2

This question stems from an answer to the following question: What's the most fundamental definition of temperature?

The issue here is how to define the temperature in a system not in thermal or thermodynamic equilibrium. Is there a fundamental definition beyond just calculating the second velocity moment in the population rest frame?

I found two other questions that are somewhat related:
How is temperature defined in non-equilibrium?
Temperature out of thermodynamic equilibrium
However, I think they limit the scope to just non-equilibrium systems. I am asking a slightly more general question where we cannot assume thermodynamic or thermal equilibrium. An example would be the plasma in the solar wind, which is a weakly collisional, ionized gas. There are multiple ion species, none of which at the same "temperature." There are even multiple electron populations (e.g., see https://iopscience.iop.org/article/10.3847/1538-4365/ab22bd).

I know that one cannot expect a single "temperature" in such systems. However, is there a fundamental definition of the temperature for each population in such systems or is the mean kinetic energy in the population rest frame as good as we can hope to expect?

honeste_vivere
  • 15,592

2 Answers2

2

An average energy is not enough to assign a temperature to a subsystem. There should also be a distribution of energies or velocities that is reasonably thermal.

So a solid bar with a temperature gradient does not have one or two temperatures.

But when the thermal coupling between the subsystems is much weaker than within the subsystems, one can assign different temperatures. Examples are nuclei that can be cooled to nanokelvins in a solid. Or conduction electrons in solids that can be heated by high-power picosecond lasers to temperatures far higher than the temperature of the lattice.

Or indeed plasmas but I do not know much about those. Of course it is always possible to talk about "1 million kelvin electrons" to indicate energy even when it is not really justified to talk about a temperature.

  • So the tricky part is that in something like the solar wind, two different electron populations can stream "through each other" for nearly an astronomical unit with only a single collision per particle or so. That is, the collision rate is so low that there can exist multiple populations and their velocity distributions are not Maxwellian. Thus, why I resort to mean kinetic energy in the population rest frame but I do not do this for all electrons. I separate the populations before taking this velocity moment approach. Does that make more sense? – honeste_vivere Jul 29 '20 at 16:07
  • 1
    @honeste_vivere That makes good sense to me. But I am not at all aware of the conventions in that field. –  Jul 29 '20 at 18:16
1

Note that the proportional relationship between temperature and mean particle kinetic energy (or velocity moment, as you write here) is a result, not a precondition, of the entropy-related definition in my answer to the linked question. It's convenient and simple to understand that temperature should be proportional to internal energy, so that comes first in the pedagogy, but it's not really right. Temperature is less about internal energy content than about a system's willingness to give that energy away.

When I was a graduate student I used a neutron polarizer that was made of polarized helium-3 nuclei. This was a glass cell filled with helium-3 and rubidium vapor, illuminated by a polarized laser. The someone-else's-thesis explanation goes like this: the laser would rapidly polarize all of the rubidium electrons, rare collisions would transfer the polarization from the rubidium electrons to the rubidium nuclei, and different rare collisions would transfer the polarization from the rubidium nuclei to the helium nuclei. The helium polarization would eventually reach a thermal distribution whose "spin temperature" had much more to do with the laser and the magnetic field than with the oven that prevented the rubidium vapor from condensing. The spin temperature of a polarized gas is cold, since the unpolarized state is the high-temperature limit, and the spin degree of freedom is nearly completely decoupled from the translational kinetic energy of the very same nucleus.

A possible answer to your question is that, if you have a system in which the statistical assumptions of thermodynamics don't apply, temperature may just be a not-useful way to think about energy transfer. Compare to quantum mechanics, where newcomers are very interested in whether such-and-such is a wave or a particle, and the expert answer is "neither." If you were interacting with a slow-pitch baseball machine in a batting cage, "temperature" is not a useful way to describe the interaction between the batter and the baseballs. The same may be true for a noninteracting, monochromatic particle beam, including for a parcel of the solar wind. If the timescale during which you interact with a system is much faster than the timescale for internal interactions within the system, your interactions aren't changing its entropy and internal energy in the correlated way that we refer to as "temperature."

This isn't quite an answer, but it got too out of control for the comment box.

rob
  • 89,569
  • Yes, there are all sorts of nit-picky arguments between folks at meetings about how one defines temperature and whether it has physical meaning. If it does have physical meaning, how do we interpret it? The list goes on and on. I tend to lean toward mean kinetic energy in bulk flow rest frame because it does describe the spread of energy, if done correctly, but I agree it is not thermodynamic temperature. Unfortunately, many in my field try to pigeon hole non-equilibrium stuff into thermodynamics. – honeste_vivere Jul 30 '20 at 18:29