<p>The photosphere of a star at 2700 K is still highly conductive, due to the ionisation of metals like sodium and potassium. Even if the photospheres were pure hydrogen, the ionisation fraction would still be <span class="math-container">$\sim 10^{-10}$</span>, which is a lot of free electrons.</p>
<p>Note that <span class="math-container">$k_BT$</span> does not have to reach anything like the ionisation energy before ionisation can occur. That is because the average energy of photons is actually about <span class="math-container">$3k_BT$</span> in a Planck distribution, the distribution itself has high energy tails and, there are a lot of photons.</p>
<p>Another key factor is that it is relatively easy to maintain a small ionisation fraction in a sparse plasma (like the photosphere of a star) because whilst the ionisation rate is proportional to the number density of atoms, the recombination rate depends on the product of the number densities of free electrons and ions (both of which would be much smaller). This is governed by the <a href="https://en.wikipedia.org/wiki/Saha_ionization_equation" rel="nofollow noreferrer">Saha equation</a> for the equilibrium ionisation fraction.</p>
<p>The cooler temperatures and lack of ionisation in even cooler (M9 and L-type) brown dwarf atmospheres <em>has</em> been suggested as a reason for their lack of magnetic activity and X-ray emitting coronae (see picture below from <a href="https://ui.adsabs.harvard.edu/abs/2010ApJ...709..332B/abstract" rel="nofollow noreferrer">Berger et al. 2010</a>). The basic idea is similar to what you are suggesting in your question - if the photosphere is neutral then it cannot couple to the magnetic field effectively. This means that photospheric motions, either turbulent or due to some systeamtic differential rotation do not stress the magnetic fields that emerge from the star and there is this no energy release above the photosphere that might accelerate particles or heat a plasma (e.g., <a href="https://iopscience.iop.org/article/10.1086/339911/fulltext/" rel="nofollow noreferrer">Mohanty et al. 2002</a>). The issue is not settled because there still seems to be plenty of radio activity caused by magnetic flaring in these cooler objects (Berger et al. 2010).</p>
<p><a href="https://i.sstatic.net/k2d16.jpg" rel="nofollow noreferrer"><img src="https://i.sstatic.net/k2d16.jpg" alt="X-ray activity in M and L-type objects" /></a></p>
<p><em>X-ray luminosity of a sample of cool M and L-type dwarfs (the latter will be brown dwarfs, not stars). There appears to be a break at around spectral type M9 (<span class="math-container">$T_{\rm eff} \simeq 2200$</span> K) which <em>might</em> be caused by lack of ionisation in their photospheres. (From Berger et al. 2010).</em></p>

CC BY-SA 4.0
2024-04-10T21:10:58.717Z
*"Note that $k_BT$ does not have to reach anything like the ionisation energy before ionisation can occur. That is because the average energy of photons is actually about $3k_BT$ in a Planck distribution"*  Only far UV photons would be able to ionize anything.
Thomas
CC BY-SA 4.0
2024-04-10T23:16:15.217Z
@Thomas After claiming that it takes temperatures of 60,000K to ionise sodium, are you actually disputing the veracity of the Saha equation?
ProfRob
CC BY-SA 4.0
2024-04-13T15:59:35.203Z
Is the 2200K "break" due to Hydrogen no longer ionizing or due to sodium/potassium no longer ionizing? Or do we not know?
Kevin Kostlan
CC BY-SA 4.0
2024-04-10T06:46:37.117Z
2024-04-10T23:17:47.060Z
2024-04-10T23:17:47.060Z
ProfRob
<p>The average gravitational potential and kinetic (via the virial theorem) energy of a particle in the solar plasma is of the order of 1 keV (corresponding to a temperature of about <span class="math-container">$10^7 K$</span>). The particle energies for a red dwarf will not be much lower as both the mass and radius are reduced by about the same factor compared to the Sun and the gravitational potential energy <span class="math-container">$\propto M/R$</span> . The mass/radius ratio would have to be less than about a 1/100 of that of the Sun for the plasma energy be less than the ionization energy of 13.6 eV (Jupiter just about falls below this limit, with a hydrogen atom at its surface having a gravitational energy of about 10 eV).  The photospheric temperature is pretty much irrelevant for this. The photosphere is just a thin layer where the plasma density is small enough so that bound atoms can form and which is cooled by inelastic collisions. A small amount of high energy particles will be able to penetrate the photosphere from below though, especially in regions where it is thinner.</p>

