# What are the six degrees of freedom of the atoms in a solid?

A monoatomic ideal gas has heat capacity $C_v=1.5$ which comes from the three translational degrees of freedom. For solids at high temperature, $C_v=3$, implying six degrees of freedom.

What are those six degrees of freedom? I do not know much about how solids work, but I suppose translation and rotation do not contribute. What then? Vibration?

## 4 Answers

For a degree of freedom whose energy is quadratic in just momentum (but flat in position, or flat with hard walls), the average energy classically is $kT/2$. That is the basic equipartition theorem for an ideal gas. However a lesser known result is that a classical degree of freedom with energy quadratic in both momentum and position has an average energy of $kT$. The atoms in a solid are in some sense each in a 3-way harmonic oscillator (this is the Einstein model) and hence one has $3NkT$ energy, i.e. $3Nk$ heat capacity.(†)

To understand this intuitively you should of course derive the equipartition theorem for yourself. But, basically, by having energy also quadratic in position you make the lower energy states less common; not only does the low energy require a small momentum, but also a particular position. By increasing energy, more and more positions become available. In contrast with a flat potential the position can always take on any value and so a low energy state only needs the momentum to be small.

So if you were to imagine each atom in a solid as instead as being inside its own little box with hard walls, then such a model would only give $3Nk/2$ heat capacity.

(†) Okay, actually the atoms are all coupled together however when you look at it this way, you can't so simply talk about the separate contributions of individual atoms anymore. Looking at these whole vibrations gives you phonons and the Debye model. Basically though, all of the atomic harmonic oscillators mix together into various modes, but of course the number of modes remains the same as the original number of individual oscillators. But, each mode is itself a harmonic oscillator so you get the $3Nk$ heat capacity at high temperature.(‡)

(‡) Actually, only $3(N-\frac{1}{2})k$ since three of the collective modes do not oscillate but rather correspond to the linear motion of the whole block of material. So, those three modes each give only $kT/2$.

The six degrees of freedom are indeed, as you supposed, vibrational. Just like there are three translational degrees of freedom, each for one spatial direction, there are two (the number of normal modes) vibrational degrees of freedom per direction. This makes for a total of $3*2=6$ degrees of freedom per atom.

Response to comment; relating normal modes and degrees of freedom

An oscillator can vibrate in many different ways, but they're all a superposition of its normal modes. In this sense, it's somewhat analogous to, for instance, the eigenstates of the hamiltonian in problems in QM. For instance, there are two spin states (up and down) for an electron, and it can only be in a superposition of those two. Similarly, a system with two normal modes can only be in a superposition of these two 'basis vibrations'. Thus, the number of normal modes is equal to the number of degrees of freedom.

• I'm sorry. I don't get what you mean with "normal modes". Doesn't the number of modes depend on the number of particles you have in the system? – The Quantum Physicist Oct 14 '13 at 12:41
• @TheQuantumPhysicist 6 would be the number of degrees of freedom per atom. Is that what you mean? – Andreas Oct 14 '13 at 12:44
• Does this mean that the number of DOF depends on the lattice structure of the solid, so that it isn't the same for all solids? – Andreas Oct 14 '13 at 12:46
• I don't know what the relation is between the number of modes and the number of degrees of freedom. Indeed I find no relation at all. I would need a kind explanation from @Danu – The Quantum Physicist Oct 14 '13 at 12:49
• @TheQuantumPhysicist The number of degrees of freedom of a solid scales with the size of the solid. However, most people talk about the degrees of freedom per atom, which is 6. – Jahan Claes Aug 22 '15 at 22:13

There are 3 directions of vibration: namely the components $x$, $y$ and $z$ and each of these directions has 2 degrees of freedom; one of potential energy and one of kinetic energy. So in total there are $3\times 2 = 6$ degrees of freedom.

degree of freedom = number of ways to move around freely

In general there are three axix to move around for any particular particle or atom or molecule : x axix , y axix and z axix

In solid, the particle/molecule/atom cannot move freely as it is congested by surrounding atoms compared to gases.

However, they are free to move along those x, y and z axis with the help of their energy or the neighbours' energy. Let me explain this: The atom's personal energy can be thought as its potential energy which can be used to move along any of those three axis (x,y or z). This means the potential energy of the atom can be used to move along any of the three axis by pushing the neighbouring atoms along those axis. Therefore, it has 3 directions or degrees or axis to move freely by using its own energy. Hence, 3 ways of moving freely by using its personal energy.

The rest of the 3 ways or degree which makes an atom move freely in solid is my getting hit by the neighbouring atoms via any of those 3 axis (x,y or z).

So, in solids atom can move freely in those three axis by either pushing neighbouring atoms or by being pushed by neighbouring atoms along those x, y or z axis.

3 axis to push other atoms + the same 3 axis to get pushed by other atoms = 6 ways to move ( Freedom ).

• I think your definition of "degrees of freedom" is wrong. You consider as "ways to move", but this is not precise enough to derive anything. I could kick a ball, throw it, let it fall... etc etc, those different "ways of moving" are not different degrees of freedom. Thus I think that your answer is incorrect. – Frotaur Jun 1 '18 at 12:13