Photon walk in stars with convection

Question

I'm having trouble figuring this out. I've read, that when photons are created via nuclear processes inside a star, it can take about 1 million years for photons to actually reach the surface of a star, due to the "random walk" motion inside the stars interior. But in our Sun, the core is radiative, and the outer layers are convective. Does that mean, that when the photons reach the bottom of the convective layer, they can't go any further, or are they somehow transported by convection, so they don't do their "random walk"?

Don't know if my question is clear. I'm just confused about what happens with the photons when there is a change from radiative to convective zones.

Answer 1

No, it doesn't mean that the photons don't go any further. It means that when the temperature gradient inside the star reaches a threshold, the gas becomes convectively unstable. Heat is transferred more efficiently by moving parcels of gas than by transferring photons from hotter regions to cooler regions.

So, the photons continue to diffuse outwards, but the heat transfer becomes dominated by convective motions.

Answer 2

Don't think about individual single photons bouncing around. The energy (heat) diffuses slowly, exchanging energy with neighboring particles by various means (smashing into each other and radiation that only makes it as far as the next particle before being absorbed). Different photons are emitted and absorbed over and over, as well as physical collisions between particles exchanging kenetic energy.

Light doesn't travel through opaque material. But the heat can.

At the end, the hot matter loses energy due to radiation that escapes to space, cooling the material at the surface.

As for why matter at any temperature will emit black-body radiation, that's another question.

In the interior, hot matter shines on adjecent hot matter, which shines back, so nothing changes. The cooler surface provides a gradiant so heat will move in that direction in the long run.

Answer 3

The gas in a star is emitting light by blackbody radiation at the same rate as it's absorbing emitted radiation so it's temperature varies extremely slowly with time. The blackbody radiation increases the rate of diffusion of heat in a star. See Does nature really follow the heat equation?. That diffusion of heat is what people call photon random walk. No individual photon actually random walks. Each part of the gas absorbs photons and then emits new photons with a continuous range of energies. It can't possibly be emitting the same photons as it absorbed because the rate that it emits photons is affected by its temperature, not by the number of photons it absorbed.

The gas at the top of the outer layer is cooler than the gas at the bottom of it so it emits radiation at a lower energy. Also, because it's cooler, it sinks so it gets surrounded by hotter gas and absorbs the radiation from it heating up in the process. Since there's no convection further down, heat is much slower to diffuse by radiation. I'm not sure why there's no convection further down. Maybe, the pressure at the top of the inner core is already high enough to generate heat by nuclear fusion so there's no convection.