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Bose-Einstein Gases. Chapter 18. 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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Bose-Einstein Gases Chapter 18
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.
The wavelength spectrum of blackbody radiation energy for three temperatures: T1>T2>T3
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. (For phonon gas, N is not independent!) • The Lagrange multiplier α that was determined by ΣNJ = N becomes 0, so that e- a= e-0 = 1 6. Photons are bosons of spin 1 and thus obey Bose-Einstein Statistics
For a continuous spectrum of energy The energy of a photon is hv So: Recall that f(v)dv is the number of quantum states with frequency v in the range v to v + d 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)
The energy within the range v to v+ dv equals the number of photons within such a range times the energy of each photon:
The above is the Planck radiation formula, which gives the energy per unit frequency. • When expressed in terms of the wavelength Since
Hereμ(λ)is the energy per unit wavelength The Stephan-Boltamann Law states that the total radiation energy is proportional to T4
The total energy can be calculated Setting one has The integral has a value of with
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 this is the Stephan-Boltzmann Law with the Stephan-Boltzmann constant.
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 Displacement Law
For long wavelengths: (Rayleigh- Jeans formula) For short wavelength: Sun’s Surface T≈ 6000K, thus
Sketch of Planck’s law, Wien’s law and the Rayleigh-Jeans law.
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.
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:
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
Substitute • Entropy
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,
18.1 a) calculate the total electromagnetic energy inside an oven of volume heated to a temperature of 400 F • Solution Use equation 18.8
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: