理学院物理系沈嵘
1
5.1 Phonon heat capacity
5.2 Experimental Determination of Phonon
Spectrum
5.3 Anharmonic Effects
Chapter 5 Crystal Thermodynamics
2
5.1 phonon heat capacityI. Density of states
Consider an atomic chain consisting of N lattice sites:
lL
lNa
q pp 22 ==
where l is an integer
NaL = —the length of the chain
the q-points are regularly arranged in a 1d chain, with
distance to each other.Lp2
5.1 phonon heat capacity
3
For large L, the distance becomes
small, and the q-points are nearly
continuous; each q-value (a single
point) represents a vabration
mode.
For an arbitrary distance dq,number
of modes dn in between is:
qLn d2
dp
=
q is related to frequence w via the dispersion relation, we can
thus get the # of vibration modes in the range of .( )www d, +
5.1 phonon heat capacity
4
1. DefinitionDensity of modes (states): # of vibration modes per unit frequency interval
( )w
wddng =
2. D.O.S. of an atomic chain
( ) ng dd =ww
qLn d2
dp
=
( ) qLg d2
dp
ww =
5.1 phonon heat capacity
5
( ) ÷ø
öçè
æ=q
Lgdd
2w
pw
for a given w value, there exist two symmetric q vlues:
( ) ÷ø
öçè
æ=q
Lgddw
pw
dispersion:2
2 qaM
sin÷øö
çèæ=bw
222
dd qaa
mqcos÷
øö
çèæ=bw
22qaa
m cosw=
Mmbw 4=in which,
Mb4
5.1 phonon heat capacity
6
( )
2
2qaa
Lgm coswp
w =
2
cos2m
Nqapw
=
2 2
2
m
Np w w
=-
is finite,0=•w ( )wg
,mww =• ( ) ¥®wg,mww >• ( ) 0=wg
5.1 phonon heat capacity
7
3. Diatomic chain
for each branch ( )q+w ( )q-w
corresponds: ( ) ( ),g gw w+ -
( ) ( ) ( )www -+ += ggg
4. General crystalsq-points with the same frequencies
consititutes closed contours, or a
isosurface.
5.1 phonon heat capacity
8
( ) ww =qj!
( )d dj q q! !w w w+ = +
Count # of all modes between these two isosurfaces: jnd
Equal. ( ) ww djg Þ determine ( )wjg
Sum over all branches, one gets the total density of states:
( ) ( )å=j
jgg ww
For a given branch j
two isoenergetic contours/isosurfaces
5.1 phonon heat capacity
9
Example:A cube with length L
According to periodic boundary condition,
ii
ii aN
lq p2=L
lip2
=
volume per q-point:32÷øö
çèæ
Lp
A unit volume in q-point space contains:
3
3
82 ppVL
=÷øö
çèæ
allowed q
V——volume of the crystall
5.1 phonon heat capacity
10
Isosurfaces are spheres, with radius q. The # of modes in this sphere:
33 2
34
÷øö
çèæ=
Lqn pp
( )3
3 34
2qV p
p=
For q, modes between q ~ q+dq:
( )qqVn d3
34
2d 23
pp
=
( )qqV d4
22
3 pp=
5.1 phonon heat capacity
11
( )( )
qqVg jj d42d 23 ppww =
( ) 23d
4d2( )
jj
Vg qqw
w pp
æ ö= ç ÷
è ø
qj
ddw
can be obtained from dispersion relation
( )( ) ( )33
33
22
11
221111
ppr VNN
bN
bN
bN
q dr
=W=W
=÷ø
öçè
æ´×
=!!!
For arbitrary isosurface, and any branch, vibration modes are
uniformly distributed in space. Corresponding DOS is:q!
