30 Photons and internal motions

Size: px

Start display at page:

Download "30 Photons and internal motions"

Julie Cannon
5 years ago
Views:

1 3 Photons and internal motions 353 Summary Radiation field is understood as a collection of quantized harmonic oscillators. The resultant Planck s radiation formula gives a finite energy density of radiation field. The internal degrees of freedom have (often well-separated) characteristic energy scales. Key words photon gas, Planck s radiation formula, ultraviolet catastrophe, Stefan- Boltzmann law, internal degrees of freedom, rotational partition functions The reader should be able to: Derive Planck s formula. Recognize clearly the main features of Planck s formula. Itemize internal degrees of freedom of a molecule and telling their energy scales (in K). Sketch the molecular ideal gas specific heat as a function of T. 3.1 Quantization of harmonic degrees of freedom Photons and phonons are obtained through quantization of the systems that can be described as a collection of harmonic oscillators. 1 Possible energy levels for the i-th mode whose angular frequency is ω 2 i are (n + 1/2)ω i, where n =, 1, 2, (A.29). The canonical partition function of a system with modes {ω i } is given by (cf. Section 23) Z(β) = ( ) e β(ni+1/2)ωi, (3.1) i n i= since no modes interact with each other. Here, the product is over all the modes. The sum in the parentheses gives the canonical partition function for a single harmonic oscillator, which we have already computed (Section 23). The canonical partition 1 That is, the system whose Hamiltonian is quadratic in canonical coordinates (quantum mechanically in the corresponding operators). 2 What mode means was explained in 2.7.

2 354 Photons and internal motions function (3.1) may be rewritten as: [ ( Z(β) = (e βωi/2)] where i Ξ(β, ) = i i n i e βniωi ) = [ i ( e βωi/2)] Ξ(β, ), (3.2) ( ) e βnωi = (1 e βωi ) 1, (3.3) n= i which may be obtained from the definition of the grand partition function (see (28.16) and (28.17)) by setting ε i = ω i, and µ =. As long as we consider a single system, the total zero-point energy of the system i ω i/2 is constant and may be ignored by shifting the energy origin. 3 Therefore, the canonical partition function of the system consisting of harmonic modes (or equivalently, consisting of photons or phonons) may be written as the bosonic grand partition function with a zero chemical potential Ξ BE (β, ) (recall 28.6), regarding each mode ω i as a one particle state energy. From this observation, the reader should immediately recognize that T dependence of various thermodynamic quantities can be computed easily (or dimensional analytic approaches allow us to guess many T -dependent behaviors). 3.2 Warning: grand partition function with µ = is only a gimmick 4 The thermodynamic potential for the system consisting of photons or phonons is the Helmholtz free energy A whose independent variables are T and V. The expected number n i of phonons (photons) of mode i is completely determined by the temperature T and the volume V. Notice that we do not have any more handle like µ to modify the expectation value. If we set formally µ =, then da = SdT P dv, so we have A = P V (see Section 27 around the Gibbs-Duhem equation 27.4). That is, our observation log Z(β) = log Ξ(β, ) is consistent with thermodynamics. Thus, we may consistently describe systems consisting of phonons and photons in terms of the grand partition function for non-interacting bosons with zero chemical potential (as long as the zero-point energy is constant). However, do not understand this relation to indicate that the chemical potentials of photons and phonons are indeed zero; actually they cannot be defined. The 3 Warning: zero-point energies can shift However, if the system is deformed or chemical reactions occur, the system zero-point energy can change, so we must go back to the original formula with the total zero-point energy and take into account its contribution. For the electromagnetic field, the change of the total zero-point energy may be observed as force. This is the Casimir effect. 4 This is emphasized by Akira Shimizu. There is a report on the Einstein condensation of photons (Klaers, J. et al. (211). Bose-Einstein condensation of photons in an optical microcavity, Nature 468, ). In this case, the experiment is performed in an artificial environment in which the number of photons mimics a conserved quantity.

