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Chapter 18 Bose-Einstein Gases
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  • Chapter 18 Bose-Einstein Gases

  • 18.1 Blackbody RadiationThe 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 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. NJ = N does not apply.

    The Lagrange multiplier that was determined by NJ = N becomes 0 (i.e. the chemical potential 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 calculated with hvSo:

    Recall that g(v)dv is the number of quantum states with frequency v in the range v to v + dv

  • 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) g(v)

  • 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:

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

  • Here U() is the energy per unit wavelength

    The Stephan-Boltzmann 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

    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 Wiens isplacement Law

  • For long wavelengths:

    (Rayleigh- Jeans formula)For short wavelength:

    Suns Surface T 6000K, thus

  • Sketch of Plancks law, Wiens law and the Rayleigh-Jeans law.

  • Using

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

  • 18.2 Properties of a Photon GasThe number of photons having frequencies between v and v + d v 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 1 m3 heated to a temperature of 400 FSolution 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:

  • 18.3 BoseEinstein CondensationA 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:

  • First Goal: Analyzing how the chemical potential varies with temperature T.

    Choosing the ground state energy to be ZERO! At T = 0 all N Bosons will be in the ground state. must be zero at T = 0 is slightly less than zero at non zero, low temperature.

  • At high temperature, in the classical limit of a dilute gas, M B distribution applies:

    In chapter 14:

    Thus

  • Example: one kilomole of 4He at STP

    = -12.43The average energy of an ideal monatomic gas atom is

    Confirming the validity of the dilute gas assumption.

  • From chapter 12:

    There is a significant flow in the above equation (discussion )