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Suppose that I have a monatomic gas sample consisting of $N$ atoms (e.g., $N$ argon atoms); thus there are no vibrations or rotations. How many degrees of freedom does the system have?

  • Does the system have $3N$ degrees of freedom? Each of the $N$ atoms can translate in $x$, $y$, and $z$, so $3N$ degrees of freedom seems reasonable. This seems to be supported by this website on molecular dynamics (MD), which states: "If there are $N$ atoms and $N_c$ internal constraints, then the number of degrees of freedom is $N_f = 3N - N_c$." It seems that a monatomic gas such as argon has no internal constraints, so the website seems to be suggesting $3N$ degrees of freedom.
  • Does the system have $3N - 3$ degrees of freedom? Page 64 of Frenkel & Smit's Understanding Molecular Simulation: From Algorithms to Applications (2nd edition) seems to suggest that the system has $3N - 3$ degrees of freedom: "In practice, we would measure the total kinetic energy of the system and divide this by the number of degrees of freedom $N_f$ ($ = 3N - 3$ for a system of $N$ particles with fixed total momentum)." The only thing I am unsure of is whether a monatomic gas in general has fixed total momentum: probably not, unless there is solely center-of-mass motion.
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  • $\begingroup$ First thing that you need to understand is that $N$ is number of atoms in the molecule and not in the whole sample. $\endgroup$
    – Cheeku
    Commented Mar 10, 2013 at 16:14
  • $\begingroup$ @Cheeku So each atom in my monatomic system has only translations (three of them). But I would like to compute, for example, the average kinetic energy of the entire system: $\langle K \rangle = 3N_f kT/2$, in which I think $N_f$ is indeed the number of degrees of freedom of the entire $N$ atoms/molecules, not just one of them. $\endgroup$
    – Andrew
    Commented Mar 10, 2013 at 16:20
  • $\begingroup$ $N_f$ is just the number of molecules. Each of them has the energy $3kT/2$, hence the expression. Degrees of freedom is 3 $\endgroup$
    – Cheeku
    Commented Mar 10, 2013 at 16:21

2 Answers 2

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Assuming that the $N$ argon atoms are being treated as a classical system, then there are 3 degrees of freedom per atom. That assumes that we are neglecting electronic degrees of freedom, which is OK since one needs a fairly high temperature to thermally excite the electrons in argon.

Now if there are $N$ atoms, then there are $3N$ degrees of freedom. No if ands or buts. However, when doing computer simulations it is very common to make the center of mass of the system of atoms stationary. The center of mass can always be separated out from the other coordinates of the system and so can its coordinates and momentum. In this case there are $3N-3$ remaining degrees of freedom.

The reason why this is often done in computer simulations is that what is interesting are the motions inside the gas, and not the bulk motions of the gas as a whole. More importantly, the center of mass will remain fixed and the momentum of the center of mass will be zero. To see this understand that there are no forces acting upon the center of mass.

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Since the gas is monoatomic, it has no internal constraints. It has $3N$ degrees of freedom. I think that the $-3$ part comes from the constraint that the center of mass is at rest. This is a choice that you can make.

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  • $\begingroup$ Thanks for your time. What if the center of mass is not at rest? Then would the system have $3N - 3$ degrees of freedom instead of $3N$? $\endgroup$
    – Andrew
    Commented Mar 10, 2013 at 16:24
  • $\begingroup$ Yes. I must admit that I haven't seen this constraint before, I think it's rather uncommon to assume that. $\endgroup$
    – zonksoft
    Commented Mar 10, 2013 at 16:33

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