The Space of Physical Frameworks (Part 5)

Azimuth 2024-10-02

In Part 4, I presented a nifty result supporting my claim that classical statistical mechanics reduces to thermodynamics when Boltzmann’s constant $k$ approaches zero. I used a lot of physics jargon to explain why I care about this result. I also used some math jargon to carry out my argument.

This may have been off-putting. But to understand the result, you only need to know calculus! So this time I’ll state it without all the surrounding rhetoric, and then illustrate it with an example.

At the end, I’ll talk about the physical meaning of it all.

The main result

Here it is:

Main Result. Suppose $S \colon [0,\infty) \to \mathbb{R}$ is a concave function with continuous second derivative. Suppose that for some $C > 0$ the quantity $C E - S(E)$ has a unique minimum as a function of $E,$ and $S'' < 0$ at that minimum. Then for this value of $C$ we have

$\displaystyle{ \lim_{k \to 0} - k \ln \int_0^\infty e^{-(C E - S(E))/k} \; d E \quad = \quad \inf_E \left(C E - S(E)\right) }$

when $k$ approaches $0$ from above.

Let’s make up names for the things on both sides of the equation:

$\displaystyle{ \Psi_k(C) = - k \ln \int_0^\infty e^{-(C E - S(E))/k} \; d E }$

and

$\displaystyle{ \Psi(C) = \inf_E \left(C E - S(E)\right) }$

So, the result says

$\displaystyle{ \lim_{k \to 0} \Psi_k(C) = \Psi(C) }$

when the hypotheses hold.

An example

Let’s do an example! Let’s try

$S(E) = \alpha \ln E$

where $\alpha$ is any positive number. This meets the hypotheses of the result for all $C > 0.$

So, let’s show

$\displaystyle{ \lim_{k \to 0} \Psi_k(C) = \Psi(C) }$

in this case. Of course, this follows from what I’m calling the ‘main result’—but working through this example really helped me. While struggling with it, I discovered some confusion in my thinking! Only after straightening this out and fixing Part 4 of my series was I able to get this example to work.

I’ll say more about the physics of this example at the end of this post. For now, let’s just do the math.

First let’s evaluate the left hand side, $\lim_{k \to 0} \Psi_k(C).$ By definition we have

$\displaystyle{ \Psi_k(C) = - k \ln \int_0^\infty e^{-(C E - S(E))/k} \; d E }$

Plugging in $S(E) = \alpha \ln E$ we get this:

$\begin{array}{ccl} \Psi_k(C) &=& \displaystyle{ - k \ln \int_0^\infty e^{-(C E - \alpha \ln E)/k} \; d E } \\ \\ &=& \displaystyle{ - k \ln \int_0^\infty e^{-C E/k} \; E^{\alpha/k} \; d E } \end{array}$

To evaluate this integral we can use the formula

$\displaystyle{ \int_0^\infty e^{-x} \, x^n \; d x = n! }$

which makes sense for all positive numbers, even ones that aren’t integers, if we interpret $n!$ using the gamma function. So, let’s use this change of variables:

$x = C E/k , \qquad E = k x/C$

We get

$\begin{array}{ccl} \Psi_k(C) &=& \displaystyle{ - k \ln \int_0^\infty e^{-C E/k} \; E^{\alpha/k} \; d E } \\ \\ &=& \displaystyle{ - k \ln \int_0^\infty e^{-x} \; \left( \frac{k x}{C} \right)^{\alpha/k} \, \frac{k d x}{C} } \\ \\ &=& \displaystyle{ - k \ln \left( \left( \frac{k}{C} \right)^{\frac{\alpha}{k} + 1} \int_0^\infty e^{-x} x^{\alpha/k} \, d x \right) } \\ \\ &=& \displaystyle{ - k \left(\frac{\alpha}{k} + 1\right) \ln \frac{k}{C} - k \ln \left(\frac{\alpha}{k}\right)! } \\ \\ &=& \displaystyle{ -\left(\alpha + k \right) \ln \frac{k}{C} - k \ln \left(\frac{\alpha}{k}\right)! } \end{array}$

Now let’s take the $k \to 0$ limit. For this we need Stirling’s formula

$\displaystyle{ n! \sim \sqrt{2 \pi n} \left(\frac{n}{e}\right)^n }$

or more precisely this version:

$\displaystyle{ \ln n! = n \ln \frac{n}{e} + O(\ln n) }$

Let’s compute:

$\begin{array}{l} \lim_{k \to 0} \Psi_k(C) = \displaystyle{ \lim_{k \to 0} \left( -\left(\alpha + k \right) \ln \frac{k}{C} - k \ln \left(\frac{\alpha}{k}\right)! \right) } \\ \\ = \displaystyle{ -\lim_{k \to 0} \left(\alpha \ln \frac{k}{C} \; + \; k \ln \left(\frac{\alpha}{k}\right)! \right) } \\ \\ = \displaystyle{ -\lim_{k \to 0} \left(\alpha \ln \frac{k}{C} \; + \; k \left(\frac{\alpha}{k} \ln \frac{\alpha}{e k} + O\left(\ln \frac{\alpha}{k} \right) \right) \right) } \\ \\ = \displaystyle{ -\alpha \ln \frac{k}{C} - \alpha \ln \frac{\alpha}{e k} } \\ \\ = \displaystyle{ \alpha \ln \frac{e C}{\alpha} } \\ \\ \end{array}$

