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Import converted Year 1 Semester 2 lecture notes from 443937a7d37850a3c57dfa863f35e80d07f2ccbc

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<b>Lecture 16: Statistical Physics III: Atmosphere and Boltzmann Factors</b><br>
Thu 12 Mar 2026 13:00
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[[Year 1 Sem 2/LC Temperature and Matter/Lecture 15|← Previous lecture]] · [[Year 1 Sem 2/LC Temperature and Matter|Module]] · [[Year 1 Sem 2/LC Temperature and Matter/Lecture 17|Next lecture →]]

Recap from last lecture:

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We derived: <div class="pn-import-equation"><nowiki>$$\rho_N(h) = \rho_N(0) \exp \left(- \frac{mgh}{k_B T}\right)$$</nowiki></div> <div class="pn-import-equation"><nowiki>$$P(h) = P(0) \exp \left(- \frac{mgh}{k_B T}\right)$$</nowiki></div>

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

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<li>

Since <nowiki>$mgh / k_B T$</nowiki> is being passed into an exponential, it must be dimensionless. Hence <nowiki>$k_B T / mg$</nowiki> must have dimensions of <nowiki>$h$</nowiki>.

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<li>

This leads to the scale height <nowiki>$h_0$</nowiki> where <nowiki>$P(h_0) = P(0) / e$</nowiki>

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== Atmosphere Continued ==

We have established that the number density (number of fluid molecules per unit volume), at height 0, <nowiki>$\rho_N(0)$</nowiki> can be related to atmospheric pressure (<nowiki>$P(0)$</nowiki>) by: <div class="pn-import-equation"><nowiki>$$\rho_N(0) = \frac{P(0)}{k_B T}$$</nowiki></div>

To determine the total number of gas molecules in the slab of atmosphere with cross-sectional area <nowiki>$A$</nowiki>, we can integrate number density across all <nowiki>$h$</nowiki>: <div class="pn-import-equation"><nowiki>$$\begin{aligned} N &= A \int_{0}^{\infty} \rho_N(h) \, dh \\ &= A \int_{0}^{\infty} \rho_N(0) \exp\left(- \frac{mgh}{k_B T}\right) \, dh \\ &= A \rho_N(0) \int_{0}^{\infty} \exp\left(- \frac{mgh}{k_B T}\right) \, dh \\ &= A \rho_N(0) \left[- \frac{k_B T}{mg} \exp \left(- \frac{mhg}{k_B T}\right)\right]_0^\infty \\ &= -A \rho_N(0) \frac{k_B T}{mg} \left[0 - 1\right] \\ N &= A \rho_N(0) \frac{k_B T}{mg} \\ \implies \rho_N(0) &= \frac{Nmg}{A k_B T} \end{aligned}$$</nowiki></div>

And using: <div class="pn-import-equation"><nowiki>$$h_0 = \frac{k_B T}{mg}$$</nowiki></div> <div class="pn-import-equation"><nowiki>$$\rho_N(0) = \frac{N}{A h_0}$$</nowiki></div>

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

We further define a new quantity: <div class="pn-import-equation"><nowiki>$$\Pr(h) = \frac{A \rho_N(h)}{N}$$</nowiki></div>

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

=== Probability and Normalisation ===

We start from: <div class="pn-import-equation"><nowiki>$$\Pr(h) = \frac{A \rho_N(h)}{N}$$</nowiki></div> <div class="pn-import-equation"><nowiki>$$h_0 = \frac{k_B T}{mg}$$</nowiki></div> <div class="pn-import-equation"><nowiki>$$\rho_N(h) = \rho_N(0) \exp(-mgh / k_B T)$$</nowiki></div>

And substitute our previous results for <nowiki>$\rho_N(0)$</nowiki> and <nowiki>$\rho_N(h)$</nowiki> in: <div class="pn-import-equation"><nowiki>$$\Pr(h) = \frac{A}{N} \times \rho_N(0) \exp(-mgh / k_B T)$$</nowiki></div> <div class="pn-import-equation"><nowiki>$$\Pr(h) = \frac{A}{N} \times \frac{N}{A h_0} \exp(-mhg / k_B T)$$</nowiki></div> <div class="pn-import-equation"><nowiki>$$\Pr(h) = \frac{mg}{k_B T} \exp(-mhg / k_B T)$$</nowiki></div>

As this is a probability, we need to check that it is normalised: <div class="pn-import-equation"><nowiki>$$\int_{0}^{\infty} \Pr(h) \, dh$$</nowiki></div> <div class="pn-import-equation"><nowiki>$$= \int_{0}^{\infty} \frac{A \rho_N(h)}{N} \, dh$$</nowiki></div> <div class="pn-import-equation"><nowiki>$$\frac{A}{N}\int_{0}^{\infty} \rho_N(h) \, dh$$</nowiki></div> <div class="pn-import-equation"><nowiki>$$\int_{0}^{\infty} \Pr(h) \, dh = \frac{1}{N} \left(A \int_{0}^{\infty} \rho_N(h) \, dh \right)$$</nowiki></div>

And as previously derived: <div class="pn-import-equation"><nowiki>$$A \int_{0}^{\infty} \rho_N(h) \, dh = N$$</nowiki></div>

So:

<div class="pn-import-equation"><nowiki>$$\int_{0}^{\infty} \Pr(h) \, dh = \frac{1}{N} \left(A \int_{0}^{\infty} \rho_N(h) \, dh \right) = \frac{N}{N} = 1$$</nowiki></div>

So <nowiki>$\Pr(h)$</nowiki> is normalised, as required!

