thermodynamics - How is the relationship of the value $kT$ and a degree of freedom derived?

Sources that discuss the derivation of the Maxwell-Boltzmann Statistics end up with two unknown constants ($\alpha$ and $\beta$) through the Lagrange Multipliers, of which $\alpha$ is derived by normalizing an integrand containing the derived Maxwell-Boltzmann probability distribution formula.

However, $\beta$ is approached differently by introducing a completely different information; saying that on average, a particle has $\frac{3}{2}kT$ energy (translational). Equalizing this with the average energy per particle according to the derived Maxwell-Boltzmann formula: $$\int^\infty_0 2 \cdot \sqrt{\frac{E}{\pi}} \cdot \beta^{1.5} \cdot e^{-\beta E} \cdot E \cdot dE = \frac{3}{2}kT$$ Which shows that $\beta = \frac{1}{kT}$.

However, it is not explained how $kT$ itself is derived. How did this value get associated with a degree of freedom in the first place? Is this derived experimentally by measuring the amount of energy needed for a system to reach a certain temperature $T$ and somehow knowing the number of moles and the number of degrees of freedom for a particle in that system?

Answer

So presumably your derivation looks like this, that in the discrete case we have a set of states $A$ and a probability variable for each state $p_a$ such that $\sum_{a\in A}p_a = 1$ and an energy of each state $E_a$ such that the average energy is fixed, $\sum_{a\in A} p_a E_a = \eta$. The goal is to maximize $s(\{p_a\})=-\sum_a p_a \ln p_a$ subject to those constraints and then with Lagrange multipliers we institute two parameters, let's call them $\beta$ and $\gamma$, so that we instead minimize this constrained entropy, $$\bar s(\{p_a\}) = \sum_{a \in A} \big(-p_a \ln p_a - \beta p_a E_a - \gamma p_a\big).$$ We then would find that $$\frac{\partial \bar s}{\partial p_a} = -1 - \ln p_a - \beta E_a - \gamma = 0$$ and thus that $$p_a = \frac{1}{Z}~e^{-\beta E_a}$$ for some $Z = e^{1 + \gamma}$ that enforces the normalization of the distribution, $Z = \sum_a e^{-\beta E_a}.$

We call this function $Z = Z(\beta)$ the partition function and recognize it specially because, for example, if we were to want to calculate $\sum_a p_a E_a = \eta$ we can now do it simply by looking at $$\eta=-\frac\partial{\partial \beta}\ln Z = -\frac{1}{Z}\frac{\partial Z}{\partial \beta} = -\frac1Z\sum_a e^{-\beta E_a}\cdot(-E_a) = \sum_a p_a E_a,$$ so that in its functional form it contains a great deal more information than you might have expected out of a simple normalization constant. Similarly $$s = -\sum_a p_a \ln p_a = +\sum_a p_a~(\beta E_a + \ln Z) = \ln Z + \beta \eta = \left(1 - \beta \frac{\partial}{\partial \beta}\right)\ln Z.$$

Now you are asking about the details of why we say that this parameter $\beta = 1/k_\text B T,$ where $T$ is the absolute temperature. Suppose we give a little bit of energy $\delta\eta$ to the system and allow it to again return to equilibrium. Since $Z = Z(\beta)$ we know that this requires somehow changing $\beta$ with some $\delta\beta.$ We can then work out that $$\delta Z = \sum_a e^{-\beta E_a} (-E_a) \delta \beta = -Z~\eta~\delta\beta$$ and thus that $$\delta s = \delta(\ln Z + \eta\beta) = \frac{\delta Z}{Z} + \eta~\delta\beta + \beta~\delta\eta = \beta ~\delta \eta.$$ We don't need to work out the details about $\delta\beta$ because it simply cancels in this expression.

So now if you imagine two such systems trying to exchange one packet of energy $\delta \eta$ you can see that the energy will flow from system 1 into system 2 spontaneously if the total entropy increases, $$\delta s = \delta s_1 + \delta s_2 = \beta_1~(-\delta\eta) + \beta_2~\delta \eta = (\beta_2 - \beta_1)\delta \eta,$$so the criterion for this is simply $\beta_2 > \beta_1$, and you might therefore say that $\beta$ measures some "coldness" of a system, with energy spontaneously flowing into a colder system from a warmer system.

So it turns out that far from $\beta$ being some sort of instance-specific parameter, we can use thermal contact to compare the $\beta$ factors between two otherwise-thermalized objects and it therefore represents some sort of universal property akin to temperature that we can use to describe thermal heat flows.

