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Fermi gas predictions for heat capacity and susceptibility

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First let's ignore that electrons in metals are a degenerate Fermi gas by ignoring Fermi stastistics. We then have a classical gas of electrons (in the sense that Maxwell-Boltzmann statistics applies). The internal energy would then be given by:

E = 3/2 N k T

where N is the total number of electrons in the metal.

The heat capacity is thus given by:

Cv = 3/2 N k (1)

The susceptibility can be computed as follows. The energy of an electron in a magnetic field H is:

E = - mu dot H

where mu is the magnetic moment of the electron, which is proportional to the spin:

mu = g mu_0 s

Here s is the spin of the electron mu_0 is the Bohr magneton and g is the so-called g-factor.

Now, if we denote the magitude of the electron magnetic moment by |mu|, then in a magnetic field the electron has two energy eigenstates with energies plus or minus |mu| |H|. The plus sign corresponds to the magnetic moment pointing in the opposite direction of H. The partition function for the spin part of the electron in a magnetic field is thus given by:

Z = exp(beta H mu) + exp(-beta H mu)

where for notational convenience mu and H now stand for |mu| and |H|.

The probability that mu points in the direction of H is thus:

P_{+} = exp(beta H mu) /[exp(beta H mu) + exp(-beta H mu) ]

And the probability that mu points in the direction opposite to H is:

P_{-} = exp(-beta H mu) /[exp(beta H mu) + exp(-beta H mu)]

The ...

Solution Summary

A detailed explanation from first principles is given.