1

CHAPTER 18

BOSE-EINSTEIN GASES

2

Math Note: Solid Angle

An element of solid angle is defined by d 3r

adrd

O

n̂

unit normal vector ndaad ˆ

add by Oat subtended angle solid

R

rn ˆˆ RdaadrrdaadrRr ˆˆ

223 R

dad

R

da

R

Rdad

2

02

0

2

02

0

sin

R

dRdR

R

dad

sphere

This gives 4

r

3

Now consider an infinitesimal area dA

2333 r

cosdA

r

cos)dA(r

r

n̂)dA(r

r

adrd

This is the solid angle subtended by dA at some point O

What fraction is this of the total solid angle at O?

2r4

cosdAd

We will use this result later in this chapter

4

Blackbody radiation:A blackbody is a perfect absorber, that is, it absorbs all of the

electromagnetic radiation incident upon it. We are interested in the radiation from a blackbody held at a certain temperature. A good approximation is cavity radiation. An enclosed volume is held at a fixed temperature and a small hole is drilled into the cavity. All radiation falling on the hole will be absorbed and so the radiation emitted from the hole is a good approximation to that emitted from a blackbody, providing that the hole is small enough so that it does not distort the spectrum of radiation in the cavity. An important problem for classical physics in the 19th century was to calculate the spectrum of radiation emitted from the cavity. Attempts to do so were spectacularly unsuccessful. This led to the development of quantum mechanics. Blackbody radiation is very important in the subject of astrophysics.

We are considering photons, which have s=1 and hence are bosons and we must use Bose-Einstein statistics. In a cavity photons are being constantly emitted and absorbed by the walls and so the

5

number of particles is not constant. A photon gas differs from a molecular gas in this important aspect. However energy must be conserved.

The chemical potential appears in the BE distribution. What is its value? It is set equal to zero! I will try to give, without going into any detail, several reasons why this is reasonable.

Firstly, since the number of photons is not constant, we must remove this constraint condition when using the method of Lagrange Undetermined Multipliers. We introduced the constraint as idN

To remove the constraint, we set and the above term in

the distribution becomes and the chemical potential does not appear.

0

This resulted in a distribution with the term ie

kTi

i ee

6

Secondly, at equilibrium the Helmholtz potential must achieve its minimum value. In moving towards equilibrium the number N will take on the value that minimizes F, for a fixed T, V.That is, we must have

0,

VTN

F

But, as we showed earlier, for a single speciesVT

N

F

,

Hence the chemical potential is zero.

Thirdly, the condition for diffusive equilibrium isConsider the reaction (similar to a chemical reaction) which describes the emission and absorption of photons by an electron:

AB

eeFor equilibrium, so ee 0

7

Setting the chemical potential equal to zero then results in the BE distribution

1

1)()(

)(

kTe

g

Nf

This is the number of photons per energy state.For photons and it is more usual to work with h

)1(1e

1

)(g

)(N)(f

kT

h

dg )( is the number of quantum states in the frequency range ),( d

Earlier (Ch.12) we used quantum mechanics and the concept of a quantum-number space to calculate the density of states. The calculation of is very similar. However the photon does not have a mass, unlike a particle, so the previous result cannot be directly used, although the approach is identical.

dg )(

(PHOTONS!!)

8

dc

Vdg 2

3

8)(

The derivation is given in the textbook. The result is

9

Now consider the energy . In the frequency range , )(u ),( d du )( can be obtained by multiplying the number of photons in

this range by the energy of each photon.

hdfghdNdu ])()([])([)(

h

e

d

c

Vdu

kT

h

1

8)( 2

3

1

8)(

3

3kT

h

e

d

c

Vhdu

Planck radiation formula

The wavelength is the variable measured, so we write the formula in terms of the wavelength.

[from eqn. (1)]

10

dc

ddc

cddc22

1

Planck’s formula then becomes

1

8)(

2

3

3kT

hc

e

dcc

c

Vhdu

1

8)(

5kT

hc

e

dVhcdu

)(u is the energy per unit wavelength. We can plot this function for various temperatures.

