Something that has bothered me for a while regards the interpretation of chemical potential for different statistics. While I understand its meaning in metals (and its relation with the Fermi surface), I cannot quite relate this definition with the thermodynamic chemical potential, defined as the change in energy of the system when one particle is added to it (or according to Wolfram Demonstrations, " It can be interpreted, for example, as the ability of the system to perform phase transitions or chemical reactions, or its tendency to diffuse").

1) Are those concepts (thermodynamic vs Fermi-Dirac chemical potential) related or they should be thought as different things?

2) Am I missing something trivial or this cited demonstration is misleading? It is mentioned there that the x coordinate corresponds to (E-u). Shouldn't the function blow up for x = 0 independent of temperature, then? I do not understand the shift for negative values as temperature increases.


1 Answer 1


Of course they are related, the Fermi-Dirac chemical potential is defined to enforce the exclusion principle inside a thermodynamic framework rather than having to apply it with a separate mechanism.

Consider three systems, an ideal gas, a van der Waals type gas, and a Fermi gas. At a fixed volume and energy, ask "How much energy is required to add another particle?"

  • In an ideal gas the presence of a new particle has no effect on the resident particles.

  • In the van der Waals case, introducing a new particle pushed the particles on average closer together, and so adds to the total potential energy, but the new particle can be introduced with little or no kinetic energy. The amount of energy required (i.e. the chemical potential) depends on the density and the strength of the van der Waals force.

  • In the Fermi case, if the low lying states are filled then you can not introduce a new particle with low KE, but must introduce it with a lot of energy (how much being set by the current level of the Fermi surface). The chemical potential is at least the level of the Fermi surface.

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    $\begingroup$ Nice answer. The only point I'd make is that, about ideal gases, your point is not absolutely true: to introduce a new particle in such system one would have to change the energy by an amount that depends on the equation of state of the system. That can sound useless but it is necessary to understand the process of nucleation and growth in hard-sphere systems, though. Thanks! $\endgroup$
    – PFD
    Commented Mar 27, 2011 at 16:12
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    $\begingroup$ Yes, strictly speaking, the statement should be "How much energy is required to add another particle, such that the system is still in equilibrium?" To add a particle to the ideal gas in equilibrium, one has to pay the average kinetic energy, so the ideal gas still has a finite chemical potential. $\endgroup$ Commented Apr 1, 2013 at 15:09

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