What Determines the Equilibrium Separation Between Atoms in a Lattice?

[acherentia]

Homework Statement [/b]


Debye considered atoms to oscillate from 0 up to a nu max. It is explained further in the text that the complication (i.e., not all atoms oscillating at same frequency as shown in Einstein's formula) is accounted for, by averaging over all the frequencies present.

How did he measure the frequencies up to the maximum frequency that was present so as to get the average?

 

[G01]

You can theoretically compute the average frequency without physically knowing which modes are excited and which aren't. The argument is statistical.

To compute the average frequency, $<f>$ we would compute the weighted average:

$$<f>=\Sigma_k f_k P(f_k)$$

That is, we take the frequency of mode, f_k, and multiply that by the probability that the mode is excited. We do that for every possible mode, and them add up all the resulting values. This will give us the average frequency.

The possible frequencies form a continuous set, so our sum becomes an integral:

$$<f>=\int_0^_f_{max}} f P(f) df$$

The probability of a given mode being excited is given by the Bose-Einstein distribution:

$$P(f)=\frac{1}{e^{hf/kT}-1}$$More detail will be given in any thermal physics or statistical mechanics text (Schroeder, for example)

More info on the web:

http://hyperphysics.phy-astr.gsu.edu/hbase/solids/phonon.html

 

[acherentia]

are you saying that even nu max is directly obtained from the formulas below? this is most of my concern how is the upper frequency limit set?

thank you btw

You can theoretically compute the average frequency without physically knowing which modes are excited and which aren't. The argument is statistical.

To compute the average frequency, $<f>$ we would compute the weighted average:

$$<f>=\Sigma_k f_k P(f_k)$$

That is, we take the frequency of mode, f_k, and multiply that by the probability that the mode is excited. We do that for every possible mode, and them add up all the resulting values. This will give us the average frequency.

The possible frequencies form a continuous set, so our sum becomes an integral:

$$<f>=\int_0^_f_{max}} f P(f) df$$

The probability of a given mode being excited is given by the Bose-Einstein distribution:

$$P(f)=\frac{1}{e^{hf/kT}-1}$$


More detail will be given in any thermal physics or statistical mechanics text (Schroeder, for example)

More info on the web:

http://hyperphysics.phy-astr.gsu.edu/hbase/solids/phonon.html

 

[acherentia]

on wikipedia I found a reasonable explanation for $$\lambda$$ minim i.e. $$\nu$$ maxim. How to get to $$\nu$$ maxim was what actually puzzled me in the beginning:

There is a minimum possible wavelength, given by twice the equilibrium separation a between atoms. As we shall see in the following sections, any wavelength shorter than this can be mapped onto a wavelength longer than 2a, due to the periodicity of the lattice.

so my question now is,

What defines equilibrium separation between atoms?