3 355 Photons and internal motions relation is only a mathematical formal relation that is useful sometimes Expectation number of photons The µ = boson analogy tells us that the average number of photons of a harmonic mode with angular frequency ω is given by (as we know, cf. (23.13) and a footnote 1* there) 1 n = e βω 1. (3.4) 3.4 Internal energy of photon systems The photon contribution to the internal energy of a system may be computed just as we did for the Debye model (Section 23). We need the density of states (i.e., photon spectrum = the distribution of the frequencies of the modes) D(ω). The internal energy of all the photons is given by E = n(ω) ω = modes ω dω D(ω) e βω 1. (3.5) This is the internal energy without the contribution of zero-point energy. A standard way to obtain the density of states D(ω) is to study the wave equation governing the electromagnetic waves, but here we use our usual shortcut (28.13). The dispersion relation for photons is ε = c p = ω, so ω D(ω )dω = V h 3 d 3 p. (3.6) p ω/c Differentiating the above equality, we obtain exactly as in 23.1 D(ω) = 4πV ( ) 2 ω h 3 c c = V ω2 2π 2 c 3. (3.7) Photons have two polarization directions, so the actual density of the modes is this formula Planck s distribution, or radiation formula The internal energy de ω and the number dn ω of photons in [ω, ω + dω) in a box of volume V are given by ω de ω = 2D(ω) e βω dω, (3.8) 1 1 dn ω = 2D(ω) e βω dω. (3.9) 1 The factor 2 comes from the polarization states (i.e., D here is given by (3.7)). 5 When chemical potential is definable Intuitively speaking, chemical potential may be defined only for particles we can pick up. More precisely speaking, if no (conserved) charge of some kind (say, electric charge, baryon number) is associated with the particle, its chemical potential is a dubious concept.

4 356 Photons and internal motions Fig. 3.1 Therefore, the energy density u(t, ω) at temperature T due to the photons with the angular frequencies in [ω, ω + dω) reads u(t, ω) = This is Planck s radiation formula (19 6 ) (Fig. 3.1). ω2 ω π 2 c 3 e βω 1. (3.1) Classical electrodynamics gives the Rayleigh-Jeans formula (3.12) (gray); this is the result of equipartition of energy. The density is not integrable due to the contributions from the UV modes (the total energy diverges). Wien reached (3.13) empirically (dotted curve). Planck initially guessed his formula (black) by interpolation of Wien s formula and another empirically known result: u(t, ω) T for small frequencies. Although the latter agrees with the Rayleigh-Jeans formula, it was theoretically derived only after Planck s formula. 3.6 Salient features of Planck s law It is important to know some qualitative features of Planck s formula (3.1): (i) Planck s law can explain why the spectrum blue-shifts as temperature increases: The peak position is 2.82k B T corresponding to the wavelength λ = /T nm. 7 (ii) The total energy density u(t ) = E/V of a radiation field at temperature T is finite. u(t ) is obtained by integration: u(t ) = dω u(t, ω). (3.11) With Planck s law (3.1) this is always finite (we will study this in the next entry). (iii) In the classical limit, we get u(t, ω) = k BT ω 2 π 2 c 3 ( = 2D(ω)k B T ), (3.12) which is the formula obtained by classical physics (i.e., the equipartition of energy; recall the last portion of 2.6) called the Rayleigh-Jeans formula. Upon integration, the classical limit gives an infinite u(t ). This divergence is obviously due to the contribution from the high frequency modes. Thus, this difficulty is called the ultraviolet catastrophe, which destroyed classical physics. (iv) In the high frequency limit ω k B T Planck s law (3.1) goes to u(t, ω) k BT π 2 c 3 ω2 e βω, (3.13) which was empirically proposed by Wien (1896) (indicating clearly the existence of 6 [19: The Paris World Exhibition; Boxer Rebellion; de Vries rediscovers Mendel s Laws; Sibelius Finlandia premiered; Evans begins to unearth Knossos.] 7 The Sun The surface temperature of the Sun is about 58 K, and its radiation roughly follows Planck s law of T = 55 K with the peak wavelength about 525 nm (green).

