# What is the viscosity of Sun surface?

It should be possible to compute value from hydrodynamics of Solar prominence. What is the current estimate of Solar surface viscosity?

• Please be more precise. What do you mean by "surface"? Commented Oct 16, 2017 at 23:04
• This is an astrophysics question, not a physics question; it would be better asked on astronomy.stackexchange.com . Commented Oct 17, 2017 at 2:05
• Yes, I am asking about the surface of photosphere. Commented Oct 18, 2017 at 12:51

What is the viscosity of Sun surface?

By "solar surface" I am going to assume you mean the photosphere. In that case, then the question is whether you are asking about dynamic ($\mu$) or kinematic ($\nu$) viscosity.

According to a paper by Cowley [1990], the relevant parameters for the photospheric surface are: T ~ 4700 K, $\rho$ ~ 4.9 x 10-9 g/cm3, and ne ~ 1.1 x 1011 cm3.

We know that the following is an okay approximation $\mu \propto \rho \ \lambda \ V_{T}$, where $\lambda$ is the mean free path and $V_{T}$ is the thermal speed. One can approximate $\lambda \propto (n_{e} \ \sigma)^{-1}$, where $\sigma$ is the collisional cross-section (here Coulomb collisions).

From Table 2 in Cowley [1990], we find that $\mu$ ~ 4.1 x 10-4 g cm-1 s-1 and since we know $\rho$ we can see that $\nu$ = $\mu/\rho$ ~ 8.4 x 104 cm2 s-1.

What is the current estimate of Solar surface viscosity?

I imagine things have not changed much other than more accurate density and temperature constraints for various regions. However, any model that attempts to make things more accurate will quickly become nearly impossible without simulations (e.g., adding multiple ion species with energy-dependent cross-sections is not a trivial task when they have different temperatures and densities).

• "ρ ~ 4.9 x 10^-9 kG/L " Didn't realize that photosphere is so thin. Seems like ideal gas would give good approximation. Thank you. Commented Oct 18, 2017 at 12:59
• @Stepan - To be fair to the photosphere, it gets denser as one goes deeper rather quickly (i.e., exponentially). ;) Commented Oct 18, 2017 at 15:14