Download Part IV - TTU Physics

Phonons:
Quantum Mechanics of Lattice Vibrations
What is a Phonon?
• We’ve seen that the physics of lattice vibrations in a
crystalline solid reduces to a CLASSICAL normal mode
problem.
• The goal of the entire discussion has been to find
the normal mode vibrational frequencies of the solid.
• In the harmonic approximation, this is achieved by first
writing the solid’s vibrational energy as a system of coupled
simple harmonic oscillators & then finding the classical
normal mode frequencies & ion displacements for that system.
• Given the results of these classical normal mode calculations,
in order to treat some properties of the solid,
it is necessary to QUANTIZE
these normal modes.
• These quantized normal modes of vibration are called
PHONONS
• PHONONS are massless quantum mechanical particles
which have no classical analogue. They behave like
particles in momentum space or k space.
• Phonons are one example of many like this in many areas of
physics. Such quantum mechanical particles are often called
“Quasiparticles”
Examples of other Quasiparticles:
•
•
•
•
•
Photons: Quantized Normal Modes of electromagnetic waves.
Rotons: Quantized Normal Modes of molecular rotational excitations.
Magnons: Quantized Normal Modes of magnetic excitations in solids
Excitons: Quantized Normal Modes of electron-hole pairs
Polaritons: Quantized Normal Modes of electric polarization
excitations in solids
+ Many Others!!!
Comparison of Phonons & Photons
PHONONS
• Quantized normal modes
of lattice vibrations. The
energies & momenta of
phonons are quantized
E phonon 
p phonon
h s

h


Phonon wavelength:
λphonon ≈ a0 ≈ 10-10 m
Comparison of Phonons & Photons
PHONONS
• Quantized normal modes
of lattice vibrations. The
energies & momenta of
phonons are quantized
E phonon 
p phonon
h s

h


PHOTONS
• Quantized normal modes
of electromagnetic waves.
The energies & momenta
of photons are quantized
E photon 
hc
p photon 
h


Phonon wavelength:
Photon wavelength: (visible)
λphonon ≈ a0 ≈ 10-10 m
λphoton ≈ 10-6 m
Quantum Mechanical Simple Harmonic Oscillator
• Quantum mechanical results for a simple harmonic
oscillator with classical frequency ω:
• The energy is quantized:
 1
 n   n   n = 0,1,2,3,..

The energy ε
levels are
equally spaced!




