Talk:Degrees of freedom (physics and chemistry)

This article seems wrong in asserting that translation consists of one degree of freedom — it actually consists of 3 (in 3-dimensional space). Likewise rotation consists of 2 degrees of freedom. The article also asserts that each degree of freedom corresponds to an average energy of kT — but it's actually kT/2 (see Equipartition theorem). I'll leave the revision to someone versed in physics, but added a warning for the mean while. --Tromer 20:56, 15 October 2005 (UTC)

The previous version didn't actually explicitly say that translation was one DOF in itself, but it seemed like that's what he meant so I edited it and added a list of all the DOF for the system of the two-atom molecule, as well as an illustration. I actually had a lecture in exactly this example today, except we used a oxygen molecule instead, and the lecturer explained after being asked about it that there is only one degree of freedom in the rotation. I think it's a littlebit strange myself, but I guess he's right. I fixed the kT/2 after reading this discussion. I'll leave for someone to proof-read it and remove the warning. --josteinaj 23:30, 22 October 2005 (UTC)

If one considers rotation in Cartesian coordiantes, then it seems as if rotation has two DOF, as a rotation about the z-axis means a change in x and y coordiantes. However, examined in polar coordinates, this rotation is just changing the phi coordinate, i.e. one DOF. Since both lines of reasoning follow the same logic, and since the second is the simplest answer, I assert that it (the second answer, one DOF for rotation) is the best answer. Moreover, DOF are based on the minimum number of degrees required to represent a change. Again, in the case of rotation, this is one DOF. -bbrueggert 15:33, 25 October, 2005

There are 2 rotational degrees of freedom. The total number of degrees of freedom is therefore 5 (below the temperature at which the vibrational mode is excited) or 6 (above vibrational excitation temperature). So the molar specific heat capacity of a diatomic molecule is either 5/2 R or 3 R. In fact it's not quite that simple, because the vibrational mode excitation isn't a completely sharp transition... Zargulon 22:00, 29 October 2005 (UTC)

QUADRATIC degrees of freedom

OK here's my take on all this: First you only have an energy of kT/2 per independent degree of freedom if the degree of freedom is a quadratic degree of freedom, and in the classical limit. The two restrictions are important.

The amount of internal energy per independent quadratic d.o.f is:

${\frac {1}{2}}\,k_{B}\,T$ (I can demonstrate that in the article if you wish)

Monoatomic ideal gas

For a monoatomic ideal gas of N particles you have 6N degrees of freedom (all are translational). In the classical limit, a micro-state is given by the positions and the impusions of all particles. In 3D, that's 6 d.o.fs for each atom, or 6N in total.

For each atom, the three degree of freedom corresponding to position have no energy attached to them (this is an ideal gas). They do not contribute.

For each atom, the three degrees of freedom corresponding to momentum have a quadratic energy attached to them: that's the kinetic energy $p^{2}/2m$.

We thus have 3N quadratic d.o.fs.

In the classical limit, the internal energy of an ideal gas is thus:

$U={\frac {3}{2}}\,N\,k_{B}\,T$

diatomic ideal gas

We still have 3 translational d.o.fs per moecule, like for the monoatomic ideal gas.

However we also have:

• For each molecule, 4 degrees of freedom corresponding to the rotational modes of the molecule. We can describe the orientation of the molecule by 2 angles, which are the equivalent of a position, and we can describe the rotational speeds of the molecule by two angular momentums, which are the equivalent of momentums. Molecules are free to rotate in space, so that no energy is attached to the angular d.o.fs. The rotational kinetic energy attached to each angular momentum is proportional to the square of of the 2 angular moementums, which means we have 2 quadratic rotational d.o.fs per molecule.
• For each molecule, the vibrational mode is linked to 2 degrees of freedom. One describes the linear distance between the two atoms (=position), the other describes the speed of vibration (=momentum). This time, all 2 are quadratic d.o.fs, because kinetic energy is quadratic as usual, and the valence link between the two atoms can be quite adequately described as having a linear elastic response, which means a quadratic potential energy. This means 2 vibrational quadratic rotational d.o.f per molecule.

The total number of quadratic d.o.fs per molecule is 3+2+2 = 7.

In the classical limit, we have:

$U={\frac {7}{2}}N\,k_{B}\,T$

Now this works only in the classical limit, which explains why you have 3/2 which becomes 5/2 at some temperature which then only becomes 7/2 at a higher temperature. The reason is that translational d.o.fs most generally reach the classical limit at a lower temperature than rotational degrees of freedom, which in turn reach the classical limit at a lower temperature than vibrational degrees of freedom. I will go back to this in more details later.

This decisive explanation was brought to you by ThorinMuglindir 13:34, 31 October 2005 (UTC)

who's the strongest?

