I haven't understood the mechanism by which energy from the core of a star reaches its outer layers. Primarily, heat is transfered via convection or radiation.

For stars of nearly one solar mass, radiation dominates in the inner regions because convection needs a high temperature gradient, and their cores aren't hot enough. Now, the wikipedia page says-

The outer portion of solar mass stars is cool enough that hydrogen is neutral and thus transparent to ultraviolet photons, so convection dominates.

I find this reasoning kind of counter intuitive.If the outer regions are transparent to ultraviolet photons, why does convection dominate?

A similar argument is presented in the case of high mass stars.

In the outer portion of the star, the temperature gradient is shallower but the temperature is high enough that the hydrogen is nearly fully ionized, so the star remains opaque to ultraviolet radiation. Thus, massive stars have a radiative envelope.

The above reasoning is not convincing enough. I would be grateful if someone could help me out.

  • $\begingroup$ I think they mean that if the star is transparent to UV, the UV radiation will escape the star and thus cannot heat the outer parts of it. Therefore convection must do the job. $\endgroup$
    – Gnorkx
    Commented Jan 8, 2017 at 14:58
  • $\begingroup$ When the radiation escapes, it must pass through the outer layers and thereby will cause heating. $\endgroup$
    – P_RS
    Commented Jan 8, 2017 at 15:00
  • 3
    $\begingroup$ Only if the radiation is absorbed in the outer layers will it cause heating. Not if it passes through. $\endgroup$ Commented Jan 8, 2017 at 15:29
  • $\begingroup$ No you must get your basics correct,I guess. The mean free path of photons inside a star is very short. It is impossible for a photon to fly out without encountering any particle. $\endgroup$
    – P_RS
    Commented Jan 8, 2017 at 15:44

1 Answer 1


Whether a region becomes convective or not depends (roughly) on whether the temperature gradient exceeds the adiabatic temperature gradient. If it does so, then a convective instability is set up whereby a rising (adiabatic) parcel of gas experiences a continuing upward force.

In a radiative zone the temperature gradient is directly proportional to the opacity of the gas, $\kappa$ $$ \frac{dT}{dr} \propto \frac{\kappa \rho}{T^3} F_{\rm rad},$$ where $\rho$ is the density and $F_{\rm rad} = L/4\pi r^2$ is the radiation flux emerging from a star of luminosity $L$ at radius $r$. The interior opacity of a star can be approximated by Kramer's law: $\kappa \propto \rho T^{-7/2}$.

There are a lot of competing effects going on here, but in general at a given temperature, the outer parts of a 1 solar mass star have a lower energy flux, but the density is a lot higher, than the outer parts of a high-mass star. The net outcome is that the outer part of a solar-mass star is unstable to convection, but the temperature gradient in the outer part of a high-mass star remains below the adiabatic temperature gradient.

I find the wikipedia explanation baffling. Only the very outer skin of a star is cool enough for hydrogen to remain neutral so this is not referring to the bulk convection zone.

  • $\begingroup$ Since the temperature of a star decreases as we move outwards the center, the temperature gradients are always negative. When you say "Whether a region becomes convective or not depends (roughly) on whether the temperature gradient exceeds the adiabatic temperature gradient". Does it means $\frac{dT}{dr} > \left(\frac{dT}{dr}\right)_{\textrm{ad}}$ or $\left|\frac{dT}{dr}\right| > \left|\frac{dT}{dr}\right|_{\textrm{ad}}$? $\endgroup$
    – Stefano
    Commented Jul 6, 2017 at 13:04
  • 1
    $\begingroup$ @Stefano The second one. $\endgroup$
    – ProfRob
    Commented Jul 6, 2017 at 14:59

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