Einstein solid

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Einstein solid is a model of a solid based on two assumptions:

Each atom in the lattice is a 3D quantum harmonic oscillator
Atoms do not interact with each another

While the first assumption is quite accurate, the second is not. If atoms did not interact with one another, sound waves would not propagate through solids.

[edit] Heat Capacity

Heat Capacity of an Einstein Solid as a function of temperature. Experimental value of 3Nk is recovered at high temperatures.

Heat capacity is, perhaps, one of the most important properties of solids and can be expressed like so

$C_V = \left({\partial U\over\partial T}\right)_V$

$T$ , the temperature of the system, can be found in from entropy

${1\over T} = {\partial S\over\partial U}$

To find the entropy consider a solid made of $N$ quantum harmonic oscillators (hereafter SHOs). Because $x$ , $y$ , and $z$ modes of an SHO are not coupled (they don't interact, really -- they don't), one can work with $3 N$ oscillators instead.

$N^{\prime} = 3N$

Possible energies of an SHO are given by

$E_n = \hbar\omega\left(n+{1\over2}\right)$

or, in other words, the energy levels are evenly spaced and one can define a quantum of energy

$\epsilon = \hbar\omega$

which is the smallest and only amount by which the energy of SHO can be incremented. Next, we must compute the multiplicity of the system. That is, compute the number of ways to distribute $q$ quanta of energy among $N^{\prime}$ SHOs. This task becomes simpler if one thinks of distributing $q$ pebbles over $N^{\prime}$ boxes

or separating stacks of pebbles with $N^{\prime}-1$ partitions

or arranging $q$ pebbles and $N^{\prime}-1$ partitions

The last picture is the most telling. The number of arrangements of $n$ objects is $n!$ . So the number of possible arrangements of $q$ pebbles and $N^{\prime}-1$ partitions is $\left(q+N^{\prime}-1\right)!$ . However, if partition #2 and partition #5 trade places, no one would notice. The same argument goes for quanta. To obtain the number of possible distinguishable arrangements one has to divide the total number of arrangements by the number of indistinguishable arrangements. There are $q!$ identical quanta arrangements, and $(N^{\prime}-1)!$ identical partition arrangements. Therefore, multiplicity of the system is given by

$\Omega = {\left(q+N^{\prime}-1\right)!\over q! (N^{\prime}-1)!}$

which, as mentioned before, is the number of ways to deposit $q$ quanta of energy into $N^{\prime}-1$ oscillators. Entropy of the system has the form

$S/k = \ln\Omega = \ln{\left(q+N^{\prime}-1\right)!\over q! (N^{\prime}-1)!}$

$N^{\prime}$ is a huge number -- subtracting one from it has no overall effect whatsoever

$S/k \approx \ln{\left(q+N^{\prime}\right)!\over q! N^{\prime}!}$

With the help of Stirling's approximation, entropy can be simplified

$S/k \approx \left(q+N^{\prime}\right)\ln\left(q+N^{\prime}\right)-N^{\prime}\ln N^{\prime}-q\ln q$

Total energy of the solid is given by

$U = {N^{\prime}\epsilon\over2} + q\epsilon$

We are now ready to compute the temperature

${1\over T} = {\partial S\over\partial U} = {\partial S\over\partial q}{dq\over dU} = {1\over\epsilon}{\partial S\over\partial q} = {k\over\epsilon} \ln\left(1+N^{\prime}/q\right)$

Inverting this formula to find $U$

$U = {N^{\prime}\epsilon\over2} + {N^{\prime}\epsilon\over e^{\epsilon/kT}-1}$

Differentiating with respect to temperature to find $C V$

$C_V = {\partial U\over\partial T} = {N^{\prime}\epsilon^2\over k T^2}{e^{\epsilon/kT}\over \left(e^{\epsilon/kT}-1\right)^2}$

$C_V = 3Nk\left({\epsilon\over k T}\right)^2{e^{\epsilon/kT}\over \left(e^{\epsilon/kT}-1\right)^2}$

While Einstein model of the solid predicts the heat capacity accurately at high temperatures, it noticeably deviates from experimental values at low temperatures. See Debye model for accurate low-temperature heat capacity calculation.

[edit] Heat Capacity (alternative derivation)

Heat capacity can be obtained much quicker through the use of partition function of an SHO.

$Z = \sum_{n=0}^{\infty} e^{-E_n/kT}$

where

$E_n = \epsilon\left(n+{1\over2}\right)$

substituting this into the partition function formula yields

$Z = \sum_{n=0}^{\infty} e^{-\epsilon\left(n+1/2\right)/kT} = e^{-\epsilon/2kT} \sum_{n=0}^{\infty} e^{-n\epsilon/kT}=e^{-\epsilon/2kT} \sum_{n=0}^{\infty} \left(e^{-\epsilon/kT}\right)^n =$

$= {e^{-\epsilon/2kT}\over 1-e^{-\epsilon/kT}} = {1\over e^{\epsilon/2kT}-e^{-\epsilon/2kT}} = {1\over 2 \sinh\left({\epsilon\over 2kT}\right)}$

This is the partition function of one SHO. Because, statistically, heat capacity, energy, and entropy of the solid are equally distributed among its atoms (SHOs), we can work with this partition function to obtain those quantities and then simply multiply them by $N^{\prime}$ to get the total. Next, let's compute the average energy of each oscillator

$\langle E\rangle = u = -{1\over Z}\partial_{\beta}Z$

where

$\beta = {1\over kT}$

therefore

$u = -2 \sinh\left({\epsilon\over 2kT}\right){-\cosh\left({\epsilon\over 2kT}\right)\over 2 \sinh^2\left({\epsilon\over 2kT}\right)}{\epsilon\over2} = {\epsilon\over2}\coth\left({\epsilon\over 2kT}\right)$

Heat capacity of one oscillator is then

$C_V = {\partial U\over\partial T} = -{\epsilon\over2} {1\over \sinh^2\left({\epsilon\over 2kT}\right)}\left(-{\epsilon\over 2kT^2}\right) = k \left({\epsilon\over 2 k T}\right)^2 {1\over \sinh^2\left({\epsilon\over 2kT}\right)}$