Year 1 Sem 2/LC Temperature and Matter/Lecture 16

Lecture 16: Statistical Physics III: Atmosphere and Boltzmann Factors
Thu 12 Mar 2026 13:00

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Recap from last lecture:

• We derived:
  $$\rho_N(h) = \rho_N(0) \exp \left(- \frac{mgh}{k_B T}\right)$$
  $$P(h) = P(0) \exp \left(- \frac{mgh}{k_B T}\right)$$

  To give the number density and pressure as an exponential function of height in the atmosphere.

• Since $mgh / k_B T$ is being passed into an exponential, it must be dimensionless. Hence $k_B T / mg$ must have dimensions of $h$.
• This leads to the scale height $h_0$ where $P(h_0) = P(0) / e$

Atmosphere Continued

We have established that the number density (number of fluid molecules per unit volume), at height 0, $\rho_N(0)$ can be related to atmospheric pressure ($P(0)$) by:

$$\rho_N(0) = \frac{P(0)}{k_B T}$$

To determine the total number of gas molecules in the slab of atmosphere with cross-sectional area $A$, we can integrate number density across all $h$:

$$\begin{aligned} N &= A \int_{0}^{\infty} \rho_N(h) \, dh \\ &= A \int_{0}^{\infty} \rho_N(0) \exp\left(- \frac{mgh}{k_B T}\right) \, dh \\ &= A \rho_N(0) \int_{0}^{\infty} \exp\left(- \frac{mgh}{k_B T}\right) \, dh \\ &= A \rho_N(0) \left[- \frac{k_B T}{mg} \exp \left(- \frac{mhg}{k_B T}\right)\right]_0^\infty \\ &= -A \rho_N(0) \frac{k_B T}{mg} \left[0 - 1\right] \\ N &= A \rho_N(0) \frac{k_B T}{mg} \\ \implies \rho_N(0) &= \frac{Nmg}{A k_B T} \end{aligned}$$

And using:

$$h_0 = \frac{k_B T}{mg}$$

$$\rho_N(0) = \frac{N}{A h_0}$$

This has a physical interpretation. If you took all $N$ molecules and distributed them in a volume $Ah_0$, i.e. our slab of cross-sectional area $A$ with $h = h_0$, the average density of this would be equal to $\rho_N(0)$. This defines $h_0$ additionally as the average height of an equivalent uniform atmosphere with the same number of molecules.

We further define a new quantity:

$$\Pr(h) = \frac{A \rho_N(h)}{N}$$

Which gives the probability of (for a given specific particle) finding it at height $h$

Probability and Normalisation

We start from:

$$\Pr(h) = \frac{A \rho_N(h)}{N}$$

$$h_0 = \frac{k_B T}{mg}$$

$$\rho_N(h) = \rho_N(0) \exp(-mgh / k_B T)$$

And substitute our previous results for $\rho_N(0)$ and $\rho_N(h)$ in:

$$\Pr(h) = \frac{A}{N} \times \rho_N(0) \exp(-mgh / k_B T)$$

$$\Pr(h) = \frac{A}{N} \times \frac{N}{A h_0} \exp(-mhg / k_B T)$$

$$\Pr(h) = \frac{mg}{k_B T} \exp(-mhg / k_B T)$$

As this is a probability, we need to check that it is normalised:

$$\int_{0}^{\infty} \Pr(h) \, dh$$

$$= \int_{0}^{\infty} \frac{A \rho_N(h)}{N} \, dh$$

$$\frac{A}{N}\int_{0}^{\infty} \rho_N(h) \, dh$$

$$\int_{0}^{\infty} \Pr(h) \, dh = \frac{1}{N} \left(A \int_{0}^{\infty} \rho_N(h) \, dh \right)$$

And as previously derived:

$$A \int_{0}^{\infty} \rho_N(h) \, dh = N$$

So:

$$\int_{0}^{\infty} \Pr(h) \, dh = \frac{1}{N} \left(A \int_{0}^{\infty} \rho_N(h) \, dh \right) = \frac{N}{N} = 1$$

So $\Pr(h)$ is normalised, as required!

Boltzmann Factors

This is on the exam. We have the quantity $mgh / k_B T$, which is dimensionless. What does this quantity actually physically mean? $mgh$ is a term for the potential energy (gravitational here), and $k_B T$ is a term for thermal energy.

Boltzmann factors describe the probabilities of measuring a given energy state in a system. In general form:

$$\Pr(E_i) \propto \exp \left(-\frac{E_i}{k_B T}\right)$$

This is true for all energy states, i.e. a mass on a spring oscillating in SHM has:

$$\Pr(\text{state where extension is } x) \propto \exp \left(-\frac{0.5kx^2}{k_B T}\right)$$

From this figure, we can see that higher temperatures have a higher probability of observing higher energy states, which makes sense. We can use this to describe continuous energy distributions, or discrete ones (i.e. energy levels in atoms).

The integral across all energy states must be equal to one.

Example: Simple Quantum System

Consider a simple atom with two energy levels $E_0$ and $E_1$.

What is the probability of finding the atom at energy state $E_0$?

$$\Pr(E_0) \propto \exp \left(- \frac{E_0}{k_B T}\right)$$

Or adding a constant of proportionality:

$$\Pr(E_0) = C \exp \left(- \frac{E_0}{k_B T}\right)$$

And if both energy states have the same constant of proportionality, which is true here:

$$\Pr(E_1) = C \exp \left(- \frac{E_1}{k_B T}\right)$$

Since the probability must be normalised across all states:

$$\Pr(E_0) + \Pr(E_1) = 1$$

$$C \exp \left(- \frac{E_0}{k_B T}\right) + C \exp \left(- \frac{E_1}{k_B T}\right) = 1$$

$$1 = C \left(\exp \left(- \frac{E_0}{k_B T}\right) + \exp \left(- \frac{E_1}{k_B T}\right)\right)$$

$$C = \frac{1}{\exp(-E_0/k_B T) + \exp(-E_1 / k_B T)}$$

Is the constant of proportionality here. Hence:

$$\Pr(E_1) = \frac{\exp(-E_1 / k_B T)}{\exp(-E_0/k_B T) + \exp(-E_1 / k_B T)}$$

$$\Pr(E_0) = \frac{\exp(-E_0 / k_B T)}{\exp(-E_0/k_B T) + \exp(-E_1 / k_B T)}$$

We can now do a bit of a maths trick:

$$\Pr(E_1) = \frac{\exp(-E_1 / k_B T)}{\exp(-E_0/k_B T) + \exp(-E_1 / k_B T)} \times \frac{e^{E_1 / k_B T}}{e^{E_1 / k_B T}}$$

$$\frac{1}{\exp \left(\frac{E_1 - E_0}{k_B T}\right) + 1}$$

Which lets us start considering limits. If we consider the limit as $T \to 0$:

$$\Pr(E_1) \to 0 \text{ and } \Pr(E_0) \to 1$$

The maths is unpleasant, but we can take it as given! This implies that in the low temperature limit, the atom always sits in the non-excited state. What about the high temperature limit? We begin with a Taylor expansion:

$$\Pr(E_1) \approx \frac{1}{1 + (E_1 - E_0)/k_B T + 1}$$

Now as $T \to \infty$:

$$\Pr(E_1) \to \frac{1}{1 + 0 + 1} \implies \Pr(E_1) \to \frac{1}{2} \text{ and } \Pr(E_0) \to \frac{1}{2}$$

So in the high temperature limit, we end up with an even distribution between the two states.

Negative Temperatures

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