5 357 Photons and internal motions the energy gap corresponding to the quantum ω). 3.7 Statistical thermodynamics of black-body radiation The total energy density u(t ) of the radiation field at T is given by u(t ) = ω π 2 c 3 e βω dω = β 4 1 ω 2 We see (as seen above) which is called the Stefan-Boltzmann law. 8 (βω) 2 βω π 2 c 3 e βω d(βω). (3.14) 1 u(t ) T 4, (3.15) Since we know the T 3 -law of the phonon low temperature specific heat (the Debye theory; 23.1), this should be expected. This is understandable by counting the number of degrees of freedom (as explained in Fig. 29.1). Clearly recognize that the radiation field problem and the low temperature specific heat of crystals are the same problem. 3.8 Black-body equation of state The proportionality (3.15) was obtained purely thermodynamically by Boltzmann before the advent of quantum mechanics (as shown below). The proportionality constant contains, so it was impossible to obtain the constant theoretically before Planck. Photons may be treated as ideal bosons with µ =, so the equation of state is immediately obtained as ( 2 in front of the following integral is due to the polarization states) P V = log Ξ = 2 dε D(ε) log(1 e βε ). (3.16) k B T For 3D superrelativistic particles, D(ε) ε 2, so ε dε D(ε) = 1 εd(ε). (3.17) 3 This gives us (review what we did to derive P V = 2E/3 for the ordinary particles; recall and 28.14) P V = 1 E. (3.18) 3 Just as P V = 2E/3 is a result of pure mechanics, (3.18) is a result of pure electrodynamics, so this was known before quantum mechanics. Boltzmann started with (3.18) to obtain the Stefan-Boltzmann law 9 as follows. Since we know generally E = T S P V = T S 1 E, (3.19) 3 8 The proportionality constant can be computed as k 4 B π2 /15 3 c 3. Cf Stefan deduced the law from Tindal s data in [1879: Edison s electric light bulb; Ibsen s A Doll s House; Frege s Begriffsschrift; Einstein was born.]

6 358 Photons and internal motions ST = 4 3 E or S = 4 E 3 T. (3.2) Differentiating S wrt E under constant V and noting ( S/ E) V = 1/T, we obtain 1 T = 4 ( ) T 3T 2 E + 4 E 3T V (3.21) or 1 3T = 4 ( ) T 3T 2 E, E V (3.22) that is, under constant V de E = 4 dt T. (3.23) This implies the Stefan-Boltzmann law E T 4. Compare this with (23.3) in Let us look at the quantum effect on internal degrees of freedom in gas molecules. 3.9 Internal degrees of freedom of classical ideal gas If noninteracting particles are sufficiently dilute (µ ), we know classical ideal gas approximation is admissible. However, the internal degrees of freedom may not be handled classically, because energy gaps may be huge. We have already glimpsed at this when we discussed the gas specific heat (Section 23). Let us itemize internal degrees of freedom of a molecule: i) Each atom has a nucleus, and its ground state could have a nonzero nuclear spin. This interacts with electronic angular momentum in the atom to produce the ultrafine structure. The splitting due to this effect is very small, so for the temperature range relevant to the gas phase we may assume all the levels are energetically equal. Thus, (usually) we can simply assume that the partition function is multiplied by a constant g = degeneracy of the nuclear ground state. 1 ii) Electronic degrees of freedom has a large excitation energy (of order of ionization potential a few ev), so unless the ground state of the orbital electrons is degenerate, we may ignore it. 11 iii) If a molecule contains more than one atom, it can exhibit rotational motion. The quantum of rotational energy (Θ R below) is usually of order 1 K Nuclear spin-rotation interference in light homonuclear diatomic molecules In the case of light homonuclear diatomic molecules (e.g., H 2, D 2, T 2 ), nuclear spins could interfere with rotational degrees of freedom through quantum statistics, so we cannot take the nuclear effect simply by g. This effect is historically important in showing protons are fermions, but is not discussed in this book [See Kubo s problem book]. For not so light homonuclear diatomic molecules, usually we may ignore this effect. 11 What if the ground state is degenerate? If the ground state is degenerate, then it could have a fine structure with an energy splitting of order a few hundred K due to the spin-orbit coupling. For ground state oxygen ( 3 P 2 ) the splitting energy is about 2 K, so we cannot simply assume that all the states are equally probable nor that only the ground slate is relevant. 12 However, for H 2 it is 85.4 K. For other molecules, the rotational quantum is rather small: N 2 : 2.9 K; HCl: 15.1 K.

