<p>It does not come out quite clearly from the accepted answer that the underlying reason for an exponential height decrease of the atmospheric density is the fact that in thermodynamic equilibrium the energy distribution function is the Boltzmann distribution</p>
<p><span class="math-container">$$n(E)=n_0\cdot e^\frac{-E}{kT}$$</span></p>
<p>with <span class="math-container">$E$</span> the energy, <span class="math-container">$k$</span> the Boltzmann constant and <span class="math-container">$T$</span> the temperature.</p>
<p>Now if the particles of mass <span class="math-container">$m$</span> do work <span class="math-container">$mgh$</span> against the gravitational field <span class="math-container">$g$</span> along a distance <span class="math-container">$h$</span>, we have to replace <span class="math-container">$E$</span> with <span class="math-container">$E+mgh$</span>, so</p>
<p><span class="math-container">$$n(E,h)=n_0\cdot e^{-\frac{E+mgh}{kT}}$$</span>.</p>
<p>So it follows quite naturally from the assumption of thermodynamic equilibrium at the base of the atmosphere that the density of the particles at greater heights will be diminished by an exponential factor. If the actual density decrease deviates from a simple exponential function, then this can be due to 2 reasons: 1) there are local sources or sinks at greater height that lead to an increase/decrease of density or temperature, or 2) the energy distribution function at the base was not a strict Boltzmann distribution function in the first place. In case of the Sun, the latter option is pretty much the only realistic possibility. For high enough energies the assumption of thermodynamic equilibrium (which implies that collisions dominate all other physical processes) will indeed break down as the Coulomb collision cross section decreases like <span class="math-container">$1/E^2$</span> with increasing <span class="math-container">$E$</span>. This provides actually a straightforward possible explanation for the so called 'Coronal heating problem' (see <a href="https://www.plasmaphysics.org.uk/research/sun.htm" rel="nofollow noreferrer">my website</a> for more in this respect).</p>
<p>The scale height for charged particles is actually a factor 2 higher than for neutrals because the small mass of the electrons means they can escape from the gravitational field much more easily than the ions, which results in an electric 'plasma polarization field' which compensates for half the gravitational field on the ions (<a href="https://astronomy.stackexchange.com/questions/23319/what-are-the-experimental-limits-to-the-residual-charge-of-the-sun/44797#44797">see this SE answer</a> for more).</p>