2
Often, we consider εn as being constructed by adding n
excitation quanta of energy  to the ground state.
Ground state energy
1
E
 0  
of the oscillator.
2
If the system makes a transition from a lower energy level
to a higher energy level, it is always true that the change in
energy is an integer multiple of 
Phonon absorption
or emission.
ΔE = (n – n΄) 
n & n ΄ = integers
In complicated processes, such as phonons interacting with
electrons or photons, it is known that The Number of
Phonons is not conserved. That is, they can be created and
destroyed during such interactions.
Thermal Energy & Lattice Vibrations
• As we’ve been discussing in detail, the atoms in a
crystal vibrate about their equilibrium positions.
• This motion produces vibrational waves.
The amplitude of this vibrational motion
increases as the temperature increases.
• In a solid, the energy associated with these
vibrations is called the Thermal Energy
• A knowledge of the thermal energy is
fundamental to obtaining an understanding many
of the basic (thermodynamics) properties of solids.
Examples: Heat Capacity, Entropy,
Helmholtz Free Energy,
Equation of State, etc....
• A relevant question is how do we calculate this
thermal energy?
• Also, we would like to know how much thermal
energy is available to scatter a conduction electron
in a metal or semiconductor.
• This is important; this scattering contributes to electrical
resistance in the material.
• Most important is that this thermal energy
plays a fundamental role in determining the
Thermal Properties of a Solid
• A knowledge of how the thermal energy
changes with temperature gives an
understanding of the heat energy which is
necessary to raise the temperature of the
material.
• An important, measureable property of a solid
is it’s
Specific Heat or Heat Capacity
Lattice Vibrational Contribution
to the Heat Capacity
• The thermal energy is the dominant contribution to the
heat capacity in most solids. In non-magnetic insulators,
it is the only contribution. Some other contributions are:
Conduction Electrons in metals & semiconductors.
Magnetic ordering in magnetic materials.
• A calculation of the vibrational contribution to the
thermal energy & heat capacity of a solid has 2 parts:
1. Evaluation of the contribution of a single
vibrational mode.
2. Summation over the frequency distribution of
the modes.
Classical Theory of the Heat Capacity of Solids
• Assume that each atom is bound to its neighbors by
harmonic forces. When heated, the atoms vibrate
around their equilibrium positions as a set of coupled
harmonic oscillators. Assuming Classical (Maxwell
Boltzmann) Statistics & using the Equipartition
Theorem, the thermal average energy for a 1D
oscillator is kBT. So, the thermal average energy
per atom, regarded as a 3D oscillator, is 3 kBT, so
the energy per mole in the solid is:
<> = 3NkB T
N is Avagadro’s number,
kB is Boltzmann constant
• The classical thermal average the energy
per mole in the solid:
<> = 3NkB T
• Formally, the molar heat capacity at
constant volume, Cv , is given by the
temperature derivative of the mean energy:
d
Cv 
dT
So, classically, Cv = 3R
(R = gas constant)
This is known as the Dulong-Petit Law
Vibrational Specific Heat of Solids
cp Data at T = 298 K
The Molar Heat Capacity
Experimentally, the Dulong-Petit Law, however, is
found to be valid only at high temperatures.
15
Quantum Thermal Energy & Heat Capacity
• The Quantized Energy of a single
simple harmonic oscillator is:
 1
n   n   
 2
• First, calculate the mean thermal energy of one mode,
then, sum over modes to find the mean thermal energy
due to all modes. From the Canonical Ensemble of
Statistical Mechanics, the Mean Energy of a harmonic
oscillator & & hence of a lattice mode of frequency ω at
temperature T has the form: _
   Pn n
n
• In the Canonical Ensemble, the probability Pn of the
oscillator being in energy level n at temperature T is
proportional to:
exp( / k T )
n
B
Now, some straightforward math manipulation!
Thermal Averaged Energy: _  P 
Putting in the explicit form gives:

n
n
n
1
 
1


n


exp

n


/
k
T




B 
 
_
2
2



n 0 




 
1

exp

n


/
k
T


B 
 
2

n 0
 


(*)
The Partition Function for this problem is:
1 
z   exp[(n  )
]
2 k BT
n 0

z  e
 / 2 k BT
 e 3
 / 2 k BT
 e 5
 / 2 k BT
z  e
 / 2 k BT
(1  e 
 / k BT
 e 2
 / k BT
z  e
 / 2 k BT
(1  e 
 / k BT
) 1
 .....
 .....
The thermal averaged energy can thus be written
1 z

 k BT 2
(ln z )
z T
T
_
 e   / 2 k BT 
2 
  k BT
ln 

T  1  e   / kBT 
_
  k BT 2
    / 2 k BT
  / k BT

ln
e

ln
1

e




T
_
  

  
  / k BT
  k BT 2   

ln
1

e




T
2
k
T

T
B

 

 k B   / k BT 

 2k  k 2T 2 e
 1
_
e
2
B
B

  k BT  2 2 

  / k BT
4
k
T
2
 B
1  e

1  e


_
  k BT 2
_
Finally, the   1  
result is:
2
e

x'
(ln x) 
x
x
 / k BT
 / k BT

 / k BT
1


_
1
  
2
e
 / k BT
1
(1)
• This is the Mean Phonon Energy. The first term in (1) is
called the Zero-Point Energy. As mentioned before, even
at 0 K the atoms vibrate in the crystal & have a Zero
Point Energy. This is the minimum energy of the system.
• The thermal average number of phonons n(ω) at
temperature T is given by The Bose-Einstein (or Planck)
Distribution, & the denominator of the second term in (1)
is often written:
1
n( ) 

e
kBT
1
_
1
  
2
e

 / k BT
1
(1)
1
n( ) 

e
kBT
1
(2)
• By using (2) in (1), (1) can be rewritten:
<> = ћω[n() + ½]
(1)
• In this form, the mean energy <> looks analogous to a
quantum mechanical energy level for a simple harmonic
oscillator. That is, it looks similar to:
• So the 2nd term in the mean energy (1) is interpreted as
The number of phonons at temperature
T & frequency ω.

