Unformatted text preview: Summary of last lecture Linear Oscillator  QM and Classical treatment Equipartition theorem Einstein Model of heat capacity of crystals. Chemistry 161: Statistical Thermodynamics
Lecture 13 Thermal Properties of Crystals
Reading: Reif, Ch6 Key Concepts and Lessons: a) Heat capacity of one dimensional crystals exact solution b) Debye model of thermodynamics of crystals. One dimensional crystals: Exact solution
Crystals appear to be far more complicated than ideal gas of noninteracting molecules that we discussed so far. Indeed atoms/molecules constituting a crystal interact strongly and it delivers stability and longrange order to a crystal! However strength of interactions appears to be a simplifying factor in this case! As we discussed earlier we can view in this case molecular motions as small deviations from the equilibrium positions and view motion as harmonic, i.e. the energy 0 function can be presented as a quadratic form of deviations i (t) = qi (t) qi from equilibrium positions of molecules in crystals: U ( 1 , 2 .... 3N 6 ) 1 3N 6 = U 0,0...0 + k 2 i, j =1 ij
3N 6 ( ) i j We start from Newton's equation of motion of molecules/atoms (i.e. use classical approximation): mi i = j =1 kij j this is a set of 3n6 linear equations of motion. (6 degrees of freedom correspond to motion of the crystal as a whole and as such are irrelevant). In Harmonic motion the time dependence is `'sinusoidal'' or in other words we can state that s (t) = ei t s (where we used the same notation for time independent part of displacement coordinate, in order not to unduly complicate notations). As a result we get for equation of motions the following 3N6 linear equations determining amplitudes :
3N i =1 6 2 j ij j kij =0 One dimensional crystals: Exact solution (Cont)
The set of Newton's equations of motion is linear with no free terms. The necessary and sufficient condition that it has solutions is:
1 k m 2 11 1 det k 2 21 1 k 2 3N 6,1
2 1 k 2 12 1 k m 2 22 1 k 2 3N 6,2
2 1 k 2 1,3N 1 k 2 2,3N 1 k 2 3N
6,3N 6 6 6 2 =0 m This condition is equivalent to the polynomial equation of 3N6 power for . This equation has 3N6 solutions called normal modes. Now the motion of crystal can be presented as 3N6 simultaneous vibrations each can be treated as harmonic oscillator. We get immediately for the partition function and heat capacity of the vibrational motion of the crystal: i 2 Z crystal =
3N i =1 3N i =1 6 e 1
i kT
i e
2 kT
i Cv = 6 e
i kT 2 kT e
kT 1 One dimensional crystals: Exact solution (Cont)
kT for all i=1...3N6 Now suppose that all degrees of freedom are `'soft'' in a sense that . i In this case everything is simple: Equipartition, i.e. each degree of freedom contributes k to heat capacity and we obtain the Dulong Petitt law for heat capacity: Cv = k 3N ( 6 ) 3Nk However reality is a bit more complicated some degrees of freedom are `'soft'' and some are `'hard'' so that their vibrations are quantum frozen. The key question then is to determine the vibrational spectrum of a molecule i.e. its set of frequencies { i }. Let us consider first a one dimensional problem of set of masses m on springs with rigidity k. In this case energy function is 1 N U= f 2 i=1 ( i i+1 ) 2 f = 2 N (2
i=1 2 i 2 i i+1 ) Now we seek solution in the form of standing waves of vibration: s = Re aeiks One dimensional crystals: Exact solution (Cont)
Newton equations of motion for each atom are: 0 = aeik (s 1) f 2f + aeiks m m 2 + aeik (s+1) f m And we get after simple algebra and trigonometry:
2 = 2f 1 ik 1 e +e m 2 ( ik ) = 4f 2 k 2f 1 cos(k) = sin m m 2 we got a gem here a dispersion relation (k) !!!. Why is it a gem? Because it is easy to count normal space The point is that we can (and should) impose boundary conditions on modes in kspace than in atom motions in our crystal. E.g. that first (zeroth) and Nth atom are fixed. That gives = N =0 0 and the condition for possible k: Nk j = 2 j or k j = 2 j , j=1.....N N This condition determines the set of possible wavevectors k which are consistent with standing waves In our onedimensional crystal. One dimensional crystals: Exact solution (Cont)
We get the frequencies of normal modes as:
== 2 f sin m kj 2
j Our next step is to determine density of normal modes g( )d which tells me how many normal modes fall into the range of frequencies ( , + d ) . How do we find it? The strategy is to first count normal Modes in k space  that is easy. Then convert the count in kspace into count in omegaspace Using the dispersion relation above. Now let us carry out this program. First thing first: kspace. We define g(k) in the same way  density of normal modes in kspace. It is constant: g(k ) = 1 kspacing = ( 1 N = 2 2 /N ) Now we want to know the same in omegaspace. Us dispersion relation as follows: g ( )d = 2g(k)dk; g( ) = 2g(k) dk 2g(k) = d d dk Factor 2 appears because both k and k correspond to the same frequency omega. One dimensional crystals: Exact solution (Cont)
From the dispersion relation we get:
d = dk i.e. 1 4f 1 k = cos 2 m 2 2 1 k = 2 2 k 2
1/ 2 2 1/ 2 max cos max 1 sin 2 = 1 2 max 1
max d 1 = dk 2 ( 2 max 2 ) 1/ 2 So that: g( ) = 2 N /2 1 2 ( ( )
2 2 max ) 1/ 2 And we get for heat capacity summation over all normal modes with proper density:
2
max Cv = d g( )
0 e kT
2 kT e kT 1 =
0 d 2 N/ ( ( )
2 2 1/ 2 2 max ) e kT
2 kT e kT 1 One dimensional crystals: Exact solution (Cont)
High temperature is defined as
kT at this T all modes are fully excited and classical:
2 1 12 kT
max Cv = kN ( ) 2 2 0 g( )d
kT contribute to frequency. At low T only vibrations that have low enough frequency such as Heat capacity in this case can be approximated by the integral:
kT / Cv
A reminder what At small
g( ) is: k
0 g(
2 N /2 1 2 )d g( ) = (
+ ( )
2 2 max ) 1/ 2 it can be expanded
Cv g( ) 2N
max 1+ 1 2 2 2 max 2NkT
max N 3
3 max kT 3 At low T only first term matters and we get the low T asymptotics of heat capacity of a 1D crystal:
1D Cv T The Debye Model of heat capacity of 3D crystals
The 1dimesional example of lattice dynamics and exact calculation of TD of crystals is useful and very illuminating but it is of course limited because we live in 3D. To that end Peter Debye worked out an approximate formula that works well in most instances and gives good insight into physical behavior of solids. The Debye formula is an approximation for our old friend dispersion relation g ( ) . As before we assume that there is a wave of displacements but now this wave is in 3D: ( i kr r ,t = Ae ( ) t ) We should impose boundary conditions of zero displacement at the boundaries of the crystal. Such conditions determine possible values of k: kx = n Lx x kx = kz = Ly Lz ny nz Where n are of course integer numbers and Lx,y,z are spatial dimensions of our crystal. The counting of standing waves in k space is easy: V
3 kx k y kz = nx n y nz The Debye Model of heat capacity of 3D crystals
Converting to polar coordinate in kspace and taking into account that we are interested only in positive n's (i.e. 1/8 of the space) we get the number of normal modes between k and k+dk as: Vk 2 dk g(k )dk = 2 2
Now we have to relate and k. This is done using the relation (approximately!) where u is speed of sound. In other words: =2 = uk / 2 g ( ) 4 V d = u3 2 d This is (almost) the desired result. To make it more precise we note that there are 3 sound waves: 2 transverse and 1 longitudinal and their speed of sound may differ. Therefore a better approximation would be: g () 2 1 12 V d = 3 + 3 4 V 2d = ut ul u3 2 d The Debye Model of heat capacity of 3D crystals (Cont)
Now we define the limiting frequency (called Debye frequency) 1D case with max ) so that:
D D (pretty much as we've seen in g( )d = 3N; i.e.
0 D 3N = 4 V 1/ 3 u This relation tells us that total number of normal modes should be 3N. Then presented in a simple form: g( ) can be g ( )d = 9N
3 D 2 d >
D 0 D 0
Now we define the Debye temperature D =
3 h k D and change variables to
/T x= h kT to get: CV = 9Nk T
D D 0 (e x 4ex
x 1 ) 2 dx The Debye Model of heat capacity of 3D crystals (Cont)
The integral can be calculated approximately and in asymptotics. When extended up to infinity and:
T
D then integration can be CV = AT 3
In the opposite case when
T
D exponentials can be expanded (because x is always small) CV
(prove that: 20pt) 3Nk for soft molecular crystals lies in the range 100200K while for several metals it is in the range of 300500K. For diamond D is 1860K so that this solid should described as lowtemperature case where QM is crucial.
D The Debye Model of heat capacity of 3D crystals (Cont)
The Debye function can be defined as:
1/ x D(x) = 3x 3
0 ( e 1)
y y 4e y 2 dy Then:
T
D CV = 3NkD The concept of phonons
The picture of normal modes suggests that we can present total vibrational energy of a crystal as: E= 3N h
j =1 j 1 = nj + 2 3N j =1 h j n j + E0 It is as if we have independent particles with energies per particle and with occupancy numbers Since there are no limits on n these particles are bosons. Now we can write the basic formula for occupancy numbers of these particles (called phonons)
nj = e 1
h kT 1
E = E0 +
0 With density of states g we get as before: g( )h d e
h kT 1 Why is this a useful concept? The reason is that in more complicated situations where behavior of electrons depends on the state of nuclear lattice this factor can be treated as interaction of electrons with phonons as if they were real particles and that helps a lot in the analysis of complex situations where anharmonicity of lattice vibrations plays an important role. Next lecture:
1) Partition Functions of Polyatomic Ideal gases (Reif
9.12, McQuarrie 6.16.3. 2) Chemical Equilibrium (Reif 8.7,8.10. McQuarrie Ch9) ...
View
Full Document
 Spring '10
 Shaklovich
 normal modes, kspace, Peter Debye, Debye model

Click to edit the document details