Supernova Expansion

The supernova expansion

In a supernova explosion, initially the supernova expands freely. Then the supernova enters a new stage when in the existent gas around the star, shock waves push out the matter. This lasts for $10^4$ years. The model that describes the shock waves stage is named the Taylor-Sedov model.  Let us say that the shock waves are spherical or radius $R=R(t)$. the characteristic quantities are $E$ (the total energy at $t=0$) and the density $\rho$ of the gas between the stars.

Please use dimensional analysis to determine the dependence of $R$ on $E$ (energy), $t$ (time) and $\rho$ (density). What can you say about the time dependence of the velocity $V=dR/dt$?

Answer

We do dimensional analysis on the quantities involved:

$<R> = m$ (meter)

$<E> = J (Joule) = N*m = kg*m/s^2 *m =kg*(m/s)^2$

$<t> = s$ (second)

$<\rho> = kg/m^3$

The dimensional analysis for $R$ (radius) goes as:

$m = ( {[kg*(m/s)^2] / (kg/m^3)} *(s^2) )^{1/5}$

The relation is

$R = [(E/\rho)*t^2]^{1/5}$          (1)

Since the velocity is defined as
$V= dR/dt$   its dimensions are $<V> =m/s$

The dimensional analysis for V (speed) is

$m/s =( {[kg*(m/s)^2] / (kg/m^3)} / (s^3) )^{1/5}$

And the relation between physical quantities is

$V = [E/(ro*t^3) ]^{1/5}$

Which could be deduced also from eq. (1) by taking the first derivative

Evaluation of results:

Energy of supernova is $E =10^44 J$

(From Hyperphysics)

Time of expansion

$t = 10^4 years =3*10^{11} s$

Density of interstellar matter

$\rho =100 atoms/m^3 =10^-25 kg/m^3$

(From The Interstellar Medium)

Radius R is thus

$R = [(E/\rho)*t^2]^{1/5} =2.46*10^{18} m = 2.5 light-year$

The value of radius of expansion is reasonable.

The evaluation for the speed gives a value of

$V = [E/(\rho*t^3) ]^{1/5} =8*10^6 m/s$

Also the value of the expansion speed is believable being about 50 times less than the speed of light.