Photon (or Planck) Statistics

1. Photon (or Planck) Statistics

2. Blackbody Radiation • Experiments on EM Radiation from a Hot Body: “Black Body Radiation” • Energy loss of a hot body is attributable to the emission of electromagnetic waves from the body. • The distribution of the energy flux over the wavelength spectrum does not depend on the nature of the body but does depend on its temperature. In Quantum Electrodynamics: EM radiation is regarded as a photon gas.

3. Qualitative Sketch of the Blackbody Radiation Energy Spectrum for 3 temperatures: T1 > T2 > T3

4. Photons emitted by an atomic energy level can be absorbed at another, so The total number of photons is not constant, i.e. ΣNJ = N does not apply • Lack of Photon Number Conservationcan be shown to imply that the Chemical Potentialin the Bose-Einstein Distribution = Zero for Photons. So, photon = 0 • Photons are Bosons of Spin 1& they obey Bose-Einstein Statistics:

5. For a continuous spectrum of energy • T • The energy of each photon is  = hν, • So: • Definition: f(ν)dνnumber of photon states with frequency v in the range ν to ν + dν

6. In a photon gas, there are 2 EM polarization states, corresponding to the 2 independent directions of polarization of an EM wave. • As a result t • c speed of EM wave (light) in vacuum.

7. The energy within the frequency range v to v+ dvequals the number of photons within that range times the energy of each photon: u(ν) u(ν) So we finally have: u(ν)

8. u u(λ) • So the total photon energy at frequency  is: u(ν)dν • This is the Planck radiation formula, which gives the energy per unit frequency. • Re-expressing this in terms of the wavelength gives:

9. u(λ) • u(λ) is the energy per unit wavelength. • The Stephan-Boltzmann Lawstates that The total radiation energy is proportional toT4 • This can be derived from the above expression by integrating u() over all wavelengths , as follows:

10. u(λ) • The total energy can be calculated by integrating this over all : • Setting one has

11. The total energy of the photon gas is: From a table, the integral has a value of • So

12. The particle flux equals , where ν is the mean speed and n is the number density. In a similar way, the energy flux e can be calculated: • This is the Stephan-Boltzmann Law with the Stephan-Boltzmann constant =

13. u(λ) u(λ) • The wavelength at which is a maximum can be found by setting the derivative of equal to zero: • This is known as Wien’s Displacement Law

14. u • For long wavelengths: (Rayleigh- Jeans formula) • For short wavelengths: • Sun’s Surface T≈ 6000K, thus

15. Sketch of Planck’s law, Wien’s law & the Rayleigh-Jeans law.

16. Using • The surface temperature of earth equals 300K, which is in the infrared region • The cosmic background microwave radiation is a black body with a temperature of 2.735± 0.06 K.

17. Properties of a Photon Gas The number of photons having frequencies between v and v+ dv is The total number of photons in the cavity is determined by integrating over the infinite range of frequencies:

18. Leads to Where T is in Kelvin and V is in m3 • The mean energy of a photon • The ratio of is therefore of the order of unity. • The heat capacity

19. Substitute • Entropy

20. Therefore, both the heat capacity and the entropy increase with the third power of the temperature! • For photon gas: • It shows that F does not explicitly depend on N. Therefore,

21. Calculate the total electromagnetic energy inside an oven of volume 1 m3 heated to a temperature of 400 F • Solution

22. Show that the thermal energy of the air in the oven is a factor of approximately 1010 larger than the electromagnetic energy. • Solution: