# Smallest star radius by loss (or production) of heat

I'm looking for the smallest possible theoretical size for a spherical uniform star in rotation, after it loses some energy as heat. I'm puzzled by the following calculations.

Initially, the star has a radius $R_0$ and angular velocity $\omega_0$. Its mass $M$ is conserved, and its mechanical energy is this : $$\tag{1} E_0 = \frac{1}{2} \, I_0 \, \omega_0^2 - \frac{3 G M^2}{5 R_0} = \frac{1}{5} \, M R_0^2 \, \omega_0^2 - \frac{3 G M^2}{5 R_0}.$$ Then for some reason, the star releases (or produces) some heat $Q$ and reduces its size to $R$. Angular momentum is conserved : $$\tag{2} I_0 \, \omega_0 = I \, \omega \quad \Rightarrow \quad \omega = \frac{R_0^2}{R^2} \, \omega_0.$$ Conservation of energy (with heat released) : $E_0 = E + Q$, gives this equation : $$\tag{3} \frac{1}{5} \, M R_0^2 \, \omega_0^2 - \frac{3 G M^2}{5 R_0} = \frac{1}{5} \, M R^2 \, \Big( \frac{R_0^2}{R^2} \, \omega_0 \Big)^2 - \frac{3 G M^2}{5 R} + Q.$$ I then could do two things :

1. Isolate $R$ as a function of $R_0$, $\omega_0$ and $Q$ (equation (3) gives a second order equation with two roots), then ask which value of $Q$ gives the smallest value of $R$, i.e. $\frac{\partial R}{\partial Q} = 0$. The calculations are a bit messy, and I'm not sure of the result.
2. Express $E$ as a function of $R$ while considering $R_0$, $\omega_0$ and $Q$ as fixed parameters. Then $\frac{dE}{dR} = 0$ would minimize the final energy, but I'm not sure that considering $Q$ as fixed is making any sense. This calculation is easy though and gives this : \begin{align}\tag{4} \tilde{R} &= \frac{2 \omega_0^2 \, R_0^4}{3 G M}, & E_{\text{min}} \equiv E(\tilde{R}) = -\, \frac{9 G^2 M^3}{20 \omega_0^2 \, R_0^4}. \end{align} Notice that $\tilde{R}$ is not the minimal value of $R$, but gives the minimal value of final $E$ instead. I think this approach is wrong.

Maybe there's another way of finding the minimal value of $R$ that a star can achieve, for given $R_0$ and $\omega_0$ ($Q$ is a variable, and $M$ is conserved). Apparently, even when $Q$ is arbitrary large, conservation of energy and angular momentum implies there's a non-trivial solution to this problem, but I may be wrong.

What is the best way of doing this ?

\begin{align} T_0&=\frac{1}{2}I_0\omega_0^2\\ U_0&=-\frac{3 G M^2}{5 R_0}\\ r&=\frac{R}{R_0} \end{align}
then your equation read \begin{align} T_0+U_0&=\frac{T_0}{r^2}+\frac{U_0}{r}+Q \end{align} and it becomes much more readable: $$r(Q) = \frac{U_0\pm\sqrt{4 T_0 \left(-Q+T_0+U_0\right)+U_0^2}}{2 \left(-Q+T_0+U_0\right)}$$
The solution with minus is the one giving $r(0)=1$. Then Mathematica thinks there is no minimum of $r(Q)$ actually. I will let you confirm that!