomesTherefore, for constant $\alpha$,

The shell volume is $M_s/{\rho }_1=W/q$. Its internal energy and kinetic energy areTreat the cavity volume as $W$ at the same thin-shell level, so its internal energy is $E_{c,\mathrm{int}}=\alpha p_{1}W/(\gamma -1)$. Their sum is proportional to $R^3{V}^2$ for constant $\alpha$ and $\gamma$. Constant total energy thus requires $3+6(\alpha -1)=0$, giving $\alpha =1/2$. ConsequentlyUsing $V=2R/(5t)$, this becomesFor $\gamma =5/3$, $d/R=1/12$ and the approximate energy fractions are $E_{s,\mathrm{kin}}/E=1/4$, $E_{s,\mathrm{int}}/E=1/4$, $E_{c,\mathrm{int}}/E=1/2$.

This is an approximate normalization, not an exact treatment of finite shell thickness. If instead one subtracts the shell volume and uses $W_c=W(1-1/q)$ for the cavity, while retaining the other shell assumptions, the energy coefficient becomes $2(2\gamma -1)/[(\gamma -1)(\gamma +1{)}^2]$ and ${\xi }_0^5=75(\gamma -1)(\gamma +1{)}^2/[32\pi (2\gamma -1)]$. The difference is an order-$d/R$ geometric correction of this crude shell model; a consistent exact coefficient requires the full self-similar profiles.

In the idealized mass budget all swept gas is assigned to the shell, so the cavity inertia is neglected. The hot cavity's residual mass density is not specified, so its kinetic energy cannot be assigned a unique numerical fraction from the stated data. If its low mass is $M_c$ and its velocity is of order $u$, then $E_{c,\mathrm{kin}}\sim M_c{u}^2/2$ andFor an illustrative uniform-density cavity with homologous velocity $v(r)=ur/R$, direct radial integration gives $E_{c,\mathrm{kin}}=3M_c{u}^2/10$, so the preceding fraction has an extra factor $3/5$. The cavity can carry substantial internal energy despite negligible kinetic energy because it is hot. Neglect of its inertia is justified in the swept-mass-dominated stage, not merely by its small initial radius.

The calculation describes the adiabatic expansion stage of a supernova remnant, giving an estimate of its age or explosion energy from its radius, expansion speed and ambient mass density. For example $t\simeq 2R/(5V)$, and $E\propto {\rho }_0{R}^3{V}^2$. Such blasts inject energy and momentum into the interstellar medium and redistribute stellar ejecta. The early ejecta-dominated stage, radiatively cooled late shell, nonuniform ambient medium, finite ambient sound speed, magnetic fields, nonspherical geometry and energy transfer to cosmic rays violate different assumptions. Even in the adiabatic spherical stage, the real profiles are not a uniform thin shell and a uniform-pressure cavity, so ${\xi }_0$ here is an estimate.

Let $D=u_1>0$ denote the front speed in the stationary upstream frame. In the shock frame, the upstream and downstream velocities are $-D$ and $-D/r$. Taking $M_1\to \infty$ in the Rankine-Hugoniot conditions for a perfect gas givesThe downstream laboratory velocity is $D-D/r=2D/(\gamma +1)$, which distinguishes the gas speed from the front speed.

For the planar blast-wave energy scaling of a self-similar blast wave, integration of the total energy density over the shocked interval $0<z<Z$ gives the energy per unit area on this side:The similarity solution makes $I$ time-independent; a finite positive explosion energy requires $0<I<\infty$. Conservation of energy therefore gives $\dot{Z}=(E_{+}/({\rho }_{1}I){)}^{1/2}{Z}^{-1/2}$ for the expanding front. Integrating from $Z(0)=0$,If $E$ denotes the one-sided energy, $C=(9/(4I){)}^{1/3}$. If the released energy $E$ feeds two symmetric fronts, $E_{+}=E/2$ and $C=(9/(8I){)}^{1/3}$. In either convention the requested scaling is $Z=C(E/{\rho }_1{)}^{1/3}{t}^{2/3}$. The constant depends on the similarity profiles and the energy convention; energy conservation determines the exponent without solving those profiles. The Strong-shock Rankine-Hugoniot conditions additionally fix $f(1)=(\gamma +1)/(\gamma -1)$ and $g(1)=h(1)=2/(\gamma +1)$.

Work in the shock frame, assume a calorically perfect gas with fixed specific-heat ratio $\gamma >1$, and neglect magnetic stresses and body forces across the thin shock. The internal energy per unit mass is $e=p/[(\gamma -1)\rho ]$, and the specific enthalpy is $w=\gamma p/[(\gamma -1)\rho ]$. Integrating the local conservation laws across a thin stationary control volume makes the mass flux $\rho u$, momentum flux $p+\rho u^2$, and energy flux $u(\rho u^2/2+\gamma p/(\gamma -1))$ continuous. Hence the Rankine-Hugoniot conditions for a perfect gas areMultiplication of the last equality by $m$ gives continuity of the stated energy flux. These express that mass, momentum and energy cannot accumulate in an infinitesimally thin steady shock, although entropy production occurs on the physical compression branch. The same constant in the polytropic equation of state $p=K{\rho }^{\gamma }$ must therefore not be imposed on both sides. For flow from $z>0$ to $z<0$, both signed velocities are negative; the flux equations and ratios remain valid.

In the Strong-shock Rankine-Hugoniot conditions, $p_1/({\rho }_1{u}_1^2)=1/(\gamma M_1^2)$ tends to zero. Put $r={\rho }_2/{\rho }_1$. The first two flux balances give $u_2=u_1/r$ and $p_2={\rho }_1{u}_1^2(1-1/r)$ to leading order. The energy balance becomesDiscarding the unchanged-state root $r=1$ gives the compressive shock compression ratio and the required limits:

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

The Strong-shock Rankine-Hugoniot conditions give the compression ratiofor $\gamma =5/3$. ThusIn the shock frame, mass conservation saysUsing ${\rho }_2=4{\rho }_0$ gives

For an energy-conserving strong spherical shock wave expanding into mass density $C{r}^{-\beta }$, $0\le \beta <3$, dimensional analysis gives the displayed radius law. Neglect ambient pressure, radiation losses and gravity in this blast-dominated approximation. Write $u=\dot{R}U(\xi )$, $\rho =C{R}^{-\beta }D(\xi )$ and $p=C{R}^{-\beta }{\dot{R}}^{2}P(\xi )$, with $\xi =r/R$. The radial continuity equation, momentum equation and adiabatic pressure equation reduce toThe Strong-shock Rankine-Hugoniot conditions give $U(1)=P(1)=2/(\gamma +1)$ and $D(1)=(\gamma +1)/(\gamma -1)$. The explosion energy fixes the radius coefficient through $E=4\pi C{R}^{3-\beta }{\dot{R}}^2{\int }_0^1[D{U}^2/2+P/(\gamma -1)]{\xi }^{2}d\xi$.