CC BY-SA 4.0
2024-04-04T06:20:23.133Z
Are there any dynamic (differential, partial differential) equations that describe the evolution of a red dwarf and its state variables over time?
ayr
CC BY-SA 4.0
2024-03-05T13:41:14.290Z
The virial theorem does not apply to coronal plasma, which is not in equilibrium - obviously so in the case of flares. And if it did, you would have to include a magnetic energy term. The temperature of 1 keV is largely determined by a minimum in the radiative loss function, so that's where most of the cooling, dynamic plasma accumulates, before cooling more rapidly at lower temperatures.
ProfRob
CC BY-SA 4.0
2024-03-05T22:23:19.410Z
@ProfRob The Sun as a whole is in a (quasi)-hydrostatic equilibrium. This implies an average particle energy in the Sun of the order of 1 keV. This also matches pretty closely the energy in the solar corona and solar wind. It is only the energy in the photosphere which is considerably different from this.
Thomas
CC BY-SA 4.0
2024-03-03T18:08:01.217Z
2024-03-04T21:16:13.687Z
2024-03-04T21:16:13.687Z
Thomas
<p>Even the smallest red dwarf stars have <a href="https://www.google.com/amp/s/www.syfy.com/syfy-wire/good-news-immensely-powerful-red-dwarf-superflares-may-not-sterilize-their-planets%3famp" rel="nofollow noreferrer">amazing flares</a>. But the photosphere temperature can be as low as 2700K, which (I think) is too cool to ionize hydrogen. Even the sun at 5800K is cool enough that most of the hydrogen is not ionized.</p>
<p>The plasma must come from the corona. Which is in turn kept very hot due to resistive heating from currents induced by fluctuating magnetic fields.</p>
<p>If this plasma was removed there would be no such heating. Could this stop the star from having flares?</p>
<p>My guess is no: all it takes is a single ion in the rarefied upper atmosphere to get the process started. It would be accelerated to a very high speed and smash into neutral hydrogen, liberating more ions in a chain reaction.</p>

CC BY-SA 4.0
2024-03-02T11:33:30.967Z
Hard to know what you are asking. The stars you refer to undergoing large flares are not cool enough to have no free electrons and they don't just come from hydrogen, but from sodium and potassium. The magnetic fields in the corona are jostled by motion of coronal loop footpoints, which does require some coupling between the photosphere and the magnetic field. The stressed fields accelerate particles, release energy and drive flares. Other mechanisms have been proposed for even cooler objects (brown dwarfs).
ProfRob
CC BY-SA 4.0
2024-04-10T22:54:07.127Z
@Thomas your comments are totally incorrect and should be removed. The basic physics of ionisation equilibrium is given by the Saha equation. As an obvious example, hydrogen has an ionisation energy of 13.6 eV, but is kept sufficiently ionised at temperatures just above 3000 K ($k_B T = 0.26$ eV) to make the early universe opaque to radiation.
ProfRob
CC BY-SA 4.0
2024-04-10T23:06:13.323Z
As another example, potassium (ionisation energy 4.34 eV) is 99.9% ionised in the solar photosphere ($k_BT = 2.2$ eV).
ProfRob
CC BY-SA 4.0
2024-04-11T18:17:41.657Z
@ProfRob You would need energies of several times the ionization energy if you wanted to create a chain reaction for ionization that the OP suggested. If you just ionize to the ionization threshold you end up with zero energy electrons. Anyway, electrons are dynamically practically irrelevant due to their small mass. They can only transmit about 1/1800 of their energy in one elastic collision (and there won't be many more as they recombine quite quickly), so the ion temperature (which is what you need to increase here) would be unaffected. You really need keV ion energies in the first place.
Thomas
CC BY-SA 4.0
2024-04-11T18:20:02.900Z
@Thomas I'm taling about your incorrect statements about the temperature required for ionisation.
ProfRob
CC BY-SA 4.0
2024-04-11T20:04:27.480Z
@ProfRob I have been using the temperature as a characteristic figure for the average energy here, not as a parameter for an infinitely extended energy distribution. Yes, there are always a few particles with an energy high enough to cause ionization for temperatures significantly less than that (in fact for any temperature T>0K),  but the resulting effect on the overall dynamics of the gas will be negligible.
Thomas
CC BY-SA 4.0
2024-03-05T21:00:00.753Z
@ProfRob: Below 3000K (recombination) even an extremely dilute hydrogen gas has little ionization. But sodium takes less than half the energy to ionize, so it should ionize at even the coolest photosphere temperatures. Makes me wonder if the first M-dwarfs which formed would (after burning their Big-Bang lithium) have more trouble generating plasma and flares?
Kevin Kostlan
CC BY-SA 4.0
2024-03-05T22:37:59.543Z
Sodium has an ionization energy of about 5 eV, which corresponds  to a temperature of about $6*10^4 K$ . Below that temperature you would not ionize anything,
Thomas
CC BY-SA 4.0
2024-03-08T20:25:15.613Z
@Thomas: Water has a vaporization energy of about 0.42 eV per molecule, corresponding to 4800K. But water boils at a much lower temperature and evaporates at a meaningful rate down to ~225 K.
Kevin Kostlan
CC BY-SA 4.0
2024-03-08T21:00:10.153Z
You were talking about ionization not vaporization.
Thomas
CC BY-SA 4.0
2024-03-08T23:31:12.303Z
@Thomas: Why would the thermodynamics be fundamentally different? Both involve adding one extra particle to the gaseous phase which causes a drastic rise in entropy.
Kevin Kostlan
<stellar-astrophysics><plasma-physics>
Could a red dwarf be "deionized"?
Kevin Kostlan
# Could a red dwarf be "deionized"?

(Asked by: username_0 on 2024-03-02)

Even the smallest red dwarf stars have [amazing flares](https://www.google.com/amp/s/www.syfy.com/syfy-wire/good-news-immensely-powerful-red-dwarf-superflares-may-not-sterilize-their-planets%3famp). But the photosphere temperature can be as low as 2700K, which (I think) is too cool to ionize hydrogen. Even the sun at 5800K is cool enough that most of the hydrogen is not ionized.

The plasma must come from the corona. Which is in turn kept very hot due to resistive heating from currents induced by fluctuating magnetic fields.

If this plasma was removed there would be no such heating. Could this stop the star from having flares?

My guess is no: all it takes is a single ion in the rarefied upper atmosphere to get the process started. It would be accelerated to a very high speed and smash into neutral hydrogen, liberating more ions in a chain reaction.

**Tags:** stellar-astrophysics, plasma-physics

> **Comments:**
> **username_1**: Hard to know what you are asking. The stars you refer to undergoing large flares are not cool enough to have no free electrons and they don't just come from hydrogen, but from sodium and potassium. The magnetic fields in the corona are jostled by motion of coronal loop footpoints, which does require some coupling between the photosphere and the magnetic field. The stressed fields accelerate particles, release energy and drive flares. Other mechanisms have been proposed for even cooler objects (brown dwarfs).  
> **username_0**: @username_1: Below 3000K (recombination) even an extremely dilute hydrogen gas has little ionization. But sodium takes less than half the energy to ionize, so it should ionize at even the coolest photosphere temperatures. Makes me wonder if the first M-dwarfs which formed would (after burning their Big-Bang lithium) have more trouble generating plasma and flares?  
> **username_2**: Sodium has an ionization energy of about 5 eV, which corresponds  to a temperature of about $6*10^4 K$ . Below that temperature you would not ionize anything,  
> **username_0**: @username_2: Water has a vaporization energy of about 0.42 eV per molecule, corresponding to 4800K. But water boils at a much lower temperature and evaporates at a meaningful rate down to ~225 K.  
> **username_2**: You were talking about ionization not vaporization.  
> **username_0**: @username_2: Why would the thermodynamics be fundamentally different? Both involve adding one extra particle to the gaseous phase which causes a drastic rise in entropy.  
> **username_1**: @username_2 your comments are totally incorrect and should be removed. The basic physics of ionisation equilibrium is given by the Saha equation. As an obvious example, hydrogen has an ionisation energy of 13.6 eV, but is kept sufficiently ionised at temperatures just above 3000 K ($k_B T = 0.26$ eV) to make the early universe opaque to radiation.  
> **username_1**: As another example, potassium (ionisation energy 4.34 eV) is 99.9% ionised in the solar photosphere ($k_BT = 2.2$ eV).  
> **username_2**: @username_1 You would need energies of several times the ionization energy if you wanted to create a chain reaction for ionization that the OP suggested. If you just ionize to the ionization threshold you end up with zero energy electrons. Anyway, electrons are dynamically practically irrelevant due to their small mass. They can only transmit about 1/1800 of their energy in one elastic collision (and there won't be many more as they recombine quite quickly), so the ion temperature (which is what you need to increase here) would be unaffected. You really need keV ion energies in the first place.  
> **username_1**: @username_2 I'm taling about your incorrect statements about the temperature required for ionisation.  
> **username_2**: @username_1 I have been using the temperature as a characteristic figure for the average energy here, not as a parameter for an infinitely extended energy distribution. Yes, there are always a few particles with an energy high enough to cause ionization for temperatures significantly less than that (in fact for any temperature T>0K),  but the resulting effect on the overall dynamics of the gas will be negligible.  