1

2
kBT
Mean energy
of a harmonic
oscillator as a
function of T.
T
Consider the Low Temperature Limit:
  kBT
_
1
  
2
e
_

 / k BT
T

0,
Exponential

1
1
1
  
2
Zero Point Energy

1

2
Mean energy
of a harmonic
oscillator as a
function of T.
kBT
T
Consider the High Temperature Limit:
 
<< kBT

_
1
  
2
e
 / k BT
1
Taylor’s Series expansion of the exponential:
2
x
ex  1 x 
 ..........
2!

e
k BT

 1
k BT

1

2
Mean energy
of a harmonic
oscillator as a
function of T.
kBT
T
High Temperature Limit:

e

1

<<
  kBT
k T

k BT
B
This gives:
_
1
2
   


1
1
k BT
OR
_
Finally:
_
1
    k BT
2
  kBT

1

2
kBT
Mean energy
of a harmonic
oscillator as a
function of T.
T
High Temperature Limit:
_
<<
  kBT
 k T
B
• In this limit, the Mean Energy is independent of frequency.
• This is the classical limit because the energy steps
are very small compared with the harmonic oscillator
energy. So, the high temperature limit gives the thermal
energy of the classical 1D harmonic oscillator,
calculated with classical (Maxwell-Boltzmann) statistics.
Heat Capacity Cv
• The heat capacity Cv is found by differentiating
the average phonon energy
_
1

     / k BT
2
e
1
Cv 
Let
d

dT
 


 kB
 k BT 

e

k
k BT

e
2

1
2
kBT


Cv  k B
2
 kBT 

2

e

e
k BT
kBT

1
2

eT
 
Cv  k B   
 T  e T 1
2


2

eT
 
Cv  k B   
 T  e T 1
2



2

k
Cv
kB

2
Area =

kB
T
• The specific heat in this
approximation vanishes
exponentially at low T &
tends to the classical
value at high T.
• These features are
common to all quantum
systems; the energy tends
to the zero point-energy
at low T & to the classical
value at high T.
• The specific heat at constant volume depends on
temperature as shown qualitatively in figure below. At
high temperatures, T, Cv is close to 3R, where R is the
universal gas constant. R  2 cal/K-mole. So, at high
temperatures Cv  6 cal/K-mole.
3R
Cv
T, K
• From the figure. it can be seen
that Cv = 3R at highT
regardless of the substance.
This fact is known as the
Dulong-Petit law. This states
that the specific heat of a given
number of atoms of any solid is
independent of temperature & is
the same for all materials!
Einstein Model for the Heat Capacity
of Lattice Vibrations
• The theory of Cv(T) proposed by Einstein was the
first use of quantum theory to understand
the physics of solids.
• He made the (absurd!) assumption that all 3N vibrational
modes of a 3D solid of N atoms have the same
frequency, so that the solid has a heat capacity 3N times

eT
 
Cv  k B   
 T  e T 1
2


2
Einstein Model for Lattice Vibrations in a Solid
Cv vs T for Diamond
Einstein, Annalen der Physik 22 (4), 180 (1907)
Points:
Experiment
Curve:
Einstein Model
Prediction
• In this model, the atoms are treated as
independent oscillators, but the energy of the
oscillators is quantum mechanical.
• This refers to an isolated oscillator, but the atomic
oscillators in a solid are not isolated. They
continually exchange energy with neighboring atoms.
• Even this crude model gives the correct limit at
high temperatures: The heat capacity of the
Dulong-Petit law:
Cv = 3R
• At high temperatures, all crystalline solids have a
specific heat of 6 cal/K per mole; they require 6
calories per mole to raise their temperature 1 K.
• This agreement between observation and classical theory
breaks down if T is low. Experiments show that at room
temperature & below the specific heat of crystalline
solids is strongly temperature dependent.
• In each material, Cv
asymptotically approaches
the classical value 3R at
high T. But, at low T, Cv
decreases to zero. This
completely contradicts
T the classical result.
Cv
6
cal
Kmol

