Temperature of individual particles in kinetic theory?

The first thing that needs to be said in this discussion is the fundamental connection between statistical physics and thermodynamics. Statistical physics describes microstates and thermodynamics describes macrostates. They are connected by so called thermodynamic limit—limit of infinite size and infinite particle number.

Microscopic parameters describing particles, like kinetic energy, are not automatically equivalent to macroscopic parameters like internal energy. They are found to be equivalent, which is a nontrivial result. The connection between micro and macro world appears only after doing the thermodynamic limit.

It is also important that many quantities are not the features of a single particle, but of the system. For instance, there is no entropy assigned to a single particle like the energy or momentum is, but there is entropy of the system containing one particle, if you specify the states that can be there.

Temperature is most commonly defined in the macroscopic realm as a quantity that lets you compare the state of two different systems that are interacting only by heat transfer. It turns out that this quantity is: $$T :=\left(\frac{\partial U}{\partial S}\right)_{V,N}.$$ Stretching this definition to the microscopic realm means treating the above equation as definition and putting to it the entropy before the thermodynamic limit. It might cause some trouble—like negative temperatures (that are in fact considered for systems that undergo saturation). So you might talk about a temperature of a system containing one particle but:

1. it will be not a temperature of this particle, but of the system containing it,
2. it is a very different temperature than you would expect.

Appendix:

1) Temperature for one particle system

Let us first consider how it is usually done in two particle system ($N=2$), before the problem trivializes in the one particle case. Consider number of microstates given two particles in a box, that can have discrete energy states separated by a constant energy portion of $\epsilon$. We divide the box into $n$ virtual compartments.

Counting the microstates. When the total energy in the system is $1\epsilon$, this energy can be either on first particle, or on the second. Particles are the same, so two situations are exactly the same. This is only one possibility. At the same time, particles can fill compartments in $W=n^2$ ways, so there are $n^2$ possibilities.

For the total energy of $2\epsilon$ we will have 2 possibilities - either energy is distributed equally over both particles, or it is all on one of them. The number of total microstates is $W=2n^2$.

You can see that the number of microstates changes with energy. To calculate what the temperature of this system is, you need to express the number of microstates in terms of energy. This is simple combinatorics but it's not the part of the question. Then entropy is: $S=k_b \log(W)$.

Then you have the expression to calculate

$$\frac{1}{T}=\left(\frac{\partial S(U)}{\partial U}\right)_{V,N},$$ and this will be the expression for temperature of this system.

Let us consider number of microstates for a single particle in a box. As before, we will assume volume divided into compartments.

At energy $1\epsilon$ number of microstates is $n$. At energy $2\epsilon$ number of microstates is $n$, at $3\epsilon$ it's the same.

The number of microstates does not change when adding energy to the system. This means $\frac{1}{T}=\left(\frac{\partial S(U)}{\partial U}\right)_{V,N}$ is zero. Since it's not a limit, it means that a temperature defined this way is not infinite, it simply doesn't exist.

2) As for temperature dependence on the observer, the entropy is not dependent on the change of the observer, so the temperature isn't either, if it were then I would start wondering if the definition is right.

Is it valid to assign a temperature to individual particles within kinetic theory and then claim that the temperature of the gas is simply the average of the temperature of the molecules?

I don't believe so. Temperature is a macroscopic property of a system. We don't normally talk about the temperature of a single particle.

Thanks for your response. I understand that this is not normally done. But I am asking if there is a logical flaw with such an interpretation.

I wouldn't say there is a logical "flaw" per se. It's just that temperature is defined as a macroscopic property of an object that reflects the collective behavior (in this case average translational kinetic energy) of the multiple microscopic particles that make up an object. But consider the following single particle example at the "macroscopic" level.

I have a ball which I throw and give it translational kinetic energy with respect to the ground. The ball is now my "particle". Assuming I throw it in a vacuum (no air friction) what temperature would I assign to the ball based on the velocity I gave it? The temperature I measure on the ball is only due to the collective microscopic kinetic energies internal to the ball. The balls "internal" kinetic energy. In the absence of air friction, the external kinetic energy of the ball, which is due to the velocity of its center of mass with respect to an external (to the ball) frame of reference, has no influence on temperature that I measure on the ball.

Hope this helps.

Temperature is a valid concept for any system in contact with a thermal bath. As such, you can take any subset of your system that is much smaller than the system, and consider it to be in contact with the remainder of the system as its thermal bath. Since the entropy and temperature are related by

$$\frac{1}{T}=-\left.\frac{\partial S}{\partial E}\right|_N,$$

Then by doing the "marble and matchstick" calculation (e.g. Callen, Thermodynamics, Ch.15) and using

$$S = \log\Big(\,{\rm number\,of\,microstates}\Big),$$

it is easy to show that the "temperature" of the subsystem is just

$$ T = \frac{E_{\rm subsystem}}{N_{\rm subsystem}}$$

In particular, you are allowed to take a subsystem consisting of only one particle, in which case its temperature is just its energy, as you are suggesting. However, the concept of temperature might not be very useful in this case.