I know that a gas behaves more like an ideal gas at higher temperature, and that is very well achieved in the Sun's core. But also low pressure is needed for a gas to behave like an ideal gas, and the pressure at the Sun's core is very high. So now if I know correctly and accurately the temperature, density and pressure of the Sun and I want to calculate the average kinetic energy of a Hydrogen or a Helium ion at the center of the Sun's core, can I simply use the $\frac32kT$ law to calculate that ?

Also, if the electrons were degenerate, but I knew the temperature at different parts from a standard solar model (not by using ideal gas laws), can I still use $\frac32kT$ for the ions (not the electrons) ? Or will the electron degeneracy prevent me somehow ?

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    $\begingroup$ I don't think you could treat it as an ideal gas. The atoms in the core are separated from their electrons (we call this a plasma) and the kinetic contributions would probably have to be calculated independently. Also, since the pressure is so high, the velocity of those particles is probably huge: you'd have to use special relativity too, maybe. I'm just guessing. $\endgroup$ Jun 30, 2015 at 15:13
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    $\begingroup$ According to here, "the gas [at the centre of the sun] can be approximated as an ideal gas". $\endgroup$
    – lemon
    Jun 30, 2015 at 15:28
  • $\begingroup$ Ideal gases require (mostly) non-interacting particles and elastic collisions if they do interact. Unless you are specifically referring to the outer core, I am inclined to think that fusion (thus a loss term) is not elastic. I am also a bit concerned about the density of $\gamma$-rays emitted by fusing particles. I am not a nuclear physicist, but I am not sure whether $\gamma$-ray-particle collisions can be considered elastic. I guess I am not sure how much these terms matter, which is why I left this as a comment. $\endgroup$ Jun 30, 2015 at 17:47
  • $\begingroup$ Related question: physics.stackexchange.com/q/142417/44126 $\endgroup$
    – rob
    Jul 1, 2015 at 14:51
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    $\begingroup$ @honeste_vivere fusion is so slow and rare in a low mass main-sequence star like the sun that it is the least of your worries in deciding if the ideal gas approximation is reasonable. $\endgroup$ Feb 17, 2018 at 22:26

2 Answers 2


According to a NASA page, the density in the middle of the Sun is about 150 g/cm3. That's about 9 × 1025 protons in a 1cm3 box, or 450 million to a side, and using that spacing for a voltage calculation reveals a typical interaction energy of 65 eV or so. (If you've never seen this unit before, that is the energy used by a 1V battery to move an electron's charge from one terminal to the other. If you've never seen these calculations before, they belong to a part of physics called "classical electromagnetism.")

The same source says "The temperature at the very center of the Sun is about 15,000,000° C", which we can convert to a thermal energy of about 1.2 keV. That means that every degree of freedom has about 200 times the thermal energy as any particle-particle interaction has.

So it's not at the level where it's a great approximation (you'd want this number to be thousands or millions for that), but it is certainly at the level where it's a useful approximation, yes, since it's in the tens or hundreds range. (It also matters that the electron mass of 512 keV is probably big enough compared to that thermal energy to neglect relativity to first order.) In fact if it does deviate, those numbers are probably small enough to view it as a van der Waals gas with the usual "attractive potential" having its sign reversed or so.

  • $\begingroup$ Excuse me if this sounds stupid, but what is actually approximated when using an ideal gas assumption for the Sun ? Is it the temperature and pressure ? So if we use the currently accepted model of the Sun with the known pressure and temperature, will the value we get from (3/2kT) be correct ? My problem here is actually whether or not the (3/2kT) law can be applied assuming we have a complete model of the Sun's temperature, pressure and density. $\endgroup$ Jun 30, 2015 at 16:38
  • $\begingroup$ The ideal gas assumption states that a box contains classical particles which are not interacting: that is the approximation it makes. You get approximately-correct results when the particles interact little, compared to all of the other stuff that's going on. You can probably apply those laws for a good "rule of thumb" about overall magnitude (maybe trust it to within 10%) and better when talking about deviations ($P~dV+V~dP=(3/2)k_B(dn~T+n~dT)$ I'd trust to within few percent), but what it's leaving out is not insignificant and I'd probably be careful beyond those levels of accuracy. $\endgroup$
    – CR Drost
    Jun 30, 2015 at 17:26

The first part is handled well by Chris Drost - the kinetic particle energies are a lot larger than their interaction energies, so the gas can be considered (approximately) ideal.

The last part - yes, as long as the Coulomb energy is a lot lower than the thermal energy then the protons or He ions can be considered an ideal gas with the appropriate average kinetic energy.

The same is true whether or not the electrons are degenerate, however in white dwarfs it is possible for the ions to "freeze" when Coulomb energies reach a certain multiple ($\sim 100 kT$) of the thermal energy. The ionic "gas" then behaves more like a solid with energy $3kT$ per ion.

  • $\begingroup$ Are there some parts in the Sun where the Coulomb energy is not a lot lower than the thermal energy ? In other words, how can I calculate the Coulomb energy for different radii of the Sun ? Is it the same way Chris Drost calculated the 65 eV number, or is this different ? $\endgroup$ Jul 1, 2015 at 16:28
  • $\begingroup$ @AbanobEbrahim Coulomb energy $\sim e^2/4\pi\epsilon_0 r$, where $r$ is the ion separation. $\endgroup$
    – ProfRob
    Jul 1, 2015 at 16:42
  • $\begingroup$ I tried to calculate the Coulomb energy at the center of the core (150 $g/cm^3$) and at 0.25 of the radius (20 $g/cm^3$). The proton separation was $2.23\times$$10^{-11}$m in case of the center, and $4.47\times$$10^{-11}$m in case of 0.25 radius. The Coulomb energy turned out to be ~64 eV at center and ~32 eV at 0.25 of the radius. So are there results within the correct/expected numbers ? $\endgroup$ Jul 1, 2015 at 17:25
  • $\begingroup$ @AbanobEbrahim OK, so, 64 eV corresponds to a thermal energy temperature of $7.4\times 10^{5}$K and 32eV is half of this. So how does that compare with the temperature in the Sun?? $\endgroup$
    – ProfRob
    Jul 1, 2015 at 19:19
  • $\begingroup$ It's obvious that the temperature in the Sun is many times the Coulomb energy, I was just unsure what "a lot larger" meant. $\endgroup$ Jul 1, 2015 at 19:29

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