Jump to content

Year 1 Sem 2/LC Temperature and Matter/Lecture 18

From Ash's Lecture Notes

Lecture 18: Statistical Physics V: Maxwell-Boltzmann Distribution
Thu 19 Mar 2026 13:00

← Previous lecture · Module · Next lecture →

Recap from last lecture:

  • We derived:
    $$\Pr(v) = g(v) \left(\sqrt{\frac{m}{2 \pi k_B T}}\right)^{(3/2)} \exp\left(- \frac{mv^2}{2 k_b T}\right)$$
  • Where the degeneracy is given by the density of states:
    $$g(v) = 4 \pi v^2$$
  • Giving the full Maxwell-Boltzmann distribution:

Using Maxwell-Boltzmann

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

$$\Pr(v > 400) = \int_{400}^{\infty} \Pr(v) \, dv$$

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

$$\Pr(v > c) = \int_{c}^{\infty} \Pr(v) \, dv = \text{non zero}$$

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

Maxwell-Boltzmann Distribution for $T = 10$K. Note that the value is approaching zero for $v \approx 1000$, let alone $c$.

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

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

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:

$$\Pr(v) = \underbrace{g(v) \left(\sqrt{\frac{m}{2 \pi k_B T}}\right)^{(3/2)} v^2}_\text{quadratic term} \underbrace{\exp\left(- \frac{mv^2}{2 k_b T}\right)}_\text{exponential term}$$

This lets us look at the limiting cases:

  • As $v \to \infty$:
    $$v^2 \to \infty$$
    $$\exp \left(- \frac{mv^2}{2k_B R}\right) \to 0$$

    Hence $\Pr(\infty) \to 0$. 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.

  • As $v \to 0$:
    $$v^2 \to \infty \to 0$$
    $$\exp \left(- \frac{mv^2}{2k_B T}\right) \to 1$$

    Hence $\Pr(0) \to 0$

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

$$\frac{d \Pr(v)}{dv} = 0$$
$$\Pr(v) = 4 \pi \left(\sqrt{\frac{m}{2 \pi k_B T}}\right)^{(3/2)} v^2 \exp\left(- \frac{mv^2}{2 k_b T}\right)$$
$$\Pr(v) = \left[4 \pi \left(\sqrt{\frac{m}{2 \pi k_B T}}\right)^{(3/2)} \right] \left\{ v^2 \exp\left(- \frac{mv^2}{2 k_b T}\right) \right\}$$

And performing the product rule on the second {} terms gives:

$$\frac{d\Pr(v)}{dv} = 4 \pi \left(\sqrt{\frac{m}{2 \pi k_B T}}\right)^{(3/2)} \left[ 2v \exp\left(- \frac{mv^2}{2 k_b T}\right) \right]+ 4 \pi \left(\sqrt{\frac{m}{2 \pi k_B T}}\right)^{(3/2)} \left[ v^2 \times \frac{-2mv}{2k_B T} \exp\left(- \frac{mv^2}{2 k_b T}\right) \right] = 0$$
$$\implies \left[4 \pi \frac{m}{2 \pi k_B T}^{3/2} 2v \exp \left(- \frac{mv^2}{2 k_B T}\right)\right] \left[1 - \frac{mv^2}{2k_B T}\right]$$

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

$$1 - \frac{mv^2}{2k_B T} = 0 \implies v = \sqrt{\frac{2 k_B T}{m}}$$

For hydrogen gas at $10$K, this gives us $287$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 $x$ can be written as:

  • $\langle x\rangle$.
  • $\bar{x}$.
  • $E\{x\}$ or $\mathbb{E} \{x\}$.

We’ll use the first notation, and strictly it’s given by:

$$\langle x\rangle = \frac{1}{N} \sum_{i=1}^{N} N_i x_i = \sum_{i=1}^{N} \Pr_i(x_i) x_i$$

If we consider a histogram, i.e. measurements of velocities for particles. If we take the number of bins to infinity, we have:

$$\langle f(x)\rangle = \int_{0}^{\infty} \Pr(x) f(x) \, dx$$

For average velocity we expect a distribution which is symmetric about $0$, so we expect the average velocity to be $0$.

