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Chapter 18 Bose-Einstein Gases

Chapter 18 Bose-Einstein Gases. 18.1 Blackbody Radiation. The energy loss of a hot body is attributable to the emission of electromagnetic waves from the body.

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Chapter 18 Bose-Einstein Gases

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  1. Chapter 18Bose-Einstein Gases

  2. 18.1 Blackbody Radiation • The 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. • Electromagnetic radiation can be regarded as a photon gas.

  3. The spectrum of blackbody radiation energy for three temperatures: T1>T2>T3

  4. Photons emitted by one energy level can be absorbed at another, so the total number of photons is not constant, i.e. ΣNJ = N does not apply. • The Lagrange multiplier α that was determined by ΣNJ = N becomes 0 (i.e. the chemical potential µ becomes 0!), so that e- a= e-0 = 1 6. Photons are bosons of spin 1 and thus obey Bose-Einstein Statistics

  5. For a continuous spectrum of energy The energy of a photon is calculated with hv So: Recall that g(v)dv is the number of quantum states with frequency v in the range v to v + dv

  6. g(v) For phonon gas (chapter 16) In a photon gas, there are two states of polarization corresponding to the two independent directions of polarization of an electromagnetic wave. As a result Where c is the speed of electromagnetic wave (i.e. light)

  7. u u u u The energy within the frequency range v to v+ dv equals the number of photons within such a range times the energy of each photon:

  8. The above is the Planck radiation formula, which gives the energy per unit frequency. • When expressed in terms of the wavelength: Since

  9. u HereU(λ) is the energy per unit wavelength The Stephan-Boltzmann Law states that the total radiation energy is proportional to T4

  10. The total energy can be calculated Setting one has The integral has a value of with

  11. Particle flux equals , where is the mean speed and n is the number density. In a similar way, the energy flux e can be calculated as Assuming is the Stephan-Boltzmann Law with the Stephan-Boltzmann constant.

  12. The wavelength at which is a maximum can be found by setting the derivative of equal to zero! Or equivalently, set (minimum of ) One has is known as Wien’s isplacement Law

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

  14. Sketch of Planck’s law, Wien’s law and the Rayleigh-Jeans law.

  15. 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.

  16. 18.2 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:

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

  18. Substitute • Entropy

  19. 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 depends on N. Therefore,

  20. 18.1 a) calculate the total electromagnetic energy inside an oven of volume 1 m3 heated to a temperature of 400 F • Solution Use equation 18.8

  21. 18.1b) Show that the thermal energy of the air in the oven is a factor of approximately 1010 larger than the electromagnetic energy. • Solution:

  22. 18.3 Bose–Einstein Condensation • A gas of non-interacting particles (atoms & molecules) of relatively large mass. • The particles are assumed to comprise an ideal BE gas. • Bose – Einstein Condensation: phase transition • B – E distribution:

  23. First Goal: Analyzing how the chemical potential μ varies with temperature T. • Choosing the ground state energy to be ZERO! At T = 0 all N Bosons will be in the ground state. μmust be zero at T = 0 μ is slightly less than zero at non zero, low temperature.

  24. At high temperature, in the classical limit of a dilute gas, M – B distribution applies: In chapter 14: Thus

  25. Example: one kilomole of 4He at STP = -12.43 The average energy of an ideal monatomic gas atom is Confirming the validity of the dilute gas assumption.

  26. From chapter 12: There is a significant flow in the above equation (discussion … )

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