What would I need in addition to the mass to figure out the radius of a main sequence star?
[Physics] How to calculate the radius of a main sequence star based on mass
astrophysicsstarsstellar-evolutionstellar-physics
Related Solutions
If you have $L$ and you have $T$, then nothing more complicated than Stefan's law is required. If $T$ is the effective temperature of the star then this gives an exact answer.
$$ R = \left(\frac{L}{4\pi \sigma_B T^4}\right)^{1/2}$$, where $\sigma_B = 5.67\times 10^{-8}$ in SI units.
If on the other hand you are trying to solve the structure from first principles then you need to learn about polytropes and the solutions of the Lane-Emden equation. A star supported solely by radiation pressure can be treated as a $n=3$ polytrope, which has no analytic solution.
On p.155-162 of Clayton's "Principles of stellar evolution and nucleosynthesis" you can find a treatment using polytropes and some tables with solutions for various values of $n$. The radius of a star is $$ R = \left[ \frac{(n+1)K}{4\pi G}\right]^{1/2} \rho_c^{(1-n)/2n} \alpha_1,$$ where $\rho_c$ is the (here unknown) central density, $n=3$ and $K$ is the constant in the polytropic equation of state (the exact value of $K$ depends on what proportion of the gas pressure is due to radiation pressure) and for a $n=3$ polytrope $\alpha_1=6.9$.
The mass is given by $$ M = -4\pi \left[ \frac{(n+1)K}{4\pi G}\right]^{3/2} \rho_c^{(3-n)/2n} \alpha_1^2 \left(\frac{d\phi}{d\alpha}\right)_{\alpha_1},$$ where $-\alpha_1^2 (d\phi/d\alpha)_{\alpha_1} = 2.02$ for a $n=3$ polytrope.
In the standard model, the ratio of normal gas pressure to total pressure is $\beta$, such that $\beta=0$ for a star solely supported by radiation pressure. It can be shown that the mass of such a star is given by $$ M = 18 \frac{(1 - \beta)^{1/2}}{\mu^2 \beta^2} M_{\odot},$$ where $\mu$ is the mean number of mass units per particle. Thus if you know $M$ and the composition, this gives you $\beta$.
The value of $K$ is then given by $$ K = \left[ \frac{9N_0^{4} k_B^{4}c}{4\mu^4 \sigma_B} \frac{(1-\beta)}{\beta^4}\right]^{1/3}$$ and $N_0$ is Avogadro's number.
This value of $K$ enables you to derive $\rho_c$ from the second polytropic relation and then substitute this into the first polytropic relation to get $R$. Good luck!
The basic concept here is that of hydrostatic equilibrium.
If you consider a thin slab of material of density $\rho$ and thickness $\Delta r$ in the star. It has a pressure of $P$ below the slab and a pressure $P + \Delta P$ above the slab. The weight of the slab will be $\rho g A \Delta r $, where $A$ is the area covered by the slab and $g$ is the local value of gravity. To keep the slab in equilibrium you need to balance this weight with the force exerted upwards on the slab due to the pressure difference between the top and bottom. i.e. $$ \rho g\ A\ \Delta r = -\Delta P\ A$$ Thus $\rho g\ \Delta r = -\Delta P$ and as $\Delta r \rightarrow 0$, we can say $$ \frac{dP}{dr} = -\rho(r) g(r) = - \rho \frac{GM(<r)}{r^2},$$ where $\rho(r)$ and $g(r)$ are functions of radius within the star and $M(<r)$ is the mass contained within radius $r$.
To solve this differential equation requires a self-consistent solution of the equations of stellar structure (involving the energy generation and energy transport equations), since the pressure also depends on temperature and composition.
To make progress with this in a back-of-the-envelope kind of way then some vast simplifications are required, namely an assumption about how the density depends on radius. If we assume that the density is constant (awful, but it does give the right proportionalities) then
$$\frac{dP}{dr} = - \frac{G \rho}{r^2} \frac{4\pi}{3} \rho r^3$$
$$ \int^{0}_{P(r)} dP = - \frac{4\pi G \rho^2}{3} \int^{R}_{r} r\ dr,$$
where $P=0$ at the surface of the star where $r=R$.
$$ P(r) = \frac{2\pi G \rho^2}{3} (R^2 - r^2)$$
Then we could put this in terms of the mass of the star $M$ by noting that $\rho =3M/4\pi R^3$ $$ P(r) = \frac{2\pi G}{3}\left(\frac{3M}{4\pi R^3}\right)^2 (R^2 - r^2) = \frac{3G}{8\pi}\left(\frac{M^2}{R^4}\right)\left(1 -\frac{r^2}{R^2}\right), $$ and the central pressure (at $r=0$) would be $$P(0) = \frac{3G}{8\pi} \left(\frac{M^2}{R^4}\right)$$
The proportionality is correct here, but comparison with a real star, like the Sun reveals that whilst the average pressure is reasonable, the central pressure is a couple of orders of magnitude too low, because the density of the Sun is not constant - the pressure and density are much higher in the centre.
Best Answer
Well, you would need an accurate stellar evolution model and you would need the age, chemical composition and probably the rotation rate too.
The mass-radius relationship is an area of intense research. Masses and radii are only determined for stars in eclipsing binary systems. These could define an empirical mass-radius relationship, but stellar evolution models tell us that the radius also depends on age. eg pre main sequence stars are larger, and even whilst still on the main sequence, stars gradually get larger by tens of percent over the course of their main sequence lifetime. Practically, this means that knowing the age would be important for stars of mass of about 80% of the Sun or larger. Lower mass main sequence stars would not have departed significantly from their initial main sequence radii even if they were as old as the Galaxy.
The radius will also depend on the chemical composition - both the overall metallicity and the fractional helium abundance are important, through their effects on atmospheric opacities and mean atomic masses. This is an effect at the level of 10-20%.
Lastly, it is becoming apparent that rotation probably plays an important role. In stars less massive than the Sun it appears that rotation-induced magnetic activity inflates stars (for poorly understood reasons). In higher mass stars rotation induces interior mixing that alters internal composition gradients and changes the evolution (as well as centrifugal distortion). Again, these are phenomena that affect radii at the level of 10% or so.
If you want something rough and ready, which at least allows you to account for the effects of age and metallicity, you could interpolate the results tables of radius versus mass that are easily obtained from this website, that reports calculations from the models of Siess, Dufour & Forestini (2000).