11

)(u

3T

2T

1T

321 TTT

Temperatures can be determined with high precision using the Planck formula. An optical pyrometer is used to determine the energy spectrum from some radiating source, such as a furnace.A fit to the measured spectrum provides a determination of T.

12

Now we wish to develop the Stephan-Boltzmann Law.We begin by calculating the energy density (U/V)

0 50 1

8)(1

kT

hc

e

dhcdu

VV

U

We set 2x

dx

kT

hcd

kTx

hc

kT

hcx

0

x

5

2 1e

1

kTx

hc

x

dx

kT

hchc8

V

U

dxe

x

hc

kThc

V

Ux

0

34

18

15)()(8 4

3

4 hc

kT

V

U

(MAPLE)

13

431644

3

45

1055.7)(15

8Km

JaaT

V

UT

hc

k

V

U

We have been studying a photon gas inside a box in thermal equilibrium. We wish to discuss the photons emitted from an object.Consider a small hole in a box. Some photons will escape through this hole. The box is held at a temperature T.

T

R

A

All photons have speed c, so photons of all energies have the same probability of leaving the hole. This implies that the spectrum of the photons leaving the hole will be the same as the spectrum of the photons inside. Those photons that leave the hole must have been moving towards the hole.

14

Consider a small time interval dt. We consider a hemispherical shell, shown as green in the diagram. Let the radius be R, which depends on how long ago we looked, and the thickness be cdt.

The volume of a small element of this shell is

cdtdRRddV ))sin()(( Since we know the energy density, we can determine the energy of the photons in this volume element:

cdtddRV

UdVind ))sin(()( 2

Most of the photons in dV will not escape through the hole because they are not moving in the right direction. Given that there is no preferred direction for the photons, the fraction of the photons escaping will be the solid angle subtended by the hole at dV, divided by the total solid angle, ie, by

24cosR

A

15

Hence, for those photons in dV that escape

22

4

cos))sin((

R

AcdtddR

V

Ud

))cos()(sin(4

ddV

UAcdtd

The total energy carried by the photons leaving the hole is then

2

0

2/

0

)cos()sin(4

ddV

UAcdt

2/

0

)cos()sin(2

dV

UAcdt

cdtV

U

4

A

V

U

4

c

Adt

16

The power per unit area (or energy flux) is then V

U

4

ce

Using the expression for the energy density in the cavity:

Stephan-Boltzmann Law

The Stephan-Boltzmann Constant is 42

81067.5Km

W

4Te

Whenever we can consider a radiating body to be a blackbody, the Stephan-Boltzmann Law is applicable

4aT4

ce

17

Wien Displacement Law. As shown earlier, a plot of as a function of the wavelength displays a maximum. If we consider thePlanck formula

)(u

1

8)(

5kT

hc

e

dVhcdu

will have its maximum value when the denominator is a minimum.)(uWe can solve the problem using MAPLE (next slide)The result is giving965.4

max

kT

hc

KmT 3max 1090.2 Wien Displacement Law

For a black body, as the temperature increases, the maximum wavelength decreases and the maximum frequency increases.

18

> # Wien Displacement Law> restart:

> # denominator of the Planck formula in terms of wavelength

> f:=x->x^(-5)*(exp(x)-1):> z:=diff(f(x),x);

z := -5*(exp(x)-1)/x^6+exp(x)/x^5> solve(z=0,x);

LambertW(-5*exp(-5))+5> ans1:=evalf(%);

ans1 := 4.965114232> #ans1=hc/(lambda[max]kT)

19

An approximation can be made to the Planck law for short and for long wavelengths. Let us consider the case of long wavelengths. By long is meant In this regime we can make a good approximation to the exponential by expanding:

1kT

hc

kT

hce kT

hc

1 This gives

dVkT

du

kThcdVhc

du45

8)(

8)(

This is the expression obtained on the basis of classical theory by Raleigh and Jeans. The result does agree with the experimental results at long wavelengths, but is in total disagreement at shorter wavelengths. We compare the Rayleigh-Jeans Law with the Planck result on the next slide.

