Chapter 7: Quantum Statistics

Outline Introduction Blackbody Radiation Thermal Quantities of Photon System Radiation from Objects

Chapter 7: Quantum StatisticsPart II: Applications

X Bai

SDSMT, Physics

2013 Fall

X Bai Chapter 7: Quantum Statistics


1 IntroductionPhotons, E.M. Radiation

2 Blackbody RadiationThe Ultraviolet Catastrophe

3 Thermal Quantities of Photon SystemTotal EnergyEntropy

4 Radiation from ObjectsRadiation from a BoxRadiation from ObjectsRadiation from Objects



Bosons

Two types of bosons:(Type-1) Composite particles: contain an even number of fermions. Thenumber of these particles is conserved if the energy does not exceed thedissociation energy (in the range of MeV in the case of the nucleus - thenuclear binding energy).

(Type-2) Particles associated with a field. The most important example is thephoton. These particles are not conserved: if the total energy of the fieldchanges, particles appear and disappear. Well see that the chemical potentialof such particles is zero in equilibrium, regardless of density.

Typically, radiation emitted by a hot body, or from a laser is not in equilibrium:energy is flowing outwards and must be replenished from some source. Thefirst step towards understanding of radiation being in equilibrium with matterwas made by Gustav Kirchhoff (1824-1887, German), who considered a cavityfilled with radiation, the walls can be regarded as a heat bath for radiation.



E.M. Radiation

The walls emit and absorb E.M. waves. In equilibrium, the walls and radiationmust have the same temperature T . The energy of E.M. radiation is spreadover a range of frequencies, and we define us (ν,T )dν as the energy density(per unit volume) of the radiation with frequencies between ν and ν + dν,us (ν,T ) is the spectral energy density. The internal energy of the photon gasshould be: u(T ) =

∫∞0

us (ν,T )dν.

In equilibrium, us (ν,T ) is the same everywhere in the cavity, and is a functionof frequency and temperature only. If the cavity volume increases atT = const, the internal energy U = u(T )V also increases. The essentialdifference between the photon gas and the ideal gas of molecules:

♠ for an ideal gas, an isothermal expansion would conserve the gas energy,

♠ for the photon gas, it is the energy density which is unchanged, the numberof photons is not conserved, but proportional to volume in an isothermalchange.



E.M. Radiation in Equilibrium with Matter

A real surface absorbs only a fraction of the radiation falling on it. Theabsorptivity α is a function of ν and T ; a surface for which α(ν) = 1 for allfrequencies is called a black body.

The electromagnetic field has an infinite number of modes (standing waves) inthe cavity. The black-body radiation field is a superposition of plane waves ofdifferent frequencies. The characteristic feature of the radiation is that a modemay be excited only in units of the quantum of energy hν (similar to aharmonic oscillators): εi = (ni + 1/2)hν.

This fact leads to the concept of photons as quanta of the electromagneticfield. The state of the E.M. field is specified by the number n for each of themodes, or, in other words, by enumerating the number of photons with eachfrequency: Eγ = hν = cpγ , pγ = hν

c

According to the quantum theory of radiation, photons are massless bosons ofspin 1 (in units ~). They move with the speed of light.



Photons: Frequency and modes

T



Equilibrium in a Photon Gas

The linearity of Maxwell equations implies that the photons do not interactwith each other.

Non-linear optical phenomena are observed when a large-intensity radiationinteracts with matter.

The mechanism of establishing equilibrium in a photon gas is absorption andemission of photons by matter. Presence of a small amount of matter isessential for establishing equilibrium in the photon gas. Well treat a system ofphotons as an ideal photon gas, and, in particular, well apply the BE statisticsto this system.



Where classical physics fails

E.M. radiation is a continuous field that permeates all space. E. M. field a box:

field =∑∞ν=0(Standing Waves)ν

Each standing wave has an average thermal energy of 2 · 12kT . Very much like

infinite number of oscillators.

Efield =∑∞ν=0 kT →∞

But this infinite thermal energy never appears in nature.

