In Astrophysics, an A-Z Introduction, Page 67, Patrick Betts states that in the Sedov-Taylor solution for a blast wave, the temperature of a supernova shell decreases over time, but the internal energy always remains 72% of the original energy released in the explosion.

I can not find any direct reference of this 72% figure. Does anyone know where such a direct reference can be found in primary literature? Thank you.

REF: Betts, P., Astrophysics. Pediapress.


1 Answer 1


I could not find a direct reference to that specific figure, but the answer seems to be in the equations (31) and (32) of the Taylor (1950) seminal paper.

Taylor derived the following numerical solutions for the total kinetic energy and total "heat energy" (i.e. the internal energy for an ideal gas) evaluated across the entire blast wave:

\begin{equation} K.E. = 2 \pi (0.185) \rho_0 A^2 = 1.164 \rho_0 A^2 \end{equation}

\begin{equation} H.E. = \frac{4 \pi}{(1.4)(0.4)} (0.187) \rho_0 A^2 = 4.196 \rho_0 A^2 \end{equation}

Where $A^2$ is a constant and $\rho_0$ is the density of the undisturbed medium. The numerical constants are all relative to the adopted adiabatic index (in the equations above, $\gamma = 1.4$).

As you can see, the ratio of Eq. (31) and (32) yields $(H.E.) / (K.E. + H.E.) \sim 78\%$. I assume that injecting a different value of $\gamma$ in the calculation of the constants would allow to obtain your figure.

In fact, Chevalier (1976) does state that the Sedov-Taylor solutions yields a kinetic energy that represents either 20% (assuming $\gamma=4/3$) or 30% (assuming $\gamma=5/3$) of the total energy. I suppose that R. Chevalier might have truncated from 72% to 70%, hence your figure would be true for $\gamma=5/3$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.