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)

Taking thermo atm, and there is zero mention of vibrational freedom as a new dof. Its treated as x,y,z translational freedom. This means the max dof for any molecule should be x,y,z,rx,ry,rz (in euclidian space) giving 6 degrees. If the functions used are using a different system (noneuclidian space for example) this number may vary and perhaps vibrational has a different meaning, but in euclidian space, it is translation and rotation only.

HWE Comment: The vibrational mode of a diatomic molecule incorporates TWO degrees of freedom, one being kinetic energy of the oscillation, the other the harmonic potential energy analogous to that of a spring (which the potential energy we’ll closely resembles). The vibrational state(s) is/are excited only at high temperatures - and, depending on the molecular binding energy, the molecule may dissociate before they are fully excited. A diatomic molecule thus displays three dof at lot T, five at intermediate T, and possibly seven at high T. The transition temps depend on the molecule - the two atomic masses, the details of the PE well shape, etc.

The non-allowance of the rotation about the axis along the molecular bond in a diatomic molecule is rationalized in the simplest model by the fact that the atoms are represented by “point” particles. Points do not rotate. In fact, of course, atoms have finite, but very small, sizes. But quantum mechanics demands that the rotational KE be quantized, and this energy is inversely proportional to the rotational inertia! This rotational inertia of extremely small atoms located ON THE ROTATIONAL AXIS is near zero, which means that the first excited rotational energy is extremely large - almost always above the molecule’s dissociation energy.

END HWE COMMENT — Preceding unsigned comment added by 72.184.140.249 (talk) 02:28, 25 April 2019 (UTC)

No citation needed
Who put the "citation needed" mark on If the time evolution of the system is deterministic, where the state at one instant uniquely determines its past and future position and velocity as a function of time, such a system has six degrees of freedom.[citation needed] ? This is an easy calculation like 3+3=6, you need no citation to add up x, y, z, vx, vy & vz. Those are just the positions and velocities in 3 space dimensions, that is no big deal one would need an extra citation for. --Yukterez (talk) 02:50, 14 February 2016 (UTC)

The illustration for degrees of freedom
The illustration describing the degrees of freedom of a diatomic molecule, is confusing and the axes that describe rotation are incorrect. 91.155.192.63 (talk) 15:03, 6 November 2008 (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:
 * $$Ec = \frac{1}{2} \frac{m}{2}(V-v/2)^2+\frac{m}{2}(V+v/2)^2 $$
 * $$Ec = \frac{1}{2} m V^2+\frac{1}{2} m \frac{v^2}{4} $$
 * 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)

If Zargulon is right then just about every text book I can find is wrong. A good example for diatomic is laid out on page 60-61 of Tipler, P.A. "Modern Physics," 1992, (Worth: New York) ISBN 0-87901-088-6. Furthermore, quatitaive experimental data for H2 bear out that that there are 7 degrees of freedom total for diatomic molecules.

I have been struggling with this as well - it seems that derivations commonly presented with the EPT contradict the EPT. For instance, many sources explain that a diatomic gas has four quadratic energy types: translational, rotational, vibrational kinetic and vibrational potential. These are presented as contributing energies 3*RT/2 (trans.), 2*RT/2 (rot.), 1*RT/2 (vib.kin.) and 1*RT/2 (vib.pot.) to the total internal energy of the molecule/system (thus 7*RT/2). Since the EPT asserts that RT/2 is contributed for each quadratic D.F., I'm tempted to conclude this diatomic molecule has 7 degrees of freedom. But if the total degrees of freedom are given by DF = 3N, then for a diatomic shouldn't there be only 6 degrees of freedom? As I see it, the EPT predicts 7*RT/2 (RT/2 for each DF, 7 DFs by careful analysis of the molecule). But a straightforward application of the assumptions DF = 3N, E = DF*RT/2 yield an energy of 6*RT/2. If the internal energy of well-behaved diatomic molecules is experimentally verified to be around 7*RT/2 (at high temp to include vibrational contribution), validating the EPT, what does this say about the assumption DF = 3N? Can a molecule have more degrees of freedom than 3N? As you good folks have detailed, the discrepancy seems to lie in the vibrational degrees of freedom. The expectation DF = 3N is obeyed if the vibrational contribution is treated as a single quadratic term (contributing RT/2 per DF), with no distinction between kinetic and potential (under Zargon's mutual dependence assumption). However, if kinetic and potential vibrational energies are treated as independent (under ThorinMuglindir's statistical independence assumption), then for every vibrational degree of freedom, RT is contributed rather than RT/2. The vibrational degrees of freedom are effectively being doubled (contradicting DF = 3N), or if the DFs are not doubled then the energy being contributed per DF is RT (contradicting the EPT's assertion RT/2 per DF). If the solution really does depend on whether the molecule is stationary or in motion, then most texts seem to be assuming the molecules are not stationary (in agreement with ThorinMuglindir and doubling the vibrational DFs). I can certainly live with the assumption of molecular motion, but then isn't DFsys = 3N misleading and not general enough for model being considered? Particleinabox (talk) 01:09, 12 December 2016 (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)

Translation link
The link to translation leads to the the 'translation' refering to languages... so we have to manually go the the disambiguation page to get the physics related article..... somebody who knows how to, please fix it.


