Download Lecture ? Einstein-Debye theory

Lecture 27. Debye Model of Solids, Phonon Gas
In 1907, Einstein developed the first quantum-mechanical model of
solids that was able to qualitatively describe the low-T heat capacity
of the crystal lattice. Although this was a crucial step in the right
direction, the model was too crude.
In Einstein’s model, all oscillators are identical, and their frequencies are the same:


3N
1 3N
h
2
2
U   mri  kri   h ni  3N
2 i 1
2
i 1
ħ
1
However, Einstein’s model ignores the fact that the
atomic vibrations are coupled together: the potential
energy of an atom in the crystal depends on the
distance from its neighbors:
2
3
3N
2
1 3N
1 3N
2
U   mri   k ri  rj 
2 i 1
2 i 1, j 1
This energy is not a sum of single-particle energies. Thus, the calculation of the
partition function may look rather difficult. But a system of N coupled three-dimensional
oscillators is equivalent to a system of 3N independent one-dimensional oscillators.
the price to be paid is that the independent oscillators are not of the same frequency;
the normal modes of vibration of a solid have a wide range of frequencies. These
modes are not related to the motion of single atoms, but to the collective motion of all
atoms in the crystal – vibrational modes or sound waves.
Einstein’s Model of a Solid
In 1907 Einstein, in the first application of quantum theory to a problem other than
radiation, modeled a solid body containing N atoms as a collection of 3N harmonic
oscillators. The partition function of a single oscillator:
n



1 

  h  
Z1   exp   n    exp    n  h   exp  
 exp  h 
2
2

 

 n 0
n 0
n 0

  h 
exp  

1
2 
  h 


 cosech

1  exp   h  2
 2 
The mean energy:
E 
The oscillators are
independent of each
other, thus:
Z  Z1
N
1


1
1
  h 

ln Z1  h coth 
  h  

2
 2 
 2 exp  h   1 
This looks familiar: the same energy would have a photon of frequency .
The internal energy is not a directly measurable quantity, and instead we measure the
heat capacity:
CV T  
Limits:
dU
U
exp  h 
2
 k B  2
 U  3 N E   3 Nk B  h 
dT

exp  h   12
high T (kBT>>h):
low T (kBT<<h):
CV T   3NkB
equipartition
CV T   3NkB  h  exp   h   exp  T 
2
The Einstein model predicts much too low a heat capacity at low temperatures!
Debye’s Theory of the Heat Capacity of Solids
Debye’s model (1912) starts from the opposite point of view, treating the
solid as a continuum, i.e., the atomic structure is ignored. A continuum
has vibrational modes of arbitrary low frequencies, and at sufficiently low
T only these low frequency modes are excited. These low frequency
normal modes are simply standing sound waves.
Nobel 1936
If we quantize this elastic distortion field, similar to the
quantization of the e.-m. field, we arrive at the concept of
phonons, the quanta of this elastic field. For the thermal
phonons, the wavelength increases with decreasing T :
k B T  h  h
cs

cS – the sound velocity
hcs
6.6 10 34  2 103
7

 T  1K  ~
~
10
m  0.1m
k BT
1.4 10 34 1
These low-energy modes remain active at low temperatures when the high-frequency
modes are already frozen out. Large values of  that correspond to these modes justify
the use of a continuum model.
There is a close analogy between photons and phonons: both are “unconserved”
bosons. Distinctions: (a) the speed of propagation of phonons (~ the speed of sound
waves) is by a factor of 105 less than that for light, (b) sound waves can be longitudinal
as well as transversal, thus 3 polarizations (2 for photons), and (c) because of
discreteness of matter, there is an upper limit on the wavelength of phonons – the
interatomic distance.
Density of States in Debye Model
For a macroscopic crystal, the spectrum of sound waves is almost continuous, and we
can treat  is a continuous variable. As in the case of photons, we start with the density of
states per unit frequency g(). The number of modes per unit volume with the wave
number < k:
k3
- multiplied by 3 since a sound wave in a solid can have
G k   3
2
three polarizations (two transverse and one longitudinal).
6
ck
  s
2
1  2 


G    3
2 
6  cs 
3
dG   12 2
g   

3
d
cs
- this eq. only holds
for sufficiently low 
(= large wavelengths and the continuous approximation is valid). There is
also an upper cut-off for the frequencies (  interatomic distances), the
so-called Debye frequency D, which depends on the density n:
D
 g  d  3n  D
0
1/ 3
 3n 
 cs 

 4 
cS ~3 km/s, a~0.2 nm,
D~1013 Hz
Each normal mode is a quantized harmonic oscillator. The mean energy of each mode:
D
D
1

