Chapter 7: Quantum Statistics

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1 Part II: Applications SDSMT, Physics 2014 Fall

2 1 Introduction Photons, E.M. Radiation 2 Blackbody Radiation The Ultraviolet Catastrophe 3 Thermal Quantities of Photon System Total Energy Entropy 4 Radiation from Objects Radiation from a Box Radiation from Objects

3 Bosons Two types of bosons: (Type-1) Composite particles: contain an even number of fermions. The number of these particles is conserved if the energy does not exceed the dissociation energy (in the range of MeV in the case of the nucleus - the nuclear binding energy). (Type-2) Particles associated with a field. The most important example are photons associated with electromagnetic field. These particles are not conserved: if the total energy of the field changes, particles appear and disappear. We ll see that the chemical potential of such particles is zero in equilibrium, regardless of density. Typically, radiation emitted by a hot body or from a laser: systems not in equilibrium - energy is flowing outwards and must be replenished from some source. The first step towards understanding of radiation being in equilibrium with matter was made by Gustav Kirchhoff ( , German), who considered a cavity filled with radiation, the walls can be regarded as a heat bath for radiation.

4 E.M. Radiation The walls emit and absorb E.M. waves. In equilibrium, the walls and radiation must have the same temperature T. Field and temperature - not a crazy correlation? No. The energy of electromagnetic (E.M.) radiation is spread over a range of frequencies, and we define u s(ν, T )dν as the energy density (per unit volume) of the radiation with frequencies between ν and ν + dν, u s(ν, T ) is the spectral energy density. The internal energy of the photon gas should be: u(t ) = 0 u s(ν, T )dν. In equilibrium, u s(ν, T ) is the same everywhere in the cavity, and is a function of frequency and temperature only. If the cavity volume increases at T = const, the internal energy U = u(t )V also increases. The essential difference 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 number of photons is not conserved, but proportional to volume in an isothermal change.

5 E.M. Radiation in Equilibrium with Matter A real surface absorbs only a fraction of the radiation falling on it. The absorptivity α is a function of ν and T ; a surface for which α(ν, T ) = 1 for all frequencies is called a black body. Our current understanding is, the E.M. field has an infinite number of modes (standing waves) in the cavity. The black-body radiation field is a superposition of plane waves of different frequencies. The characteristic feature of the radiation is that a mode may be excited only in units of the quantum of energy hν (similar to a harmonic oscillators): ɛ i = (n i + 1/2)hν. This fact leads to the concept of photons as quanta of the E.M. field. The state of the E.M. field is specified by the number n for each of the modes, or, in other words, by enumerating the number of photons with each frequency: E γ = hν = cp γ, p γ = hν c According to the quantum theory of radiation, photons are massless bosons of spin 1 (in units ). They move with the speed of light.

6 Photons: Frequency and modes T

7 Equilibrium in a Photon Gas The linearity of Maxwell equations implies that the photons do not interact with each other. Non-linear optical phenomena are observed when a large-intensity radiation interacts with matter. The mechanism of establishing equilibrium in a photon gas is absorption and emission of photons by matter. Presence of a small amount of matter is essential for establishing equilibrium in the photon gas. We will treat a system of photons as an ideal photon gas, and, in particular, we will apply the Bose-Einstein statistics to this system.

8 Where classical physics fails Blackbody radiation & the ultraviolet catastrophe in the classical picture: E.M. radiation is a continuous field that permeates all space. E. M. field in a box: E.M. field = ν=0 (Standing Waves)ν Classical picture: Each standing wave has an average thermal energy of 2 1 kt, very much like infinite number of one-dimensional mechanical 2 oscillators: each oscillator has two degrees of freedom, one for kinetic energy (ẋ or p), one for potential energy (x). The total energy in the field is, E field = ν=0 kt which is obviously wrong! In quantum physics: Each standing wave is a standing mode. A single mode (with a fix ν value) is equivalent with a single-particle state.

