I was thinking that for a star to be stable, the rate of energy emittance through a shell of radius r is constant, otherwise there would be a buildup of energy which would change the temperature and hence the radius of the star. So I get $$r^2 T^4 = R_{surface}^2T_{surface}^4$$ using the Stefan Boltzmann law. But it gives the wrong results when I check for the temperature near the core of the Sun. But why? If this model is wrong, how would I get Temperature as a function of distance from centre for a star?

  • 1
    $\begingroup$ The Sun is a complicated thing - it doesn't emit only from the centre. In reality the Sun is completely opaque to EM waves at the centre and the light we measure is only emitted from the surface. $\endgroup$
    – Akerai
    Sep 27, 2019 at 11:57
  • $\begingroup$ en.wikipedia.org/wiki/Stellar_structure $\endgroup$
    – Kyle Kanos
    Sep 27, 2019 at 12:04
  • 1
    $\begingroup$ do not forget there is a lot of convection too $\endgroup$
    – user65081
    Sep 27, 2019 at 12:33

1 Answer 1


That isn't quite right for a number of reasons. Firstly you have to assume that the energy transport is only by radiation and that there is no energy generation occurring in the region where you use your formula. The Sun has an outer convection zone and energy generation takes place in the central 20% by radius.

Secondly, the flux of radiation is almost isotropic in the interior of a star so your expression for net outward flux is true at the photosphere but not true in the stellar interior.

The net flux radiated outward by a layer is actually the gradient of the flux multiplied by its effective thickness, given by the mean free path of the photons. You cannot assume that the latter is independent of radius - it changes with the gas opacity and density.

Thus outside the 20% core of the Sun, and interior to the convection zone, you can write that the net outward flux $$l \frac{d}{dr} \sigma T^4 = {\rm constant}\ ,$$ where $l$ is the photon mean free path and the constant is $L/4\pi r^2$, with $L$ the solar luminosity.


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.