== Boltzmann Factors ==

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

Boltzmann factors describe the probabilities of measuring a given energy state in a system. In general form: <div class="pn-import-equation"><nowiki>$$\Pr(E_i) \propto \exp \left(-\frac{E_i}{k_B T}\right)$$</nowiki></div>

This is true for all energy states, i.e. a mass on a spring oscillating in SHM has: <div class="pn-import-equation"><nowiki>$$\Pr(\text{state where extension is } x) \propto \exp \left(-\frac{0.5kx^2}{k_B T}\right)$$</nowiki></div>

[[File:TM-Y1S2-lec16-01.png|thumb|center|600px]]

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 <nowiki>$E_0$</nowiki> and <nowiki>$E_1$</nowiki>.

[[File:TM-Y1S2-lec16-02.png|thumb|center|600px]]

What is the probability of finding the atom at energy state <nowiki>$E_0$</nowiki>? <div class="pn-import-equation"><nowiki>$$\Pr(E_0) \propto \exp \left(- \frac{E_0}{k_B T}\right)$$</nowiki></div>

Or adding a constant of proportionality: <div class="pn-import-equation"><nowiki>$$\Pr(E_0) = C \exp \left(- \frac{E_0}{k_B T}\right)$$</nowiki></div>

And if both energy states have the same constant of proportionality, which is true here: <div class="pn-import-equation"><nowiki>$$\Pr(E_1) = C \exp \left(- \frac{E_1}{k_B T}\right)$$</nowiki></div>

Since the probability must be normalised across all states: <div class="pn-import-equation"><nowiki>$$\Pr(E_0) + \Pr(E_1) = 1$$</nowiki></div> <div class="pn-import-equation"><nowiki>$$C \exp \left(- \frac{E_0}{k_B T}\right) + C \exp \left(- \frac{E_1}{k_B T}\right) = 1$$</nowiki></div> <div class="pn-import-equation"><nowiki>$$1 = C \left(\exp \left(- \frac{E_0}{k_B T}\right) + \exp \left(- \frac{E_1}{k_B T}\right)\right)$$</nowiki></div> <div class="pn-import-equation"><nowiki>$$C = \frac{1}{\exp(-E_0/k_B T) + \exp(-E_1 / k_B T)}$$</nowiki></div>

Is the constant of proportionality here. Hence: <div class="pn-import-equation"><nowiki>$$\Pr(E_1) = \frac{\exp(-E_1 / k_B T)}{\exp(-E_0/k_B T) + \exp(-E_1 / k_B T)}$$</nowiki></div> <div class="pn-import-equation"><nowiki>$$\Pr(E_0) = \frac{\exp(-E_0 / k_B T)}{\exp(-E_0/k_B T) + \exp(-E_1 / k_B T)}$$</nowiki></div>

We can now do a bit of a maths trick: <div class="pn-import-equation"><nowiki>$$\Pr(E_1) = \frac{\exp(-E_1 / k_B T)}{\exp(-E_0/k_B T) + \exp(-E_1 / k_B T)} \times \frac{e^{E_1 / k_B T}}{e^{E_1 / k_B T}}$$</nowiki></div> <div class="pn-import-equation"><nowiki>$$\frac{1}{\exp \left(\frac{E_1 - E_0}{k_B T}\right) + 1}$$</nowiki></div>

Which lets us start considering limits. If we consider the limit as <nowiki>$T \to 0$</nowiki>: <div class="pn-import-equation"><nowiki>$$\Pr(E_1) \to 0 \text{ and } \Pr(E_0) \to 1$$</nowiki></div> 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: <div class="pn-import-equation"><nowiki>$$\Pr(E_1) \approx \frac{1}{1 + (E_1 - E_0)/k_B T + 1}$$</nowiki></div>

Now as <nowiki>$T \to \infty$</nowiki>: <div class="pn-import-equation"><nowiki>$$\Pr(E_1) \to \frac{1}{1 + 0 + 1} \implies \Pr(E_1) \to \frac{1}{2} \text{ and } \Pr(E_0) \to \frac{1}{2}$$</nowiki></div>

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

[[File:TM-Y1S2-lec16-03.png|thumb|center|600px]]

== Negative Temperatures ==

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Ash: Import converted Year 1 Semester 2 lecture notes from 443937a7d37850a3c57dfa863f35e80d07f2ccbc