kB
Cv  3R
• The Einstein model also correctly gives a
specific heat tending to zero at T  0.
• But the, temperature dependence near
T= 0 does not agree with experiment.
• By more accurately taking into account the
actual distribution of vibrational
frequencies in a solid, this can be
corrected using a model due to Peter Debye.
Thermal Energy & Heat Capacity
Debye Model
Density of States
From Quantum Mechanics, if a particle is constrained;
the energy of particle can only have discrete energy
values. It cannot increase infinitely from one value to
another. It has to go up in steps.
• These steps can be so small depending on the system
that the energy can be considered as continuous.
• This is the case of classical mechanics.
• But on atomic scale the energy can only jump by a
discrete amount from one value to another.
Definite energy levels
Steps get small
Energy is continuous
• In some cases, each particular energy level can be
associated with more than one different state (or
wavefunction )
• This energy level is said to be degenerate.
• The density of states  ( ) is the number of
discrete states per unit energy interval, and so
that the number of states between  and   d
is  ( )d 
.
There are two sets of waves for solution;
• Running waves
• Standing waves
Running waves:
0

4
L

2
L
2
L
4
L
6
L
k
These allowed k wavenumbers corresponds to the running
waves; all positive and negative values of k are allowed. By
means of periodic boundary condition
an integer
Na 2
2
2
L  Na  p   

k 
pk 
p
p
k
Na
L
Length of
the 1D chain
These allowed wavenumbers are uniformly distibuted in k at a
density of R  k  between k and k+dk.
running waves
L
 R  k  dk  dk
2
5
L
Standing waves:
0

L
2
L
k
3
L
6
L
7
L
4
L
0
3
L

L
2
L
In some cases it is more suitable to use standing waves,i.e. chain
with fixed ends. Therefore we will have an integral number of half
wavelengths in the chain;
L
n
2
2 n
n
;k 
k 
k 
2

2L
L
These are the allowed wavenumbers for standing waves; only
positive values are allowed.
2
k
p
L
for
running waves
k

L
p
for
standing waves
These allowed k’s are uniformly distributed between k and k+dk
at a density of  S (k )
 S (k )dk 
 R  k  dk 
L

dk
L
dk
2
DOS of standing wave
DOS of running wave
•The density of standing wave states is twice that of the running waves.
•However in the case of standing waves only positive values are
allowed
•Then the total number of states for both running and standing waves
will be the same in a range dk of the magnitude k
•The standing waves have the same dispersion relation as running
waves, and for a chain containing N atoms there are exactly N distinct
states with k values in the range 0 to  / a .
The density of states per unit frequency range g():
• The number of modes with frequencies  and +d
will be g()d.
• g() can be written in terms of S(k) and R(k).
dR
dn
modes with frequency from  to +d corresponds
modes with wavenumber from k to k+dk
dn  S (k )dk  g ( )d
dn   R (k )dk  g ( )d;
Choose standing waves to obtain
g ( )
g ( )  S (k )
dk
d
Let’s remember dispersion relation for 1D monoatomic lattice
4K
2 ka
 
sin
m
2
2
d
2a

dk
2
K
ka
cos
m
2
K
ka
2
sin
m
2
g ()  S (k )
1
K
ka
a
cos
m
2
1 m
1
g ( )  S (k )
a K cos  ka / 2 
sin x  cos x  1  cos x  1  sin x
2
2
2
1 m
g ( )  S (k )
a K
1
 ka 
1  sin 2  
 2
4
4
 ka 
2  ka 
cos    1  sin  
 2
 2
Multibly and divide
Let’s remember:
g ( )  S (k )
g ( ) 
1
a
2
4 K 4 K 2  ka 

sin  
m
m
 2
1
L 2
2
 a max
2
True density of states
 S (k )dk 
L

dk
L  Na
4K
 ka 
2 
sin 2  
m
 2 
4K
2
max 
m
g ( )
g ( ) 
N

m
K
2N


2
max


2 1/ 2
True density of states by
means of above equation

max
K
2
m
K

m
constant density of states
K
True DOS(density of states) tends to infinity at max  2 ,
m
since the group velocity d / dk goes to zero at this value of .
Constant density of states can be obtained by ignoring the
dispersion of sound at wavelengths comparable to atomic spacing.
The energy of lattice vibrations will then be found by
integrating the energy of single oscillator over the distribution
of vibration frequencies. Thus

1
    
2
e
0

 / kT

  g   d
1 
2N
Mean energy of a harmonic
oscillator


2
max

 for 1D
2 1/ 2
One can obtain same expression of g ( ) by means of using
running waves.
It should be better to find 3D DOS in order to compare the
results with experiment.
3D DOS
• Let’s do it first for 2D
• Then for 3D.
• Consider a crystal in the shape of 2D box with crystal lengths
of L.
ky
y

