 Solution

Let the mean particle mass be $\bar{m}=\mu m_H$, with $n_u$ counting total particles so that ${\rho }_u=n_u\bar{m}$. The ideal-gas law and the strong-shock pressure give the immediate post-shock temperatureThe robust prediction is $T_d\propto t^{-6/5}$. For a transparent numerical convention take $\mu =1$, $n_u=10^6{\mathrm{m}}^{-3}$ and $\xi =1$. ThenFor an ionized mixture with $\mu =0.6$ under the same total-particle-density convention, these normalizations become $5.13$ pc and $5.49\times 10^7$ K. The paper does not specify composition or whether its quoted density instead counts hydrogen nuclei; that convention must be stated, not hidden in a precise age.

The ambient-temperature limit of an adiabatic blast wave gives a formal matching estimate, $T_d\sim 10^4$ K, givesWith $\mu =0.6$ the age is about $1.3\times 10^6$ yr and the radius stays about 90 pc. This matching criterion already lies outside a very strong shock: $T_d=T_u$ in the strong-shock formula corresponds to ${\mathcal{M}}_s=\sqrt{16/(3\gamma )}\simeq 1.79$. A separate sonic estimate $V_s\sim c_{s,u}$ gives about $4.4\times 10^6$ yr and 133 pc for $\mu =1$; neither extrapolation is a precise transition solution. Using $\xi =1.15$ increases these inferred matching ages and radii by factors ${\xi }^{5/3}$.

In a real supernova remnant, substantial radiation losses commonly end the energy-conserving phase before these formal pressure-merging limits. The relevant comparison is $t_{\mathrm{cool}}\sim (3/2)n{k}_{B}T/(n_e{n}_i\Lambda (T))$ against the expansion time. A cooling function, composition and ionization state are needed to calculate that earlier age and radius; they are not supplied. The numerical values above answer the ambient-temperature breakdown estimate under explicit assumptions and must not be presented as a uniquely determined radiative shell-formation age.

After cooling, an interior-pressure-driven shell can approximately follow $R\propto t^{2/7}$ while pressure remains important. In the limiting momentum-conserving stage, $M_s\dot{R}\simeq \mathrm{const}$ and $M_s\propto {\rho }_u{R}^3$ give $\boxed{R\propto t^{1/4}}$. Ultimately ambient pressure and turbulent motions make the disturbance merge with the ISM. These slower stages respect the reduced energy budget rather than continuing an adiabatic strong-shock law indefinitely.