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


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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