3 photons phonons

8/13/2019 3 Photons Phonons

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Statistical and Low Temperature Physics (PHYS393)

3. Photons and phonons

Kai Hock

2011 - 2012

University of Liverpool


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Contents

3.1 Phonons

3.2 Photons

3.3 Einsteins Model

3.4 Useful Integrals

Photons and phonons 1


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3.1 Phonons



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Specific heat capacities of solids

Previously, we have looked at Einsteins model of the 3-D

simple harmonic oscillator, in which each atom in a solid is inits own potential well. The predicted heat capacity was

accurate at high temperature, but fails at low temperature.

In 1912, Peter Debye, a Dutch physicist working in Germany,

produced a theory which turns out to be very accurate both

hot and cold.



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Atoms vibrating in a solid

In Debyes model, we think of the atoms in a solid as

connected by springs. They vibrate in 3D in a complicated way.

The problem can be simplified by treating the vibration as a

superposition of waves of different frequencies.

If we suppose that the atoms are fixed at the edge of the solid,

the frequency would be quantised. A quantum of this vibration

energy is called a phonon.



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Quantised thermal waves

We can count the number of states for the phonons, like wedid for atoms in an ideal gas, or electrons in a metal.

Unlike electrons, however, the exclusion principle does not

apply to phonons.

So each energy state can be occupied by any number of

phonons.

This means that we need yet another way to find the

macrostate. This was worked out by Bose and Einstein in the1920s.



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Bose-Einsten Statistics

The method is similar to what we have used for the ideal gas.

There, we have made the assumption that there are many

more energy levels than there are particles. This has meant

that it is unlikely for two gas atoms to occupy the same state.

We do not make this assumption now, since we would be

particularly interested in the low temperature behaviour.

Since the exclusion principle does not apply to phonons, more

than one phonons can occupy the same state.



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Bose-Einsten Statistics

Suppose, in the energy bundle i, that there are gi levels and ni

phonons. The phonons are now completely free to arrangethemselves among the energy levels.

This means that we have ni+ gi objects altogether in the

bundle. There are ni of one type, and gi of the other type. The

number of possible arrangements would be

i=(ni+ gi)!

ni!gi!

The total number of arrangements for all bundles would be

obtained by multiplication of the number for every bundle:

=

i

(ni+ gi)!

ni!gi! .



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Bose-Einstein Distribution

Then we need to maximise ln using the Lagrange multiplier

method.

Using the same constraints as before on the number of

particles N and the total energy U, the method would give the

following answer:

ni

gi=

1

exp(1 2i) 1.

where 1 and 2 are the Lagrange multipliers.

This is called the Bose-Einstein distribution function. It looks

almost the same as the Fermi-Dirac distribution function,except that it has -1 in the deniminator instead of +1.

When we apply this to phonon, however, we need to make

some changes.



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Bose-Einstein Distribution

Phonons are not real particles like atoms or electrons. The

number of phonons is related to the vibration energy. The

number would increase when the temperature increases. At 0

Kelvin, there is no vibration, so there would be no phonons.

Because of this, there would be no constraint on the number N.

We really only have one constraint - that of constant energy U.

The Lagrange function is ln + U. N does not appear.

Maximising this function, we get

ni

gi

= 1

exp(i) 1

.

With the help of entropy as before, we would find

= 1

kBT.



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Phonon occupation number

Maximising Lagrange function, we have obtained

nigi =

1

exp(i/kBT) 1.

Following the same notations that we have used for ideal gas

and electrons, we rewrite:

ni as n()d and

gi as

g(

)

d.

This gives

n()d= g()d

exp(/kBT) 1

Next, we need a formula for the density of states g().

Note that although we have previously used the same symbol

g() for the particle in a 3-D box, the phonon density of states

would have a different formula.



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density of states for phonons

The vibrations in a solid is described by a wave equation. The

solution for the displacement of the atoms , can be given bythe same function

= sin kxx

in each of the x, y and z directions. Assuming that the

displacements are zero at the edge of the solid, the

wavevectors are discretised in the same way:

kx=nx

a .

So the density of states can also be given by the same formula:

g(k) = V k2

22.

However, two changes are needed from the way we have used it

for the ideal gas and the electrons.



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The first change is in the change of variable to energy .

For atoms or electrons, the energy is

= 2k2

2m .

For vibrations, the energy is given by by the same formula as

for photons:

= ,

where is the angular frequency 2f and f is the frequency.

is in turn related to k by

v =

k,

where v is the velocity.



