4
$\begingroup$

I'm trying to show that the mass-luminosity for a fully convective star is $$L \propto M^{\frac{113}{66}}$$ I know that the energy generation is via pp-chain and the opacity is due to $H^{-}$ ions such that:

$\epsilon = \epsilon_0 \rho T^3$ and $\kappa = \kappa_0\rho^{1/2}T^9$

I use the equation of stellar structure:

hydrostatic equilibrium: $P \propto M^2/R^4$

equation of state: $P \propto \rho^{5/3}$

energy generation: $L \propto R^3\rho \epsilon$

and there's obviously that $\rho \propto M/R^3$ and the ideal gas $P \propto \rho T$.

I can't use the equation of radiative transfer, because the star is convective but I instead have the equation of state, assuming the star is isentropic due to being fully convective. These together result in

$$L \propto M^7$$ which is not even close. Also in my calculation I never utilize the opacity relation, as it only really comes up in the radiative transport equation.

Any ideas where I'm going wrong?

$\endgroup$
5
  • 1
    $\begingroup$ Perhaps better asked on Astronomy SE than Physics SE. $\endgroup$ Apr 24, 2017 at 17:18
  • 1
    $\begingroup$ I would say this is really a physics problem. All this is a part of a broader picture, where equilibrium and gas dynamics is linked to different fuel cycles and plasma and degenerated electron gas properties. $\endgroup$
    – jaromrax
    Apr 25, 2017 at 7:25
  • 1
    $\begingroup$ It is a physics ploblem but I agree with @StephenG $\endgroup$
    – Stefano
    Dec 14, 2017 at 1:15
  • $\begingroup$ @DilithiumMarixt. I see that you are using 2 different equation of state. You should use only one. If the energy production is from the PP chain, I think you should use $T^4$ instead of $T^3$. And regarding the opacity, you may want to use the Kramer's opacity law. $\endgroup$
    – Stefano
    Dec 14, 2017 at 1:24
  • $\begingroup$ You could loop up Polytrope Models which can be used to model fully convective stars (and other types of EOSes) $\endgroup$ Feb 27, 2022 at 8:34

1 Answer 1

0
$\begingroup$

The bit you are missing is that the radius of the star is determined by the opacity.

Hydrostatic equilibrium yields $$P_R = \frac{GM}{R^2}\int \rho\ dr,$$ where $P_R$ is the pressure at the radius where light can escape. This in turn is related to the opacity $\kappa$ by $$\int \kappa \rho\ dr = \bar{\kappa}\int \rho\ dr = 1$$ Thus $$P_R = \frac{GM}{R^2\bar{\kappa}} \propto \frac{GM}{R^2 \rho^a T_{\rm eff}^b}, \tag*{(1)}$$ where I have substituted in $\bar{\kappa}\propto \rho_R^a T_{\rm eff}^b$.

With this equation you also have $$ L = 4\pi R^2 \sigma T_{\rm eff}^4 \tag*{(2)}$$ $$ L \propto M \rho T^c \tag*{(3)}$$

From the virial theorem and perfect gas law you also know that the interior $T \propto M/R$ and interior $\rho \propto M/R^3$.

You thus have 6 variables $L, M, R, T_{\rm eff}$, $P_R$, $\rho_R$ and 3 equations. To eliminate all but $L$ and $M$ you need another.

This comes from the interior polytropic equation of state $P \propto \rho^{\gamma}$ for a fully convective star (a polytrope with $\gamma = 5/3$ for a fully convective star). However, this proportionality only applies to a particular star - the constant of proportionality in fact depends on the mass and radius of the star in question. Here is not the place to repeat textbooks that deal with polytropes, but it can be shown that (e.g. here) $$P\rho^{-\gamma} \propto P^{1-\gamma} T^{\gamma} \propto M^{2-\gamma} R^{3\gamma -4}$$ and thus near the surface $$P_R^{1 -\gamma} T_{\rm eff}^{\gamma} \propto M^{2 -\gamma} R^{3\gamma-4} \tag*{(4)}$$

You then solve to get $L$ only as a function of $M$ and substitute in $\gamma=5/3$ and your favourite values for $a$, $b$ and $c$.

$\endgroup$

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.