L
0
+
-
-
+
+
L

L
L
kx
x
Standing wave pattern for a 2D
box
Configuration in k-space
•Let’s calculate the number of modes within a range of
wavevector k.
•Standing waves are choosen but running waves will lead
same expressions.
•Standing waves will be of the form
U  U 0 sin  k x x  sin  k y y 
• Assuming the boundary conditions of
•Vibration amplitude should vanish at edges of
x  0; y  0; x  L; y  L
Choosing
p
q
kx 
; ky 
L
L
positive integer
ky
y
+
L
0
-
-
+
+
-
L


L
L
kx
x
Standing wave pattern for a
2D box
Configuration in k-space
•The allowed k values lie on a square lattice of side
positive quadrant of k-space.
in
 /the
L
•These values will so be distributed uniformly with a density of
2
L
/

perunit area.

• This result can be extended to 3D.
L
Octant of the crystal:
kx,ky,kz(all have positive values)
The number of standing waves;
L
3
V 3
L 3
s  k  d k    d k  3 d k

 
1
 4 k 2 dk
L /
8
V 1
3
 s  k  d k  3  4 k 2 dk
 8
2
Vk
 s  k  d 3k  2 dk
2
ky
Vk 2
S  k   2
2
3
L
kz
dk
k
kx
2
Vk
•  k  
is a new density of states defined as the number of
2
2
states per unit magnitude of in 3D.This eqn can be obtained by
using running waves as well.
• (frequency) space can be related to k-space:
g   d    k  dk
g      k 
dk
d
Let’s find C at low and high temperature by means of using the
expression of
g . 
High and Low Temperature Limits
  3NkBT
•
Each of the 3N lattice
modes of a crystal
containing N atoms
d
C
dT
This result is true only if
C  3NkB
T


kB
At low T’s only lattice modes having low frequencies can be
excited from their ground states;

long 
Low frequency
sound waves
  vs k
0

k
a
vs 

k

k
1
dk 1
vs    

k
 vs
d vs
 2 
V 2 
vs  1

g   
2 2 vs
and
Vk 2 dk
g    2
2 d
at low T
vs depends on the direction and there are two transverse, one
longitudinal acoustic branch:
V2 1
V2  1 2 
g   
 g   
 3
2
3
2  3
2 vs
2  vL vT 
Velocities of sound in longitudinal and transverse direction


1
    
2
e
0

 / kT
1
    
2
e
0

  g   d
1 

 / kT
Zero point energy =  z
2
 V  1 2 
  2  3  3  d
 1  2  vL vT 
x



V  1 2  
3
   z  2  3  3      / kT
d 


2  vL vT   0  e
 1



3
V  1 2   k BT   4
  z  2  3  3 
3
2  vL vT 
15
4

e 
0

e 
3
/ kT
0

/ kT
3
1
1

d  
d 
0
 k BT 
3
 k BT  3

 x kT


B
dx
x
e 1

k BT

k BT
x
d 
k BT
dx
4 
x3
0 e x  1dx
 4 15
 1 2   kBT 
d
2
2
Cv 
 V  kB  3  3  

dT 15
v
v

 L
T 
3
at low temperature
How good is the Debye approximation at low T?
 1
d
2
2   k BT 
2
Cv 
 V  kB  3  3  

dT 15
v
v


 L
T 
The lattice heat capacity of solids
thus varies as T 3at low
temperatures; this is referred to
as the Debye T 3 law. Figure
illustrates the excellent
agreement of this prediction with
experiment for a non-magnetic
insulator. The heat capacity
vanishes more slowly than the
exponential behaviour of a
single harmonic oscillator
because the vibration spectrum
extends down to zero frequency.
3
The Debye interpolation scheme
The calculation of g ( ) is a very complicated calculation
for 3D, so it must be calculated numerically.
Debye obtained a good approximation to the resulting heat
capacity by neglecting the dispersion of the acoustic waves, i.e.
assuming
  s k
for arbitrary wavenumber. In a one dimensional crystal this is
equivalent to taking g ( ) as given by the broken line of
density of states figure rather than full curve. Debye’s
approximation gives the correct answer in either the high and
low temperature limits, and the language associated with it is
still widely used today.
The Debye approximation has two main steps:
1. Approximate the dispersion relation of any branch by a linear
extrapolation of the small k behaviour:
Einstein
approximation to
the dispersion
Debye
approximation
to the dispersion
  vk
Debye cut-off frequency  D
2. Ensure the correct number of modes by imposing a cutoff frequency  D , above which there are no modes. The
cut-off freqency is chosen to make the total number of
lattice modes correct. Since there are 3N lattice 
vibration
D
modes in a crystal having N atoms, we choose
so
that
2
D
 g ( )d  3N
g ( ) 
0
V
1
2
(

