# Debye model

1. May 23, 2015

### Wminus

Hi! I feel like I've understood none of this stuff!

A 1D chain of springs and masses modeling a chain of atoms has a dispersion relation ala $\omega$ ~ $|sin(k a /2) |$, where $k$ is the wave vector and $a$ the distance between atoms. As far as I have understood, the debye model (in 1D) approximates this dispersion relation as simply a straight line, and from that calculates the heat capacity. But why bother doing that?? Wouldn't it be more accurate to use the proper dispersion relation to calculate the internal energy of the chain of atoms as a function of temperature? And this would just carry over to 3D, right?

Why not just treat the atoms in the solid as a bunch of masses connected to each other with springs vibrating at various modes? Surely it isn't too difficult for a physicist with some grit to solve such a system? And how accurate is this mass-and-spring model anyways?

2. May 23, 2015

### mhsd91

Well, I must disagree with you on this one. It is by no means trivial to solve the equations of motion for a 3D-system of masses connected with springs. I have a background in computational physics and mathematical modeling. Without having looked to much into the details, I would guess there wouldn't even be possible to find an analytical solution, even if all masses, and all spring constants were equal.

Just to enlighten the difficulity, consider a simple cubic lattice of lattice constant $a$, and let's try 2D first. We can label each particle by $i,j$, which at rest will be positioned at $(x_i,y_j) = (i a,j a)$ respectively. There will be four neighbouring particles which are connected with springs such that the total force on our particle will be

$\sum \vec{F} = F_{north} + F_{south} + F_{west} + F_{east} = - k( \vec{r}_n + \vec{r}_s + \vec{r}_w + \vec{r}_e) = \vec{r}_{i,j}$

note that $\vec{r}_e$ is the position of particle $(i+1,j)$, $\vec{r}_w$ of $(i-1,j)$, and so on for the north and south particles.

To make this the simplest possible problem, we assume there is a finite number of $N^2$ particles, that is, $N$ particles in both the $i$ and $j$ directions. We can then treat this a a Boundory Value Problem (BVP) and demand that the end of the crystal is kept at rest: these particles are held fast.

This, actually really simple system, will result in a large set of $N^2$ equations on the form

$m \frac{d^2}{dt^2}\vec{r}_{i,j} = \sum \vec{F}$

which will written out, be a matrix equation. Morover, the number of unknowns (rows&colums of the matrix) will be doubled as the 2D-vectors will need to be to decomposed into separate directions.

I might be wrong, but I am pretty sure this will have to be solved computationally. Even though 1D models are boring, they tend to catch the qualitative behaviour of many physical systems, and are better suited as teaching material as they may be solved as exam problems with pen and paper.

3. May 23, 2015

### DrDu

The Debye model stems from a time long before the advent of computers. Of course nowadays you can calculate better dispersion relations for the phonons and calculate heat capacities etc. from it.
As a sidenote to mhsd91: Usually you would impose periodic boundary conditions and make use of the periodicity of the system using Bloch's theorem. Then at least the model of coupled springs becomes quite tractable.

4. May 23, 2015

### Wminus

Thanks for the replies. I guess I can appreciate the difficulty of the spring and mass model, but still the Debye model is completely pointless for the 1D example.. So in 1D it just linearizes the dispersion relation. What does it do in 2D? Does it turn the dispersion relation into a cone from a parabola? I read in wikipedia that debye's approximation models the system as "phonons in a box". Can't seem to understand anything more of it though, anyone care to help?

As for the computers: Surely you could still find an accurate dispersion relation back in the old days numerically? Did they really not have anything better than Debye's approximation 100 years ago?

5. May 23, 2015

### DrDu

The Debye model is not as bad as it may seem. It's goal is (among others) to calculate the heat capacity for temperatures much lower than the Debye temperature. Hence practically only the phonons of lowest energy (and k values) will contribute to the heat capacity and in this range the dispersion relation is linear in an excellent approximation. Only phonons whose wavelength is much larger than the atomic spacing contribute and thus the detailed molecular structure of the solid isn't important.
You would also not argue that one shouldn't use the index of refraction in optics to calculate a lens system but use a full fledged solution of the microscopic Maxwell equations?

In fact, the situation is not very different, say, in quantum electrodynamics (QED). QED is only an effective field theory whose range is limited to rather low energies (e.g. as compared to the Planck scale). Nevertheless, it makes very precise predictions for, say, the fine structure of atoms.

6. May 24, 2015

### Wminus

OK good points. After I slept on it everything is more clear. Thanks for the help!

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