4
$\begingroup$

Suppose I know the Luminosity $L$, temperature $T$ and Mass $M$ of star. Assuming the star is very heavy so that we can treat it to be radiation-dominated star. This would imply that pressure inside the star goes (roughly) as,

$$ P(r) = \frac{\sigma T^4}{4 \pi c r^2} $$

How can I calculate the radius of (run-of-the-mill) star $R$ by balancing the radiation pressure and gravitational pressure? For this purpose the equation of Hydrostatics may be used,

$$ \frac{dP}{dr} = - \frac{G\ m(r)}{r^2} \rho $$

but since density $\rho$ is $r$ dependent I don't know how to deal with it. It would be nice if we can simply use the idea of balancing pressures since that's what defines the main-sequence star.

$\endgroup$
  • 3
    $\begingroup$ Can you clarify your question. If you "know the luminosity, temperature and mass", then the radius is known trivially from Stefan's law. I think if you are asking how do you calculate that structure from first principles, then the answer is you need to (numerically) integrate the equations of stellar structure (for instance both temperature and of course opacity vary with $r$ too). $\endgroup$ – Rob Jeffries Apr 7 '15 at 10:03
  • $\begingroup$ physics.stackexchange.com/questions/179045/… -Related $\endgroup$ – Joshua Lin Aug 6 '15 at 7:36
3
$\begingroup$

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!

$\endgroup$
0
$\begingroup$

The overly simplified (and empirically incorrect) way is to just balance the pressure at the surface $$ P_{\text{gravity}} = \frac{Gm^2}{4\pi R^4} $$ And $$ P_{\text{radiation}} = \frac{\epsilon \sigma T_{\text{surface}}^4}{c} $$

$\endgroup$
-1
$\begingroup$

It is a nice question, however the answer is complicated : I recommend to look at Eddington standard model in http://www.astro.umass.edu/~wqd/astro640/model.pdf, there are also nuerical methods mentioned there.

$\endgroup$
  • $\begingroup$ Whilst this may theoretically answer the question, it would be preferable to include the essential parts of the answer here, and provide the link for reference. $\endgroup$ – Kyle Kanos Aug 6 '15 at 0:42

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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