In Mehran Kardar's volume 1: Statistical Physics of Particles, he introduces the Maxwell Boltzmann distribution function just after the discussion on the microcanonical ensemble as follows:

The joint probability distribution function for a microstate is $$p(\mu)=\frac{1}{\Omega(E,V,N)} \times \begin{cases}1 & q_i\in [0,V^{1/3}]\ \text{and} \sum p_i^2/2m =E \\ 0 & \text{otherwise} \end{cases} .... (4.27)$$

The above text is for the ideal gas. After this and a few paragaphs on deriving the ideal gas laws, he said this:

The unconditional probability of finding a particle with momentum $\vec p_1$ in the gas can be calculated from the joint PDF in 4.27 by integrating over all the other variables $$p(\vec p_1)=\int d^3q_1 \prod_{i=2}^{N}d^3q_id^3p_i\ p (\{ \vec p_i, \vec q_i \}) = V \frac{\Omega (E-p_1^2/2m,V,N-1)}{\Omega (E,V,N)}$$

I understand how unconditional probabilities are calculated. But I can't understand how he wrote that last expression out of thin air. Obviously, the first $V$ comes out because he integrated $d^3q_i$. But I didn't get how that $\Omega$ thing came out. I am a beginner with these things so it would be great to keep the explanation to beginning graduate level.

Thank you. Any help is appreciated.

  • 2
    $\begingroup$ How exactly has $\Omega$ been defined? I can see more or less what it has to be, but it will be easier to give a helpful answer if we know exactly how Kardar has done it $\endgroup$ Feb 28, 2022 at 18:34

1 Answer 1


Defining $$p(\mu) \equiv \frac{1}{\Omega(E,V,N)} \delta(E-H(P,Q)) \quad, $$ we find from the normalization condition

$$\int \mathrm d^3p_1\, \mathrm d^3p_2\ldots \mathrm d^3 p_N \int \mathrm d^3 q_1\,\mathrm d^3 q_2 \ldots \mathrm d^3 q_N\, p(\mu) = 1 $$ that $$ \Omega(E,V,N) = \Gamma_N(E) \, V^N \quad , $$ where \begin{align} \Gamma_N(E) &\equiv \int \mathrm d^3p_1\, \mathrm d^3p_2\ldots \mathrm d^3 p_N\, \delta\left(E-\sum\limits_{i=1}^N \frac{p_i^2}{2m}\right) \\ V^N&= \int_V \mathrm d^3 q_1\,\mathrm d^3 q_2 \ldots \mathrm d^3 q_N \quad .\end{align} Note that we treat the particles as distinguishable and omitted some constant factors. Of course, we have assumed that $H$ is the Hamiltonian of an ideal gas. Here, $\Gamma_N(E)$ is related to the surface area of a $3N$-dimensional hypersphere with radius $\propto \sqrt E$.

Eventually, this yields $$p(p_1) = \frac{V\, \Gamma_{N-1}\left(E-\frac{p_1^2}{2m}\right)\, V^{N-1}}{\Omega(E,V,N)} = V\, \frac{\Omega\left(E-\frac{p_1^2}{2m},V,N-1\right)}{\Omega(E,V,N)} $$


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