# Why diatomic molecules (ideal gas) are 1-d oscillators.

1. May 2, 2010

### kof9595995

I think I did understand this once but now I am confused. If we choose the center-of-mass frame for a diatomic molecule, it also obeys the force law F=-kr, where r=(x,y,z), so why isn't it a 3-d harmonic oscillator, like an atom in solid? I know it may have something to do with the fact that gas molecules are free to rotate, but still I can't see the essential difference.

2. May 2, 2010

### Feldoh

I'm just taking a guess here but it might be because the force only acts along a line joining the two atoms.

It might be a combination of x,y,z but it's always the same x,y,z combination. Effectively if you rotated the coordinate axes you could rotate in such a way that the atoms would only move along a single direction.

3. May 2, 2010

### kanato

^ yes, that's correct. In order to have an oscillator, you need to have a restoring force which opposes the motion. For motion along the axis connecting the atoms, this is the case. But for motion of the atoms perpendicular to the molecular axis, the force (which acts along the molecular axis) is a central force and will result in the atoms rotating around each other instead of an oscillation. For an atom in a solid, just about any displacement will result in a restoring force somehow, so you always get 3 vibrational modes for each atom in a solid.

4. May 2, 2010

### DrDu

Even in case of a (finite) solid you don't have 3N degrees of approximately harmonic oscillators but 3N-6 (three translations and 3 rotations of the center of mass). In a diatomic molecule there are 3N-5 degrees of freedom as the orientation of a linear molecule is specified already by 2 angles and not 3 as in the general case.

5. May 2, 2010

### kof9595995

I don't get it. Can't we model the diatomic molecule as two masses connected by a spring? If so, isn't it a 3-d oscillator since the spring can be oriented along any direction in space?

6. May 2, 2010

Why?

7. May 3, 2010

### DrDu

It is a one-dimensional oscillator because the vibration depends only on the change of one co-ordinate, namely the interatomic distance r.
You may see it also in another way. Assume the positions of the two atoms (assumed to have equal weight for simplicity) at equilibrium distance being given in a Cartesian co-ordinate system. You can now obtain the force matrix by differentiating the potential energy of the system twice with respect to the Cartesian co-ordinates of the atoms. Upon diagonalization of the force matrix you will find that it has 5 vanishing eigenvalues which correspond to an infinitesimal translation or rotation of the whole molecule. The remaining non-zero eigenvalue is just the force constant of the one-dimensional oscillator.

8. May 3, 2010

### kof9595995

Emm, then why is a solid atom different? why can't we do the same for solid atoms?

9. May 3, 2010

### DrDu

As I said, a solid is not really different and you can perform the same analysis. In case of a solid (or a non-linear molecule), you will get 6 vanishing eigenvalues corresponding to 3 infinitesimal translations and 3 rotations. In the case of a linear molecule, there will only be two eigenvalues corresponding to rotations.

10. May 3, 2010

### kof9595995

So in fact we can treat solid atoms as 1-d oscillators?

11. May 3, 2010

### DrDu

No, you can treat them as (3N-6)/N dimensional oscillators if you are really up to.

12. May 3, 2010

### D H

Staff Emeritus
This is incorrect.

A diatomic molecule modeled as atoms connected by a spring has seven degrees of freedom: 3 for translation, 2 for rotation, and 2 more for vibration (1 each for kinetic and potential energy). An isolated gas comprising N such translating/rotating/vibrating diatomic molecules has 7N-5 degrees of freedom. That -5 can be safely ignored. The kinetic theory of gases treats a translating/rotating/vibrating diatomic gas as having 7 degrees of freedom per molecule.

There are two simpler models for a diatomic molecule. One is to replace the spring with a rigid rod. The vibrational modes do not exist in this model. With this model a diatomic gas has 5 degrees of freedom per molecule. An even simpler model results by shrinking the length of the rod to zero. There is no difference between a diatomic gas and a monatomic gas in this model.

So which model is correct? In a sense, all of them. Many diatomic gases have $C_v \approx 3/2 R$ at low temperatures, $5/2 R$ at moderate temperatures, and $7/2 R$ at high temperatures. It is as if the gas operates with the simple model at low temperatures, then with the rigid connector model at moderate temperatures, and then with the spring model at high temperatures. In another sense, none of them is a better answer. This mode switching is a rather ad hoc explanation. Moreover, the transitions between low, moderate, and high temperature behavior is smooth rather than a step functions.

The spring model has a central force between the molecules. It is directed along the line connecting the molecules. A 3d oscillator would require forces normal to this line.

13. May 3, 2010

### Count Iblis

Let me just say that what D.H is writing here is completely wrong. "Degrees of freedom" are defined differently, at least in all the textbooks on Statisical Mechanics I have seen.

You don't have a "degree of freedom for potential energy" at all. You can do a simple accounting of all the kinematic degrees of freedom exactly as Dr. Du has pointed out to compute e.g. how many vibrational degrees of freedom there are. The fact that each of these degrees of freedom will contribute k T and not 1/2 k T in the high temperature limit doesn't figure in here.

