I have been designing a habitable planet, its moon, and its star for a while now.


My star is a K2V star with very strong solar wind and is about the age of the Sun. Every so often, when the star is extremely active (once every 10,000 years or so), the star forms some soft x-rays (10-35 Å) from a weak [O VII] forbidden line arising from MCWS (magnetically confined wind shocks) in my star.

The Planet

When these shocks happen, some of the x-rays hit my planet's upper atmosphere, and before being absorbed by lots of ozone (in the upper atmosphere), they hit some neon atoms, exciting them and creating a red-orange glow. I have named this red-orange glow, "psuedo-aurorae".

The planet's atmosphere is roughly composed of:

  • Nitrogen ($N_2$) - 61.5%
  • Oxygen ($O_2$) - 21%
  • Neon ($Ne$) - 15.5% (very dominant in the high atmosphere, as it is quite light)
  • Xenon ($Xe$) - 1% (since it is heavy, it's almost completely in the lower atmosphere, so it doesn't really affect the upper atmosphere much)
  • Water Vapor ($H_2O$) - 0.5%
  • Argon ($Ar$) - 0.479%
  • Carbon Dioxide ($CO_2$) - 0.02%
  • Trace - the rest (this includes the ozone)

The Numbers

Comments have told me that I need more quantitive data, so here it is:

  • Atmospheric Pressure - 0.98 atm
  • Temperature of Lower Atmosphere - 290.78ºK
  • X-ray flux - $10^{30}$ ergs (please note that this is a very rough estimate) - and please tell me if this value makes no sense and give a reasonable value if so. I'd like the "psuedo-aurorae" to be visible or even dramatic, so you can change this number. Also, I would like to keep the upper atmosphere intact and not kill everyone, so don't make it that high if you are changing the value.
  • Planet Radius - 1.03 R🜨
  • Planet Mass - 1.2 M🜨
  • Planet Gravity - ≈11.107 $\frac{m}{s}$ (≈1.132 G)

Star Info

  • Mass - 0.82 M
  • Luminosity - 0.452 L
  • Radius - 0.853 R
  • Metallicity - 0.2 (dex is ≈-0.699)
  • Star Flux - 0.055 (using the $G_{SC}$ or solar constant)
  • Age - 4.6 billion years
  • Star Coronal Composition
    • Hydrogen ($H$) - 72.3%
    • Helium ($He$) - 27.4%
    • Oxygen (including the [O VII] line; $O$ and $O^{6+}$) - 0.3%
    • Carbon ($C$) - 0.05%
    • Nitrogen ($N$) - 0.02%
    • Other Elements - 0.02% (mostly neon, magnesium, silicon, and sulfur, and there are almost no metals higher than iron present in this star)
  • Corona Temperature - $3 \cdot 10^6 ºK$ (this is higher from the peak $2 \cdot 10^6 ºK$ to form the O VII line, from the answer of this question)

Tell me if you need any more info and I will provide it to you.

The Question

I would like to know the visibility of these "pseudo-aurorae" at day and night and how visible they are. Basically, use your math skills to figure out when the aurorae are visible and if they are, how visible are they (barely, very visible, etc.).

  • $\begingroup$ In the absence of any quantitative input data, nobody can say anything quantitative. What is the atmospheric pressure on the surface of the planet? Temperature of the lower atmosphere? Is the planet about the size and mass of Earth? How strong is the flux of X-rays? (Note that on Earth, auroras produced by lots and lots of energetic electrons exciting the few oxygen atoms to be found 100 km up are visible only at night.) Given the geometry, all I can say is that if the pseudo-auroras are visible they are visible only in winter nights at high latitudes. $\endgroup$
    – AlexP
    Commented Sep 24, 2023 at 16:15
  • $\begingroup$ Just say it is bright. I get going down a rabbit hole but nobody can answer this without a lot more data. So any answer you get will be as good as just guessing. $\endgroup$
    – ErikHall
    Commented Sep 24, 2023 at 16:26
  • $\begingroup$ @AlexP I edited my question to give more info. $\endgroup$
    – Neil Iyer
    Commented Sep 24, 2023 at 17:33
  • $\begingroup$ (1) Ergs are a measure of energy. Ten million ergs make a joule. (It's a artificial word made up from en-erg-y.) A flux is the amount of energy per second and unit area. (2) One erg is about the amount of energy needed to blink an eye. One tenth millionth of an erg is one tenth millionth of the energy needed to blink an eye. $\endgroup$
    – AlexP
    Commented Sep 24, 2023 at 17:40
  • $\begingroup$ @AlexP I made the intensity $10^{30}$ ergs, which I think is more reasonable and is much more significant. $\endgroup$
    – Neil Iyer
    Commented Sep 24, 2023 at 22:25

