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Import converted Year 1 Semester 2 lecture notes from 443937a7d37850a3c57dfa863f35e80d07f2ccbc

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<b>Lecture 18: Statistical Physics V: Maxwell-Boltzmann Distribution</b><br>
Thu 19 Mar 2026 13:00
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[[Year 1 Sem 2/LC Temperature and Matter/Lecture 17|← Previous lecture]] · [[Year 1 Sem 2/LC Temperature and Matter|Module]] · [[Year 1 Sem 2/LC Temperature and Matter/Lecture 19|Next lecture →]]

Recap from last lecture:

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We derived: <div class="pn-import-equation"><nowiki>$$\Pr(v) = g(v) \left(\sqrt{\frac{m}{2 \pi k_B T}}\right)^{(3/2)} \exp\left(- \frac{mv^2}{2 k_b T}\right)$$</nowiki></div>

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Where the degeneracy is given by the density of states: <div class="pn-import-equation"><nowiki>$$g(v) = 4 \pi v^2$$</nowiki></div>

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<li>

Giving the full Maxwell-Boltzmann distribution:

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== Using Maxwell-Boltzmann ==

What is the probability of finding a particle with speed in excess of 400m/s?

<div class="pn-import-equation"><nowiki>$$\Pr(v > 400) = \int_{400}^{\infty} \Pr(v) \, dv$$</nowiki></div>

Using this distribution, what is the probability of finding a particle in excess of the speed of light?

<div class="pn-import-equation"><nowiki>$$\Pr(v > c) = \int_{c}^{\infty} \Pr(v) \, dv = \text{non zero}$$</nowiki></div>

Which is non-physical. Luckily, by the time we get close to <nowiki>$c$</nowiki>, the exponential term becomes so so small that the probability is effectively infinitesimal.

[[File:TM-Y1S2-lec18-01.png|thumb|center|800px|Maxwell-Boltzmann Distribution for <nowiki>$T = 10$</nowiki>K. Note that the value is approaching zero for <nowiki>$v \approx 1000$</nowiki>, let alone <nowiki>$c$</nowiki>.]]

As the temperature increases, the peak decreases in amplitude and shifts towards higher energy states:

[[File:TM-Y1S2-lec18-02.png|thumb|center|800px]]

Effectively, the probability of occupying higher energy states increases with increasing temperature. What about changing tem

[[File:TM-Y1S2-lec18-03.png|thumb|center|600px]]

This has the opposite effect, which comes from the ratio in the Boltzmann factor.

== Most Probable Speed ==

There are two very distinct components to the MBD, when examined componentwise:

<div class="pn-import-equation"><nowiki>$$\Pr(v) = \underbrace{g(v) \left(\sqrt{\frac{m}{2 \pi k_B T}}\right)^{(3/2)} v^2}_\text{quadratic term} \underbrace{\exp\left(- \frac{mv^2}{2 k_b T}\right)}_\text{exponential term}$$</nowiki></div>

This lets us look at the limiting cases:

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<li>

As <nowiki>$v \to \infty$</nowiki>: <div class="pn-import-equation"><nowiki>$$v^2 \to \infty$$</nowiki></div> <div class="pn-import-equation"><nowiki>$$\exp \left(- \frac{mv^2}{2k_B R}\right) \to 0$$</nowiki></div>

Hence <nowiki>$\Pr(\infty) \to 0$</nowiki>. Showing that the product of an infinite limit and the zero limit would require L’Hopital, but effectively the exponential decays faster than the quadratic grows and wins out.

</li>
<li>

As <nowiki>$v \to 0$</nowiki>: <div class="pn-import-equation"><nowiki>$$v^2 \to \infty \to 0$$</nowiki></div> <div class="pn-import-equation"><nowiki>$$\exp \left(- \frac{mv^2}{2k_B T}\right) \to 1$$</nowiki></div>

Hence <nowiki>$\Pr(0) \to 0$</nowiki>

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</ul>

As we have <nowiki>$\Pr(0) \to 0$</nowiki> and <nowiki>$\Pr(\infty) \to 0$</nowiki>, this gives us a curve with a central peak, representing the most probable speed. This speed is give by:

<div class="pn-import-equation"><nowiki>$$\frac{d \Pr(v)}{dv} = 0$$</nowiki></div> <div class="pn-import-equation"><nowiki>$$\Pr(v) = 4 \pi \left(\sqrt{\frac{m}{2 \pi k_B T}}\right)^{(3/2)} v^2 \exp\left(- \frac{mv^2}{2 k_b T}\right)$$</nowiki></div>

