It is very simple for monoatomic gas $3/2R$, and the gas-question also gives an good approach for this theme; Why is molar specific heat at constant volume of a monatomic ideal gas a constant?

So this is about degrees of freedom. Dulong-Petit law is also good source to start with.

So does the fact, that plasma is consisted from charged particles, influence on it's degrees of freedom and thus to it's heat capacity?

At least the magnetism caused by moving charges seems to have more limited degrees of freedom, what it comed to right-hand rule.

And this predicts to me that plasma should have smaller heat capacity than gas. But I don't find any clear existing source where this aspect would have been propely discussed.

  • $\begingroup$ These sources have been proposed me elswhere; nrl.navy.mil/ppd/sites/www.nrl.navy.mil.ppd/files/pdfs/… and thermopedia.com/content/1033 and "Introduction to Plasma Physics: A graduate level course" -Richard Fitzpatrick, but I haven't found the answer. $\endgroup$
    – Jokela
    Apr 23, 2017 at 20:33
  • $\begingroup$ Please don't capitalize random words. In English, we capitalize only the first word of each sentence, and proper nouns. $\endgroup$
    – DanielSank
    Apr 23, 2017 at 22:14
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    $\begingroup$ Question is too broad. Plasma encompasses many different compositions, densities, temperatures, ionisation states, equilibrium vs non-equilibrium states etc. etc. So there isn't "the answer". You calculate how the energy density depends on temperature for specific cases. $\endgroup$
    – ProfRob
    Apr 23, 2017 at 22:38
  • $\begingroup$ The heat capacity can be higher or lower. There is no straightforward answer. $\endgroup$
    – ProfRob
    Apr 29, 2017 at 22:23
  • $\begingroup$ @RobJeffries Could you please make a short answer about your comments. "No answer" can also be "the answer" $\endgroup$
    – Jokela
    Apr 30, 2017 at 15:37

1 Answer 1


Firstly, some definitions. A plasma consists of various "species" of particle, such as electrons or nuclei (I will not discuss electron-positron or quark-gluon plasmas here, but you can probably infer many of the details). Furthermore, there may be different species of nuclei, such as different elements (e.g. Helium or Hydrogen) and different charge states of a given element (e.g. He 1+ or He 2+). In general, plasmas may be out of thermodynamic equilibrium and hence the different species may have different temperatures.

Let me also define what is typically meant by a heat capacity. A heat capacity is the amount of thermal energy required to raise the temperature of an amount of substance. For convenience, I will use the heat capacity per unit volume as follows: $\varepsilon$ is the amount of thermal energy per unit volume at a given temperature. Therefore we are looking for a heat capacity of the form $\varepsilon=C(T)$, where $C$ is some mathematical function. It is straightforward to then get to a heat capacity per unit mass or per particle.

As you have pointed out, a non-ionized gas of particles has a heat capacity as follows (note I am using constants more common to plasma physics): $$\varepsilon=\frac{3}{2}n k_B T \tag{1}$$

However, some species may be fermions (as in the case of electrons) with a heat capacity defined through the following relation:

$$\varepsilon_{fermion}=4\pi\left(\frac{2m_e c^2}{h^2 c^2}\right)^{3/2}\int_0^\infty \frac{x^{3/2}}{\exp[(x-\mu)/T]+1} dx \tag{2}$$

Here the physical constants have their usual meaning, $n_{fermion}$ is the density of the particles and the chemical potential $\mu=\mu(n_{fermion},T)$. Please read the appropriate literature on the chemical potential and the Fermi-Dirac distribution. These formulas arise from the quantum nature of fermions (only one allowed per quantum state). If this looks complicated, please bear in mind that at high temperatures or low densities, Equation (2) reduces to Equation (1).

Similarly, the species of boson in a plasma satisfy the following equation:

$$\varepsilon_{boson}=4\pi\left(\frac{2m c^2}{h^2 c^2}\right)^{3/2}\int_0^\infty \frac{x^{3/2}}{\exp[(x-\mu)/T]-1} dx$$

This is from the Bose-Einstein distribution.