20

)(u

Raleigh-Jeans The result is obviously not correct for the radiation at short wavelengths.As )(0 u

This was called the“ultraviolet catastrophe” and attempts to obtain a formula valid at all wavelengths led to the development of quantum mechanics.

21

22

The Sun: The solar constant is the amount of radiation that the Earth receives from the Sun. The value is (above atm)2/1370 mW

Using the Earth-Sun distance the Sun’s luminosity is . The Sun’s surface area is

)10150( 6 kmW26109.3 218101.6 m

Using the Stephan-Boltzmann Law 4Te 4

428

218

26

1067.5101.6109.3

TKm

W

m

W

KT 3108.5

We can now use the Wien Displacement Law to determine the wavelength at which is a maximum. )(u

KmT 3max 1090.2 m

K

Km 73

3

max 1000.5108.5

1090.2

nm500max This is near the middle of the visible spectrum.

( =power/area)e

23

At equilibrium a body must emit as much energy as it absorbs. If a body is not black, it will reflect some of the photons incident on it.Consider photons of a particular energy and suppose that for every10 photons incident on the body, 9 are absorbed and 1 is reflected.The body will then emit only 9 photons at this energy and so a total of 10 photons will move away from the object. Let represent the fraction of photons incident on a body that is absorbed by the body. This is also the fraction of photons emitted, when compared with a blackbody.

The symbol represents the emissivity of the body. For a blackbody, and for a perfect reflector 1

0It is clear that a good reflector is a poor emitter. Experimentally it is found that so the spectrum of a body with is not the same as that of a blackbody. If we take some weighted average over all relevant wavelengths, we can write

)( 1

4Te

24

Now let us consider the Earth. We begin by considering the Earth to be a blackbody. How much power is absorbed from the Sun? This is the solar constant multiplied by the cross-sectional area of the Earth as viewed from the source (the Sun). So the answer is [(solar constant) ] 2RThe power that the Earth emits is 424 TR At equilibrium )1(TR4R)constantsolar ( 422

KTTm

W279)1067.5(41370 48

2

(Stefan-Boltzmann constant)

Considering the crudeness of the calculation, this is amazingly close to the measured average value of 288K. {Of course, the temperature of the atmosphere is a function of the height above the surface. The value of 288K is for the atmosphere at the surface of the Earth.}

25

(from clouds )In reality the Earth is far from a blackbody. About 30% of

the sunlight is reflected. Taking this reflection into account, only 70% of the radiation incident on the Earth (the solar constant) is absorbed. Multiplying the solar constant in equation (1) by 0.7 gives T=255K. This correction worsens the agreement with the measured value.But wait, you might say. Since the Earth is not a blackbody,

one should include the emissivity of the Earth on the RHS of equation (1). This is also not correct! The Sun and the Earth emit radiation in different regions of the spectrum because of their vastly different temperatures. The surface of the Earth emits in the infrared and, in fact, is a very efficient emitter ( ) for most of the wavelengths emitted. What is going on?

1

26

We must consider the carbon dioxide and water vapor in the atmosphere. Let us devise a very crude (!)model (refinements later) in which the atmosphere is a layer some distance above the surface of the Earth. This layer absorbs infrared radiation. Hence we have the following situation:

Earth’s Surface

Sunlight

atmosphere

infrared (IEarth)

{Important molecules in the air for absorption in the infrared are carbon dioxide (about 350ppm),water,ozone and methane.}

Equal(each IAt)

(IS)

(0.7 IS)

(0.3 IS)

27

In this model all the energy radiated by the Earth’s surface is absorbed by the layer of atmosphere. The atmosphere then radiates this energy, returning half to the Earth. Hence, for energy balance: (I represents energy)

IS=0.3IS+IAt 0.7Is=IAt IEarth=0.7IS+IAt

IEarth=2(0.7IS)

242 R)constantsolar (7.02TR4

This gives T=303K This is a bit high, but we have made several approximations, chiefly that the atmosphere is a layer some distance above the surface of the Earth. The absorption of the radiation by the atmosphere is called the Greenhouse Effect.

28

Note!