)D3  3 N
2
3
3
6
vL vT
g ( ) 
9N
D3
V 1 2
( 3  3)
2
2 vL vT

V 1 2 D 2
(  )  d  3 N
2 2 vL3 vT3 0
V
1
2
3N
9N
(

)

3

2 2 vL3 vT3
D3
D3
2
g ( ) /  2

The lattice thermal energy is
E  (
0
becomes
and,
9N
E 3
D
1

   / kBT )g ( )d
2
e
1
D
D
D
3


1

9
N

3
2
0 ( 2   e  / kBT  1) d  D3  0 2 d  0 e  / kBT  1d 


9
9N
E  N D  3
8
D
D
 3 d
e
0
/ k BT
1
First term is the estimate of the zero point energy, and all T
dependence is in the second term. The heat capacity is obtained by
differentiating above eqn wrt temperature.
C
The heat capacity is
9
9N
E  N D  3
8
D
D

0
 d
e  / k BT  1
3
dE
dT
dE 9 N
CD   3
dT D
D

0
 4 e  / k BT
d
2
2
kBT  e  / kBT  1
2
Let’s convert this complicated integral into an expression for the
specific heat changing variables to
x

k BT
d
kT

dx
and define the Debye temperature

D 
D
kB
kT
x
The Debye prediction for the lattice specific heat

dE 9 N kBT  kBT  
CD 
 3

 
2 
dT D

  kBT 
4
 T 
CD  9 Nk B 

 D 
where
D 
2
3  /T
D
D
kB

0
D / T

0
x 4e x
 e 1
x
x 4e x
e
x
 1
2
2
dx
dx
How does C D limit at high and low temperatures?
High temperature
x is always small
T
D
x2
x3
e  1 x 


2!
3!
x
x 4 (1  x)
2



x
2
2
2
x
x
1

x

1

 e  1 
x 4e x
T
x 4 (1  x)
 T 
 D  CD  9 Nk B 


 D
3  /T
D

0
x 2 dx  3Nk B
How does C D limit at high and low temperatures?
Low temperature
D
T
For low temperature the upper limit of the integral is infinite; the
integral is then a known integral of
.
4 4 /15
T
 T 
 D  CD  9 Nk B 


 D
3  /T
D

0
x 4e x
e
x
 1
2
dx
We obtain the Debye T 3 law in the form
CD 
12 Nk B  T 


5

 D
4
3
Lattice heat capacity due to Debye interpolation scheme
 T 
CD  9 Nk B 


Figure shows the heat capacity  D 
3 D / T

0
x 4e x
e
x
between the two limits of high and
low T as predicted by the Debye
interpolation formula.
 1
2
dx
C
3 Nk B T
1
Because it is exact in both high and low T limits
the Debye formula gives quite a good
representation of the heat capacity of most
solids, even though the actual phonon-density of
states curve may differ appreciably from the
Debye assumption.
Lattice heat capacity of a solid as
predicted by the Debye interpolation
scheme
1
T / D
Debye frequency and Debye temperature scale with the velocity of sound in the solid.
So solids with low densities and large elastic moduli have high  D . Values of  Dfor
various solids is given in table. Debye energy D can be used to estimate the
maximum phonon energy in a solid.
Solid
D ( K )
Ar
Na
Cs
Fe
Cu
Pb
C
KCl
93
158
38
457
343
105
2230
235
Cv vs T for Diamond
Points:
Experiment
Curve:
Einstein Model
Prediction
Vibrational Density of States (Aluminum)
Solid Curve:
From X-Ray
Experiment
Dashed Curve:
Debye
Approximation
Debye Density of States
Cv vs T
Solid Curve:
Debye
Approximation
Dashed Curve:
Einstein Model