CC BY-SA 4.0
2023-06-25T17:59:40.877Z
2023-06-25T17:59:40.877Z
Thomas
<p>The exponential decrease in density comes out naturally whenever you have a gas in hydrostatic equilibrium. The scale height <span class="math-container">$H$</span> is then given by the balance between the kinetic energy of the particles due to thermal motion, <span class="math-container">$kT$</span>, and the gravitational energy of the particles, <span class="math-container">$mg$</span>. This is often a good approximation, both in planetary and stellar atmospheres, and even in galaxies.
That is,
<span class="math-container">$$
H = \frac{kT}{mg}
$$</span>
where <span class="math-container">$k$</span> is Boltszmann's constant, <span class="math-container">$T$</span> is the temperature, <span class="math-container">$m$</span> is the average mass of the particles, and
<span class="math-container">$$
g = \frac{GM}{r^2}
$$</span>
with <span class="math-container">$G$</span> the gravitational constant, and <span class="math-container">$M$</span> the mass inside the radius <span class="math-container">$r$</span>.</p>
<p>On the surface of our Sun, <span class="math-container">$g$</span> works out to <span class="math-container">$274\,\mathrm{m}\,\mathrm{s}^{-2}$</span>, 27 times as high as on Earth.</p>
<p>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 <em>mean molecular mass</em> — i.e. the mass in terms of the hydrogen mass — is <span class="math-container">$\mu \simeq 0.6$</span>, whereas for a fully neutral gas it is (e.g. <a href="http://adsabs.harvard.edu/abs/1996ima..book.....C" rel="nofollow noreferrer">Carroll &amp; Ostlie 1996</a>)
<span class="math-container">$$
\mu \simeq \frac{1}{X + Y/4 + Z/15.5} \simeq 1.25,
$$</span>
where <span class="math-container">$X$</span>, <span class="math-container">$Y$</span>, and <span class="math-container">$Z$</span> 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.</p>
<p>Taking the average mass of a particle to be roughly equal to the proton mass <span class="math-container">$m_p$</span> (i.e. setting <span class="math-container">$\mu=1$</span>), and taking the temperature to be <span class="math-container">$T = 5770\,\mathrm{K}$</span>, the scale height is thus
<span class="math-container">$$
H_\odot = \frac{kTR^2_\odot}{GM_\odot m_p} \simeq 170\,\mathrm{km}.
$$</span>
With <span class="math-container">$\mu=0.6$</span> you'd get <span class="math-container">$H\simeq290\,\mathrm{km}$</span>, while <span class="math-container">$\mu=1.25$</span> yields <span class="math-container">$H = 140\,\mathrm{km}$</span>.</p>
<h3>Realistic density profiles </h3>
<p>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 <a href="http://www2.gi.alaska.edu/%7Ejvo/P472G/SolarAtmosphere.pdf" rel="nofollow noreferrer">these lecture notes</a> where the yellow curve shows the number density profile in the Sun's atmosphere.</p>
<p><a href="https://i.sstatic.net/Jdj4h.png" rel="nofollow noreferrer"><img src="https://i.sstatic.net/Jdj4h.png" alt="n_prof" /></a></p>
<p>Extracting the data and plotting on a log-linear scale shows reasonable agreement with an exponential decrease of scale height <span class="math-container">$H = 140\,\mathrm{km}$</span>:</p>
<p><a href="https://i.sstatic.net/HJkNQ.png" rel="nofollow noreferrer"><img src="https://i.sstatic.net/HJkNQ.png" alt="n" /></a></p>

CC BY-SA 4.0
2019-04-04T10:12:04.883Z
Where does the 0.59 come from? Also, can you really ignore magnetic effects?
Steve Linton
CC BY-SA 4.0
2019-04-04T11:25:29.660Z
@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!
pela
CC BY-SA 4.0
2019-04-04T11:28:20.877Z
@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.
pela
CC BY-SA 4.0
2019-04-04T11:38:54.220Z
Looks great! Roughly exponential over 7 orders of magnitude (~18 scale heights)! That's more than I expected, though I guess nature loves exponentials.
uhoh
CC BY-SA 4.0
2019-04-04T12:19:24.273Z
@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.
pela
CC BY-SA 4.0
2023-06-24T13:53:28.053Z
I don't know why the scale height is given a dimension km/s here. This would be a velocity not a height.
Thomas
CC BY-SA 4.0
2023-06-24T15:20:35.340Z
@Thomas that was obviously a typo; above it has also been given in correct units, km.
planetmaker
CC BY-SA 4.0
2023-06-24T19:27:17.903Z
Thanks for catching that @Thomas!
pela
CC BY-SA 4.0
2023-06-25T08:36:28.973Z
@pela It is still incorrect in the diagram.
Thomas
CC BY-SA 4.0
2023-06-25T20:16:27.040Z
@Thomas Ah sorry, I'll fix.
pela
CC BY-SA 4.0
2019-04-04T06:43:03.493Z
2023-06-25T20:17:39.900Z
2023-06-25T20:17:39.900Z
pela
<p>Much of the Earth's atmosphere has a scale height <span class="math-container">$h$</span> of roughly 7 to 8 kilometers such that the local density varies as <span class="math-container">$\exp(-(r-r_0)/h)$</span> where r is the radius vector and <span class="math-container">$r_0$</span> would be some reference radius above the surface of the Earth. </p>

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

<p>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.</p>

<p>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?</p>

CC BY-SA 4.0
2019-04-04T13:07:32.700Z
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.
AtmosphericPrisonEscape
<the-sun><stellar-atmospheres>
Does the Sun's atmosphere have a scale height?
uhoh
# Does the Sun's atmosphere have a scale height?