The purpose of the previous discussion was to indicate the use of some of the formulae that we developed and to introduce theGreenhouse Effect. The model used is incapable of addressing the topic of global warming. To discuss this issue we would need to include important effects not taken into consideration by our exceedingly crude model. Of great importance is the fact that not all of theradiation emitted from the Earth’s surface is absorbed by the cloud cover. There are certain wavelength “windows” through which surface radiation goes directly into outer space. In fact roughly 10% of the energy radiated from the surface is not absorbed by the atmosphere. As CO2 and certain other pollutants increase in the atmosphere, the widths of these windows decrease and more energy is absorbed by the atmosphere which results in a rise of the Earth’s surface temperature.

29

Some Comments: Snow, like most terrestrial bodies, is nearly a black body.

This is true in spite of the fact that it reflects about 94% of radiation in the visible part (small part) of the spectrum. It absorbs almost all infrared radiation. In particular snow emits in the microwave region. This fact is used to remotely sense snow-packs from satellites.

Human skin is also an approximate black body. All of us are black in the infrared (from 0.7 to 1000 ) m

The problem of energy balance is quite complicated and the next slide gives you some idea of the nature of the problem.Notice that the solar constant is given as 342 W/m2 which is different from the value we used (1370 W/m2). The reason for the difference is that the solar radiation intercepted by the Earth is determined by its cross section (π R2), but as the Earth rotates, this energy is distributed over the entire surface (4π R2), thus a factor of 4 enters.

30

31

Cosmic Background Radiation: The most spectacular example of a photon gas is the radiation that fills the universe, with a thermal spectrum corresponding to K002.0728.2T This is radiation left over when the universe consisted of ionized gas and radiation and they were strongly coupled. At that time the temperature was much higher (several thousand K), but the expansion of the universe has resulted in a red-shift of the spectrum. The energy spectrum peaks at about

eVkT 4106.682.2 This corresponds to a wavelength of about 1 mm, which is in the far infrared. This infrared radiation cannot penetrate the atmosphere but wavelengths in the microwave region (about 1 cm) of the spectrum can be detected at the surface of the Earth . To obtain the complete spectrum, measurements are made from satellites. (See Figure 18.3 of the textbook)

32

Properties of a photon gas. We wish to calculate the entropy of a photon gas. The easiest way is to first calculate the specific heat.We have:

43

45

)(158

Thc

k

V

U

The specific heat is then

33

45

)(15

32TV

hc

k

T

UC

V

V

Vhc

kTkCV

35

1532

The entropy is then :

dTThc

VkdT

T

CS

TTV

0

23

45

0 )(1532

35

4532

hc

kTkVS

(Note strong dependence on T.Again, as T increases, S increases)

(energy density in a cavity)

33

We can now calculate the pressure. As discussed much earlier, we can obtain thermodynamic variables from the thermodynamic potentials.

From the diagram on the Fact Sheet, F=U-TS

43

54

3

5

)(4532

)(158

kTVhc

kTVhc

F

VThc

kkTV

hcF 4

3

454

3

5

)(158

31

)(1538

VThc

kF 4

3

45

)(158

31

We can now obtain the pressure from the Helmholtz Function:

34

NTV

FP

,

43

5

)()(15

831

kThc

P

V

UP

31

(Photon Gas)

Note that this is not the pressure of the gas in the enclosure.

(reciprocity relationship)

35

Bose-Einstein Condensation:So far, in this chapter, we have been considering photons, and

so the total number of particles is not fixed. Now we consider Bose particles or systems, such as atoms with integer spin. In this case the number of particles remains fixed and this has important consequences. Unlike the case for photons, the chemical potential is not zero and we will need to determine its dependence on various parameters.

We now consider an ideal Bose gas and concentrate on low temperatures. A common convention, which we shall adopt, is to choose the zero of energy to be at the ground state.

For a Bose gas00

1

1)()(

)(/)(

kTeg

Nf

Consider the ground state, which is not degenerate.