A direct implication of the last point is the potential existence of thermometers. A thermometer is just a well-known system which we can use to tell you a value for $\beta$ without disturbing that $\beta$ by all that much.

One such thermometer would simply be an ideal gas thermometer. If the energy is independent of position in such a thermometer (i.e. gravity is negligible here) then we basically wish to parcel out velocity space into a bunch of discrete chunks so that a given atom has the state $p_{(v_x,v_y,v_z)} = \frac1Z e^{-\beta\frac12 m (v_x^2 + v_y^2 + v_z^2)}.$ We can see that in the limit of small chunks we have some sort of Gaussian integral, $$Z_1 = \alpha \int_{-\infty}^\infty dv~e^{-\beta \frac 12 m v^2} = \sqrt{2\pi\alpha^2\over m\beta},\\Z = Z_1^3.$$

We thus have for the single molecule that $\eta = -Z'/Z = \frac32 \beta^{-1},$ and so for a bunch of molecules $N$ that $E = N \eta = \frac32 N \beta^{-1}.$

From the kinetic theory it is known that $E = \frac32 n R T.$ (This comes purely from considering the momentum imparted to a ceiling of a piston and the time between collisions with that ceiling giving $E= \frac32 PV$ where the 1/2 comes from a kinetic energy prefactor while the 3 comes from the 3 dimensions of space, see comments below.) Thus this ideal gas thermometer is measuring $\beta^{-1} = (n/N)~R T.$ If we simply define that $k_\text B = (n/N) R$ then we have your resulting expression, that $$\beta = 1/(k_\text B T).$$

Note that this is a stronger result than it first appears, because coldness and temperature are such broad properties. There is an amplification where if this holds for any thermometer it must hold for all thermometers. So simply connecting the kinetic theory of gases to the statistical interpretation of temperature means that we must have one of two results:

1. The statistical mechanism of heat transport is not the only one typically at play in real physical systems coming to equilibrium, or


2. $\beta = 1/k_\text B T$ for everything.

If there is an empirical side to this, then, it is the rejection of (1). (And, to a lesser extent, the fact that we suspect $n/N$ to be some constant is also an empirical observation.) In physicists’ experience with statistical mechanics they have never needed to introduce any other mechanism; it has always been sufficient that spontaneous heat flow can be understood as the result of an overall system transferring into a more likely state by a transfer of energy.

The general relationship can now be derived. Above the 3 comes from the exponent in $Z_1^3$ which is our major hint. We take two places where energy can live (degrees of freedom) $A+B$ with their own energy contributions $E^{A,B}$ and then discover that $$Z_{A+B}=\sum_a\sum_be^{-\beta E^A_a-\beta E^B_b}=Z_AZ_B,$$ but both $\eta$ and $s$ have been expressed as derivatives of $\ln Z$ so must be additive: energy and entropy must sum over these degrees of freedom.

Now take any one continuous degree of freedom $q$ and attempt to calculate the average value of $ q\frac{\partial H}{\partial q}, $ where H is a Hamiltonian function g giving a total energy, to find:$$ \begin{align} \left\langle q\frac{\partial H}{\partial q}\right\rangle &= \frac1{Z_Q} \int_{-\infty}^\infty dq~e^{-\beta H}~ q~\frac{\partial H}{\partial q}\\ &= \frac1{Z_Q} \left[q~\frac{-e^{-\beta H}}{\beta}\right]_{-\infty}^\infty + \frac1{Z_Q}\int_{-\infty}^\infty dq~\frac{e^{-\beta H}}{\beta} \end{align} $$ by integration by parts. The boundary term looks suspiciously ignorable for most nice Hamiltonians that grow faster than logarithmically in energy towards $+\infty$ or that have one fixed bound that can be placed at $q=0$ or so, while the latter integral is just $Z_Q/\beta$, so we derive that in general $$\left\langle q\frac{\partial H}{\partial q}\right\rangle = \beta^{-1} = k_\text B T.$$ If a term contributes quadratically to the Hamiltonian $H=\alpha q^2$ then the expression on the left will be twice that contribution $\left\langle q\frac{\partial H}{\partial q}\right\rangle=\langle 2\alpha q^2 \rangle= k_\text B T$ and so we see the average energy occupation being $\eta_Q =\frac1n k_\text B T$ for a Hamiltonian that goes like $q^n.$