Let $N_1$ be number of particles in volume $V_1$ with momenta and coordinates $(p_1, q_1)$ and $N_2$ be particles with momenta and coordinates $(p_2, q_2)$ in $V_2$, if $E_1, E_2$ be the energies of subsystems satisfying $E< E_1 + E_2 < E + 2 \Delta$ with $\Delta << E$.

My book says that

Obviously the volume of the region of $\Gamma$ space that corresponds to conditions with total energy lying between $E_1 + E_2$ and $E_1 + E_2 + 2 \Delta$ is $\Gamma(E_1) \times \Gamma(E_2)$

Why is the volume of phase space multiplied? Shouldn't it be added?


Let's consider simpler examples.

Suppose first we have a particle moving in one dimension $q$ withe corresponding canonical momentum $p$. Suppose, additionally, that the particle is constrained to have positions in the interval $[q_0,q_0+Q]$ and momenta in the interval $[p_0, p_0 + P]$. Then what is the volume of the particle's phase space? Well, the phase space is a rectangle in $(q,p)$ space measuring $Q$-by-$P$, so its volume is the product $QP$; the volume for the two degrees of freedom are multiplied to get the total phase volume.

Now consider two particles with the same constraints as the single particle above. Each particle will then be able to move in a square of area $QP$, but that means that the system as a whole is allowed to move in the four-dimensional "hypercube" with volume $(QP)^2$. The mathematical way to think about this is that the total phase space consists of all $4$-tuples $(q_1, q_2, p_1, p_2)$ such that $q_1\in[q_0,q_0+Q]$,$q_2\in[q_0,q_0+Q]$, $p_1\in[p_0, p_0 + P]$, and $p_2\in[p_0, p_0 + P]$. The volume of this region of phase space is $$ \int_{q_0}^{q_0+Q}\int_{q_0}^{q_0+Q}\int_{p_0}^{p_0+P}\int_{p_0}^{p_0+P} dq_1 dq_2 dp_1 dp_2 = Q\cdot Q\cdot P\cdot P $$ So again we see that the phase volumes for independent degrees of freedom multiply.

In general, for $N_1$ particles in subsystem $1$ and $N_2$ particles in subsystem $2$, when the subsystems are independent, the volume of phase space will be an integral over an appropriate region $R$ in the phase space of the total system that splits into a product of integrals over the phase space of each independent subsystem because the phase space of the combined system is the Cartesian product $R=R_1\times R_2$ of the phase spaces for each subsystem; \begin{align} \int_{R=R_1\times R_2} d^{N_1}q_1\,d^{N_2}q_2\,d^{N_1}p_1\,d^{N_2}p_2 = \int_{R_1} d^{N_1}q_1 d^{N_1}p_1 \int_{R_2} d^{N_2}q_2 d^{N_2}p_2 \end{align}

| cite | improve this answer | |
  • $\begingroup$ thanks, i get your point, i was being foolish. to put simply, each particle is assigned $(p,q)$, and the volume of phase space is a hyperspace formed by each independent coordinates for each particle. thanks (y) $\endgroup$ – hasExams Aug 22 '13 at 16:55
  • $\begingroup$ @hasExams Yes, I think your simpler statement might even be more clear than my answer :). $\endgroup$ – joshphysics Aug 22 '13 at 17:00

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.