5.1 phonon heat capacity
12
( )( )3d d2
Vn w w wp
= ´ +频率 和 等频面间的体积
( )3d d d
2 S
Vn S qp
^= ò
dS--surface element,
dq^--distance between two surfaces
( )qq jqj ww Ñ= ^dd
( )qjqwÑ ——gradient along direction normal to the surface
(volume between and ) w ww d+
5.1 phonon heat capacity
13
( ) ( )3dd d
2j
S q j
V Snq
wwp
=Ñò
( )( ) ( )3
d dd 2
jj S q j
n V Sgq
ww wp
= =Ñò
Total Density of States:
( ) ( )å=j
jgg ww
( ) PNg 3d =ò wwwhich satifies:
5.1 phonon heat capacity
14
II. Heat Capacity1. lattice heat capacity
Specifici heat of solid is defined: V
V TC ÷
øö
çè涶
=e
: average intermal energy of solid, including energies of
lattice vibration and electron motions
e
According to classical theory, every degree of freedom
corresponds to ave. energy , with kinetic energy
and potential energy; suppose there are N atoms, the
total energy is .
TkB2TkB
2TkB
TNkB3=e
5.1 phonon heat capacity
15
suppose N=NA (1 mole of atoms)
therefore mole capacity is:
3V A BV
C N kTe¶æ ö= =ç ÷¶è ø
which has nothing to do with specific material or any temperature dependence.——Dulong-Petit’s law
5.1 phonon heat capacity
16
Experiments show,CV vialates the Dulong-Petit’s
law.
This demonstrates that
the classical theory of
equipartition of energy is
no longer valid, and the
quantum theory of crystal
vibration is needed.
5.1 phonon heat capacity
17
lattice vibrations are quantized, at temperature T, the
virbration with frequency w has energy:
we !÷øö
çèæ +=
21n
average phonon #: ( ) ( ) 11
-= Tkqi Bie
qn !"w
Ignore the zero point energy:12w!
D.O.S , satisfy:( )g w
In which, w m----max angular frequency, N----# of primitive unit cell
5.1 phonon heat capacity
18
Average Energy:
( )ò -=m
Bg
e Tkw
w wwwe
0d
1!!
=÷øö
çè涶
=V
V TC e
( )2
20d
( 1)
Bm
B
k T
B k TB
ek gk T e
ww
w
w w wæ öç ÷ -è ø
ò!
!
!
Heat Capacity:
Þ calculate the phonon heat capacity( )g w
5.1 phonon heat capacity
19
2. Einstein ModelSuppose all the vibration modes are with the same frequency wE .
Average energy of the crystal:
where N is # of primitive cells (PN is # of atoms)
heat capacity:
=÷øö
çè涶
=V
V TC e ( )2
2
13
-÷ø
öçè
æTk
Tk
B
EB
BE
BE
ee
TkNPk
w
ww!
!!
5.1 phonon heat capacity
20
Einstein Temperture: EΘ EBE Θk=w!
( )22
13
-÷øö
çèæ=
TΘ
TΘE
BVE
E
ee
TΘPNkC
Determination of :Select a proper value, such that the
calculated specific heat agrees with the experimental data,
in an extensive range of temperatures.
EΘ
,EΘT >>• at high temperature
( ) ( )22Θ221
1 TTΘTΘTΘ
EEE
E
eeee
--=
-
2
2
22
1÷ø
öçè
æ=
÷øö
çèæ
»EE Θ
T
TΘ
5.1 phonon heat capacity
21
BE
EBV PNkΘ
TTΘPNkC 33
22
=÷ø
öçè
æ÷øö
çèæ=
agrees with Dulong-Petit’s law.
,EΘT TΘEe TΘTΘ EE ee »-1
TΘEBV
EeTΘPNkC -÷
øö
çèæ=
2
3
E
Bk TEB
B
PNk ek T
2
3ww -æ ö
= ç ÷è ø
!
!
low temperature
5.1 phonon heat capacity
22
0®T
0®VC
asymptotic behavior (from experiments):
3T
Einstein model oversimplified the difference in frequenceof various lattice waves (vibration modes)
5.1 phonon heat capacity
23
3. Debye model
Regard Braivais lattices as isotropic continuous media, and the
crystal vibration an elastic wave with longitudinal and
transverse waves presumably moving in the same velocity.