Aside from the contribution of the thermal energy discussed above, the other main heat capacity for plasmas are the ionization energies of ions. There is an energy $E_i$ required to ionize an electron from an atom. For hydrogen, this is called the Rydberg energy $E_i=\mathrm{Ry}=13.6$ eV. For other elements, there are varying ionization energies, most of which are determined experimentally or using complicated computer codes. In any case, the resulting heat capacity arises as the sum of the density of each ionization stage multiplied by the ionization energy:

$$\varepsilon_{ion}=\sum_i E_i n_{ion} \tag{3}$$

Here $n_{ion}$ is the density of each type and charge of ion, such as H$^{1+}$, Fe$^{20+}$ etc. Note that the fractions of each charge state are strong functions of both total density and temperature and hence rapidly change based on plasma conditions. For example, at low temperatures an iron plasma may be almost completely neutral, but at a high temperature it may consist of significant fractions of Fe$^{19+}$, Fe$^{20+}$ and Fe$^{21+}$. These can be solved for using the Saha-Boltzmann equation or a collisional-radiative model.

I cover most of what I say in my PhD thesis, which is freely available here and which contains further references should you be interested: http://etheses.whiterose.ac.uk/13357/ All the formulas are explained in detail there. Chapter 3 and possibly 4 are the most relevant. Also note that within there, I use plasma physics units conventions (explained in the Appendix) where $k_B=1$, so temperatures are in electronvolts.

A brief example: Hydrogen atoms at a density of $10^{20}$ cm$^{-3}$ are 60% ionized. This means that there are $6\times 10^{19}$ cm$^{-3}$ electrons and H$^{1+}$ ions and $4\times 10^{19}$ cm$^{-3}$ neutral H atoms. We can ignore quantum effects and use Equation (1) for each of the three species. Then we use Equation (3) with only one term in the sum, the above $n_{\mathrm{H}^{1+}}$ and $E_{\mathrm{H}^{1+}}=\mathrm{Ry}$.

So in summary, a plasma would have a higher heat capacity than an equivalent un-ionized gas (if that were possible), since there are more particles (free electrons) and an energy cost of ionization.

  • $\begingroup$ Does the fact that the ionized gas is coupled to the photon gas also boost the heat capacity? $\endgroup$ Apr 29, 2017 at 22:15
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    $\begingroup$ @SeanE.Lake Heat capacities require some sort of equilibrium. If you're considering a plasma in Thermodynamic Equilibrium (background photon distribution at the same temperature as the plasma) , then yes. Can take a while to establish and not often seen in the lab, though - more usual to have Local Thermodynamic Equilibrium where photons are not a black-body distribution. $\endgroup$ Apr 30, 2017 at 14:45
  • $\begingroup$ Thanks for the Answer. I need to study your thesis further too. So if Plasma have higher heat capacity than un-ioninzed gas, then it also must have more degrees of freedom. This is against my observation. I was expexting it would be less. Do you have any ideas explanations to this through degrees of freedoms? I upvoted your answer, so the bounty will go to you, though -as said- I don't quite agree with this answer. $\endgroup$
    – Jokela
    Apr 30, 2017 at 17:32
  • $\begingroup$ @JokelaTurbine Well, as I mentioned at the end: more "particles" means more degrees of freedom. In one case, there is a neutral hydrogen atom with 3 spatial degrees of freedom (electron and nucleus move as a single unit), in the other there is one electron and one ion with 6 spatial degrees of freedom. $\endgroup$ Apr 30, 2017 at 19:01
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    $\begingroup$ @H.H. You are right that I forgot to change the subscript when I copy-pasted the equation. I have now fixed this. However, the integral itself has one very important difference between the bosons and fermions. (The +/- sign) $\endgroup$ Dec 6, 2020 at 11:14

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