1

1/)(0

0

kTeN

36

At low temperatures, the number of bosons in the ground state will be large, so the term must be slightly greater than unity.We expand this exponential and keep only the first two terms.

kTe /)( 0

kTe kT /)(1 0/)( 0

To this approximation

0/)(0

1

10

kT

eN

kT

Since the ground state energy is zero:

kT

N0

this requires that the chemical potential be negative. Further, atT=0, we must have (Note that μ depends on N0)0

When the temperature is very small, the number of particles in the ground state is large and so the magnitude of the chemical potential must be small.

37

Example: Consider N0=1023 and T=1K

eV1062.810

)K1(K

eV1062.8

N

kTkTN 28

23

25

0

0

(For comparison, the average energy of a molecule of an ideal monatomic gas 273K is 0.035eV.)

38

An obvious way to attempt to determine the chemical potential is to make use of the fact that N remains constant. We now return to the distribution function and we will replace sums by integrals.

)1(1

1)()(

)(/)(

kTeg

Nf

0

)()()()()()()()( dgfNdgfdNgfN

We have already calculated the density of states (chapter 12):

dVhmdg 32/324)(

)2(1

22

0/)(

2/3

2

kTe

d

h

mVN

This expression is not correct! The ground state has been omitted.For 0)(,0 dg

Generally this would not matter as the occupation of the ground state would be negligible. However, for a Bose gas, this is not true at low temperatures.

(See note on next slide)

39

Note: When most of the states populated are very closely spaced, the sum can legitimately replaced by an integral. However we are going to consider temperatures at which there are a significant number of bosons in the ground state. In such a case the low-lying states are not very closely spaced and we might be concerned that the integral does not properly take these states into consideration. We do not have time to pursue this problem, but the result of a more detailed analysis indicates that our approach is an excellent approximation.

40

We will correct the formula by adding in the number of particles in the ground state in equation (2). Equation (2) gives the number of particles in the excited states and we will label this number

We now have with equation (2) written as

)( 0N

)( EXNEXNNN 0

)2(1

22

0/)(

2/3

2b

e

d

h

mVN

kTEX

At very low temperatures and, as we discussed earlier,0NN

0

//

0/0

111

1

1

1N

eeNe

N kTkT

kT

But is large so 0N 1/ kTe

Making this approximation equation (2b) becomes

0/

2/3

2 1

22

kTEXe

d

h

mVN

(Of course this will not be true at higher temperatures.)

41

Set dxkTdkTxkT

x

0

2/3

2 1

22

xEXe

dxx

h

mkTVN

2612.2

10

xe

dxxMAPLE gives

)3(2

612.22/3

2

h

mkTVNEX

Notice that if we had ignored the ground state term and simply had , then equation (3) would be nonsense because it would give

NNEX 2/3TN

However there is a temperature at which that is, there are no particles in the ground state

NNEX 00 N

We introduce the Bose Temperature BT

42

orh

mkTVN B )4(

2612.2

2/3

2

3/22

612.22

V

N

mk

hTB

At 00 NTT B2/3

00 1)4()3(

11

B

EXEX

T

T

N

N

N

N

N

NN

N

N

Example: Typical value for the Bose Temperature.33423 104.221002.6 mVinHeofatoms

3/2

33

23

2327

234

104.22612.21002.6

)1038.1)(1066.14(2

)1063.6(

mKJ

kg

sJTB

KTB 036.0 At atmospheric pressure liquefies at 4.21K well above the Bose Temperature.

He4

NOTE!!!!!

43

Is it possible to form a gas with all its atoms in the ground state (the Bose-Einstein Condensation)? We must cool a gas to a very low temperature. However substances become solids at these low temperatures. For example Rb is a solid at room temperature. Many physicists thought that thermodynamics precluded the possibility of producing a gas at a temperature low enough to obtain a BEC. In fact a BEC was exhibited using Rb. It was possible to create the vapor in a non-equilibrium state. A supersaturated-vapor state was produced which persists for a long time (10’s of seconds). Eventually the vapor solidifies.

