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I was reading up on basic properties of plasmas and came across this description in the Wikipedia on how to convert between eV and Kelvin units to describe the amount of energy of the electrons:

The SI unit of temperature is the kelvin (K), but using the above relation the electron temperature is often expressed in terms of the energy unit electronvolt (eV). Each kelvin (1 K) corresponds to 8.617 333 262...×10−5 eV; this factor is the ratio of the Boltzmann constant to the elementary charge.[5] Each eV is equivalent to 11,605 kelvins, which can be calculated by the relation $⟨ E ⟩ = k_{\text{B}} T$.

This last equation has me confused. Why do we not write instead

$$⟨ E ⟩ = \frac32 k_{\text{B}} T$$

after all an electron has three translational degrees of freedom? What am I missing?

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    $\begingroup$ It doesn’t help that the second-to-last quoted sentence is completely wrong. $\endgroup$
    – tparker
    Mar 29, 2021 at 13:03

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You are mixing two things:

  1. What is the average kinetic energy of the system in $n$-dimensions? Thermodynamics states that $$\bar E = \frac{f}{2}k_B T$$ where $f$ is the number of degree of freedom.
  2. How do we convert between the energy unit "Joule" and the temperature unit "Kelvin"? Since $k_B T$ has the dimension of an energy, we use $T = E/k_B$ to express an energy in terms of Kelvin, see here.

Using this convention, we do not have to think about the number of degrees of freedom. Furthermore, we are even allowed to associate a temperature with a system, which does not have a temperature. E.g. consider a single electron, which obviously has no temperature. It's perfectly fine to state that the electron has a kinetic energy of $1K \approx 1.38e-23 J$. This is analog to expressing an energy in terms of $eV$ -- even if the particle is not electrically charged.

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  • $\begingroup$ Ok, thanks, I did get things mixed up. It's like defining a rotational or vibrational T then. But then that "T" derived from the electron energy has to be accompanied by a qualifier, it is the "electron temperature". So another question, just to get this out of the way and unless I should post this separately: how to make clear that this is not a real thermodynamic T? For instance, in discussion of plasmas, how does one know that the T refers to a "real" T as opposed to an energy-equivalent T? One just has to read between lines? $\endgroup$
    – Buck Thorn
    Mar 28, 2021 at 15:47
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    $\begingroup$ Consistent wording is not particular strong in physics. Hence, you have to learn the slang of plasma physicists. I did my phd in ultra-cold atomic physics, and our standard was the "real temperature" -- the $T$ which we entered in the distribution fct. If, instead, somebody liked to express the average energy as a temperature, he/she just stated this explicitly: "The average kin. energy is ...K". I reckon that "the plasma temperature is $10^5K$" has a common meaning in plasma physics. Unfortunately, I can neither say what it is nor if the factor $3/2$ might be important. Sorry. $\endgroup$
    – Semoi
    Mar 28, 2021 at 16:43
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Why do we set the number of degrees of freedom of an electron in a gas to 2 rather than 3?

In a magnetized plasma, most electrons are said to be gyrotropic -- their velocities can be decomposed into parallel and perpendicular (with respect to the background, quasi-static magnetic field), where the latter is assumed to be azimuthally symmetric. Thus, there is really only two degrees of translational freedom.

Temperature as an energy instead of the thermodynamic quantity we all know and love.

Plasmas are almost never in equilibrium so temperature will not have the same meaning as a thermodynamic system. Regardless of this, it's often the case that when discussing temperature folks will keep $k_{B}$ (= 1.38064852 x 10-23 J K-1) and $T$ together (or even absorb $k_{B}$ into $T$) so that what is called temperature is really an energy.

Generally one finds the second velocity moment of the velocity distribution function (e.g., see https://physics.stackexchange.com/a/218643/59023) of the particle species of interest, multiplied by some constants, to get the pressure tensor. Then one takes one-third the trace divided by the number density to get what most plasma physicists will call the temperature of the plasma.

Conversion between electron volts and kelvin.

We know that 1 eV = 1.6021766208 x 10-19 J and we know that J and K are just a constant apart. So it's not difficult to show that 1 eV ~ 11,604.5 K.

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  • $\begingroup$ I thought the explanation could be that a degree of freedom is lost because of the strong interactions between the particles resulting in increased order (as in a liquid crystal maybe) or restrained motion as in a harmonic oscillator, but could not explain this. What do you think the Wikipedia means with its statement, though? Do you think it is also mixing up (like I seem to have done) the two different interpretations of $fk_BT$, one equipartition and the other a conversion factor (like frequency in wavenumbers)? $\endgroup$
    – Buck Thorn
    Mar 29, 2021 at 13:41
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    $\begingroup$ Generally thermal pressure in a plasma is defined as $P = n \ k_{B} \ T$ for similar reasons. I tend to find attempts to pigeonhole plasmas into thermodynamic concepts to be misleading at best. Most plasmas are not even close to equilibrium so it doesn't make much sense to try and force those concepts onto plasmas. As for the conversion between kelvin and eV, that doesn't depend upon degrees of freedom. That's just a unit conversion. $\endgroup$
    – honeste_vivere
    Mar 29, 2021 at 14:10

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