Volume of the crystal V, for elastic wave in an isotropic media
quw = qdd uw =
In each branch, # of vibration modes in the range of q~q+ dq
( )qqVn d4
2d 23 pp
=
5.1 phonon heat capacity
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# of vibration modes in the range of w ~ w + dw :
( )w
up
pd4
2d
2
3qVn = w
uw
pd
2 32
2V=
For each q, there corresponds 3 elastic waves (1 longitudinal,
2 transverse modes).
Therefore, # of modes in w ~ w + dw :
wwup
d2
3d 232Vn =
( ) wwup
ww d2
3dd 232Vng ==
5.1 phonon heat capacity
25
( ) 23223 wup
w Vg =
ò ×-=m
B
Ve Tk
w
w wwupwe
0
232 d2
31!
!
ò -=m
BTkeV w
w ww
up 0
3
32 d123
!
!
( )ò -÷øö
çè
æ×=
m
B
B
Tk
Tk
BBV
ee
TkkVC
w
w
w
wwwup 0 2
22
3 d123
!
!!
Average Energy:
Heat Capacity:
Density of States:
5.1 phonon heat capacity
26
For N atoms, 3N canonical frequencies:
( )òm g
www
0d NVm 3d
23 2
320== ò wwup
w
upw31
26÷ø
öçè
æ=V
Nm
Tkx
B
w!=
Tkx
B
mm
w!=
( )ò -÷øö
çè
æ=mx
x
xB
BV xe
xeTkVkC0 2
43
2 d123
up !
set
5.1 phonon heat capacity
27
upw31
26÷ø
öçè
æ===QV
NTkTk
xT BB
mm
D !!
Debye Temperature: DΘB
mD k
Θ w!=
xe
xΘTTNk
TΘ
xD
BD d
19
0
33
ò -÷øö
çè
æ=e
ò -×÷øö
çè
æ= m
x
x
x
DBV xe
xeΘT
VNVkC
0 2
432
2 d)1(6
23 p
p
ò -×÷øö
çè
æ=
TΘ
x
x
DBV
D xe
xeΘTNkC
0 2
43
d)1(
9
5.1 phonon heat capacity
28
Debye temperature is a characteristic temperature of the system
DT Q>>• xe x +» 1
TNkT
TTNk BDD
B 3319
33
=÷øö
çèæQ÷
ø
öçè
æQ
=e
BV
V NkTC 3=÷
øö
çè涶
=e
tends to classical limit
5.1 phonon heat capacity
29
DΘT
301.from elastic const. and wave velocity2.from specific heat dataD
Θ ofion determinat
5.1 phonon heat capacity
31
4. Comparison of Einstein and Debye Models
optical branch---Einstein model
acoustic branch---Debye modelCrystal Vibration
5.1 phonon heat capacity
32
Einstein Model:
Approximate all branches as a straight line, ignore the
effects of low energy phonons, and treat all lattice waves as
optical modes.
Debye Model:
For long wavelength acoustic waves, effective model at low
temperatures.
5.1 phonon heat capacity
33
We also adopt the combination of the two models—— hybridized model
( ) ( ) ( )www DE ggg +=
5.1 phonon heat capacity
34
5.2 Experimental Determination of Phonon Spectrum
The dispersion relation between frequency and wave vector
——vibration sepctrum of crystal, phonon spectrum
When photons or neutrons are shot into crystals, they can
exchange energies with the lattice, and excite/annihilate
phonons—inelastic scattering.
Inelastic X-Ray scattering, neutron scattering, light scattering...
5.2 experimental determination of phonon spectrum
light scattered by acoustic phonon——Brillouin Scattering
light scattered by optic phonon——Raman Scattering.
Raman spectrum is very important tool for investigating
microstructure in condensed matter.
35
5.2 experimental determination of phonon spectrum
36
90K, Na crystal, determined dispersion relation along three directions
5.2 experimental determination of phonon spectrum
37
5.3 Anharmonic Effects
Cannot convey energy, nor establish thermal equilibrium.
Harmonic Approximation:
Force on the atom is proportional to its displacement, potential
energy is kept up to d 2.
Crystal vibration can be discribed as a series of
independent harmonic oscillators, which do not interact
with each other, nor exchange energy with each other.
