According to Boltzmann statistics the number of particles $n$ with a certain energy level $E_j$ in case of discrete energy levels is: $$n(E_j) = \frac{N}{Z} \cdot e^{-\frac{E_j}{k_B T}} \cdot g_j$$ Where $Z$ is the partition function, $N$ the total number of particles and $g_j$ the degeneracy of energy level $E_j$.

The energy levels $E_j$ in this case are usually only reserved for the translational energy consisting of 3 dimensional translational degrees of freedom. The formula $n(E_j)$ is then further converted into a continuous equation $f(E)$ with energy being treated continuously, giving the Maxwell-Boltzmann Distribution for gases. Multiplying $f(E)$ with the energy $E$ and dividing it by $N$ and then integrating it over infinity should give the average energy per particle which is $\frac{3}{2}k_BT$.

However, is it possible to treat $E_j$ from the start as an energy level also containing rotational energy and vibrational energy and then convert it into an MB Distribution containing all these degrees of freedom? If so, what would the MB Distributio formula look like and would deriving the average energy in this case yield $\frac{5}{2}k_BT$?

  • 1
    $\begingroup$ You mean the way all statistical mechanics books do to compare the heat capacity of monoatomic to diatomic gases? $\endgroup$
    – Jon Custer
    Commented Jun 27, 2019 at 13:10
  • $\begingroup$ @JonCuster I was wondering how the MB Distribution formula would look like if all degrees of freedom are taken into account instead of just the 3 translational ones. $\endgroup$
    – Phy
    Commented Jun 27, 2019 at 20:01

2 Answers 2


Yes, so long as the temperature is high enough that the quantum discreteness of the energy levels can be ignored.

In classical statistical mechanics, it is possible to prove that any degree of freedom which contributes quadratically to the energy (i.e., kinetic energy, rotational energy, elastic potential energy) will contribute a factor of $\frac12 k T$ to the average energy of the system. The proof requires you to take into account the energy of every possible state of the system, and since we're thinking about things classically, this means that you have to integrate over all of the states of the system.

In quantum statistical mechanics, though, this integral is replaced by a discrete sum. If the quantum states are "close enough" in energy, then this sum is approximately equal to the integral you would perform in the classical case (think of taking a Riemann sum where the intervals are very narrow.) But if the quantum states are widely spaced, then the sum and the integral can give wildly different results (think of taking a Riemann sum where the intervals are very wide.)

The general rule for when the energy states are "close enough" is that the difference in the energy levels must be significantly smaller than the thermal energy scale $kT$. An example of this can be seen in the heat capacity for molecular hydrogen: enter image description here At low temperature, the spacing between the rotational energy states and the vibrational energy states is much greater than $kT$, and so H2 behaves like a monoatomic gas. At room temperature, $kT$ is smaller than the rotational spacing, but still larger than the vibrational spacing, and so the rotational modes contribute an extra $\frac{1}{2} kT$ each to the heat capacity. And at even higher temperatures, $kT$ is large enough that the vibrational modes start to "turn on" as well. Unfortunately, H2 dissociates into atomic hydrogen before these vibrational modes contribute their full $k T$ ($\frac{1}{2}kT$ for kinetic energy, $\frac{1}{2}kT$ for potential) to the heat capacity.

Another example of this is the ultraviolet catastrophe. In classical physics, a mode of the electromagnetic field in a cavity contributes quadratically to the energy (since the energy is proportional to the square of the amplitude of the field.) Thus, classically, the energy of each mode should be $\frac{1}{2} kT$, and since there are an infinite number of modes, there should be an infinite amount of electromagnetic energy in a cavity at any finite temperature. The paradox was resolved by Planck, by assuming (ad hoc) that the energy in these vibrational modes could only occur in discrete packets of energy $hf$; so the highest-frequency modes have a wide spacing compared to $kT$ and don't behave classically.

  • $\begingroup$ Thanks for your extensive explanation. My question is more what the formula for the MB Distribution would look like if all these degrees of freedom, including rotational and vibrational modes, are taken into account. $\endgroup$
    – Phy
    Commented Jun 27, 2019 at 18:59

Yes, it is perfectly valid to combine translational and rotational degrees of freedom in a system and get the corresponding distribution of states.

The case you are describing, the diatomic gas with translational, rotational and vibrational DoF, has the following energy:

$E = E_{tras} + E_{rot} + E_{vib} =\sum_i \frac{p_i^2}{2m} + \frac{\hbar^2}{2I} j_i(j_i+1)+\hbar \omega (n_i+\frac1{2})$.

You can identify the three terms as the kinetic energy, rotational energy and vibrational energy. Notice that we are using quantum mechanics formalism, which is the formalism that makes the correct precision.

The Maxwell Boltzmann distribution is $P(E_i) = \frac1{Z} e^{-\beta E_i}$, whit the energies written above.

Since the three terms of the energy is independent, we can factorize the partition function into three contributions:

$Z = Z_{tras}Z_{rot}Z_{vib},$

$Z_{tras} = \int \frac{d^{3N} p d^{3N} q}{h^{3N}}e^{-\sum_i \beta p_i^2 /2m}=\left(\frac{V}{\lambda^3}\right)^N$,

$Z_{rot} = \sum_{\{j_i\}} (2j+1)e^{- \sum_i \beta \frac{\hbar^2}{2I} j_i(j_i+1)} $,

$Z_{vib} =\sum_{\{j_i\}} e^{- \sum_i \beta\hbar\omega (n_i + \frac1{2})}= \left(\frac1{2\sinh(\beta \hbar \omega/2 )} \right)^N$.

Notice that we have included the degeneracy of the different rotational states, $g_j = 2j+1$. You can check that, for sufficiently high temperature, $Z_{rot} \approx \left(\frac{2I}{\beta \hbar \omega}\right)^N$ and $Z_{vib} \approx \left(1/\beta \hbar \omega \right)^N$. Thus, multiplying the three partition functions and doing some algebra we get:

$C_v = - \frac{\partial}{\partial T}\left(\frac{\partial \log Z}{\partial \beta}\right) = \frac{7}{2} Nk$.

So as you see, combining discrete and continuous degrees of freedom is perfectly valid. Notice that we do not have the factor 5/2 that you mention, but a factor 7/2. This factor 5/2 arises when considering only translational and rotational DoF. In other words, to find $C_v = 5/2kT $, you must tune the temperature so that the rotational degrees of freedom get excited but the vibrational ones don't. In other words, you must take $T\to\infty$ in $Z_{rot}$ but $T\to 0$ in $Z_{vib}$.

I hope this clarifies yous question.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.