Much of the Earth's atmosphere has a scale height $h$ of roughly 7 to 8 kilometers such that the local density varies as $\exp(-(r-r_0)/h)$ where r is the radius vector and $r_0$ would be some reference radius above the surface of the Earth.

This works approximately, at least up to 50 km (six or seven scale heights).

The derivation of this approximation is based on a simple gas at constant temperature, and those probably don't apply to the Sun's atmosphere very well.

Nonetheless, are there regions of the Sun's atmosphere where density varies approximately exponentially in such a way that a scale height characterization would work over the range of at least a few scale heights?

  • $\begingroup$ You can define the isothermal scaleheight anywhere, in any gas. For the stellar atmosphere $H$ will then be a function of radius $H(r)$. Only if it then is constant on a non-zero domain, you can integrate that and get the exponential behaviour. $\endgroup$ – AtmosphericPrisonEscape Apr 4 at 13:07

The exponential decrease in density comes out naturally whenever you have a gas in hydrostatic equilibrium. The scale height $H$ is then given by the balance between the kinetic energy of the particles due to thermal motion, $kT$, and the gravitational energy of the particles, $mg$. This is often a good approximation, both in planetary and stellar atmospheres, and even in galaxies. That is, $$ H = \frac{kT}{mg} $$ where $k$ is Boltszmann's constant, $T$ is the temperature, $m$ is the average mass of the particles, and $$ g = \frac{GM}{r^2} $$ with $G$ the gravitational constant, and $M$ the mass inside the radius $r$.

On the surface of our Sun, $g$ works out to $274\,\mathrm{m}\,\mathrm{s}^{-2}$, 27 times as high as on Earth.

The average particle mass depends weakly on metallicity, and mostly on the ionization state of the gas, since the small mass of free electrons compared to that of atoms pulls the average down. For a fully ionized gas, the mean molecular mass — i.e. the mass in terms of the hydrogen mass — is $\mu \simeq 0.6$, whereas for a fully neutral gas it is (e.g. Carroll & Ostlie 1996) $$ \mu \simeq \frac{1}{X + Y/4 + Z/15.5} \simeq 1.25, $$ where $X$, $Y$, and $Z$ are the mass fractions of hydrogen, helium, and metals, respectively. I mistakenly wrote initially that the gas is fully ionized, but that's not true; it is only partially ionized, and the hydrogen is largely neutral.

Taking the average mass of a particle to be roughly equal to the proton mass $m_p$ (i.e. setting $\mu=1$), and taking the temperature to be $T = 5770\,\mathrm{K}$, the scale height is thus $$ H_\odot = \frac{kTR_\odot}{GM_\odot m_p} \simeq 170\,\mathrm{km}. $$ With $\mu=0.6$ you'd get $H\simeq290\,\mathrm{km}$, while $\mu=1.25$ yields $H = 140\,\mathrm{km}$.

Realistic density profiles

The above calculation are quite basic, assuming a completely isotropic Sun. But observations and more realistic models, both 1D and 3D, do indeed predict exponential density profiles, although with quite large variations across the surface (according to a Solar physicist colleague down the hall). I found this model from these lecture notes where the yellow curve shows the number density profile in the Sun's atmosphere.


Extracting the data and plotting on a log-linear scale shows reasonable agreement with an exponential decrease of scale height $H = 140\,\mathrm{km}\,\mathrm{s}^{-1}$:


  • $\begingroup$ Where does the 0.59 come from? Also, can you really ignore magnetic effects? $\endgroup$ – Steve Linton Apr 4 at 10:12
  • 1
    $\begingroup$ @uhoh Okay I'm sorry, I underestimated your prior knowledge :) I also overestimated my own, because realistic models of stellar atmospheres are really not my expertise. But luckily, several Solar physicists are located roughly 30 meters from me, and they 1) confirm that an exponential decrease in density is applicable here, but 2) that there are quite large variations across the surface of the Sun. They pointed me however to some references that show that average profiles are indeed exponential. I'll edit! $\endgroup$ – pela Apr 4 at 11:25
  • $\begingroup$ @SteveLinton The 0.59 is the mean molecular mass for a fully ionized gas: 1/µ ~ 2X + 3Y/4 + Z/2. But as I wrote now, the gas isn't fully ionized, so that's a bad approximation. As for magnetic fields, 3D models that incorporate this still result in exponential profiles, on average, but of course do influence the result quantitatively. $\endgroup$ – pela Apr 4 at 11:28
  • $\begingroup$ Looks great! Roughly exponential over 7 orders of magnitude (~18 scale heights)! That's more than I expected, though I guess nature loves exponentials. $\endgroup$ – uhoh Apr 4 at 11:38
  • 1
    $\begingroup$ @uhoh Yes it looks surprisingly nice. I should say that I arbitrarily normalized the exp() to coincide with the data at the first data point, which is at r ~ 200 km. Also, r = 0 is defined as where the optical depth is unity for light at λ = 500 nm, which is also a bit arbitrary. If you normalize somewhere else, slightly different values of H would fit better. But overall it's a pretty good model, I'd say. $\endgroup$ – pela Apr 4 at 12:19

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.