1. Lecture 28 — The Planck Distribution Chapter 8, Monday March 24 th Before Planck: Wien and Rayleigh-Jeans The ultraviolet catastrophe The Planck distribution Reading: All of chapter 8 (pages 160 - 185) Homework 8 due Mon. Mar. 31st Assigned problems, Ch. 8: 2, 6, 8, 10, 12

2. Black-body spectrum Wien's displacement Law m T = constant = 2.898 × 10  3 m.K, or m  T  m T = constant = 2.898 × 10  3 m.K, or m  T  Found empirically by Joseph Stefan (1879); later calculated by Boltzmann  = 5.6705 × 10  8 W.m  2.K  4. Stefan-Boltzmann Law Power per unit area radiated by black-bodyR =  T 4 Rayleigh-Jeans equation u( )d = (# modes in cavity in range d ) × (average energy of modes) Define spectral energy distribution such that u( )d is the fraction of energy per unit volume in the cavity with wavelengths in the range to + d. Density of modes 

3. Wien, Rayleigh-Jeans and Planck distributions Wilhelm Carl Werner Otto Fritz Franz Wien

4. The ultraviolet catastrophe There are serious flaws in the reasoning by Rayleigh and Jeans Furthermore, the result does not agree with experiment Even worse, it predicts an infinite energy density as  0! (This was termed the ultraviolet catastrophe at the time by Paul Ehrenfest) Agreement between theory and experiment is only to be found at very long wavelengths. The problem is that statistics predict an infinite number of modes as 0; classical kinetic theory ascribes an energy k B T to each of these modes!

5. Planck's law (quantization of light energy) In fact, no classical physical law could have accounted for measured blackbody spectra The problem is clearly connected with u( )  , as  0 Planck found an empirical formula that fit the data, and then made appropriate changes to the classical calculation so as to obtain the desired result, which was non-classical. Max Planck, and others, had no way of knowing whether the calculation of the number of modes in the cavity, or the average energy per mode (i.e. kinetic theory), was the problem. It turned out to be the latter. The problem boils down to the fact that there is no connection between the energy and the frequency of an oscillator in classical physics, i.e. there exists a continuum of energy states that are available for a harmonic oscillator of any given frequency. Classically, one can think of such an oscillator as performing larger and larger amplitude oscillations as its energy increases.

6. Planck's law (quantization of light energy) N distinguishable oscillators in the walls of the cavity M indistinguishable energy elements (quanta)  so that U N = M  W N,M N = 1  1, 1 N = 2   1, 2, 1 N = 3    1, 3, 3, 1 N = 4     1, 4, 6, 4, 1 Pascal’s triangle

7. Planck's law (quantization of light energy) N distinguishable oscillators in the walls of the cavity M indistinguishable energy elements (quanta)  so that U N = M 

8. Maxwell-Boltzmann statistics Define energy distribution function: Then, This is simply the result that Rayleigh and others used, i.e. the average energy of a classical harmonic oscillator is k B T, regardless of its frequency. Planck postulated that the energies of harmonic oscillators could only take on discrete values equal to multiples of a fundamental energy  = h, where is the frequency of the harmonic oscillator, i.e. 0, , 2 , 3 , etc.... Then,U n = n  nh = 0, 1, 2... Where n is the number of modes excited with frequency. Although Planck knew of no physical reason for doing this, he is credited with the birth of quantum mechanics.

9. The new quantum statistics Replace the continuous integrals with a discrete sums: Solving these equations together, one obtains: Multiplying by D( ), to give.... This is Planck's law

10. n = 0, 1, 2... (quantum number)  = natural frequency of vibration  Vibrational energy levels for diatomic molecules Energy