1 Answer 1


Your pseudo-aurorae will be visible as a sky glow

Since the primary source of your aurorae are X-rays, and they are mere photons with power of 350-1240 eV according to wavelength of 35-10 Angstrom provided, they do not behave as accelerated electrons that cause aurorae here on Earth. No night-side lighting, that's for sure, although the presense of "normal" aurorae is not excluded, since your star still can eject mass and produce magnetic acceleration for its electrons so that they would reach your planet and travel by magnetic lines. If true, normal aurorae would also exhibit orange colors in its components, but to achieve this, the electrons should reach 18.7 eV energy upon reaching the upper atmosphere layers to excite neon atoms (see Wikipedia, the neon part), but this is way below observed values of 1.5 keV minimum median value of Earth's pulsating aurorae, which are a subclass of normal ones.

However, this energy is enough for ionization, making electrons fly across the ionosphere that would cause secondary aurora, provided your planet has a magnetic field of strength comparable to Earth's.

The X-rays are enough to excite and ionize atoms in atmosphere

As we know, atoms can only absorb energy in certain portions, this also applies to photons, if a neon atom gets hit by a photon that has 18.7 eV energy or higher, and has interfered with it, that photon would lose 18.7 eV (or more, as ionization is also a thing, and it requires more energy) and get re-emitted at a random angle, likely retaining enough energy to ionize or excite several other atoms or molecules. Eventually that photon would not cause any more excitation or just would be emitted towards outer space. But since everything would get ionized or excited, sometimes multiple times, as top ionization energy among atmospheric components of your planet that could be spent for an ionization event by an X-ray photon of your star is Neon 9+ at 1195 eV, followed by fully ionized O and N at 871 and 667 eV respectively, you can expect that every possible ionization and excitation state attainable by either compound (N2, O2, N, O, Ne, Xe, H2O, H, Ar) would eventually be reached, thus the entire set of visible spectrum would be produced in the atmosphere following this X-ray emission by the star.

How light would it be

You have provided a "random" 1e30 erg as an emission value, but the problem here is that it's not described whether it's just the total energy of the star flash transformed to X-rays, or it's the portion of that energy that hits your planet. But given that 1e30 erg is 1e23 J, and hitting a planet with this would just vaporize its outer layers, I assume that this amount is what has been emitted by the star. To compare, Sun emits 3.846e26 J/s, and your star, given your flux rate of 0.055 of that, would emit 2.115e25 J/s, making the flash only provide a 0.5% flux increase if it lasts 1 second. From the linked article on MCWS, the mass ejection is directed, so this event of yours should also produce a CME, aka normal aurora some decent tie after the glow. I don't exactly understand the restrictions on the MCWS confinement ratio for your star to ever produce O[VII] forbidden line emission, yet the diagram 1, right in this PDF displays that X-ray emission region is not largely dependent on confinement factor, and this emission is still undirected, meaning the planet is only hit by a fraction of this energy.

The fraction is equal to the planet's solid angle when viewed from the point of emission (which is not right at the star, or right at its surface, and the radius of the emission point, listed as Alfven radius, depends on confinement factor at 1/4 power), that is determined by the star's goldilocks zone radius for given surface temperature. This calc displays Goldilocks zone ranging from 0.396 to 0.74 AU, with 290K temperature located at about 3/4 of this range, or at 0.654 AU or roughly at 1e11 m from the star. Since the star radius is listed as 8.55e8 m, and the Alfven radius has such small dependence on confinement, thus RA/R* does not exceed 10 even for the stars with strongest fields, assuming the center of X-ray emission to be at 2.0 the star radius would do. The exact values depend on star matter loss to the star wind and the magnetic field strength (the smaller is the loss, the farther is the emission point, so having the star losing mass quickly, aka very strong wind, is detrimental to the possibility of it having MCWS!), also there's a radius cap to Kepler corotation radius, the shocks can only happen if the Alfven radius is smaller than this, the Kepler radius depends on angular speed of the star's own rotation RK = W^(−2/3)R*. The actual solid angle for a planet of 6561 km radius when viewed from 9.8e10 m is 1/2231000 steradian, meaning the planet receives 1/4*PI\2231000 flux from the emission, or for a 1e30 erg total emission, the planet will be hit by X-rays of 3.567e22 erg, or 3.567e15 J. Note this is for the surface, the atmosphere would get a little more, as its radius is greater than the planet's.