5.3 Anharmonic effects
38
Realistic Crystal:
ü Strictly, the interaction between atoms are not completely
harmonic, the crystal vibration modes are not completely
independent, but can instead talk to each other.
ü Phonon-phonon interaction exists, which enable them to
exchange energies. A phonon with a specific frequency could
be transfered into another phonon with different frequency,
and the distribution of phonons achieves thermal equilibrium
after a period of time.
ü Anharmonic terms accounts for the existence of thermal
equilibrium of crystals.
5.3 Anharmonic effects
39
I. Lattice Free Energy and Equation of State
lattice free energy :
1. thermodynamic relationsTSUF -=
TVFP ÷øö
çè涶
-=
VTFS ÷øö
çè涶
-=
VVV T
FTTSTC ÷
ø
öçè
涶
-=÷øö
çè涶
= 22
( )VT
FTFTSFTU ÷øö
çè涶
-=+=
pressure:
entropy:
energy:
heat capacity:
5.3 Anharmonic effects
40
2. Free Energy( ) ZTkVUF B ln-=
Z —phonon partition function :
å -= TkBieZ e
lattice free energy:
U(V) —crystal cohesive energy at T = 0 , a function of volume V, independent of temperature T
For lattice wave wi: å¥
=
÷øö
çèæ +-
=0
21
i
B
ii
n
Tkn
i eZw!
i
B
i
B
k T
k T
e
e
12
1
w
w
-
-=
-
!
!
Partition functionfor all vibration modes: ÕÕ -
-
-==
iTk
Tk
ii Bi
Bi
eeZZ w
w
!
!
1
2
5.3 Anharmonic effects
phonon contribution
41
( ) ( )i Bk TiBi B
F U V k T ek T
1 ln 12
ww -ì üé ù= + - - - -í ýê úë ûî þ
å !!
( ) ( )å úûù
êëé -++= -
i
TkBi
BieTkVU ww !! 1ln21
Sum over all branches and values (modes).q!
3. Equation of StatesNonlinear vibration changes the volume and frequency wi, wi is a function of V.
TVFP ÷øö
çè涶
-=Ve
eVU i
iTk
Tk
Tbi
Bi
dd
121 w
w
w
å úû
ùêë
é-
+-÷øö
çè涶
-= --
!
!!
!
5.3 Anharmonic effects
42
VeVU i
iTki
iT
bi ddln
121 www wå úû
ùêëé
-+-÷
øö
çè涶
-=!
!!
VVVU i
ii
T lndlnd1 weå-÷
øö
çè涶
-=
121
-+= Tk
iii Bie w
wwe!
!!in which:
Vi
lndlnd wg -=set ——Gruneisen parameter,
independent of wi, related with nonlinear vibration.
VVUP eg+-=
dd
5.3 Anharmonic effects
43
II. Thermal ExpansionVolume changes with T, under a
fixed pressure.
1. Qualitative analysisGiven a symmetric potential
around equilibrium position, then
Realistic potential curve is not strictly a parabola, steeper on the
left and smoother on the right side. Atoms tend to move
rightwards when vibration amplitudes increase.
Þ equilibrium position independent of amplituide of vibration or
temperature.
5.3 Anharmonic effects
44
Anharmonic potential is non-symmetric, inducing some
“repulsive” interactions when atoms vibrate, accounting for
the themal expansion phenomenon.
2. quantitative analysisvolume expansion
coefficient: VP TP
KTV
V ÷øö
çè涶
=÷øö
çè涶
=11
0
a
linear expansion
coefficient: Pl T
ll ÷ø
öçè涶
=0
1a
isotropic cubic system:V
l TP
K ÷øö
çè涶
==313aa
K is bulk modulus
5.3 Anharmonic effects
TVPVK ÷øö
çè涶
= 0-
45
According to equation of state:
VVUP eg+-=
dd
For most solids, volume change is small, expand around
equilibrium V0: VU
dd
( ) !+÷ø
öçè
æ-+÷
øö
çèæ=
00
2
2
0 dd
dd
dd
VV VUVV
VU
VU
First term vanishes, retain up to the second term:
úúû
ù
êêë
é÷ø
öçè
æ-=
0
2
2
00
0
dd
dd
VVUV
VVV
VU
0
0
VVVK -=
5.3 Anharmonic effects
46
VVVVKP eg+--=0
0
0=PVKV
VV eg=
-
0
0
thermal expansion is the relation between V and T at P = 0,
0d VVV -=
VC
KTV
VV
P
ga =÷øö
çè涶
=0
1VC
KV
l 3ga =
Grüneisen's law, thermal expansion of solids is proportional to the specific heat (per volume).