 * What you need to do is to find translation in the code and change it translation.
 * To explain: text you would like to appear as a link change to text you would like to appear as a link Rex the first 18:30, 5 April 2006 (UTC)

Que-when an ideal diatomic gas is heated at constant pressure,fraction of heat energy supplied which increase the internal energy of gas is?? (a)2/5   (b)3/5    (c)3/7   (D)5/7  —Preceding unsigned comment added by 122.173.66.178 (talk) 11:05, 13 February 2010 (UTC)

Degrees of freedom of a system of N independent particles
The article states that it's equal to 3N - 1, but shouldn't be 3N? One for each coordinate in the 3D space? Juanlu001 (talk) 15:52, 14 February 2012 (UTC)

AFAIK N particles require 3N DOF to describe their positions. Changed the article. — Preceding unsigned comment added by Myrikhan (talk • contribs) 02:58, 18 February 2012 (UTC)

History
How old is the concept of degrees of freedom? In his treatise on Heat, Planck used the term thermodynamic probability (an integer) when he seems to mean degrees of freedom (page 120, Theory of Heat, Masius translation). If the term degrees of freedom had currency then, one would think he would have used that term.Entropier (talk) 01:05, 20 March 2013 (UTC)

In statistics, the term wasn't used until 1922, in Ronald Fisher's work on chi-squares ("On the Interpretation of χ2 from Contingency Tables"). That was 8 years after Planck's Theory of Heat. Maybe the term spilled over from there? Mjsmith5 (talk) 13:31, 26 October 2015 (UTC)

Error in diagram
The diagram File:Degrees_of_freedom_(diatomic_molecule).png is incorrect; specifically the top of the three. It suggests three left-right degrees of freedom: two individual particle vibrational degrees, and one translational (centre of mass), whereas these are dependent and only constitute two degrees of freedom. I would suggest trimming this part from the diagram. — Quondum 16:35, 7 August 2013 (UTC)

Scalar number
In the definition it is said that DOF is a scalar number. Shouldn't it be natural number? Or are there cases I'm not aware of? Strasburger (talk) 11:23, 10 March 2014 (UTC)

DEFINITION: For clarity please explain the following statement which appears in the first paragraph of the Definition Section: "On the other hand, a system with an extended object that may rotate or vibrate can have more than six degrees of freedom. A force on the particle that depends only upon time and the particle's position and velocity fits this description." It is not clear exactly which description this fits, although I'd assume it's the more than six degree of freedom system, however if this assumption is correct ... then it would be very helpful if there was more explanation as to how a system with only time, position and velocity fits the more than six degree of freedom criteria. Thanks — Preceding unsigned comment added by 97.125.80.108 (talk) 16:59, 7 June 2015 (UTC)

Should this be renamed
Should this article be renamed as "Degrees of freedom (thermodynamics)". I think that would be a more accurate rubric. This article is clearly about the meaning of degrees of freedom in thermodynamics and kinetic gas theory. Moreover, there is a distinct article with name "Degrees of freedom (mechanics)", but also mechanics is a branch of physics. Indeed, degrees of freedom are the same thing both in mechanics and in thermodynamics, and their role in thermodynamis is just one example of what they mean in statistics, but there are clearly practical reasons to handle them distinctly in these fields. -KLS (talk) 08:20, 3 November 2015 (UTC)


 * I completely agree. Thermodynamics is indeed an overlap between physics and chemistry, but "degrees of freedom" in this context is specific to thermodynamics and not relevant to the myriad other subfields contained under "physics and chemistry".Particleinabox (talk) 01:13, 12 December 2016 (UTC)

Citation?
"So, if the time evolution of the system is deterministic (where the state at one instant uniquely determines its past and future position and velocity as a function of time), such a system has six degrees of freedom." Is this statement obvious, and if so, how so? If not, then we should remove if it unless it is backed by a good source. SpiralSource (talk) 20:15, 7 April 2024 (UTC)