In quantum physics:Each standing wave is a standing mode. A single mode (with a fix ν value) isequivalent with a single-particle state.



The solution by Max Planck

The oscillator cannot have any amount of energy but the allowed ones like:εn = 0, 1hf , 2hf , ...

People understand this better after the quantum physics was established.

When εn = 0, 1hf , 2hf , ..., the partition function for single oscillator becomes:Z =

∑∞n=0 e−nhf /kT =

∑∞n=0 e−nhf β = 1

1−e−nhfβ

The average energy is therefore,

E = − 1

Z

∂Z

∂β=

hf

ehf /kT − 1(1)

Therefore, when energy comes in units of ”hf”, the mean number of units of energy

in the oscillator is

nPlank =1

ehf /kT − 1(2)

This is the so-called Plank Distribution. The ”units of energy” correspond to”photons”



The solution by Max Planck - cnt.

Since photons are Bosons, the mean number of photons at certain energyshould follow the B.E. distribution,nphoton = 1

e(ε−µ)/kT−1

Comparing with nPlank = 1

ehf /kT−1, one can see µphoton = 0.

Does this make sense? Let’s take a look at the how photons can be created ordestroyed.

e ↔ e + γ

At equilibrium, the γ disappearing rate is the same with its appearing rate, thechemical potential in this case is:

µe = µe + µγ .

This means, µγ = 0.



E.M. Radiation in a ”Box” - Photon Gas

To obtain the bulk property, we need to consider photons in all modes: Totalenergy, total number of photons, etc.Taking a cubic box of photons (E.M. radiation) for example, we have:

1-d potential well:

λ =2L

np =

h

2Ln (3)

photon energy:

ε = pc =hc

2Ln (4)

3-d box:

ε =hc

2L

√n2

x + n2y + n2

z =hnc

2L(5)

Now, we can take the same steps as we didi before to calculate the total energyof the E.M. radiation.



E.M. Radiation in a ”Box” - Photon Gas - cnt.Total E.M. energy:

U = 2∑

nx

∑ny

∑nz

εnPl (ε)

”2” for two polarizations.

U =∑

nx ,ny ,nz

hcn

L

1

ehcn/2LkT − 1(6)

U =

∫ ∞0

dn

∫ π/2

0

dθ

∫ π/2

0

dφ·

n2sinθhcn

L

1

ehcn/2LkT − 1(7)

U =π

2

∫ ∞0

n2 hcn

L

1

ehcn/2LkT − 1dn

(8)

Change variable ε = hcn2L

: n = 2Lεhc

, dε = hc2L

dn



E.M. Radiation in a ”Box” - Photon Gas - cnt.

U = π

∫ ∞0

n2 hcn

2L

1

ehcn/2LkT − 1dn (9)

U = π

∫ ∞0

(2L

hc

)2

ε2 · ε · 1

eεβ − 1· 2L

hcdε (10)

U = π

∫ ∞0

(2L

hc

)3

ε3 · 1

eεβ − 1dε (11)

U

V=

∫ ∞0

8π

(hc)3· ε3

eεβ − 1dε (12)

The energy density per unit photon energy, or the spectrum of the photons:

u(ε) =8π

(hc)3· ε3

eε/kT − 1(13)



E.M. Radiation in a ”Box” - Photon Gas - cnt.



E.M. Radiation in a ”Box” - Total Energy

U

V=

∫ ∞0

8π

(hc)3· ε3

eεβ − 1dε (14)

U

Vε=kTx

=8π(kT )4

(hc)3·∫ ∞0

x3

ex − 1dx (15)

U

V=

8π(kT )4

(hc)3· π

4

15=

8π5(kT )4

15(hc)3. (16)



The Plank Spectrum

In terms of the dimensionless variable x = ε/kT

Peak x = 2.82 Or Ε=2.82 kT

By code: F7_19.C



In-class exercise: The Plank Spectrum

Exercise 07-06: Peak of the Plank Spectrum (Problem 7.37):Prove the peak of the Plank Spectrum is at x = 2.82.