The atoms in the ground state are called the condensate. Such a condensate was first formed with Rb vapor at KT 7103.1

( Good descriptive article: Carl Wieman AJP 64 pp 847-855 (1996))

We can plot the number of atoms in the condensate and in excited states as a function of temperature:

44

N

0N EXN

BT

T

0 1 2We can also determine the chemical potential as a function of the temperature (a hand-in problem).

0 1 2

BT

T0

-0.5

-1.0

BkT

0

45

The BEC has been observed in Rb, Na, Li and HThe condensate is a strange material:

It is a macroscopic system in a single quantum state. Such amacroscopic system behaves in a non-classical way.

All the atoms have the identical set of quantum numbersand so they have no individual identities. The systembehaves as a “superatom”.

The transition to the condensate is non-intuitive. The averageseparation of the atoms is of the order of

and so their interactions are minute, yet they allcondense to the ground state. It comes about, not by

some interaction force, but by quantum statistics applicable to identical particles. (Completely non- classical.)

0410 a

Note: Even though photons are bosons, there is nothing in a system of photons equivalent to a BEC. The difference is that the constraint on the number of atoms is not present for photons.

46

Properties of He4

The phase diagram for this isotope of He is as follows:

40

20

0

P(atm)

0 2 4 6T(K)

CPvapor

line solid

liquid He I liquidHe II

2.18K 5.25K

Above the critical point (5.25K), cannot exist as a liquid.2.18K<T<5.25K vapor compresses to liquidT<2.18K vapor compresses to liquid

He4

IHe4

IIHe4

The two liquid phases coexist along the line

point -

47

He is the only element that remains a liquid down to absolute zero.(It can be solidified by the application of sufficient pressure). Solid cannot exist at pressures less than 25 atm. Solid cannot co-exist with vapor under any conditions.

There are two triple points:

He4 He4

solidIIHeIHe ,, 44

vapor,IIHe,IHe:point 44The specific heat differs radically in the two liquid phases. In the phase it behaves like a normal liquid until near the λ pointIHe4

liquid liquidIIHe4 IHe4

2.18KT(K)

0 1 2 3

4

2

0

Nk

CV

48

(The shape of this curve suggested the name )point

IIHe4 is called a superfluid. It has high thermal conductivity(millions of times that of 4He I and hundreds of times that of copper) and almost zero viscosity. An object falling through a perfect superfluid will experience no resistance:- it is as if the fluid were not there. (This is no longer true after a certain speed is achieved.)

49

N

0N EXN

BT

T

0 1 2

superfluid component

normal fluid component

Two-fluid model:

IIHe4 is considered to consist of a normal fluid and a superfluid

The proportion of the superfluid is zero at the and increases to unity at T=0. In liquid He the force between atoms is weak and, to a first approximation, one might consider it to be a Bose Gas.

line

(Experimentally liquid 4He has many properties expected of a classical gas.)

50

The two-fluid model associates the superfluid to be the condensate and the normal fluid to be due to the atoms in excited states. As T decreases below the point

the number of atoms in the condensate increases and the He becomes more of a superfluid. Hence the superfluidity is associated with the start of a BEC at the

However a detailed picture is somewhat more complicated:- a better calculation takes into account the weak interatomic forces.

point

If we can treat the liquid as a gas, a calculation of the Bose temperature (valid only for gases) should give a result close to the .This is indeed so*.point

* Students will be, I am sure, anxious to calculate TB. The density of liquid 4HE II is about kmole/m106.27 33

51

Let us briefly consider He3

40

20

0

P(atm)

0 1 2 3T(K)

CPvapor

solid

liquid

He3

(Interesting dip!)

52

On this phase diagram we do not see a superfluid phase. This is as expected, since is a fermion and hence there should not be a BEC. However, for T<3mK there are three distinct superfluid phases, which are not shown on the phase diagram. How can this be if we associate superfluidity with a BE condensate? The explanation is very similar to the explanation of superconductivity in solids. In superconductors electrons pair up (Cooper pairs) to form a boson and it is the properties of these pairs that give rise to superconductivity. In the same fashion it is energetically favorable for two atoms to form a pair which act as a unit. Although each atom has spin ½, the pair has zero spin and is therefore a boson.

He3

He3