Statistical Mechanics - Physics at Oregon State University

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104 CHAPTER 5. FERMIONS AND BOSONS and get F (T, N, V ) = N 0 N 0 N Other thermodynamic variables. The entropy follows from 0 EF (N ′ )dN ′ = 3 5 NEF E −1 F (N ′ )dN ′ = 3NE −1 F µ(T, V, N ′ )dN ′ ≈ 3 5 NEF 1 − 5π2 2 kBT 12 EF S(T, N, V ) = − ∂F ≈ ∂T V,N Nπ2 kBT kB 2 EF (5.81) (5.82) (5.83) (5.84) and indeed goes to zero if the temperature goes to zero. The internal energy follows from: U(T, N, V ) = F + T S ≈ 3 5 NEF 1 + 5π2 2 kBT (5.85) 12 At zero temperature we have U = 3 5 NEF . Hence the average energy per particle is less than the Fermi energy, as expected, since the Fermi energy is the energy of the occupied orbital with the largest energy. The average is not half the Fermi energy, however, since the number of states with a certain energy is increasing with energy. There are more orbitals with an energy larger than half the Fermi energy than there are with an energy less than half the Fermi energy. The heat capacity at constant volume is: ∂U CV (T, V, N) = ∂T V,N EF π ≈ NkB 2 kBT 2 EF (5.86) First of all, we notice that the heat capacity goes to zero in a linear fashion, as is required by the third law of thermodynamics. This was not true for the ideal gas, and we clearly have now an improved theory that is valid for low temperature. The ratio of the Fermi-Dirac heat capacity to the ideal heat capacity is: CV C ideal V = π2 3 kBT EF ≪ 1 (5.87) which seem to indicate that only a small portion of the particles participate in excitation processes. That is indeed true, and will be further explored in solid state physics.
5.1. FERMIONS IN A BOX. 105 Experiments in solid state physics are able to measure the heat capacity at low temperature. From these measurements on finds a value for the Fermi energy. The density of the electrons can also be measured, from lattice spacing data. Therefore, we can compare the two parts of the equation EF = 2 2M (3nπ2 ) 2 3 . In general, we find that the left hand side is not equal to the right hand side. The only available parameter is the mass M of the particles. Apparently, the mass has changed from the free particle mass. This is reasonable, since there are many body interactions. In order to move one electron we need to move others out of the way, too, and hence a certain applied force results in a smaller acceleration. In other words, the mass seems larger. In regular solids this enhancement is a factor between 1 and 10, but in special cases it can be 1000 or more. There are also cases where the enhancement is actually less than one, and lattice effects help to accelerate the electron even more than normal. 2 − Since we also know that EF ∝ V 3 we find for the pressure ∂F p(T, V, N) = − ∂V T,N from which we obtain the Gibbs energy: G(T, V, N) = F + pV ≈ NEF ≈ 2 NEF 1 + 5 V 5π2 2 kBT 12 EF 1 − π2 12 2 kBT EF (5.88) (5.89) which shows that also here G = µN, as expected. Finally, we can also calculate the grand energy: Ω(T, V, N) = F − µN ≈ − 2 5 NEF 1 + 5π2 2 kBT (5.90) 12 which is indeed −pV as expected from the Euler equation in thermodynamics. Now we can go back to the original form for the grand energy derived in this section. Remember that the grand energy had a simple linear dependence on volume? This gives: ∂Ω p = − = − ∂V T,µ Ω (5.91) V indeed. Therefore, the simple volume dependence in the grand energy was required by thermodynamics! Surprisingly, even at T = 0 the pressure of a Fermi gas is greater than zero. This is in contrast to the ideal gas. The origin of this pressure is the Pauli principle which makes that particles of the same spin avoid each other. For example, in a metal we have n ≈ 10 29 m −3 and the Fermi energy is a few eV or about 10 −18 J, leading to a pressure of about 10 5 atm! So why do the conduction electrons stay within a metal? The answer is, of course, the large EF
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Statistical Mechanics Henri J.F. Ja
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VIII INTRODUCTION Introduction. The
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