1
h g  
E  h  




U

g

E
T
,

d


U

and

0
0
0 exp  h   1d
 2 exp  h   1
is the total energy per unit volume. The U0 term comes from the zero-point motion of atoms.
It reduces the cohesive energy of the solid (the zero point motion in helium is sufficient to
prevent solidification at any T at normal pressure), but since it does not depend on T, it
does not contribute to C. Note that we ignored this term for phonons, where it is . In QED,
this unobservable term is swept under the rug by the process known as renormalization.
The Heat Capacity in Debye’s Model
At low temperatures, we can choose the upper limit as  (the high-frequency modes
are not excited, the energy is too low). How low should be T :
hcs
6.6 10 34  2 103
T 

~ 1000 K
 23
10
k B a 1.4 10 110
h g  
4 5 k BT 
U  U0  
d  U 0 
3
exp  h   1
5 h 3 cs
0

4
dU 16 5 k B
3
C

T
3
dT
5 h 3 cs
4
The low-T heat capacity:
Thus, Debye’s model predicts that in the limit of sufficiently
low T, the heat capacity due to vibrations of the crystal lattice
(in a metal electrons also contribute to C) must vary as T3,
and not as 2exp(- h/kBT), as in Einstein’s model. (Roughly
speaking, the number of phonons ~ T3, their average energy
is proportional to T).
At high temperatures, all the modes are excited (the
number of phonons does not increase any more), and the
heat capacity approaches the equipartition limit, C=3NkB .
D
  2
U  U0  A 
d  U 0  3NkBT

0
Debye Temperature
The material-specific parameter is
the sound speed. If the temperature
is properly normalized, the data for
different materials collapse onto a
universal dependence:
4
 T 
16 5 k B
12 4
3


C
T

Nk
B
3
5

5 h 3cs
 D
The normalization
factor is called the
Debye temperature:
3
1/ 3
hc  3n 
D  s 

k B  4 
It is related to the maximum frequency D, the Debye
frequency:
1/ 3
D
 3n 
 cs 

4



k BTD  h D
The higher the sound speed and the density of ions,
the higher the Debye temperature. However, the real
phonon spectra are very complicated, and D is better
to treat as an experimental fitting parameter.
Problem (blackbody radiation)
The spectrum of Sun, plotted as a function of energy, peaks at a photon energy of
1.4 eV. The spectrum for Sirius A, plotted as a function of energy, peaks at a
photon energy of 2.4 eV. The luminosity of Sirius A (the total power emitted by its
surface) is by a factor of 24 greater than the luminosity of the Sun. How does the
radius of Sirius A compare to the Sun’s radius?
L  4R  T
2
4
R
L
T2
The temperature, according to Wien’s law, is
proportional to the energy that corresponds to the peak
of the photon distribution.
LSirius / LSun
RSirius
24


 1.67
RSun TSirius / TSun 2 2.4 / 1.42
Final 2006 (blackbody radiation)
The frequency peak in the spectral density of radiation for a certain distant star
is at 1.7 x 1014 Hz. The star is at a distance of 1.9 x 1017 m away from the Earth
and the energy flux of its radiation as measured on Earth is 3.5x10-5 W/m2.
a)
b)
c)
d)
(5) What is the surface temperature of the star?
(5) What is the total power emitted by 1 m2 of the surface of the star?
(5) What is the total electromagnetic power emitted by the star?
(5) What is the radius of the star?
h
6.6 10 34 1.7 1014
T

 3000 K
 23
2.7 k B
1.38 10  2.7
(a)


(b)
J   T 4  5.7 10 8 W / K 4  m 2  3000 K 4  4.6 106 W / m 2
(c)
power  4 r 2 J r   4  1.9 1017 m2  3.5 105W / m2  1.6 1031W
(d)

 
4 R J RS  4 r J r 
2
S
2
4

J r 
3.5 10 5
17
11
RS  r
 1.9 10 m 

5
.
2

10
m
6
J RS
4.6 10
 
Problem 2006 (blackbody radiation)
The cosmic microwave background radiation (CMBR) has a temperature of
approximately 2.7 K.
(a) (5) What wavelength λmax (in m) corresponds to the maximum spectral density
u(λ,T) of the cosmic background radiation?
(b) (5) What frequency max (in Hz) corresponds to the maximum spectral density
u(,T) of the cosmic background radiation?
(c) (5) Do the maxima u(λ,T) and u(,T) correspond to the same photon energy? If
not, why?
(d) (15) What is approximately the number of CMBR photons hitting the earth per
second per square meter [i.e. photons/(s·m2)]?
(a)
(b)
(c)
max
hc
6.6 1034  3 108
3



1
.
1

10
m  1.1mm
 23
5 k BT 5 1.38 10  2.7
 max
k BT 2.8 1.38 10 23  2.7
11
 2.8


1
.
58

10
Hz
34
h
6.6 10
hc
max
 1.8 10  22 J
h max  1.04 10 22 J
The maxima u(λ,T) and u(,T) do not correspond to the same photon energy. As 
increases, the frequency range included in unit wavelength interval increases as 2,
moving the peak to shorter wavelengths:
u  , T d  u  , T d
c
8 hc
 d




u

,
T

 d
2 
5
1
 hc 
  1
exp 

k
T
 B 
Problem 2006 (blackbody radiation) cont.
(d)