9 The solution by Max Planck The quantum oscillator cannot have any amount of energy but the allowed ones like: ɛ n = 0, 1hν, 2hν,... People understand this better after the quantum physics was established. When ɛ n = 0, 1hν, 2hν,..., the partition function for single oscillator becomes: Z = n=0 e nhν/kt = n=0 e nhνβ = The average energy is therefore, Ē = 1 Z Z β = hν e hν/kt e nhνβ Therefore, when energy comes in units of hν, the mean number of units of energy in the oscillator is n Plank = 1 e hν/kt 1 This is the so-called Plank Distribution. The units of energy correspond to photons (1) (2)

10 The solution by Max Planck - cnt. Since photons are Bosons, the mean number of photons at certain energy should follow the B.E. distribution, n photon = 1 e (ɛ µ)/kt 1 Comparing with n Plank = 1 e hν/kt 1, one can see µ photon = 0. Does this make sense? Let s take a look at the how photons can be created or destroyed. e e + γ At equilibrium, the γ disappearing rate is the same with its appearing rate, the chemical potential in this case is: µ e = µ e + µ γ. This means, µ γ = 0.

11 E.M. Radiation in a Box - Photon Gas To obtain the bulk property, we need to consider photons in all modes: Total energy, total number of photons, etc. Taking a cubic box of photons (E.M. radiation) for example, we have: 1-d potential well: λ = 2L p = h n 2L n (3) photon energy: ɛ = pc = hc 2L n (4) 3-d box: ɛ = hc 2L n 2 x + n 2 y + n 2 z = hnc 2L Now, we can take the same steps as we didi before to calculate the total energy of the E.M. radiation. (5)

12 E.M. Radiation in a Box - Photon Gas - cnt. Total E.M. energy: U = 2 ɛ n Pl (ɛ) n x n y n z 2 for two polarizations. U = hcn 1 L e hcn/2lkt 1 n x,n y,n z U = 0 dn π/2 0 dθ π/2 0 dφ (6) n 2 sinθ hcn 1 (7) L e hcn/2lkt 1 U = π n 2 hcn L e hcn/2lkt 1 dn (8) Change variable ɛ = hcn : n = 2Lɛ hc, dɛ = dn 2L hc 2L

13 E.M. Radiation in a Box - Photon Gas - cnt. U = π U = π U = π U V = n 2 hcn 1 dn (9) 2L e hcn/2lkt 1 ( ) 2 2L ɛ 2 1 ɛ hc e ɛβ 1 2L dɛ (10) hc ( ) 3 2L ɛ 3 1 dɛ (11) hc e ɛβ 1 8π (hc) 3 ɛ 3 dɛ (12) e ɛβ 1 The energy density per unit photon energy, or the spectrum of the photons: u(ɛ) = 8π (hc) ɛ 3 3 e ɛ/kt 1 (13)

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

15 E.M. Radiation in a Box - Total Energy U V = 8π 0 (hc) ɛ 3 dɛ (14) 3 e ɛβ 1 U ɛ=ktx 8π(kT )4 x 3 = dx (15) V (hc) 3 e x 1 U V = 8π(kT )4 (hc) 3 0 π4 15 = 8π5 (kt ) 4 15(hc) 3. (16)

16 The Plank Spectrum In terms of the dimensionless variable x = ɛ/kt By code: F7_19.C Peak x = 2.82 Or Ε=2.82 kt

17 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.

18 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/ɛ to derive the photon spectrum as a function of wavelength. Eq. (12): U V = 0 8π (hc) 3 ɛ 3 e ɛ/kt 1 dɛ

19 Total Energy Eq.(15) [Eq. (7.85) in the textbook] U x 3 dx V 0 e x 1 U V = 8π(kT )4 (hc) 3 = 8π(kT )4 (hc) 3 U = 8π5 (kt ) 4 V 15(hc) 3 π4 15 Energy density is proportional to T 4. Heat capacity: C V = ( U T )V = 32π5 k(kt ) 3 15(hc) 3 V = 32π5 k 4 V 15(hc) 3 T 3. Note: The integration x 3 π4 dx = 0 e x 1 15 x n dx = Γ(n + 1)ζ(n + 1), ζ(n) is the Riemann zeta function. 0 e x 1 x n dx = ( ) e x +1 2 Γ(n + 1)ζ(n + 1). n See more in Appendix B.