14. May 3, 2010

### D H

Staff Emeritus
The kinetic theory explanation is a simple freshman-level physics description or mid-level thermo course, Count, rather than the description one runs into in an upper level undergrad / lower-level graduate course on statistical physics.

I checked my freshman physics book before writing my response. You can also something very similar to what I wrote in the wikipedia article on heat capacity, http://en.wikipedia.org/wiki/Heat_capacity. In particular, see the sections entitled "Degrees of freedom", "Example of a changing specific heat capacity in a diatomic gas" and "Diatomic gas".

The fact is that on a simplistic level many diatomic gases do go through exactly those modes of operation described above. At low temperatures Cv is about 3/2R, the same result as obtained by the basic kinetic theory. Raise the temperature and Cv will remain more or less constant for a while but then will start rising. It will level out at about 5/2 R and will once again remain more or less constant with increasing temperature for a while. Then it will start rising toward 7/2 R. Many gases dissociate before reaching 7/2 R (hydrogen, for example).

Regarding hydrogen: Below 75 K, Cv for hydrogen is about 3/2 R. It is about 5/2 R between 250 and 750 K. Above 750 K Cv rises towards 7/2 R as mentioned above.

Last edited: May 3, 2010
15. May 3, 2010

### Gokul43201

Staff Emeritus
While there is some genuine disagreement here on counting the number of vibrational modes, there also appears to be some miscommunication.

For instance, Dr Du's equation: DoF = 3N - 5, is presumably for counting the vibrational degrees of freedom for a linear molecule with N atoms (and not for counting the total degrees of freedom for a diatomic gas with N molecules).

16. May 3, 2010

### Count Iblis

Ok, but the rationale for only looking at the "kinematic" degrees of freedom should be clear.

If you have a two particle system that can form a bound state, so there exists an interaction potential, then you always have that potential energy term in the Hamiltonian. So, at very high temperatures you have two almost non-interacting particles, but in reality there is still some interaction between them.

The whole point of the notion of DoF is that you want to consider how many independent variables you need to fully describe the system. For two particles you have 6 coordinates to consider. You can then break that down in different ways, like:

6 = 1 (vibration) + 3 (translation) + 2 (rotation)

But note here that you can consider two particles that are far away from each other so that you can ignore any interaction and then describe the motion at any time in terms of the motion of the center of mass, an instantaneous rotation (which conserves the distance between the particles) and a motion that changes the distance between the two particles.

17. May 3, 2010

### D H

Staff Emeritus
An argument to the contrary: Looking only at kinematics to describe the behavior of a high temperature diatomic gas requires some mighty wide wavings of the hand to explain why the constant volume specific heat for such a gas is 7/2 R, and even mightier hand waving to make this jibe with the equipartition theorem. For example, this wonderful (wonderfully bad) bit from the previously cited wikipedia article,
Each rotational and translational degree of freedom will contribute R/2 in the total molar heat capacity of the gas. Each vibrational mode will contribute R to the total molar heat capacity, however. This is because for each vibrational mode, there is a potential and kinetic energy component. Both the potential and kinetic components will contribute R/2 to the total molar heat capacity of the gas. Therefore, a diatomic molecule would be expected to have a molar constant-volume heat capacity of Cv=3/2R+R+R=7/2R.​
Umm, wait. I thought that we are only supposed to be looking at kinetic energy! So why then does this article talk about potential energy at all?

Saying there are six degrees of freedom does not jibe with the equipartition theorem and with experimental results that gases at high temperatures have a constant volume specific heat of about 7/2R. On the other hand, counting vibrational potential and kinetic energy as two degree of freedom does jibe with the equipartition theorem. Seven degrees of freedom (three translational + two rotational + two vibrational), each contributing R/2 to Cv results in Cv=7/2R. Another way to look at vibration is that vibrational phase and magnitude are independent degrees of freedom.

I tried looking for a plot of a specific heat versus temperature for a diatomic gas. I found one, and right on this very site. See [post=2696738]this post[/post]. Note: In that post, Stewart is, for the most part, talking about the constant pressure specific heat Cp rather than Cv. Hence the multipliers of 5/2, 7/2, and 9/2 in that post rather than the 3/2, 5/2, 7/2 multipliers for Cv.

18. May 3, 2010

### kof9595995

To make things clearer, i need to ask this question:
1.If i connect a mass with a spring, to a fixed point in 3-D space, is it a 3-D oscillator?
2.If so, and if we choose the center-of-mass frame of a linear diatomic molecule, what is the difference with 1. ？

19. May 3, 2010

### D H

Staff Emeritus
No. Whatever the disagreements are between some of us on vibrational modes, we are clear on vibrations being central forces.

20. May 3, 2010

### DrDu

Ok, I now see what you mean. In a 2 or 3 dimensional oscillator, the equilibrium position about e.g. an atom in a lattice oscillates is zero, while the atoms in the molecule oscillate about a non-zero equilibrium value r. Hence in one case the potential has a single minimum while in the other, the potential has the so called mexican hat shape where there is a whole rim of degenerate minima.