I have encountered this old question and since I was bored I figured I use statisical mechanics to derive the relationship $E=\frac{3}{2}kT$ for a single particle in an ideal gas without using outside help. As a disclaimer, it has been a while since I had statistical mechanics courses, so I am a bit rusty. I have encountered some strange results I cannot explain and am curious about.

The problem: Using microcanonical ensemble for a single particle gives $E=(1/2)kT$. Using canonical ensemble for a single particle gives $E=(3/2)kT$.

Using canonical enseble for $N$ particles, the energy per particle $E/N$ is $E/N=(3/2)kT$. Using microcanonical ensemble for $N$ particles, the energy per particles is $E/N=(3/2-1/N)kT$. Clearly the latter goes to $(3/2)kT$ in the $N\rightarrow\infty$ limit.


  • An ideal gas is noninteracting, so there should be no difference between using a single-particle ensemble and a multiparticle ensemble, right? Why is it that the canonical ensemble gives the same results in both cases, but the microcanonical is different?

  • Why do the microcanonical and canonical ensembles give different results, aside from the $N\rightarrow\infty$ limit? In particular, the microcanonical ensemble, while probably the furtherst removed from experiment, is the "purest" ensemble from a theoretical standpoint, the one that comes directly from the postulates of statistical mechanics. Is the formula $E=(3/2)kT$ valid only when $T$ is fixed? Or what?

The calculations I have used are presented below:

Ideal gas using microcanonical ensemble:

Assume we have an ideal gas in a box of volume $V$, at energy $E$ (at an uncertainty $(E,E+\delta E)$), and since the particles of an ideal gas are noninteracting anyways, we consider $N=1$.

The number of microstates is taken to be the phase volume between $E$ and $E+\delta E$, where $\delta E$ is a first-order "infinitesimal".

This can be calculated by calculating $\Omega(E'<E)$ the number of microstates whose energy is less than $E$. This is the following phase volume: $$ \Omega(E'<E)=\int_Vd^3x\int_Bd^3p, $$ where $B$ is the ball in momentum space given by $B=\{(p_x,p_y,p_z):\ p^2<2mE\}$, so $$ \Omega(E'<E)=Vg(2mE)^{3/2}, $$ where $g$ is an irrelevant constant. The number of microstates (of energy $E$) is then $$ \Omega(E,E+\delta E)=\frac{\partial}{\partial E}\Omega(E'<E)\delta E=\frac{3}{2}gV(2mE)^{1/2}2m\delta E=3gVm(2mE)^{1/2}\delta E. $$

The entropy is $S=k\ln\Omega$ so $$ S=k\left[\ln(3gVm\delta E)+\frac{1}{2}\ln(2m)+\frac{1}{2}\ln E\right], $$ the reciprocal of temperature is then $$ \frac{1}{T}=\frac{\partial S}{\partial E}=\frac{k}{2}\frac{1}{E}, $$ from which we get $$ E=\frac{1}{2}kT. $$

This is missing a factor of 3. Using a canonical ensemble on the other hand gives the proper answer:

Ideal gas using canonical ensemble:

We now consider the ideal gas fixed at temperature $T$, at volume $V$ and at particle number $N=1$.

The canonical partition function is $$ Z=V\left(\frac{2m\pi}{\beta}\right)^{3/2}, $$ the expectation value of energy is $$ E=-\frac{\partial}{\partial\beta}\ln Z=\frac{\partial}{\partial\beta}\frac{3}{2}\ln\beta=\frac{3}{2}\frac{1}{\beta}=\frac{3}{2}kT, $$ which is the result I expect.

Now, I have no idea why the microcanonical ensemble gives only the third of the expected result but I considered using an $N$-particle gas instead of a single particle gas. The canonical ensemble gave $E=(3N/2)kT$, so the energy of a single particle is $E/N=(3/2)kT$, the same as before, so I will not do this calculation here. The microcanonical ensemble was different however.

$N$-particle ideal gas using microcanonical ensemble:

The number of microstates whose energy is less than $E$ is now given as $$\Omega(E'<E)=V^Ng(2mE)^{3N/2},$$ where $g$ is a different constant, but still irrelevant. The number of microstates (of energy $E$) is then $$\Omega(E,E+\delta E)=\frac{\partial\Omega(E'<E)}{\partial E}\delta E=3NmgV^N(2mE)^{(3N/2)-1}\delta E,$$ the entropy is $$S=k\ln\Omega=k\left[\left(\frac{3N}{2}-1\right)\ln(2mE)+c\right],$$ the reciprocal temperature is $$\frac{1}{T}=\frac{\partial S}{\partial E}=k\left(\frac{3N}{2}-1\right)\frac{1}{E}, $$ $$ E=\left(\frac{3N}{2}-1\right)kT. $$ The energy for a single particle is then $$ \frac{E}{N}=\left(\frac{3}{2}-\frac{1}{N}\right)kT. $$ In the limit $N\rightarrow\infty$, this reduces to $E=(3/2)kT$.

  • $\begingroup$ The $3/2$ and $1/2$ derive from the degrees of freedom of the system, do they not? It is perfectly okay to have $E = 1/2 kT$ for a 1D ideal gas. It's equivalent to saying that each DoF has $1/2 kT$. $\endgroup$ Commented Jan 9, 2017 at 20:34
  • $\begingroup$ @honeste_vivere That's true but the specification of the system was the same for both ensembles, so they should yield the same answer. More precisely, I used three dimensions for both calculations. $\endgroup$ Commented Jan 9, 2017 at 20:40

1 Answer 1


You cannot use the microcanonical ensemble one particle at a time. The reason is that the energies of the particles are not independent; if one particle has some of the energy, it lowers the total energy available to the other particles. (The canonical ensemble exists to get around this complication.) Moreover, while the particles are "noninteracting" in an ideal sense, there must be some weak interactions in order for the distributions of velocities to thermalize.

As a further illustration of the failure of applying the microcanonical ensemble one particle at a time, note that it predicts that the average energy is in fact the exact energy for every particle. This is clearly not a description of the correct distribution function.


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.