With my contribution the factual accuracy of this article is not going to be disputed for long (and yes, I have well deserved bragging a bit about it). Let's just put a redlink to attract those physics and mathematics editors who are after my edits. Hopefully this will quicken the process.192.54.193.37 13:44, 31 October 2005 (UTC). Of couse that was me, I just couldn't keep logged in, for some reasonThorinMuglindir 13:50, 31 October 2005 (UTC)

debate: one or two vibrational dofs?

There is one vibrational quadratic degree of freedom in the diatomic molecule, so the high temperature limit for the molar heat capacity is 3R. Zargulon 18:24, 31 October 2005 (UTC)

I don't think so. There is one quadratic degree of freedom linked to the kinetic energy, as usual, but then there is another that is linked to the energetic cost of pulling on the link between the two atoms. I'll try to break it down completely when I have time.ThorinMuglindir 19:20, 31 October 2005 (UTC)

Don't waste your time.. there really is only one vibrational degree of freedom. Zargulon 23:11, 31 October 2005 (UTC)

there are two quadratic vibrational degrees of freedom. The momentum (or equivqlently the speed at which the molecule vibrates is also a degree of freedom. Let us define the following vectors:

• v1 the speed vector of atom 1,
• v2 the speed of atom 2,
• V the speed of the center of mass of the molecule,
• v the derivative of the vector joining atom 1 and 2

We have:

• v1 = V - v/2
• v2 = V +v/2

So that the kintic energy of the molecule is:

$Ec={\frac {1}{2}}{\frac {m}{2}}(V-v/2)^{2}+{\frac {m}{2}}(V+v/2)^{2}$

$Ec={\frac {1}{2}}mV^{2}+{\frac {1}{2}}m{\frac {v^{2}}{4}}$

The 3 components of V are the 3 quadratic translational dofs. . The 2 components of v perpendicular to V are the 2 quadratic rotational dofs. The component of v that is paralell to V is 1 of the 2 quadratic vibrational dofs. The potential energy associated to the bond is the other vibrational dof, and it can be considered quadratic to a very good approximation (7th and last dof).

Note that in classical stat mech, the microstate of a 1D oscillator is described by a position and a speed. Those are quadratic dofs

that was my breakdown. It justifies the expression of the energy of the molecule at the start of the article.ThorinMuglindir 21:11, 1 November 2005 (UTC)

The two vibrational terms which Thorin has identified are mutually dependent through the oscillator equation of motion. There is one vibrational degree of freedom, making 6 in total. Zargulon 23:08, 1 November 2005 (UTC)

Statistical mechanics don't try to solve the equation of motion. These two degrees of freedom are characterized as independent because, as the system thermally fluctuates between its micro-states, the quantities $l$ and ${\dot {l}}$ are statistically independent. In the ensemble average that yields internal energy, they contribute as indedenpent terms, kT/2 each, or kT in total for a vibrational mode.ThorinMuglindir 01:55, 2 November 2005 (UTC)

This should solve the disagreement. Let us consider a system of N harmonic oscillators that fluctuate thermally (in 3D). For me, with 3 vibrational modes accounting for 6 quadratic independent dofs per harmonic oscillator, in the classical limit, the internal energy of the system is 3NkT = 3nRT. Zargulon, what is the value of this internal energy for you?ThorinMuglindir 01:55, 2 November 2005 (UTC)

It depends on whether the oscillators are stationary.. if not, then you are right. Zargulon 07:53, 2 November 2005 (UTC)

OK, you win Zargulon 08:22, 2 November 2005 (UTC)

Suggest moving material to: equipartition theorem

Can I suggest that it would be useful to have a separate article on the equipartition of energy? IMO it is an important enough idea that it ought to have an article of its own.

Then, can I suggest that most of the discussion of "independent" degrees of freedom would be best moved to that article, with only a 5-line summary and a pointer remaining here?

I think that would also make this article a lot more focussed and less intimidating. Jheald 10:43, 7 November 2005 (UTC)

Aha. I see the article does exist, as equipartition theorem. Can I suggest most of the thermodynamic material be moved there. Jheald 11:08, 7 November 2005 (UTC)

by transfer that will make the equipartion theorem article "intimidating." And you would have to redefine degrees of freedom over there. Besides there is still something that lacks to this article, it's the description of the breakdown of set of quadratic degrees of freedom to independent degrees of freedom. This is actually my main objective in this article, I have just been demonstrating the equipartition theorem passing by because I had all the tools necessary to do it at this stage. I unfortunately have no time to start that until next week.ThorinMuglindir 22:14, 7 November 2005 (UTC)

I object the change mostly because the tools developped here are not only useful for the equipartition theorem. It's useful in the quantum case, and it's also useful for (quantum and classical) mechanics. For example phonons in solids are evidenced by this breakdown of dofs I am speaking about. If you want to quanticize a system of coupled harmonic oscillators, like phonons, what you do is you first break them down into independent dofs, then do the quantum mech part on the set of independent harmonic oscillators that you get. So that at some stage this article could also be moved to the phonon article, or to the quantum harmonic oscillator article that there must be somewhere.ThorinMuglindir 22:14, 7 November 2005 (UTC)