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So in this answer:

Folks figured out thermodynamics before statistical mechanics. In particular, we had thermometers. People measured the "hotness" of stuff by looking at the height of a liquid in a thermometer. The height of a thermometer reading was the definition of temperature; no relation to energy.

What happens, if I invoke collision dynamics from Chapman–Enskog theory? And measure the mass diffusivity (along with every other variable) to measure the temperature?

$$D = \frac{AT^{3/2}}{p \sigma^2_{12} \Omega} \sqrt{1/M_1 + 1/M_2} \tag{1}$$

where $D$ is the diffusion coefficient ($cm^2/s$), $A$ is an empirical coefficient equal to $1.859 × 10^{-3} \frac{{A}^{2}\cdot {cm}^{2}}{K^{3/2}\cdot s}$, index the two kinds of molecules present in the gaseous mixture, $T$ is the absolute temperature ($K$), $M$ is the molar mass (g/mol), p is the pressure (atm), $\sigma_{12} $ but usually of order $1$ and is the average collision diameter and $\Omega$ is the collision integral. (See here for the equation)

I suspect this argument now automatically fails:

[a] Note that if temperature had dimensions of energy then under this definition entropy would have been dimensionless (as it "should" be).
[c] Note again how $k_b$ and $T$ show up together.

More explicitly if I follow the prescription where $x \to p D$:

$$ L(p D) = \ln (\Omega) + T(E_0 - E(p D)) \tag{2}$$

we see an implicit assumption is present that the product of mass diffusivity and pressure must be a function of energy (I've never heard of this noether invariant of time translational invariance). This is not true and has to be shown valid in this case! (if not every).

In fact I will go further and prove it cannot be the case heuristically. Using Fick's law (which is a step down from Chapman's collisional dynamics)*:

$$ S = \int \sum_i \frac{{p_i }^2}{2m} dt \approx \frac{N}{2} m \bar{v} l \propto D \tag{3}$$

where $S$ is the action, $p_i$ is the momentum, $m$ is mass, $N$ is number of particles, $\bar{v}$ is the average velocity and $l$ is the mean free path. Also $ D = \bar{v} l $ There is no mention of pressure above.

Does Chapman–Enskog theory provide a way to measure temperature independent of energy? Is this answer still valid?

More Anonymous
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2 Answers2

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This question and your other recent question are ultimately about the systems of units (like SI or CGS) and not about temperature. In most systems of units the majority of quantities are expressed in terms of a few basic units, even if sometimes they are given special names like Joules/ergs for energy or Hz for frequency. The number of such basic units may vary - e.g., one does not have to define an independent unit for charge or temperature - and there is actually a price to pay for defining extra basic units - in terms of having to introduce additional constants, like Boltzmann constant $k_B$, vacuum permittivity $\varepsilon_0$ and so on. Moreover, one could even get away with less units than the basic length, mass and time - e.g., in spectroscopy frequency is often measured in inverse centimeters. Sometimes quantities may have rather surprising units when expressed in terms of basic ones - like capacitance in CGS system being measured in centimeters.

Another approach is to define natural units by expressing basic units in terms of basic constants or even by demanding that some or all the basic constants are equal to $1$ (see Planck units.)

Getting closer to the argument given in the OP: here one redefines the temperature in terms of other dimensional quantities - shifting the problem to defining the unites of these quantities. It does give the impression that the unit of temperature is fixed by those of other units... but we could always define a different unit and introduce a dimensional proportionality constant - but such a constant is our arbitrary choice, there is nothing fundamental about it.

Let me also bring up something that was lost in jumping from phenomenological thermodynamics to Chapman-Enskog - the equilibrium statistical physics, which dispenses with the collisions using quite general arguments - the collisions matter for establishing the thermodynamic equilibrium, but they can be neglected in description of this equilibrium, e.g., in terms of canonical/Boltzmann ensemble - the temperature is already present there, and have unambiguously energy units (e.g., defining the average kinetic energy of molecules in a gas, see equipartition theorem and this answer). BBGKY hierarchy, Boltzmann equation, and Chapman-Enskog arise when we try to describe how this equilibrium is achieved. It is even somewhat illogical to define temperature - which is a property of en equilibrium state - in terms of non-equilibrium theory.

Roger V.
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I did not understand all of them, but I will make an ad hoc link:$$D=\bar{v}l=\bar{v}\frac{kT}{\bar{F}}$$ with:$\;\;\bar{F}l_{x}=kT\;\;,\;\;\bar{P}=\frac{N\bar{F}}{l_{y}l_{z}}\;\;,\;\;\bar{P}V=NkT\;\;\;\;$(*)

so :$$D=\frac{ \bar{v} NkT}{\bar{P}S}=\frac{2}{3}\frac{\bar{v}}{\bar{P}S}\bar{E}$$

For the coefficient of self-diffusion, we find in (*) (rations: 46,47,48): $$D\propto \frac{1}{n}\propto \frac{1}{\bar{P}}\;\;\;\;,T=cst$$ $$D\propto T^{3/2}\;\;\;\;,\bar{P}=cst$$

(*) For example in Berkeley physics course, volume V (Statistical physics)

The Tiler
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