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

$$\Pr(v) = 4 \pi \left(\sqrt{\frac{m}{2 \pi k_B T}}\right)^{(3/2)} v^2 \exp\left(- \frac{mv^2}{2 k_b T}\right)$$
$$\langle v\rangle = \int_{0}^{\infty} v \Pr(v) \, dv$$

Letting $a = 4 \pi \left(\frac{m}{2 \pi k_B T}\right)^{3/2}$ and $b = \frac{m}{k_B T}$ gives:

$$\langle v\rangle = a \int_{0}^{\infty} v^3 e^{-bv^2} \, dv$$

And making the substitution $u = v^2 \implies du = 2v dv$:

$$\langle v\rangle = \frac{a}{2}\int_{0}^{\infty} u e^{-bu} \, du$$

Which is integrable by parts:

$$\langle v\rangle = \frac{a}{2b^2} = \frac{4 \pi \left(\frac{m}{2 \pi k_B T}\right)^{3/2}}{2 \left(\frac{m}{k_B T}\right)^2}$$
$$= \sqrt{\frac{8 k_B T}{\pi m}} = \frac{2}{\sqrt{\pi}} v_\text{most probable}$$

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

Average Kinetic Energy

Since:

$$\langle E_k\rangle = \frac{1}{2}mv^2 \implies \langle E_k\rangle = \frac{1}{2}m \langle v^2\rangle$$

We need $\langle v^2 \rangle$:

$$\langle v^2 \rangle = \int_{0}^{\infty} v^2 \Pr(v) \, dv = a \int_{0}^{\infty} v^4 e^{-bv^2/2} \, dv$$

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

$$\langle v^2 \rangle = a \int_{0}^{\infty} v^4 e^{-\alpha v^2} \, dv$$

We apply Feynman’s trick with $n = 2$:

$$\int_{0}^{\infty} e^{-\alpha x^2} x^{2n} \, dx = \frac{1}{2}\left(-\frac{\partial}{\partial \alpha}\right)^n \sqrt{\frac{\pi}{\alpha}}$$
$$-\frac{\partial}{\partial \alpha}\sqrt{\frac{\pi}{\alpha}} = \frac{1}{2}\sqrt{\pi}\,\alpha^{-3/2}$$
$$\left(-\frac{\partial}{\partial \alpha}\right)^2\sqrt{\frac{\pi}{\alpha}} = -\frac{\partial}{\partial \alpha}\left[\frac{1}{2}\sqrt{\pi}\,\alpha^{-3/2}\right] = \frac{3}{4}\sqrt{\pi}\,\alpha^{-5/2}$$

Hence:

$$\int_{0}^{\infty} v^4 e^{-\alpha v^2} \, dv = \frac{3}{8}\sqrt{\pi}\,\alpha^{-5/2}$$

Substituting back with $\alpha = \frac{m}{2k_B T}$:

$$\langle v^2 \rangle = 4\pi\left(\frac{m}{2\pi k_B T}\right)^{3/2} \cdot \frac{3}{8}\sqrt{\pi} \left(\frac{m}{2k_B T}\right)^{-5/2}$$
$$= \frac{3\pi^{3/2}}{2} \cdot \frac{m^{3/2}}{(2\pi)^{3/2}(k_B T)^{3/2}} \cdot \frac{(2k_B T)^{5/2}}{m^{5/2}} = \frac{3}{2} \cdot 2 \cdot \frac{k_B T}{m} = \frac{3k_B T}{m}$$

Therefore:

$$\langle E_k \rangle = \frac{1}{2}m\langle v^2 \rangle = \frac{1}{2}m \cdot \frac{3k_B T}{m} = \boxed{\frac{3}{2}k_B T}$$

This lets us define the RMS (root mean square) speed:

$$v_\text{rms} = \sqrt{\langle v^2 \rangle} = \sqrt{\frac{3k_B T}{m}} = \sqrt{\frac{3}{2}}\, v_\text{most probable}$$

Summarising the three characteristic speeds of the distribution:

$$v_\text{most probable} = \sqrt{\frac{2k_B T}{m}} < \langle v \rangle = \sqrt{\frac{8k_B T}{\pi m}} < v_\text{rms} = \sqrt{\frac{3k_B T}{m}}$$

← Previous lecture · Module · Next lecture →

Retrieved from "https://notes.ashstewart.uk/index.php?title=Year_1_Sem_2/LC_Temperature_and_Matter/Lecture_18&oldid=418"