Past exam of the astrophysics course of the University of Oxford / 2015 / 4 / a / Solution

Write $R=C{E}^a{\rho }_0^b{t}^d$ for a uniform ambient mass density. Equating mass, length and time dimensions gives $a+b=0$, $2a-3b=1$ and $-2a+d=0$. Thus the Sedov-Taylor blast wave hasThe dimensionless coefficient depends on the gas's adiabatic index and on whether $R$ denotes the shock radius. Solving the similarity profile for a monatomic ideal gas gives $\xi \simeq 1.15$, so taking it as unity is an order-of-magnitude approximation rather than an extra dimensional result.

Timed photographs of the 1945 atomic explosion provide a direct scale test: the measured fireball radius follows $R\propto t^{2/5}$ over the relevant blast-wave interval, and inserting the independently established explosion energy and atmospheric density gives a coefficient of order unity. The published photographic analysis tested the radius-time law and used its normalization to infer an explosion energy of the correct nuclear-test scale. This illustrates both the success and the limits of dimensional reasoning: the power law is fixed dimensionally, whereas the numerical coefficient requires a profile calculation or measured normalization.^[ref]

Neglecting the exterior temperature requires a strong shock, ${\mathcal{M}}_s=V_s/c_{s,u}\gg 1$, or ${\rho }_0{V}_s^2\gg p_u$. The swept-up mass must dominate the initial ejecta for this energy-conserving similarity stage, while radiative losses, gravity, ambient gradients and continuing energy input must be unimportant. A very hot exterior or a late weak disturbance introduces a sound-speed scale and invalidates the cold-upstream similarity assumptions.