I wonder however, how much of this would be spent for ionization and how much would hit the ground. Let's assume 50% is spent on ionization, this would be 1.784e15 J. If ignoring molecular ionization expenditure (which I think should not be ignored, as it'll take about 5% of the energy to dissociate N2 into N than to single-ionize N atoms), and averaging single ionization energies across the atmospheric composition, we'll have 14.53\*0.61+13.61\*0.21+21.56\*0.155+12.12\*0.01 = 15.1844 eV per atom, or 1465 kJ/mol of the mixed single-atom gas. Thus, at most 1.2177e9 mols of air in the atmosphere would become ionized over the flash length (which is yet undetermined, but feel free to use flux in J/s instead of plain J, then this would be in second). But, as ions and electrons would re-combine, they would emit lower-frequency light everywhere, making the atmosphere glowing with light both to space and to the surface, for a yet indetermined amount of time that depends on the lifespan of ionized particles at different heights (I can't find the papers that would display data on these), but the glow will recede exponentially after the flash would end. But for a rough estimate of the induced luminosity, we can assume that 50% of the energy used for ionization will be emitted towards the surface, and about another 50% would be emitted as visible light instead of being UV or harder. This amount of energy might be compared to what the sky disperses from sunlight in Rayleigh scattering of blue and more energized wavelength.

I have found this question that states the amount of light being scattered by Earth atmosphere is about a quarter of what comes in, majorly in low wavelengths due to Rayleigh scattering, and Wikipedia states the value of 23%, thus an estimation of your planet's sky scattering energy be about 0.23\*2.115e25/4\*PI\*2231000 = 1.173e17 J/s, thus even if only half of this would hit the ground, it would still be about two orders of magnitude more light than what's caused by the flash on the star. Thus the glow would be there, but it would be pretty faint, and dominated by green+blue hues of ionized O and N, although orange component should be there as well.

Regarding X-ray damage, the initial flux is IMHO way too big, 3.567e15 J over the planet is as much as 26.38 J/m^2 (initial flow), so a human exposed to this flash will receive about 4 Gy absorbed dose multipled by percentage of X-ray passed through the atmosphere and by percentage of absorption, and most of this absorbed dose would be received by the brain and skull. 4 Gy is quite enough for a human to die of acute radiation syndrome, athough should they have long and bushy hair or a helmet on, the effects would be greatly lessened. Also probably you would be able to adjust the atmosphere to absorb more incident X-rays, by increasing its pressure at sea level for example, this would result in brighter glow and smaller X-ray damage to life sitting in the open under sunlight.

Overall, it looks like your better source of aurora would be secondary ionized particles and electrons produced by the flash, as the amount of energy to make the sky glow would be too destructive to the life on the planet surface, or you need to devise conditions that would allow no less than 99% incoming x-ray absorption, allowing you to increase the flux and thus brightness of the ionized sky.

  • $\begingroup$ Thank you for the detailed answer, and I think I could probably use those secondary ionized particles and electrons and along with a smaller amount of the x-rays, this could create a brighter glow. For the radiation problem, I have a thick atmosphere (from the xenon) and that could help absorb some more x-rays, and I have also increased the pressure to 1.13 atm. Finally, I have increased the amount of trace ozone in the upper atmosphere to cope with these x-rays. I am also thinking about evolving my life to make them have more hair (as you said). $\endgroup$
    – Neil Iyer
    Commented Sep 26, 2023 at 14:50
  • $\begingroup$ In fact, such flashes might not be as rare as one in 10kY, after all the mechanism looks like this happens almost constantly for O-class stars, so your star being way weaker energy wise might have such flashes about 1-2 times per year, with the planet being under its influence about half the time. With this constraint more hair that's also X-ray opaque will definitely be beneficial for life that's travelling under the daylight sky. $\endgroup$
    – Vesper
    Commented Sep 26, 2023 at 17:38
  • $\begingroup$ Then, maybe it doesn't have as much activity usually, and it could have weaker x-ray emissions most of the time and some strong ones (like the $10^{30}$ ergs one every once in a while). And for the hair thing, I'll definitely change that. $\endgroup$
    – Neil Iyer
    Commented Sep 26, 2023 at 19:08

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