7 359 Photons and internal motions iv) Also such a molecule can vibrate. The vibrational quantum (Θ V order 1 K. 13 below) is of Fig Rotation and vibration Notice that there is a wide temperature range, including room temperature, where we can ignore vibrational excitations and can treat rotation classically (Fig. 3.2). Thus, equipartition of energy applied to translational and rotational degrees of freedom can explain the specific heat of many gases. C v /Nk B T 7/2 5/2 3/2 vibrational rotational translational ΘR RT Θ V The constant volume specific heat. RT means room temperature. Θ R is the rotational energy quantum (in K) and Θ V the vibrational energy quantum (in K). The hump in the rotational contribution can only be found by actual computation (no simple physical reason for it). The Hamiltonian for the internal degrees of freedom for a diatomic molecule reads H = 1 ( 2I Ĵ 2 + ω ˆn + 1 ), (3.24) 2 where I is the moment of inertia, Ĵ the total angular momentum operator and ˆn the phonon number operator. Therefore, the partition function for the internal degrees of freedom reads z i = z r z v : the rotational contribution is 14 z r = (2J + 1)e (Θ R/T )J(J+1), (3.25) J= with Θ R = 2 /2k B I and the vibrational contribution is z v = T e (Θ V /T )(n+1/2). (3.26) n= with Θ V = ω/k B. If the temperature is sufficiently low, then z r 1 + 3e 2Θ R/T. (3.27) 13 N K; O 2 : 226 K; H 2 : 61 K. 14 A rigid rotor is a symmetric top that cannot rotate around its symmetry axis. This restricted eigenspace is characterized by the total angular momentum quantum number J with the rotational energy 2 J(J + 1)/2I which is 2J + 1-tuple degenerate.

8 36 Photons and internal motions The contribution of rotation to specific heat is ( ) 2 ΘR C rot 3Nk B e 2ΘR/T. (3.28) T For T Θ R, we may approximate the summation by integration (Large Js contribute, so we may approximate J J + 1): z r 2 djje J 2 (Θ R /T ) = T Θ R. (3.29) This gives the rotational specific heat, but it is more easily obtained by the equipartition of energy, because the rotational energy is a quadratic form. Thus, C rot = k B (per molecule) in the high temperature limit for a diatomic molecule (because two orthogonal rotations are allowed; each degree of freedom classically contributes k B /2. Recall (2.25)). The vibrational contribution has already been studied in Section 23. Q3.1 [Einstein s A and B] Let us consider a system with two energy levels with energy (ground state) and ε (> ) (excited state). If many such systems are maintained at temperature T, we know the occupation ratio of these states must be given by a Boltzmann factor e βε. Now, let us assume that the system is also interacting with the electromagnetic field (radiation field) of frequency ν = ε/h with radiation energy density ρ (= the number density of the photons of energy hν; no other field interacts with the system because the energy exchange is through photons of energy hν). 15 Interacting with the radiation field, the system may be excited at the rate ρb n, where n is the number of the systems in their ground state and B a positive constant (imagine a collisional reaction between the atom and the photon). It is also de-excited at the rate ρb n ε, where n ε is the number of the systems in their excited state and B a positive constant. It is also known that there is the so-called spontaneous radiation : even if there exists no radiation field, a photon is emitted spontaneously from the excited state with the rate An ε, where A is a positive constant. If T is very large and the intensity of the radiation field is very strong, then n ε n, and the spontaneous radiation may be ignored relative to the induced transitions by radiation, so we must assume B = B = B, a common constant. In equilibrium there must be a transition balance between going up and going down. From this balance derive Planck s radiation formula ρ 1/(e βhν 1). Soln. The equilibrium condition must be (up rate = down rate) This implies Bρn = (A + Bρ)n e. (3.3) ρ = A/B e βhν 1. (3.31) 15 We follow the argument by Einstein in Einstein, A., (1917). Zur Quantentheorie der Strahlung, Phys. Z. 18, [To the quantum theory of radiation]. [1917: The Russian Revolution; the Balfour declaration]

9 PART III ELEMENTS OF PHASE TRANSITION

Physics 607 Exam 2. ( ) = 1, Γ( z +1) = zγ( z) x n e x2 dx = 1. e x2

Physics 607 Exam 2. ( ) = 1, Γ( z +1) = zγ( z) x n e x2 dx = 1. e x2 Physics 607 Exam Please be well-organized, and show all significant steps clearly in all problems. You are graded on your work, so please do not just write down answers with no explanation! Do all your