Fermi Temperature and Black-body radiation

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My understanding is that the Fermi temperature is a measure of the energy of a system at its lowest energy state. This suggests that at the Fermi temperature is a minimum temperature where the system can't radiate away any more energy.

If this were a physical temperature it seems the system should radiate black body radiation. But then where would this energy come from?

Rather than a physical temperature, would it be better to see this as a cutoff, like static friction, representing the cutoff where the lowest energy state is at thermal equilibrium with its surroundings, but above which excited states will be required?
 

DrClaude

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This suggests that at the Fermi temperature is a minimum temperature where the system can't radiate away any more energy.
That's incorrect. The Fermi temperature is simply the conversion of the Fermi energy to units of temperature, with the Fermi energy based on the T = 0 distribution of fermions (degenerate Fermi gas). It is not the temperature at which the "lowest energy state" is attained.

The usefulness of the Fermi temperature is that it serves as a cutoff for the quantum behavior of the Fermi gas. Below TF, the gas is quantum while above TF it can be considered as a classical gas.
 

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