<div class="pn-import-equation"><nowiki>$$\Pr(v) = \left[4 \pi \left(\sqrt{\frac{m}{2 \pi k_B T}}\right)^{(3/2)} \right] \left\{ v^2 \exp\left(- \frac{mv^2}{2 k_b T}\right) \right\}$$</nowiki></div>

And performing the product rule on the second {} terms gives: <div class="pn-import-equation"><nowiki>$$\frac{d\Pr(v)}{dv} = 4 \pi \left(\sqrt{\frac{m}{2 \pi k_B T}}\right)^{(3/2)} \left[ 2v \exp\left(- \frac{mv^2}{2 k_b T}\right) \right]+ 4 \pi \left(\sqrt{\frac{m}{2 \pi k_B T}}\right)^{(3/2)} \left[ v^2 \times \frac{-2mv}{2k_B T} \exp\left(- \frac{mv^2}{2 k_b T}\right) \right] = 0$$</nowiki></div> <div class="pn-import-equation"><nowiki>$$\implies \left[4 \pi \frac{m}{2 \pi k_B T}^{3/2} 2v \exp \left(- \frac{mv^2}{2 k_B T}\right)\right] \left[1 - \frac{mv^2}{2k_B T}\right]$$</nowiki></div>

The first term can only equal zero in the trivial cases where <nowiki>$v = 0$</nowiki> or <nowiki>$v = \infty$</nowiki>. If we only consider non-trivial cases, we’re left with:

<div class="pn-import-equation"><nowiki>$$1 - \frac{mv^2}{2k_B T} = 0 \implies v = \sqrt{\frac{2 k_B T}{m}}$$</nowiki></div>

For hydrogen gas at <nowiki>$10$</nowiki>K, this gives us <nowiki>$287$</nowiki>m/s. Does this mean we need to consider supersonic effects (sonic booms) for hotter scenarios?

Luckily, no. With random motion in a gas, there’s no obvious direction for pressure to build up in. The effects of gas molecules travelling in lots od different directions cancels itself out nicely.

== Averages ==

The average value of <nowiki>$x$</nowiki> can be written as:

<ul>
<li>

<nowiki>$\langle x\rangle$</nowiki>.

</li>
<li>

<nowiki>$\bar{x}$</nowiki>.

</li>
<li>

<nowiki>$E\{x\}$</nowiki> or <nowiki>$\mathbb{E} \{x\}$</nowiki>.

</li>
</ul>

We’ll use the first notation, and strictly it’s given by: <div class="pn-import-equation"><nowiki>$$\langle x\rangle = \frac{1}{N} \sum_{i=1}^{N} N_i x_i = \sum_{i=1}^{N} \Pr_i(x_i) x_i$$</nowiki></div> If we consider a histogram, i.e. measurements of velocities for particles. If we take the number of bins to infinity, we have: <div class="pn-import-equation"><nowiki>$$\langle f(x)\rangle = \int_{0}^{\infty} \Pr(x) f(x) \, dx$$</nowiki></div>

For average <b>velocity</b> we expect a distribution which is symmetric about <nowiki>$0$</nowiki>, so we expect the average velocity to be <nowiki>$0$</nowiki>.

Speed however is given by the Maxwell-Boltzmann distribution and is not symmetric.

<div class="pn-import-equation"><nowiki>$$\Pr(v) = 4 \pi \left(\sqrt{\frac{m}{2 \pi k_B T}}\right)^{(3/2)} v^2 \exp\left(- \frac{mv^2}{2 k_b T}\right)$$</nowiki></div> <div class="pn-import-equation"><nowiki>$$\langle v\rangle = \int_{0}^{\infty} v \Pr(v) \, dv$$</nowiki></div>

Letting <nowiki>$a = 4 \pi \left(\frac{m}{2 \pi k_B T}\right)^{3/2}$</nowiki> and <nowiki>$b = \frac{m}{k_B T}$</nowiki> gives: <div class="pn-import-equation"><nowiki>$$\langle v\rangle = a \int_{0}^{\infty} v^3 e^{-bv^2} \, dv$$</nowiki></div>

And making the substitution <nowiki>$u = v^2 \implies du = 2v dv$</nowiki>: <div class="pn-import-equation"><nowiki>$$\langle v\rangle = \frac{a}{2}\int_{0}^{\infty} u e^{-bu} \, du$$</nowiki></div>

Which is integrable by parts: <div class="pn-import-equation"><nowiki>$$\langle v\rangle = \frac{a}{2b^2} = \frac{4 \pi \left(\frac{m}{2 \pi k_B T}\right)^{3/2}}{2 \left(\frac{m}{k_B T}\right)^2}$$</nowiki></div> <div class="pn-import-equation"><nowiki>$$= \sqrt{\frac{8 k_B T}{\pi m}} = \frac{2}{\sqrt{\pi}} v_\text{most probable}$$</nowiki></div>

This gives an average speed which is just a little higher compared to the most probable speed.