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The variable of angular frequency is often used instead of

energy . For the change of variable, we then use

=vk.

We can now do the change of variable using

g()d =g(k)dk.Substituting

g(k) =V k2

22,

we get the density of states in :

g() = V 2

22v3.



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About polarisation

The second change is the related to the polarisation. Recall

that for electrons, we have to multiply by 2 because of the 2spin states.

For vibration, a state is a particular frequency, for a particular

set of wavevectors (kx, ky, kz).

Each state correspond to a wave, and a wave in a solid can

have 2 transverse polarisations, and 1 longitudinal polarisation.

So we have to multiply by 3 to get the correct density of state:

g()d = 3V 2d

22v3



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The Debye Frequency

There is one more practical point to note. In a solid, there is a

limit to the highest phonon frequency.

This is related to the fact that the wavelength cannot possibly

be shorter than the distance between atoms.

Let this highest frequency be D.

In a solid with N identical atoms, there are 3N energy states, or

normal modes. This is when we count every possible frequency,

and every possible direction of vibration for each frequency.

This can be proven mathematically, but we shall skip that. We

can find D by integrating the density of states: D0

g()d = D

0

3V 2d

22v3 = 3N



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The Debye Frequency

This equation D0

g()d = D

0

3V 2d

22v3 = 3N

can then be solved to find the Debye frequency. Integrating

gives

V 3D22v3

= 3N

Solving for D gives the Debye frequency:

D

= 6N 2v3

V

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The phonon distribution

As we have shown earlier, the distribution of phonons is given

by

n()d= g()d

exp(/kBT) 1

We shall express this in terms of frequency as well. The

phonon energy is given by

= .

This comes from assuming that the atoms vibrate in a

harmonic potential, so that the energy is quantised in steps of

.

In terms of the frequency , the phonon distribution is then

given by

n()d = g()d

exp(/kBT) 1.



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Debyes theory for Heat capacity of solid

The phonon distribution is given by

n()d = g()dexp(/kBT) 1

.

We can multiply by to get the energy at this frequency

interval:

n()d = g()d

exp(/kBT) 1.

The total energy is then

U= D

0n()d=

D0

g()d

exp(/kBT) 1.

We have found earlier that the density of states is:

g() = 3V 2

22v3.

Substituting into the integral for the total energy above ...



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Debyes theory for Heat capacity of solid

We get:

U = 3V

22v3

D0

3d

exp(/kBT) 1

At low temperature, it can be shown that this is proportional to

T4. This would imply that the heat capacity C is proportionalto T3. (This has been mentioned in the topic on the specific

heat of electrons in metal.)

At high temperatures, the heat capacity tends to 3N kB.



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Is verified at high temperature

In 1912, Debye measured the high temperature behaviour of

copper.

He showed that the heat capacity did tend to 3N kB, as he had

predicted.



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3.2 Photons



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Radiation from a very hot object

When an object gets very hot, it can give out light.

The temperature of a volcano lava flow can be estimated by

observing its color. The result agrees well with the measured

temperatures of lava flows at about 1,000 to 1,200 deg C.

http://en.wikipedia.org/wiki/Black_body



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An ideal black body

The amount of radiation emitted depends on the nature of theobject - its colour, whether it is smooth or rough, etc.

An ideal black body is one that absorbs all the radiation thatfalls on it.

No real material can do this. Soot is about the best, absorbing

all but 3%.

In 1859 Gustav Kirchhoff had a good idea: a small hole in theside of a large box is an excellent absorber, since any radiationthat goes through the hole bounces around inside, a lot gettingabsorbed on each bounce, and has little chance of ever gettingout again.

http://galileo.phys.virginia.edu/classes/252/black_body_radiation.html

The energy of the absorbed radiation reaches equilibriumamong different frequencies in the cavity. This results in acharacteristic spectrum that would be emitted again throughthe hole.



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Black Body Spectrum

http://en.wikipedia.org/wiki/Black_body

As the temperature decreases, the peak of the black-body

radiation curve moves to lower intensities and longer

wavelengths. The black-body radiation graph is also compared

with the classical model of Rayleigh and Jeans.



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Photons in a box

We can derive a formula for the black body radiation by

considering the energies of a free photon gas in a box

(following Kirchhoffs idea).

Photons arise from oscillations in electromagnetic fields, which

is also governed by a wave equation.

The same ideas for phonons can be used, leading to the same

density of states:

g() = V 2

22

c3

,

where the speed v is now the speed of light c. This is the

formula before considering polarisations.



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Photons in a box

Like phonons, the photon number is not fixed. We get fewphotons when the box is cold, and more photons when it is hot.

