# The monoatomic ideal gas

We now choose a specific system in order to do some explicit computations: the monoatomic ideal gas. This is a system composed of free point particles that don't interact and therefore has the following Hamiltonian:

${\mathcal {H}}(\mathbb {P} )=\sum _{i=1}^{N}{\frac {{\vec {p}}_{i}{}^{2}}{2m}}$ The volume occupied by the system in phase space is thus:
{\begin{aligned}\Omega (E,V,N)=\underbrace {\int _{{\vec {q}}_{i}\in V}\prod _{i=1}^{N}d{\vec {q}}_{i}} _{V^{N}}\int \prod _{i=1}^{N}d{\vec {p}}_{i}\delta \left(\sum _{i=1}^{N}{\frac {{\vec {p}}_{i}{}^{2}}{2m}}-E\right)=\\=V^{N}\int \prod _{i=1}^{N}d{\vec {p}}_{i}2m\cdot \delta \left(\sum _{i=1}^{N}{\vec {p}}_{i}{}^{2}-2mE\right)\end{aligned}} where we have used the known property of the Dirac $\delta$ function:
$\delta (\lambda x)={\frac {\delta (x)}{|\lambda |}}\qquad \lambda \in \mathbb {R}$ Now, the sum $\sum _{i}{\vec {p}}_{i}{}^{2}$ is the square of the distance of the point $\mathbb {P}$ from the origin in momentum space; this means that the $\delta$ selects a spherical surface of radius $R={\sqrt {2mE}}$ in momentum space. Therefore:
$\Omega (E,V,N)=V^{N}2m\cdot {\hat {\Omega }}_{3N}(2mE)$ where we have denoted with ${\hat {\Omega }}_{\ell }(R^{2})$ the surface of an $\ell$ -dimensional hypersphere of radius $R$ , namely:
${\hat {\Omega }}_{3N}(R^{2})=\int \prod _{i=1}^{N}d{\vec {p}}_{i}\delta \left(\sum _{i=1}^{N}{\vec {p}}_{i}{}^{2}-R^{2}\right)$ Let us define in general:
${\hat {\Omega }}_{\ell }(R^{2})=\int \delta ({\vec {p}}_{\text{ }}{}^{2}-R^{2})d^{\ell }p$ where ${\vec {p}}$ is an $\ell$ -dimensional vector. Using the properties of the $\delta$ function we can write:
${\hat {\Omega }}_{\ell }(R^{2})=\int {\frac {\delta (|{\vec {p}}|-R)+\delta (|{\vec {p}}|+R)}{2|{\vec {p}}|}}d{\vec {p}}=\int {\frac {\delta (|{\vec {p}}|-R)}{2R}}d{\vec {p}}$ where we have neglected the contribution of $\delta (|{\vec {p}}|+R)$ since $|{\vec {p}}|>0$ . Now, since (as it can be shown in distribution theory) the Dirac $\delta$ is the derivative of the Heaviside step function $\Theta$ , we have:
${\hat {\Omega }}_{\ell }(R^{2})={\frac {1}{2R}}{\frac {d}{dR}}\underbrace {\int \Theta (R-|{\vec {p}}|)d{\vec {p}}} _{V_{\ell }(R)}$ and $V_{\ell }(R)$ is the volume of an $\ell$ -dimensional hypersphere; its computation is shown in the appendix Volume of a hypersphere. In the end, we have:
${\hat {\Omega }}_{\ell }(R^{2})={\frac {1}{2R}}{\frac {\ell R^{\ell -1}\pi ^{\ell /2}}{\Gamma \left({\frac {\ell }{2}}+1\right)}}$ where $\Gamma$ is Euler's gamma function. Therefore:
$\Omega (E,V,N)=V^{N}2m{\frac {1}{2{\sqrt {2mE}}}}{\frac {3N\left({\sqrt {2mE}}\right)^{3N-1}\pi ^{3N/2}}{\Gamma ({\frac {3}{2}}N+1)}}$ which, after some massaging, can be rewritten as:
$\Omega (E,V,N)={\frac {3N}{2E}}V^{N}{\frac {(2\pi mE)^{{\frac {3}{2}}N}}{\Gamma \left({\frac {3}{2}}N+1\right)}}$ If we now compute the probability density in configuration space, i.e. we integrate $\rho$ over the momenta, we get:

$\rho (\mathbb {Q} )=\int d\mathbb {P} \rho (\mathbb {Q} ,\mathbb {P} )=\int d\mathbb {P} {\frac {\delta ({\mathcal {H}}-E)}{\Omega (E,V,N)}}=\int d\mathbb {P} {\frac {1}{\Omega (E,V,N)}}\delta \left({\frac {\mathbb {P} ^{2}}{2m}}-E\right)={\frac {1}{V^{N}}}$ which is intuitively clear since no forces act on the particles and thus they can be found in any point of our system with equal probability. If we now think of our system as a cubic box and mentally divide it into a right and a left half we will have that the probability to find the $i$ -th particle in one of these two halves is $1/2$ :
$p({\vec {q}}_{i}\in V/2)=\int _{{\vec {q}}_{i}\in V/2}d{\vec {q}}_{i}\rho (\mathbb {Q} )={\frac {1}{V^{N}}}V^{N-1}{\frac {V}{2}}={\frac {1}{2}}$ and so the probability that all the particles are in the same half of the system is:
$p(\mathbb {Q} \in V/2)={\frac {1}{2^{N}}}$ This is exactly what we have already seen in The microcanonical ensemble.

1. It could be legitimately argued that strictly speaking such a system will never reach an equilibrium: since the particles don't interact nor collide their velocities will always be equal to the initial ones, whatever they were; for example if in the initial state the kinetic energy of the system is held by only one particle (all the others being still), that particle will always be the only one moving and all the remaining ones will always stay still. We can solve this problem in two ways: we can either suppose that some interaction between the particle exists and such that it is sufficient to establish an equilibrium but weak enough to be neglected, or we can suppose that the particles exchange energy with the walls of the system through the collisions and that the walls "give back" this energy to the other particles when they impact them.
2. Remember that it is defined as:
$\Gamma (z)=\int _{0}^{+\infty }t^{z-1}e^{-t}dt$ and if $n\in \mathbb {Z}$ , then $\Gamma (n)=(n-1)!$ .