In Blundell and Blundell's Thermal physics book there is a chapter on the explanation of the Boltzmann distribution. It is explained that for canonical ensemble with a reservoir at temperature $T$ in equilibrium with the system, the probability that the system is at energy $\epsilon$ is proportional to: $$ e^{-\frac{\epsilon}{k_B T}} $$ which is called the Boltzmann factor.

Then, when considering velocity distributions in monatomic gases they consider a single particle acting as a system and the rest of the energy as a reservoir, so it should follow the Boltzmann distribution in terms of energy. However, the velocity distribution in a fixed direction itself is said to be proportional to a Boltzmann factor too. This time, since the relevant energy is $E = \frac{1}{2}mv_x^2$ the distribution becomes: $$ g(v_x) = C e^{-\frac{mv_x^2}{2k_B T}}$$ What I do not understand is why the distribution is known to follow this factor. It is supposed to apply to the energy of the particle, but why does it apply to the velocity distribution too?

  • $\begingroup$ I don't understand your last sentence "It is supposed to apply to the energy of the particle, but why does it apply to the energy distribution too?". It is the probability of finding a particle with that energy, or equivalently the distribution of particle energies. If you know the probability of a energy of a single particle, you then know the distribution of all particles' energies. $\endgroup$
    – dllahr
    Jan 1, 2022 at 18:58
  • 1
    $\begingroup$ Sorry, I meant to write velocity, thanks for pointing it out $\endgroup$
    – Johnn.27
    Jan 1, 2022 at 19:43

1 Answer 1


The Boltzmann distribution applies to each component of the energy.

For instance, in your monoatomic gas can be either in the fundamental electronic state or in an excited state an energy $E_{exc}$ above the fundamental one, in addition to a kinetic energy


then the distribution function for an atom in the excited state will be proportional to

$$e^{-\frac {E_{exc}+E_k} {k_B T}}$$

while that in the fundamental will just be

$$e^{-\frac {E_k} {k_B T}}$$

If you ignore the excitation state, and consider only the component of the velocity along $x$, that is, you add all the atoms with a given $v_x$ whatever the degree of excitation or the values of $v_y$ and $v_z$ are, of course then the only relevant part is

$$e^{-\frac {mv_x^2} {2k_B T}}$$

  • $\begingroup$ So, please correct me if I am wrong, I understand that we use the correspondence between energy and speed and thus, since we know what the distribution for a particle with energy 1/2mv^2 looks like this means the exact same distribution applies for a particle with velocity v? (And then put this idea into components x, y, z) $\endgroup$
    – Johnn.27
    Jan 1, 2022 at 19:45
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    $\begingroup$ @Nick.25 Exactly. The exponential of a sum is the product of the exponentials. Each contribution to the energy provides an exponential, if you are only interested in that component. The full distribution is the product of all contribution, the exponential of the total energy. $\endgroup$
    – Alfred
    Jan 1, 2022 at 20:06
  • $\begingroup$ I was having trouble realizing that the distribution for speed v is equivalent to that of energy 1/2 mv^2. Thanks! $\endgroup$
    – Johnn.27
    Jan 1, 2022 at 21:59

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