I’ve had trouble understanding exactly why there is not more mixing of plasma at the core of the sun with the outer layers.

I understand the difference between the radiative zone and the convective zone.

However, even $without$ convection, a plasma should mix at a fairly high degree with everything around it.

So there must be some other physics of high density plasma that keeps it from mixing.

I see no reason to think of this plasma at the core as effectively being some kind of solid, a gel, or even a liquid? It’s always described as a gas, which implies very free mixing with its surroundings.

Maybe someone can answer why, to some order of magnitude, a helium ion in the core does not, on average, migrate 1 million miles outside the core over the course of a billion years or so?

Regardless of temperature gradient or pressure gradient, it seems that helium ions at the core should diffuse to outer layers. I did read one reference to Swartzschild and “negative buoyancy”. It seems to me that the force of negative buoyancy would have to be very strong in order to keep heavier ions from diffusing to outer layers.

EDIT 1: ok I did some more googling and finally found this paper from 1977 on diffusion. It’s going to take me a few days to digest it though :)


It refers to a diffusion rate of ~1000 km every 5 billion years. Now I have to figure out why that estimate is so fantastically slow.

Such a low diffusion rate also seems absolutely ludicrous given that high density tokamak plasma experiments report diffusion rates on the order of 100 meters^2/second. https://www.osti.gov/servlets/purl/6637207

Experiments cite a number 9 or 10 orders of magnitude higher than the rate they “think” is going on inside of stars. Everything I’ve read sounds like handwaving. Physicists don’t seem to know what is going on in the core of stars.

EDIT 2: Here is a paper that discusses at much length, and with much humility, atomic diffusion processes in the core of stars. https://arxiv.org/abs/1707.07454 page65. They reiterate many times that there seems to be unknown process(es) that inhibit diffusion, based on studies gathered from many many different sources, encompassing just about every species of star.

  • $\begingroup$ Let us continue this discussion in chat. $\endgroup$ – PM 2Ring May 10 at 2:16
  • $\begingroup$ "that high density tokamak plasma experiments report diffusion rates on the order of 100 meters^2/second." cannot compare the two, what about the very strong gravitational well of the Sun , the higher (with respect to the center) the diffusion the deeper the well ? $\endgroup$ – anna v May 10 at 4:01

The process of atomic diffusion does not proceed quite as you describe. In order to increase (for example) the He abundance at the surface of the Sun, it is not just necessary for He to make it from the core to the surface, but also that the reverse process is slower.

Of course in a plasma with large concentration gradients, that is likely to happen; but what you are missing is that gravitational acceleration will make the He concentrate towards the centre of the Sun. The overall diffusion rate must take both effects into account. There is also the issue of differential radiative acceleration, although in the Sun this is subordinate to the effects of concentration gradients and gravity.

As to whether individual He nuclei make it from the core of the Sun to the surface, I'm sure they do, but that is not how a diffusion rate is calculated, since more He nuclei travel in the opposite direction.

The comparison with the Tokamak plasma is not apt. Although, there is a (smaller) gravitational potential gradient, the plasma density in a Tokamak is about 12 orders of magnitude lower than at the centre of the Sun, whereas the temperatures (and particle velocities) are an order of magnitude (factor of a few) higher. Since diffusion along concentration gradients is ultimately set by the mean free path between collisions, which varies as the inverse of particle number density in an ideal gas, then diffusion will be much faster in a Tokamak, by a handwaving 13 orders of magnitude.

As to your last paragraphs - why is mixing in stars difficult to model? (a) The mixing you are talking about is not difficult to model. (b) In real stars, simple atomic diffusion is only important in radiative zones and is unimportant in convective zones. (c) Even in radiative zones, there are other mixing processes that can occur, connected with gravity waves and rotational shear. It is these processes that are poorly understood and modelled. They are inherently 3-dimensional, but 3D modelling with sufficient time and spatial resolution to follow microscopic processes over an object the size of a star for 10 billion years is currently impossible. It is also these processes (along with convection) that scramble the effects of "normal" diffusion.