20 Entropy of Photon Gas δs = δq T S(T ) = T 0 S(T ) = T 0 C V (T T dt 32π 5 k(kt ) 3 15(hc) 3 V 1 T dt S(T ) = 32π5 k 4 15(hc) 3 V T 0 T 2 dt S(T ) = 32π5 k 4 15(hc) 3 V 1 3 T 3 S(T ) = 32π5 k 4 45(hc) 3 VT 3 S(T ) = 32π5 45 V in the last step, σ = 2π5 15 ( ) 3 kt k = 16 hc 3c σvt 3, k 4 h 3 c 2, called the Stefan-Boltzmann constant.

21 Radiation from a Box We have obtained the radiation inside a Box: U (7.83): = 8π ɛ 3 dɛ V 0 (hc) 3 e ɛβ 1 (7.84): u(ɛ) = 8π (hc) 3 (7.86): U V = 8π(kT )4 (hc) 3 ɛ 3 e ɛ/kt 1 π4 15 = 8π5 (kt ) 4 15(hc) 3 What is the emission from the box? dv = Rdθ Rsinθdφ dr dv = Rdθ Rsinθdφ cdt Energy in dv : U dv = V ucdtr2 sinθdθdφ Assuming photons are moving all directions with equal probability (equilibrium): dω = dacosθ R 2 Probability of photons moving into da: P = 1 dacosθ dω = 4π 4πR 2

22 Radiation from a Box - cnt. Energy escape from the box through da is: de(θ, φ, dt) = u dacosθ 4πR 2 de(θ, φ, dt) = ucdtr 2 sinθdθdφ dacosθ 4πR 2 de(θ, φ, dt) = dacosθ u cdtsinθdθdφ 4π de = c 2π dφ π/2 dθsinθ cosθ dadt 0 0 4π de = c 2π 1 1 8π5 (kt ) 4 dadt 2 4π 15 (hc) 3 de = c 8π5 (kt ) 4 = σt 4 dadt 4 15 (hc) 3 in which σ = 2π5 15 8π5 15 (kt ) 4 (hc) 3 k 4 h 3 c 2 = W /(m 2 K 4 ). This is called the Stefan s Law, discovered in 1879 from experiments. σ is the Stefan-Boltzmann constant Or, The emission power per unit area = de dadt = c 4 U V.

23 Radiation from Objects - How to Think about this? Radiation from object means no reflection happens (the black body radiation) although in reality reflection always exists. A thought experiment to demonstrate that a perfectly black surface emits radiation identical to that emitted by a hole in a box of thermal photons: T box = T black surface ; Area box = Area black surface ; Each absorbs the same fraction of the other s radiation. By the 2nd Law of Thermodynamics, we must have: Uemitted Box = U black body emitted

24 Radiation from Objects - A body with reflection Assume the body has a coefficient of absorption of e (the fraction of photons being absorbed), the number of absorbed photons should be: N Body abs = en Box emit Ideal black body: e = 1. Ideal reflector: e = 0. e is also called emissivity of the material. A good reflector must be a poor emitter, and vice versa. In order to be at equilibrium state, the total power emitted by the object must be: power = eσat 4. (17) where A is the surface area of the emitter. Again, σ is the Stefan-Boltzmann constant. Please read The Sun and the Earth on page in the textbook.

25 Two examples Star spectrum and its surface temperature, emission power, and its size.: Black Hole as a Black Body Problem Interesting problem.

Chapter 7: Quantum Statistics

Chapter 7: Quantum Statistics Part II: Applications SDSMT, Physics 2013 Fall 1 Introduction Photons, E.M. Radiation 2 Blackbody Radiation The Ultraviolet Catastrophe 3 Thermal Quantities of Photon System Total Energy Entropy 4 Radiation