== Average Kinetic Energy ==

Since: <div class="pn-import-equation"><nowiki>$$\langle E_k\rangle = \frac{1}{2}mv^2 \implies \langle E_k\rangle = \frac{1}{2}m \langle v^2\rangle$$</nowiki></div>

We need <nowiki>$\langle v^2 \rangle$</nowiki>: <div class="pn-import-equation"><nowiki>$$\langle v^2 \rangle = \int_{0}^{\infty} v^2 \Pr(v) \, dv = a \int_{0}^{\infty} v^4 e^{-bv^2/2} \, dv$$</nowiki></div>

Where we reuse <nowiki>$a = 4\pi\left(\frac{m}{2\pi k_B T}\right)^{3/2}$</nowiki> and let <nowiki>$\alpha = \frac{m}{2k_B T}$</nowiki>, so the exponent becomes <nowiki>$-\alpha v^2$</nowiki>:

<div class="pn-import-equation"><nowiki>$$\langle v^2 \rangle = a \int_{0}^{\infty} v^4 e^{-\alpha v^2} \, dv$$</nowiki></div>

We apply Feynman’s trick with <nowiki>$n = 2$</nowiki>:

<div class="pn-import-equation"><nowiki>$$\int_{0}^{\infty} e^{-\alpha x^2} x^{2n} \, dx = \frac{1}{2}\left(-\frac{\partial}{\partial \alpha}\right)^n \sqrt{\frac{\pi}{\alpha}}$$</nowiki></div>

<div class="pn-import-equation"><nowiki>$$-\frac{\partial}{\partial \alpha}\sqrt{\frac{\pi}{\alpha}} = \frac{1}{2}\sqrt{\pi}\,\alpha^{-3/2}$$</nowiki></div> <div class="pn-import-equation"><nowiki>$$\left(-\frac{\partial}{\partial \alpha}\right)^2\sqrt{\frac{\pi}{\alpha}} = -\frac{\partial}{\partial \alpha}\left[\frac{1}{2}\sqrt{\pi}\,\alpha^{-3/2}\right] = \frac{3}{4}\sqrt{\pi}\,\alpha^{-5/2}$$</nowiki></div>

Hence: <div class="pn-import-equation"><nowiki>$$\int_{0}^{\infty} v^4 e^{-\alpha v^2} \, dv = \frac{3}{8}\sqrt{\pi}\,\alpha^{-5/2}$$</nowiki></div>

Substituting back with <nowiki>$\alpha = \frac{m}{2k_B T}$</nowiki>:

<div class="pn-import-equation"><nowiki>$$\langle v^2 \rangle = 4\pi\left(\frac{m}{2\pi k_B T}\right)^{3/2} \cdot \frac{3}{8}\sqrt{\pi} \left(\frac{m}{2k_B T}\right)^{-5/2}$$</nowiki></div>

<div class="pn-import-equation"><nowiki>$$= \frac{3\pi^{3/2}}{2} \cdot \frac{m^{3/2}}{(2\pi)^{3/2}(k_B T)^{3/2}} \cdot \frac{(2k_B T)^{5/2}}{m^{5/2}} = \frac{3}{2} \cdot 2 \cdot \frac{k_B T}{m} = \frac{3k_B T}{m}$$</nowiki></div>

Therefore:

<div class="pn-import-equation"><nowiki>$$\langle E_k \rangle = \frac{1}{2}m\langle v^2 \rangle = \frac{1}{2}m \cdot \frac{3k_B T}{m} = \boxed{\frac{3}{2}k_B T}$$</nowiki></div>

This lets us define the RMS (root mean square) speed: <div class="pn-import-equation"><nowiki>$$v_\text{rms} = \sqrt{\langle v^2 \rangle} = \sqrt{\frac{3k_B T}{m}} = \sqrt{\frac{3}{2}}\, v_\text{most probable}$$</nowiki></div>

Summarising the three characteristic speeds of the distribution: <div class="pn-import-equation"><nowiki>$$v_\text{most probable} = \sqrt{\frac{2k_B T}{m}} < \langle v \rangle = \sqrt{\frac{8k_B T}{\pi m}} < v_\text{rms} = \sqrt{\frac{3k_B T}{m}}$$</nowiki></div>

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