Past exam of the mathematics course of the University of Cambridge / 2007 / iii / Paper 65 / 1

In the energy-conserving blast regime, neglect ambient pressure, radiative losses, ejecta mass and gravity compared with the explosion. The cold ambient ideal gas has no sound speed scale, and its power-law mass density has no preferred radius. The point explosion introduces only its energy $E$ and elapsed time $t$. This scale-free setting supports a self-similar blast wave; extra physics would limit the time interval over which that description applies.

The mass swept up to radius $R$ isIt is finite because $\beta <3$. With no energy loss, the kinetic energy and internal energy both scale as $C{R}^{3-\beta }{\dot{R}}^2$. Therefore $R^{3-\beta }{\dot{R}}^2$ is constant and integration givesThe dimensionless coefficient $A$ depends on the adiabatic exponent and ambient exponent through the similarity profiles. Indeed, $[E/C]=L^{5-\beta }{T}^{-2}$, which independently verifies the scaling. The shock wave decelerates for $0<\beta <3$, since its time exponent lies between $2/5$ and $1$.

Write $V=\dot{R}$, $a=2/(5-\beta )$ andHere $u$ is the outward radial velocity; $U,D,P$ are dimensionless. The relationsare sufficient for the reduction. In smooth post-shock gas the spherical continuity equation, Euler equations and adiabatic pressure equation areFor example, the mass density material derivative is $C{R}^{-\beta }(V/R)[(U-\xi )D^{\prime }-\beta D]$, and the velocity material derivative is $(V^2/R)[(U-\xi )U^{\prime }+qU]$. After cancellation of the dimensional factors the spherical blast wave in a power-law ambient density obeysPrimes here mean $d/d\xi$. In the pressure equation the scale derivative $2q-\beta =-3$ explains the constant term. Once the profiles are found, their finite energy integral fixes $A$ throughNo global constant in $p/{\rho }^{\gamma }$ is assumed: the shock wave generates different entropies in shells swept up at different times.

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.

Put $U=b\xi$ and $D=d\xi$. The shock wave conditions fix $b=2/(\gamma +1)$ and $d=(\gamma +1)/(\gamma -1)$. The continuity equation then reduces toso a linear mass density profile requiresMomentum requires $P^{\prime }=-db(b-1+q){\xi }^2$. At this exponent $q=-2(\gamma -1)/(\gamma +1)$, hence integration gives $P=2{\xi }^3/(\gamma +1)+P_0$. Substitution in the pressure equation makes the cubic coefficient vanish because $(\gamma +1)b=2$, while its constant term requires $(3\gamma b-3)P_0=0$. For $\gamma >1$ this implies $P_0=0$. Therefore all three equations and all shock wave conditions hold for the linear-profile spherical blast wave,The prescribed range $0<\beta <3$ restricts this family to $1<\gamma <7$.

For $\gamma =5/3$ one has $\beta =2$, $a=2/3$, $U=3\xi /4$, $D=4\xi$ and $P=3{\xi }^3/4$. The energy integral isConsequently $E=(3\pi /2)CR{\dot{R}}^2=(2\pi /3)C{R}^3/t^2$, which fixes the radius and dimensional fields completely:The kinetic energy and internal energy integrands are each $9{\xi }^5/8$, so each accounts for half the explosion energy. Integrating the mass density also gives the correct swept-up mass $4\pi CR$.