EINSTEIN SOLID

'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.

Contents

Heat capacity

Heat capacity (alternative derivation)

Heat capacity

The heat capacity of an object is defined as
:$C\_V\; =\; left(\{partial\; Uoverpartial\; T\}$
ight)_V.
$T$, the temperature of the system, can be found from the entropy
:$\{1over\; T\}\; =\; \{partial\; Soverpartial\; U\}$
To find the entropy consider a solid made of $N$ atoms, each of which has 3 degrees of freedom. So there are $3N$ quantum harmonic oscillators (hereafter SHOs).
:$N^\{prime\}\; =\; 3N$
Possible energies of an SHO are given by
:$E\_n\; =\; hbaromegaleft(n+\{1over2\}$
ight)
or, in other words, the energy levels are evenly spaced and one can define a ''quantum'' of energy
:
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 $left(q+N^\{prime\}-1$
ight)!. 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
:$Omega\; =\; \{left(q+N^\{prime\}-1$
ight)!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
:$S/k\; =\; lnOmega\; =\; ln\{left(q+N^\{prime\}-1$
ight)!over q! (N^{prime}-1)!}.
$N^\{prime\}$ is a huge number—subtracting one from it has no overall effect whatsoever:
:
ight)!over q! N^{prime}!}
With the help of Stirling's approximation, entropy can be simplified:
:
ight)lnleft(q+N^{prime}
ight)-N^{prime}ln N^{prime}-qln q.
Total energy of the solid is given by
:
We are now ready to compute the temperature
:
ight)
Inverting this formula to find ''U'':
:
Differentiating with respect to temperature to find $C\_V$:
:
ight)^2}
or

:
ight)^2{e^{arepsilon/kT}over left(e^{arepsilon/kT}-1
ight)^2}.

Although 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.

Heat capacity (alternative derivation)

Heat capacity can be obtained much more quickly through the use of partition function of an SHO.
:$Z\; =\; sum\_\{n=0\}^\{infty\}\; e^\{-E\_n/kT\}$
where
:
ight)
substituting this into the partition function formula yields
:$$
egin{align}
Z & {} = sum_{n=0}^{infty} e^{-arepsilonleft(n+1/2
ight)/kT} = e^{-arepsilon/2kT} sum_{n=0}^{infty} e^{-narepsilon/kT}=e^{-arepsilon/2kT} sum_{n=0}^{infty} left(e^{-arepsilon/kT}
ight)^n \
& {} = {e^{-arepsilon/2kT}over 1-e^{-arepsilon/kT}} = {1over e^{arepsilon/2kT}-e^{-arepsilon/2kT}} = {1over 2 sinhleft({arepsilonover 2kT}
ight)}.
end{align}

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
:$langle\; E$
angle = u = -{1over Z}partial_{eta}Z
where
:
Therefore
:
ight){-coshleft({arepsilonover 2kT}
ight)over 2 sinh^2left({arepsilonover 2kT}
ight)}{arepsilonover2} = {arepsilonover2}cothleft({arepsilonover 2kT}
ight).
Heat capacity of ''one'' oscillator is then
:
ight)}left(-{arepsilonover 2kT^2}
ight) = k left({arepsilonover 2 k T}
ight)^2 {1over sinh^2left({arepsilonover 2kT}
ight)}.
Heat capacity of the entire solid is given by $C\_V\; =\; 3NC\_V$:

:
ight)^2 {1over sinh^2left({arepsilonover 2kT}
ight)}.

which is algebraically identical to the formula derived in the previous section.