For an ideal gas, the multiplicity of an ideal gas with $N$ molecules in Cartesian space is $$\Omega_{\text{space}}=\Big(\frac{V}{(\Delta x)^3}\Big)^N.$$ This is pretty intuitive, because we are simply dividing the total volume, i.e. $V^N$, by the total number of states (or spaces for states), i.e. ($\Delta x)^{3N}$. However, the multiplicity in the momentum space is $$\Omega_{\text{mom}}=\frac{V_{\text{mom}}}{(\Delta p)^{3N}}.$$

Why do we not take the entire expression to the power of $N$ for the momentum space?


I think this is actually a common misunderstanding of what is the definition of $V_\text{mom}$ for the case of $N$ particles. $V_\text{mom}$ is in fact the surface area of a $3N$-dimensional hypersphere with radius given by $\sqrt{2mE}=p$, where $E$ is the average energy per particle and $p$ is the modulus of the associated momentum in $3N$ dimensions. In other words, as you go from one particle to $N$ particles in the calculation of the multiplicity, the definition of $V_\text{mom}$ includes the $N$. We do not do the same for the real space volume because it is simply more natural to think in terms of the actual volume $V$ in $3D$ without changing its definition only because you now have more particles: the definition of the $V$ is independent of the number of particles $N$; this is not true for the volume $V_\text{mom}$.

Edition: Let me explain a bit more. In the case where $N=1$, we can say that the real volume multiplicity of one particle is $V/V_{particle}$. However, if we have $N$ particles, then we need to consider the number of possible states in real space volume of each particle and do the combinatorics. That's why we have the power $N$.

For the momentum space "volume" (which is actually the hypersurface area of a $3N$-dimensional sphere since most of the particles have energy not very far from $E$,and then most of the possible states in momentum are points of that surface determined by $E$), we have $V_\text{mom} \propto (\sqrt{2mE})^{3N-1}$ (because the area of a hypershpere in $D$ dimensions is proportional to the radius to the power $D-1$). Notice that $p^2=\sum_{i=1}^{N}(p_{ix}^2+p_{iy}^2+p_{iz}^2)$.

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  • $\begingroup$ Are you saying that it should be $V_{\text{mom}}^N$ and not $V^N$? $\endgroup$ – ODP May 7 '16 at 21:08
  • $\begingroup$ No, I am saying that the $N$ that you're looking for in $V_\text{mom}$ is actually already there, inside the definition of $V_\text{mom}$. $\endgroup$ – pscm May 7 '16 at 21:10
  • $\begingroup$ @ODP, please have a look at the new piece of answer I included. I am happy to explain more details if you need. $\endgroup$ – pscm May 7 '16 at 21:37

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