J   TCMBR  5.7 108 W / K 4  m2  2.7 K 4  3 106W / m2
4
4
uT 
8 5 k BT  hc 
4



k BT  2.7 k BT
3
3
N
15hc  8 k BT   2.4 15  2.4
4
3
W 
J 2 
3 106
m 
 photons 

16 photons
N



3

10

2
 J 
2.7 1.38 1023  2.7
s  m2
 sm 
Problem (blackbody radiation)
The planet Jupiter is a distance of 7.78 x 1011 meters from the Sun and is of radius
7.15 x 107 meters. Assume that Jupiter is a blackbody even though this is not
entirely correct. Recall that the solar power output of Sun is 4 x 1026 W.
a) What is the energy flux of the Sun's radiation at the distance of Jupiter's orbit?
PSun
4 1026W
J

 52.6W / m2
2
2
4Rorbit 4 7.78 1011 m


b) Recognizing that Sun's energy falls on the geometric disk presented to the Sun
by the planet, what is the total power incident on Jupiter from the Sun?


2
PJupiter  J  RJupiter
 52.6W / m2   7.15 107 m  8.4 1017W
2
c) Jupiter rotates at a rather rapid rate (one revolution per 0.4 earth days) and
therefore all portions of the planet absorb energy from the Sun. Hence all
portions of the surface of this planet radiate energy outward. On the basis
of this information find the surface temperature of Jupiter.
2
4
PJupiter  4RJupiter
TJupiter
 PJupiter
TJupiter  
 4R 2 
Jupiter

1/ 4




1/ 4


8.4 1017W


2
7
8
4 2
 4 7.15 10 m 5.76 10 W / K m 


 123K
d) Estimates of the surface temperature of Jupiter indicate that it is clearly above
500 Kelvins. On the basis of your answer to c) what might you conclude
about this planet?
There must be some other source of energy to produce a surface temperature of
500 K.
Problem (blackbody radiation)
Tungsten has an emissivity of 0.3 at high temperatures. Tungsten filaments operate
at a temperature of 4800 K.
a) At what frequency does a tungsten filament radiate the most energy?
b) What is the power/unit surface emitted by the tungsten filament?
c) If the surface area of a tungsten filament is 0.01 cm2 what is the power output of
the bulb?
d) What is the energy flux of radiation emitted by the filament 2 meters from the
bulb?
e) What fraction of radiation incident on a tungsten filament is reflected?
[Answers: a) 2.88 x 1014 Hz, b) 9.03 x106 W/m2, c) 9.03 watts, d) 0.180 watts, and e)
70%.]
Problem (radiation pressure)
(a)
(b)
(c)
(d)
At what temperature will the pressure of the photon gas be equal to 105 Pa (=
one bar )?
At what temperature will the pressure of the photon gas be equal to 10-5 Pa?
The temperature at the Sun’s center is 107 K. What is the pressure of the
radiation? Compare it to the pressure of the gas at the center of the Sun, which
is 1011 bar.
Calculate the pressure of the Sun’s radiation on the Earth’s surface, given that
the power of the radiation reaching earth from the Sun is 1400 W/m2.
4 4
P
T
3c
(b)
(c)
1/ 4
 3cP 
T 

 4 
P=10-5
Pa
 3  3 108  1 105
(a) P=105 Pa T  
8
 4  5.7 10
 3  3 108  1 10 5
T  
8
 4  5.7 10
 
4  5.7 108  107
P
3  3 108
1/ 4



 1.4  105 K
1/ 4



 450 K
4
 2.5 1012 Pa  2.5 107 bar
- at this T, the pressure of radiation is still negligible in the balance for mechanical equilibrium
(d)
P
1
4
u T  u T   I
3
c
P
4I
4 1400

 6 10 6 Pa
8
3c 3  3 10
Problem (blackbody radiation)
The black body temperature is 3000K. Find the power emitted by this black body
within the wavelength interval  = 1 nm near the maximum of the spectrum of
blackbody radiation.
c
Jd  u  , T d
4
hc
8 hc
1
max 
u  , T d  5
d
5 k BT

 hc 
  1
exp 
  k BT 
the power emitted by a unit area in all directions:
near the maximum:
c
c 8 hc5k BT 
1
2 c 5k BT 
1
J d    u  , T d 
d


d
5
4




4
4
exp
5

1
exp
5

1
hc 
hc 
5
J d  

2  3 108  5 1.38 10  23  3000
6.6 10
34
 3 108

4
5

5

1
110 9  3.1103 W/m 2
exp 5  1