(Asked by: username_0 on 2019-04-04)

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?

**Tags:** the-sun, stellar-atmospheres
> **Comments:**
> **username_1**: 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.  

---

## 2 answers

### Answer by username_3 on 2019-04-04. Score: 7 | Accepted answer

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](http://adsabs.harvard.edu/abs/1996ima..book.....C))
$$
\mu \simeq \frac{1}{X + Y/4 + Z/15.5} \simeq 1.25,
$$
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
$$
H\_\odot = \frac{kTR^2\_\odot}{GM\_\odot m\_p} \simeq 170\,\mathrm{km}.
$$
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](http://www2.gi.alaska.edu/%7Ejvo/P472G/SolarAtmosphere.pdf) where the yellow curve shows the number density profile in the Sun's atmosphere.

[![n_prof](https://i.sstatic.net/Jdj4h.png)](https://i.sstatic.net/Jdj4h.png)

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

[![n](https://i.sstatic.net/HJkNQ.png)](https://i.sstatic.net/HJkNQ.png)

> **Comments:**
> **username_4**: Where does the 0.59 come from? Also, can you really ignore magnetic effects?  
> **username_3**: @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!  
> **username_3**: @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.  
> **username_0**: Looks great! Roughly exponential over 7 orders of magnitude (~18 scale heights)! That's more than I expected, though I guess nature loves exponentials.  
> **username_3**: @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.  
> **username_2**: I don't know why the scale height is given a dimension km/s here. This would be a velocity not a height.  
> **username_5**: @Thomas that was obviously a typo; above it has also been given in correct units, km.  
> **username_3**: Thanks for catching that @Thomas!  
> **username_2**: @pela It is still incorrect in the diagram.  
> **username_3**: @Thomas Ah sorry, I'll fix.  

---

### Answer by username_2 on 2023-06-25. Score: 1

It does not come out quite clearly from the accepted answer that the underlying reason for an exponential height decrease of the atmospheric density is the fact that in thermodynamic equilibrium the energy distribution function is the Boltzmann distribution

$$n(E)=n\_0\cdot e^\frac{-E}{kT}$$

with $E$ the energy, $k$ the Boltzmann constant and $T$ the temperature.

Now if the particles of mass $m$ do work $mgh$ against the gravitational field $g$ along a distance $h$, we have to replace $E$ with $E+mgh$, so

$$n(E,h)=n\_0\cdot e^{-\frac{E+mgh}{kT}}$$.

So it follows quite naturally from the assumption of thermodynamic equilibrium at the base of the atmosphere that the density of the particles at greater heights will be diminished by an exponential factor. If the actual density decrease deviates from a simple exponential function, then this can be due to 2 reasons: 1) there are local sources or sinks at greater height that lead to an increase/decrease of density or temperature, or 2) the energy distribution function at the base was not a strict Boltzmann distribution function in the first place. In case of the Sun, the latter option is pretty much the only realistic possibility. For high enough energies the assumption of thermodynamic equilibrium (which implies that collisions dominate all other physical processes) will indeed break down as the Coulomb collision cross section decreases like $1/E^2$ with increasing $E$. This provides actually a straightforward possible explanation for the so called 'Coronal heating problem' (see [my website](https://www.plasmaphysics.org.uk/research/sun.htm) for more in this respect).

The scale height for charged particles is actually a factor 2 higher than for neutrals because the small mass of the electrons means they can escape from the gravitational field much more easily than the ions, which results in an electric 'plasma polarization field' which compensates for half the gravitational field on the ions ([see this SE answer](https://astronomy.stackexchange.com/questions/23319/what-are-the-experimental-limits-to-the-residual-charge-of-the-sun/44797#44797) for more).


---