Back of the envelope time. Let's treat the diffusion problem as a random walk, where a He nucleus collides with other nuclei and changes direction randomly.

Let's say the speed of the He nuclei is $v\sim (k_BT/4m_u)^{1/2}$. Let's also argue that the effective interaction radius with a proton (the dominant species) will be where the Coulomb energy equals the kinetic energy. $$ \frac{2e^2}{4\pi \epsilon_0 r} \simeq k_BT$$ and the cross-section is $$\sigma = \pi r^2 \simeq \frac{e^4}{4\pi \epsilon_0^2 (k_BT)^2}.$$

If the density of the plasma is $\rho$, then the number density of protons is $n \sim \rho/m_u$ and the mean free-path of a He nucleus is $$ l \sim \frac{1}{n\sigma} = \left(\frac{m_u}{\rho}\right)\left(\frac{4\pi \epsilon_0^2 (k_BT)^2}{e^4}\right)$$

Now if we say the He nucleus has to move a distance $R$ radially, then a random walk means that rather than taking $R/l$ steps, it actually takes $(R/l)^2$ steps, each of which takes a time $l/v$.

Thus the total time for a He nucleus to diffuse a distance $R$ is $$\tau \sim \left(\frac{R}{l}\right)^2 \left(\frac{l}{v}\right) = \left(\frac{4m_u}{k_BT}\right)^{1/2} \left(\frac{R^2\rho}{m_u}\right) \left( \frac{e^4}{4\pi \epsilon_0^2 (k_BT)^2}\right) $$ $$\tau \sim \frac{2R^2\rho e^4}{4\pi \epsilon_0^2 m_u^{1/2} (k_BT)^{5/2}} = 2\times 10^{14} \left(\frac{\rho}{10^5 {\rm kg/m}^3}\right)\left(\frac{T}{10^7 {\rm K}}\right)^{-5/2}\left(\frac{R}{R_{\odot}}\right)^2\ {\rm years} $$

I believe this is the same approach as your 1977 reference which arrives at 1000 km/billion years.

This basic "molecuar diffusion" away from the core (the effect you are talking about in your question) is a very slow process indeed, although it does speed up further away from the core, since $\rho T^{-5/2}$ decreases. It is also much faster in more massive main sequence stars with lower interior densities and higher interior temperatures.

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  • $\begingroup$ Ok. Thanks for clearing up the tokamak comparison. That’s helpful. I’m still not 100% convinced that basic atomic diffusion is so slow, but I need to study the Burgers equations more first before asking more questions. As for gravitation effects, consensus seems to be that heavy elements are diffused throughout the star. But again, I need to study first. $\endgroup$ – Keith Knauber May 10 at 15:00
  • $\begingroup$ @keithknauber "consensus seems to be...". The heavy elements a star is born with are spread throughout the star. That's because protostars are convective and diffusion is slow! The products of nuclear burning in the core are certainly not. That's why the Sun will only burn about 10% of its H. $\endgroup$ – Rob Jeffries May 10 at 17:09
  • $\begingroup$ Yes I understand intuitively why convection is a more powerful process in protostars, and in old stars in the convective layer. I still don’t buy handwaving type arguments on why basic “concentration diffusion” appears to be so weak. Still studying, and searching for examples of calculations based on the formulas. There’s lots of formulas but few examples: one paper from 1977 says “diffusion goes like Temperature to the 14th power”, then backpedals and says it goes like temp to the 8th power. $\endgroup$ – Keith Knauber May 10 at 18:42
  • $\begingroup$ Hmm. I suspect that this might be a big clue. physics.stackexchange.com/questions/110563/… $\endgroup$ – Keith Knauber May 11 at 2:34
  • $\begingroup$ Ok. Your “back of the envelope” discussion was super helpful! $\endgroup$ – Keith Knauber May 11 at 3:11

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