---

## 2 answers

### Answer by username_1 on 2024-04-10. Score: 0 | Accepted answer

The photosphere of a star at 2700 K is still highly conductive, due to the ionisation of metals like sodium and potassium. Even if the photospheres were pure hydrogen, the ionisation fraction would still be $\sim 10^{-10}$, which is a lot of free electrons.

Note that $k\_BT$ does not have to reach anything like the ionisation energy before ionisation can occur. That is because the average energy of photons is actually about $3k\_BT$ in a Planck distribution, the distribution itself has high energy tails and, there are a lot of photons.

Another key factor is that it is relatively easy to maintain a small ionisation fraction in a sparse plasma (like the photosphere of a star) because whilst the ionisation rate is proportional to the number density of atoms, the recombination rate depends on the product of the number densities of free electrons and ions (both of which would be much smaller). This is governed by the [Saha equation](https://en.wikipedia.org/wiki/Saha_ionization_equation) for the equilibrium ionisation fraction.

The cooler temperatures and lack of ionisation in even cooler (M9 and L-type) brown dwarf atmospheres *has* been suggested as a reason for their lack of magnetic activity and X-ray emitting coronae (see picture below from [Berger et al. 2010](https://ui.adsabs.harvard.edu/abs/2010ApJ...709..332B/abstract)). The basic idea is similar to what you are suggesting in your question - if the photosphere is neutral then it cannot couple to the magnetic field effectively. This means that photospheric motions, either turbulent or due to some systeamtic differential rotation do not stress the magnetic fields that emerge from the star and there is this no energy release above the photosphere that might accelerate particles or heat a plasma (e.g., [Mohanty et al. 2002](https://iopscience.iop.org/article/10.1086/339911/fulltext/)). The issue is not settled because there still seems to be plenty of radio activity caused by magnetic flaring in these cooler objects (Berger et al. 2010).

[![X-ray activity in M and L-type objects](https://i.sstatic.net/k2d16.jpg)](https://i.sstatic.net/k2d16.jpg)

*X-ray luminosity of a sample of cool M and L-type dwarfs (the latter will be brown dwarfs, not stars). There appears to be a break at around spectral type M9 ($T\_{\rm eff} \simeq 2200$ K) which *might* be caused by lack of ionisation in their photospheres. (From Berger et al. 2010).*

> **Comments:**
> **username_2**: *"Note that $k_BT$ does not have to reach anything like the ionisation energy before ionisation can occur. That is because the average energy of photons is actually about $3k_BT$ in a Planck distribution"*  Only far UV photons would be able to ionize anything.  
> **username_1**: @username_2 After claiming that it takes temperatures of 60,000K to ionise sodium, are you actually disputing the veracity of the Saha equation?  
> **username_0**: Is the 2200K "break" due to Hydrogen no longer ionizing or due to sodium/potassium no longer ionizing? Or do we not know?  

---

### Answer by username_2 on 2024-03-03. Score: 1

The average gravitational potential and kinetic (via the virial theorem) energy of a particle in the solar plasma is of the order of 1 keV (corresponding to a temperature of about $10^7 K$). The particle energies for a red dwarf will not be much lower as both the mass and radius are reduced by about the same factor compared to the Sun and the gravitational potential energy $\propto M/R$ . The mass/radius ratio would have to be less than about a 1/100 of that of the Sun for the plasma energy be less than the ionization energy of 13.6 eV (Jupiter just about falls below this limit, with a hydrogen atom at its surface having a gravitational energy of about 10 eV). The photospheric temperature is pretty much irrelevant for this. The photosphere is just a thin layer where the plasma density is small enough so that bound atoms can form and which is cooled by inelastic collisions. A small amount of high energy particles will be able to penetrate the photosphere from below though, especially in regions where it is thinner.

> **Comments:**
> **username_1**: The virial theorem does not apply to coronal plasma, which is not in equilibrium - obviously so in the case of flares. And if it did, you would have to include a magnetic energy term. The temperature of 1 keV is largely determined by a minimum in the radiative loss function, so that's where most of the cooling, dynamic plasma accumulates, before cooling more rapidly at lower temperatures.  
> **username_2**: @username_1 The Sun as a whole is in a (quasi)-hydrostatic equilibrium. This implies an average particle energy in the Sun of the order of 1 keV. This also matches pretty closely the energy in the solar corona and solar wind. It is only the energy in the photosphere which is considerably different from this.  
> **username_3**: Are there any dynamic (differential, partial differential) equations that describe the evolution of a red dwarf and its state variables over time?  

---