In-class exercise: The Plank Spectrum

Exercise 07-07: Peak of the Plank Spectrum (Problem 7.39):Starting from Eq.(12), use relation λ = hc/ε to derive the photon spectrumas a function of wavelength.UV

=∫∞0

8π(hc)3· ε3

eε/kT−1dε



Total Energy

Eq.(15) [Eq. (7.85) in the textbook]UV

= 8π(kT )4

(hc)3·∫∞0

x3

ex−1dx

UV

= 8π(kT )4

(hc)3· π

4

15

UV

= 8π5(kT )4

15(hc)3

⇒ Energy density is proportional to T 4.

⇒ Heat capacity: CV =(∂U∂T

)V

= 32π5k(kT )3

15(hc)3V



Entropy of Photon Gas

δS = δQT

S(T ) =∫ T

0CV (T ′

T ′ dT ′

S(T ) =∫ T

032π5k(kT ′)3

15(hc)3V 1

T ′ dT ′

S(T ) = 32π5k4

15(hc)3V∫ T

0T ′2dT ′

S(T ) = 32π5k4

15(hc)3V 1

3T 3

S(T ) = 32π5k4

45(hc)3VT 3

S(T ) =32π5

45V

(kT

hc

)3

k

Or, using the ”Stefan-Boltzmann constant”, σ = 2π5

15k4

h3c2

S(T ) = 163cσVT 3



Radiation from a Box

We have obtained the radiation inside a Box:(7.83): U

V=∫∞0

8π(hc)3· ε3

eεβ−1dε

(7.84): u(ε) = 8π(hc)3· ε3

eε/kT−1

(7.86): UV

= 8π(kT )4

(hc)3· π

4

15= 8π5(kT )4

15(hc)3

What is the emission from the box?

dV = Rdθ · Rsinθdφ · dRdV = Rdθ · Rsinθdφ · cdtEnergy in dV :UV

dV = ucdtR2sinθdθdφAssuming photons are moving all directionswith equal probability:dΩ = dAcosθ

R2

Probability of photons moving into dA:P = 1

4πdΩ = dAcosθ

4πR2

R dV

θ dA

R dΩ

θ dA



Radiation from a Box - cnt.

Energy escape from the box through dA is:dE(θ, φ, dt) = u · dAcosθ

4πR2

dE(θ, φ, dt) = ucdtR2sinθdθdφ · dAcosθ4πR2

dE(θ, φ, dt) = dAcosθ4π· u · cdtsinθdθdφ

dEdAdt

= c∫ 2π

0dφ∫ π/20

dθsinθ cosθ4π· 8π

5

15(kT )4

(hc)3

dEdAdt

= c · 2π · 12

14π· 8π

5

15(kT )4

(hc)3

dEdAdt

= c4· 8π

5

15(kT )4

(hc)3= σT 4

in which σ = 2π5

15k4

h3c2= 5.67× 10−8 W /(m2K 4)

This is called the ”Stefan’s Law”, discovered in 1879 from experiments. σ isthe ”Stefan-Boltzmann constant”

Or,dE

dAdt= c

4· U

V



Radiation from Objects - How to Think about this?

⇒ Radiation from object means no reflection is considered, the Black BodyRadiation.⇒ In reality, reflection always exists.

Let’s first look at the Black Body Radiation.

Box of Photons

TBox = TB.B.

By 2nd Law of Thermodynamics, wemust have:

UBoxemitted = UB.B.

Emitted



Radiation from Objects - A body with reflection

Assume the body has a coefficient of absorption of e, the number of absorbedphotons should be:

NBodyabs = eNBox

emit

Ideal black body: e = 0. Ideal reflector: e = 1.

In order to be at equilibrium state, the total power emitted by the object mustbe:

power = eσAT 4 ,

in which A is the surface area of the emitter. σ is called the”Stefan-Boltzmann constant”.

e is also called ”emissivity” of the material.



Two examples

7-08. Black Hole as a Black Body:

7-09. Star spectrum and its surface temperature, emission power, andits size.:


Chapter 7: Quantum Statistics

Documents

Transcript of Chapter 7: Quantum Statistics