Unlike phonons, there is no upper limit to the frequency (no

Debye frequency). The wavelength in a solid cannot be shorter

than the distance between atoms. The electromagnetic wave ina box has no such limit.

Finally, phonons in a solid can have 3 polarisations: 2

transverse and 1 longitudinal (like sound). Photons can only

have 2: both transverse (e.g. light).



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Distribution of photon number

Like phonons, each energy state can be occupied by any

number of photons.

So it obeys Bose-Einstein statistics.

The number density is therefore given by the same formula:

n()d= g()d

exp(/kBT) 1

So the number of photons in a given frequency interval is

n()d = 2V 2d

22c3

d

exp(/kBT) 1



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Distribution of photon energy

Berlin, 1899. In the Imperial Institute of Physics and

Technology, the scientists were just trying to help the German

lighting and heating industries.

Lummer and Pringsheim measured the radiation spectrum from

a blackbody very accurately.



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Blackbody radiation

The results did not look at all like what the physicists thought

it should.

Then Max Planck came along and derived his formula for black

body radiation that agreed very well with these measurements.

And the rest is history.



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What we have learnt so far.

1. The Bose-Einstein distribution function is given by

f() = 1

exp[( )/kBT] 1.

2. For phonons and photons, the following form is used:

f() = 1

exp(/kBT) 1.

3. The density of states can take different forms:

phonons

g()d = 3V 2d

22v3

photons

g()d = 2V 2d

22v3

bosonic atoms

g() =4mV

h3 (2m)1/2



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What we have learnt so far.

electrons

g() = 2 4mVh3

(2m)1/2

classical ideal gas

g() =4mV

h3 (2m)1/2

3. At low temperature, the specific heat capacity for the Debyemodel tends to T3. At high temperature, it tends to 3N kB.

4. The energy distribution in a photon gas is given by:

u() = n() =V3

2c31

exp(/kBT) 1.

This also describes the black body radiation spectrum.



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3.3 Einsteins model



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Einsteins model for solids

The Simple Harmonic Oscillator is one of the early attempts to

understand the heat capacity of solids using quantum

mechanics.

It was proposed by Einstein. The idea is to think of - to model

- a solid as atoms vibrating independently of one another other.

Glazer and Wark (2001)

Each atom sits in its own potential well. It could undergo

simple harmonic oscillation in all 3 directions x, y and z.

Vibration in any one direction is independent of vibration in the

other two directions.



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The 1-D case

From quantum mechanics, we know that the energy for a 1-D

simple harmonic oscillator is

n= (n + 1/2)h.

where n is an integer, and is the frequency.

http://hyperphysics.phy-astr.gsu.edu/hbase/quantum/hosc.html



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The equipartition theorem

We have seen that at high temperature, the heat capacity

approaches 3N kB.

This can be understood this using the equipartition theorem,

which states that:

The mean value of each independent quadratic term in the

energy is equal to

1

2kBT.

In our case, quadratic term refers to the kinetic energy12mv

2, or the potential energy of the harmonic oscillator, 12kx2.

We can prove this by finding, for example, the mean kinetic

energy in the Boltzmann distribution:

12mv

2 exp(12mv2/kBT)dv

exp(12mv

2/kBT)dv=

1

2kBT



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At low temperature

At very low temperature, the atoms are mostly in the ground

state, with some in the first excited state. Only these two levels

are important, so its behaviour is similar to a spin 1/2 salt.

Suppose there are N atoms, with n in the excited state. The

total energy is approximately

U = (N n)h2 + n3h2 .

n is small, so N n is close to N and n is given by the

Boltzmann factor:

U Nh

2

+ exp(h/kBT)3h

2

.

The heat capacity than has an exponential form:

C=dU

dT =

h

kBT2

exp(h/kBT)3h

2 .



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The heat capacity

We can sketch the graph for the heat capacity predicted by

Einsteins model. It looks like this:

As we have just shown, the heat capacity approaches 3N kB at

high T.

This agrees with the empirical Law of Dulong and Petit, who

measured many solids at room temperature in 1819.



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3.4 Useful Integrals



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Useful Integrals

For Debye model, to find heat capacity at low temperature;

and for black body radiation, to find Stefans law:

0

x3

ex 1dx=

4

15.

For liquid helium-4, later in the course:-

to find condensation temperature of Bose Einstein condensate:

0

x1/2

ex 1dx= 2.315

to the find heat capacity:

0

x3/2

ex 1dx= 1.783


3 photons phonons

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