Now let’s compare the right hand side of the main result. By definition this is

$\displaystyle{ \Psi(C) = \inf_E \left(C E - S(E)\right) }$

Plugging in $S(E) = \alpha \ln E$ we get this:

$\displaystyle{ \Psi(C) = \inf_E \left(C E - \alpha \ln E \right) }$

Since $\alpha > 0,$ the function $C E - \alpha \ln E$ has strictly positive second derivative as a function of $E$ wherever it’s defined, namely $E > 0.$ It also has a unique minimum. To find this minimum we solve

$\displaystyle{ \frac{d}{d E} \left(C E - \alpha \ln E \right) = 0 }$

which gives

$\displaystyle{ C - \frac{\alpha}{E} = 0 }$

$\displaystyle{ E = \frac{\alpha}{C} }$

We thus have

$\displaystyle{ \Psi(C) = C \cdot \frac{\alpha}{C} - \alpha \ln \frac{\alpha}{C} = \alpha\left(1 - \ln \frac{\alpha}{C} \right) }$

$\displaystyle{ \Psi(C) = \alpha \ln \frac{e C}{\alpha} }$

Yes, we’ve seen this before. So we’ve verified that

$\displaystyle{ \lim_{k \to 0} \Psi_k(C) = \Psi(C) }$

in this example.

The physical meaning

We can interpret the function $S(E)$ as the entropy of a system when its energy is exactly $E.$ There are many systems whose entropy is given by the formula $S(E) = \alpha \ln E$ for various values of $\alpha.$ Examples include:

1) an $n$ -dimensional harmonic oscillator,

2) a collection of noninteracting classical particles in a box (a so-called ‘ideal gas’),

3) a collection of so-called ‘energy particles’, which are imaginary entities that have no qualities except energy, which can be any nonnegative number.

Examples 1 and 2 are worked out in my book, while I used Example 3 to prove Stirling’s formula here:

• Stirling’s formula from statistical mechanics.

So, it’s no coincidence that I used Stirling’s formula today: I’m just walking across the same bridge in the opposite direction!

For any system of this sort, its temperature is proportional to its energy. We saw this fact today when I showed the infimum in this formula:

$\Psi(C) = \inf_E \left(C E - S(E)\right)$

occurs at $E = \alpha/C.$ The quantity $C$ is the ‘coldness’ or inverse temperature $C = 1/T,$ so we’re seeing $E = \alpha T.$ This type of result fools some students into thinking temperature is always proportional to energy! That’s obviously false: think about how much more energy it take to melt an ice cube than it takes to raise the temperature of the melted ice cube by a small amount. So we’re dealing with a very special, simple sort of system here.

In equilibrium, a system at fixed coldness will choose the energy that maximizes $S(E) - C E.$ The maximum value of this quantity is called the system’s ‘free entropy’. In other words, the system’s negative free entropy is $\Psi(C) = \inf_E \left(C E - S(E)\right).$

All of my physics remarks so far concern ‘thermostatics’ as defined in Part 2. In this approach, we start with entropy as a function of energy, and define $\Psi(C)$ using a Legendre transform. Boltzmann’s constant never appears in our calculations.

But we can also follow another approach: ‘classical statistical mechanics’, as defined in Part 4. In this framework we define negative free entropy in a different way. We call $\exp(S(E)/k)$ the ‘density of states’, and we define negative free entropy using the Laplace transform of the density of states. It looks like this:

$\displaystyle{ \Psi_k(C) = - k \ln \int_0^\infty e^{-(C E - S(E))/k} \; d E }$

In this approach, Boltzmann’s constant is important. However, the punchline is that

$\displaystyle{ \lim_{k \to 0} \Psi_k(C) = \Psi(C) }$

Thus classical statistical mechanics reduces to thermostatics as $k \to 0,$ at least in this one respect.

In fact various other quantities should work the same way. We can define them in classical statistical mechanics, where they depend on Boltzmann’s constant, but as $k \to 0$ they should approach values that you could compute in a different way, using thermostatics. So, what I’m calling the ‘main result’ should be just one case of a more general result. But this one case clearly illustrates the sense in which the Laplace transform reduces to the Legendre transform as $k \to 0.$ This should be a key aspect of the overall story.

Warning

By the way, I simplified the calculations today by choosing units of energy where the so-called ‘energy width’ $w$ from last time equals 1. Thus, energy is treated as dimensionless. However, I am not treating entropy as dimensionless.