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Fixed minor grammar error.
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Paul J. Gans
Paul J. Gans
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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 isare 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.

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 is 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.

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.

Source Link
Paul J. Gans
Paul J. Gans
  • 1.4k
  • 7
  • 5

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 is 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.