5.3 Anharmonic effects
47
III. Thermal Conductivity
1. phonon scattering
Lattice waves are no longer independent due to anharmonic
effects, leading to phonon scattering.
The precess of phonon collision obeys energy conservation and
quasi-momentum conservation:
321 www !!! =+
hGqqq!!!! +=+ 321
5.3 Anharmonic effects
48
(1) Normal process
Wave vectors , are relatively small, resulting is still in
the 1st BZ, thus called normal process.
Total energy/wave vector keeps unchanged, only the energy
and quasi-momentum of the two phonons are tranfered to a
3rd one, net heat flow does not decreases, nor does its direction
change.
1q 2q 3q
If all phonon collisions are normal processes, the lattice
thermal conductivity diverges and resistivity is zero, i.e.,
normal process contributes nothing to thermal resistivity.
5.3 Anharmonic effects
49
(2) Umklapp process
1q
2q
3q
exceeds 1st BZ, due to the
peroidicity of lattice, wave vector
describes the same vibration state
as . can be moved
back to 1st BZ by adding some
reciprocal lattice vector.
q q1 2+! !
q!
hq G+!!
ü Such process is called umklapp process, a large angle
scattering. The direction of phonon movement is changed
greatly, making a shorter mean free path, and give rise to
some finite thermal resistivity.
5.3 Anharmonic effects
q q1 2+! !
50
2. Thermal ConductivitySuppose different temperatures at two ends of the rod: T1, T2
12 TT >Heat flow, following the temperature gradient, move from
the hot end to the cold one, with energy density proportional
to the temperature gradient.
xTJq d
dk-=
where k denotes thermal conductivity coefficient.
Thermal conductance: electrons (metal), phonon (insulator)
5.3 Anharmonic effects
Fourier's law
T2 T1
l Atoms are vibrating more
strongly in the hot (left) end
than those near the cold (right)
end, leading to a larger
density of phonons.
l Phonons move from left to
right, carrying heat, and flow
along the opposite direction of
temperature gradient.
ü Apply the gas molecule
dynamic theory to the phonon
gas, thermal conductivity
reads:
lCVuk 31
=
u —phonon velocity; l —mean free path, mean distance a phonon
moves between two successive collisions with other phonons.
5.3 Anharmonic effects
52
Three mechanisms:
Phonon-phonon scattering due to anharmonic couplings,
especially important at high tempertures.
(1) Collision between phonons
Mean phonon number:
High T: DΘT >>
TTke
nq
BTkq Bq
µ»-
=ww !! 1
1
Collision probability proportional to # of phonons,
Corresponding mean free path thus is inverse proportional to
temperatures:T
l 1~
5.3 Anharmonic effects
53
(2) Scattering of phonon with crystal defects
(3) scattering with the boundaries of specimen
Impurities and defects also scatter phonons, since they partially
break the lattice periodicity. The larger the mass difference and
density of impurities are, the stronger the scattering is, and l
becomes shorter.
At very low temperatures, (1,2) collisions are scarce.
Ø only few phonons exist, dilute phonon gas
Ø phonons with long wavelength at low temperatures, which
cannot be effectively scattered by objects like impurities of much
smaller size.
5.3 Anharmonic effects
54
At low temperatures, the main mechanism is the boundary
scattering. It cause some geometric effects, since the phonon has
very long wavelength, which is comparable to the size of
specimen. Mean free path l = L, independent of temperatures.
lCVuk 31
=Remenber the thermal conductivity:
At low T, thermal conductivity determined by heat capacity3~ Tk
At high T, thermal conductivity determined by mean free path l
T1~k
5.3 Anharmonic effects
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