Lecture 27 — The Planck Distribution Chapter 8, Friday March 21 st Quick review of exam 2 Black-body radiation Before Planck: Wien and Rayleigh-Jeans The ultraviolet catastrophe The Planck distribution Reading: All of chapter 8 (pages ) Homework 8 not due until Mon. Mar. 31st Assignment will be handed out on Monday

Exam 2 – question 1

Exam 2 – question 2

The Planck Distribution A. A. Michelson (late 1900s): “The grand underlying principles (of physics) have been firmly established......the future truths of physics are to be looked for in the sixth place of decimals.” Planck credited with the birth of quantum mechanics (1900) - developed the modern theory of black-body radiation - developed the modern theory of black-body radiation

Quantum nature of radiation 1st evidence from spectrum emitted by a black-body What is a black-body? An object that absorbs all incident radiation, i.e. no reflection A small hole cut into a cavity is the most popular and realistic example.  None of the incident radiation escapes What happens to this radiation? The radiation is absorbed in the walls of the cavity This causes a heating of the cavity walls Atoms in the walls of the cavity will vibrate at frequencies characteristic of the temperature of the walls These atoms then re-radiate the energy at this new characteristic frequency The emitted "thermal" radiation characterizes the equilibrium temperature of the black-body

Black-body spectrum

Black-bodies do not "reflect" any incident radiation They may re-radiate, but the emission characterizes the black-body only The emission from a black-body depends only on its temperature We (at 300 K) radiate in the infrared Objects at K start to glow At high T, objects may become white hot Wien's displacement Law m T = constant = × 10  3 m.K, or m  T  m T = constant = × 10  3 m.K, or m  T  Found empirically by Joseph Stefan (1879); later calculated by Boltzmann  = × 10  8 W.m  2.K  4. A black-body reaches thermal equilibrium when the incident radiation power is balanced by the power re-radiated, i.e. if you expose a black-body to radiation, its temperature rises until the incident and radiated powers balance. Stefan-Boltzmann Law Power per unit area radiated by black-bodyR =  T 4

Rayleigh-Jeans equation Consider the cavity as it emits blackbody radiation The power emitted from the blackbody is proportional to the radiation energy density in the cavity. One can define a 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. Then, the power emitted at a given wavelength, R( )  u( ) u( ) may be calculated in a straightforward way from classical statistical physics. u( )d = (# modes in cavity in range d ) × (average energy of modes) # of modes in cavity in range d  n  d   d # of modes in cavity in range d  n  d   d Average energy per mode is k B T, according to kinetic theory  u( ) = k B T n  k B T 

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

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!

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.

Maxwell-Boltzmann statistics Define an 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  = hf, where f is the frequency of the harmonic oscillator, i.e. 0, , 2 , 3 , etc.... Then,E n = n  nhfn = 0, 1, 2... Here, h is a fundamental constant, now known as Planck's constant. Although Planck knew of no physical reason for doing this, he is credited with the birth of quantum mechanics.

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

The results of Planck's law Note: the denominator [exp(hc/ k B t)] tends to infinity faster than the numerator (  5 ), thus resolving the infrared catastrophe, i.e. u( )  0 as  0. Note also: for very large  From a fit between Planck's law and experimental data, one obtains Planck's constant to be: h = × 10  34 J.s Planck's restriction of the available energies for radiation gets around the ultraviolet catastrophe in the following way: the short wavelength/high frequency modes are now limited in the energy they can have to either zero, or E  hf; in the calculation of the average energy, these modes with high energy are cut off by the Boltzmann factor exp(  E/k B T), i.e. these modes are rarely excited and, therefore, contribute nothing to the average energy in the limit  0.