Fluctuations

Can wild, chaotic fluctuations actually be tamed by state functions like S, T, and V? Turns out entropy and probability density are more buddy-buddy than you'd think.

Earlier we looked at Brownian motion, this unknowable random motion. So this time, here’s what we’re going to do. That Brownian motion thing — actually, not just Brownian motion, but fluctuation in general:

Can’t we express it using the state functions we’ve been dealing with all along, $T$, $V$, $p$, $S$….????

Let’s try to find the answer to that question.

First off, this thing called fluctuation, this absolute trainwreck of a physical phenomenon, feels like it’s somehow related to ’entropy’, right?? Entropy $S$ is Boltzmann's entropy, S = k_B ln Ω, so we’re going to focus on omega, the number of microstates!!!

$\Omega$ in terms of $U$? $\Omega$ in terms of $T$? $\Omega$ in terms of $p$? Stuff like that. So let’s leave the variable of $\Omega$ open and call it $x$. Just to be clear: take an isolated combined system with fixed total energy E, and let x describe an internal quantity that can fluctuate. Ω(x,E) counts the combined system’s microstates for a given x.

$\Omega$: number of microstates associated with a system characterized by parameter $x$ & energy $E$

Writing the count as Ω(x, E): S(x, E) = k_B ln Ω(x, E), so ln Ω(x, E) = S(x, E)/k_B and Ω(x, E) = e^(S(x, E)/k_B)

Oh…. but, the number of possible microstates… this is somehow related to the states the microstates occupy, i.e., the probability density function $p(x)$ must be related to $\Omega(x,E)$. Is it hooked in as a square? A cube? A square root? Nope, it’s the simplest one, plain ‘proportional’ (linear): every microstate is equally likely, so a macrostate with twice the microstates turns up twice as often.

The probability density p(x) ∝ Ω(x, E), so p(x) ∝ e^(S(x, E)/k_B)

Written out like this, it doesn’t look like nonsense. The point where the entropy is maximum would be equilibrium, and the point of maximum entropy is the biggest (highest) part of the probability density function, after all???!?

Alriiight— so let’s look at equilibrium, the point where the entropy is maximum. Say the entropy is maximum at $x = x_0$, and let’s write an approximation of $S(x,E)$ around $x \sim x_0$.

Taylor expansion of S(x, E) about x₀: S(x₀, E) + (∂S/∂x)(x − x₀) + ½(∂²S/∂x²)(x − x₀)² + …, derivatives at x = x₀

(at equilibrium)

But hey, there’s a term that flies away, right?!?!?!

The first derivative will be 0, since this is the maximum point, by the logic that it’s an extremum!!!

At the maximum the first-order term drops out: S(x, E) ≈ S(x₀, E) + ½(∂²S/∂x²)(x − x₀)² + …

And it goes liiike this.

Back to p(x) ∝ e^(S(x, E)/k_B)

I’m going to rewrite this proportionality as below.

The expansion put into the exponent: p(x) ∝ exp of (1/k_B) times (S(x₀, E) + ½(∂²S/∂x²)(x − x₀)²)

Now let’s play with it.

The exponential split into two factors, the constant e^(S(x₀, E)/k_B) in red

The red part will be a constant, so

Dropping the constant: p(x) ∝ exp((Δx)²/(2k_B) · ∂²S/∂x² at x₀), with Δx = x − x₀

Now here’s where we guess that $p(x)$ is a Gaussian function~~~

(It’s not such a wrong claim anyway, so, well;; heh; hehe)

Matching the Gaussian e^(−(Δx)²/2σ²)/(σ√(2π)) with that exponential

OK, so

Reading off 1/σ² = −(1/k_B) ∂²S/∂x² at x₀

So σ² = ⟨(Δx)²⟩ = −k_B / (∂²S/∂x² at x₀)

We found the fluctuation term near equilibrium.

And this equation is even more fun, because the conclusion it gives predicts real phenomena.

We left x open, right???? Now let U be the energy of a subsystem exchanging heat with a large reservoir, at fixed volume and particle number. The entropy being maximized is S_sub(U) + S_res(E − U). The reservoir supplies the first-order term that cancels at equilibrium; in the large-reservoir limit the curvature is −1/(T²C_V). Then

With x = U: ⟨(ΔU)²⟩ = −k_B / (∂²S/∂U²) at equilibrium

The denominator as ∂/∂U(∂S/∂U); dU = TdS − pdV gives ∂U/∂S = T, so ∂S/∂U = 1/T

Using dU/dT = C_V: ∂(1/T)/∂U = −1/(T²C_V), so ⟨(ΔU)²⟩ = k_B T² C_V

Near some continuous phase transitions, C_V can become very large and may diverge in the thermodynamic limit. That is not true of every transition, and a finite sample does not literally have infinite fluctuations.

So large C_V means large absolute energy fluctuations. The infinity in the original picture Original idealized divergence sketch: energy variance tends to infinity refers to an idealized limit, not every finite phase transition.

Totally

Hand-drawn graph of probability against U: an almost flat curve spread over everything

it’s in this kind of ‘pandemonium’ state, right?????

And in actual phase transitions, the fluctuations really are pandemonium, aren’t they??

And also,

Energy fluctuation ⟨(ΔU)²⟩ = k_B T² C_V

this equation tells us something. $U$ is an extensive variable, so it’s proportional to $N$, and $C_V$ is also an extensive variable, so it’s proportional to $N$ too. So we can also say the root-mean-square of the fluctuation scales like this: The rms fluctuation √⟨(ΔU)²⟩ ∝ √N ($\because C_V \propto N$).

That is, the ‘percentage change’ of the fluctuation is

The relative fluctuation √⟨(ΔU)²⟩ / U ∝ 1/√N

that’s what it iiis :) And this is telling us thiiis:

If the number of particles is enormous, the percentage change of the fluctuation term converges to 0!!!!!

Near a critical point, the simple C_V ∝ N scaling and the quadratic Gaussian approximation can break down. We have to check how C_V and U scale together before claiming anything about the relative fluctuation — it does not automatically go to infinity!!!~~ Urk!

Fun lol


Heads-up. The figures here are the original notes, slips and all. A few things to watch for:

  • The infinite-variance picture is an idealized critical-limit sketch. Not every phase transition has divergent C_V, finite samples have finite fluctuations, and a flat Gaussian over the entire real line would not be normalizable. Near a critical point, higher entropy terms can matter.

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