Chemical Potential Question, *

Hi everyone

I have the impression that the chemical potential mu for particles
obeying Bose-Einstein statistics is always less than zero. What about
for fermions? Are they always positive, or can they take any value?

As an example I calculate the number density for ultrarelativistic fermions:

n_f
= (g/2Pi^2)Integral[p^2/(exp[beta(p-mu)] + 1),{p,0,Infinity}]
= -(g/2Pi^2) T^3 PolyLog[3,-exp[mu/T]]

And for relativistic bosons we have

n_b
= (g/2Pi^2)Integral[p^2/(exp[beta(p-mu)] - 1),{p,0,Infinity}]
= (g/2Pi^2) T^3 PolyLog[3,exp[mu/T]]

where g is the number of spin degrees of freedom and p is the momentum.

For the case of bosons, we use the fact that the PolyLog[p,x] - if we
_define_ it as the sum of x^q/(q+1)^p from q = 0 to infinity - diverges
for |x|>1, i.e. for exp[mu/T]>1 or mu/T > 0. Then we say therefore mu/T
< 0 always. Is this the correct reason why bosons have negative mu?

What about for fermions? It seems by the same argument above mu/T < 0
always too. But I realize PolyLog can be "analytic continued" for x < -1
on the real line, so this seems to allow mu/T to take any value. So are
there any restrictions on the chemical potential of fermions?

One of the reasons I ask this is because it seems in
antiparticle-particle reactions, say e- + e+ <-> gamma + gamma if we
assume there is some kind of equilibrium (such as in the very early
universe) then I suppose we can say something like mu_e+ + mu_e- =
mu_gamma + mu_gamma = 0, because mu_gamma = 0 always. And therefore in
chemical equilibrium the chemical potentials of antiparticles are
negative of that of their particle counterparts. But what about bosons,
since mu_bosons/T < 0 always?

Thanks for helping me understand this.

Yi-Zen

I think the qth term should actually be x^(q+1)/(q+1)^p.


I wondered why is it in equilibrium, given a reaction A + B <-> C + D,
the sum of the chemical potentials on both sides are equal - mu_A + mu_B
= mu_C + mu_D? The reason given by Landau (Statistical Physics) is that
in equilibrium since by definition the particle number is constant the
free energy F = T ln[Z] is minimized with respect to the variation of
particle number - that leads directly to the above condition if one
recalls that the chemical potential of a given particle species is the
derivative of the free energy with respect to particle number of the same.

Yi-Zen

I want to understand this too, actually: why is it that if I have a
reaction A + B <-> C + D, and if this is in some kind of chemical
equilibrium - by that I assume what is meant is that the number of
particles of A, B, C and D remain constant - then mu_A + mu_B = mu_C +
mu_D? That is why is the sum of the chemical potentials equal?

Yi-Zen

Dear Yi-Zen,

the calculations you are doing are about ultra-relativistic particles (you
have set E=p) in the thermodynamic expressions. So the results you
have found only apply to the situation T >> m (or m=0)

However, it is correct that in the case of *ultra-relativistic* bosons
the PolyLog-function becomes imaginary whenever \mu_Bosons > 0.
Therefore ultra-relativistic bosons cannot have a chemical potential
which is positive. I.e. simply from microscopic statistical thermodynamics
we have the condition \mu_Boson <= 0 (for ultra-relativistic bosons).

Note that I will be assuming T>0 throughout the whole discussion. For
the thermodynamics of an ultra-relativistic gas of fermions and bosons
only the *ratio* /mu / T is important (it is the quantity that appears in
the polylog function and determines whether polylog is real, or becomes
imaginary - which is not desirable).

Now at ultra-relativistic energies you will have particle-antiparticle
pair production in abundance. On the other hand, the chemical potentials
of particle and anti-particle are necessarily opposite to each other:

\mu_particle + \mu_antiparticle = 0.

Combined with the thermodynamic result \mu_Boson <=0 this only leaves
\mu_boson = 0 for ultra-relativistic bosons. In fact, this is the chemical
potential of the photon (which quite definitely is an ultra-relativistic
boson, as m=0), but the result is more general: *Any* ultra-relativistic
(or zero rest mass) boson *must* have a chemical potential of *zero*,
at least if thermodynamic reasoning is still appropriate at
ultra-relativistic
energies.

However, this result is only true at ultra-relativistic energies. At *lower*
energies, massive bosons can even have a chemical potential that is
positive, as long as \mu < m. (See the appendix of gr-qc/0405010 for the
thermodynamics of an ideal gas of fermions and bosons with m arbitrary -
you can probably find the same results elsewhere, however, when I needed
them I was too lazy to look and therefore decided to derive the necessary
results myself - if you need to *understand* what you are doing, doing the
calculations oneself is usually quicker than looking them up ;-)

No, there aren't. For fermions the polylog-function is well defined for
*any* value of the chemical potential. The chemical potential of
ultra-relativistic fermions is not restricted. However, one still has the
restriction that fermions and anti-fermions have opposite chemical
potentials. Therefore, whenever an ultra-relativistic fermion species has
a *non-zero* chemical potential, we can distinguish between particles
and anti-particles in a thermodynamic sense.
The particle in the particle-antiparticle pair with *positive* chemical
potential would be called "normal matter", whereas its antiparticle
(with negative chemical potential) would be called "antimatter".

I would think, that you *always* can say \mu_particle + \mu_antiparticle =0,
independent of the nature of the particle. At least this is what I thought I
have learned from QFT. Or does anyone in this group disagree? This
would in fact be interesting.

Best wishes, M