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My lecturer told me that $\mu$, the Chemical potential, is zero or negative, and in the following example, mathematically it acts as a normalization constant. But is there any physical insight about why boson gas can be zero or negative?

I think it is due to the fact the photon gas can pop up from nowhere (i.e. vacuum fluctuation).

$$ f_{BE}(\varepsilon)=\dfrac{1}{e^{(\varepsilon-\mu)/(k_B T)}-1}$$

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    $\begingroup$ In the photon gas the chemical potential is always zero. For other cases, I'm not sure why your lecturer said it always has to be nonpositive. It is true that the chemical potential has to be less than the energy of the lowest level (less than the lowest $\varepsilon$). Depending on potentials and such, that lowest energy may be positive or negative. $\endgroup$ – Nanite Jan 4 '14 at 10:55
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    $\begingroup$ I recommend this paper: G. Cook and R. H. Dickerson, Understanding the chemical potential, 1995 aapt.scitation.org/doi/10.1119/1.17844 The authors explain in a simple and clear way the physical meaning of the chemical potential in various systems, including ideal Bose gas. $\endgroup$ – Rivka Jun 8 '15 at 12:55
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The chemical potential can be thought of as how accepting the system is of new particles -- how much work you have to do to stick a new particle in the system.

Since you can stick as many bosons in a given state as you want, the system is always accepting of new particles. At worst, you have to do zero work to add a boson (corresponding to $\mu=0$), and often the system is happy to take in a new particle (corresponding to $\mu<0$).

In contrast, you can only put one fermion in a given state. If you have a fermion with a certain energy, and you want to add it to a system where the state of that energy is already occupied, the system has to play musical chairs to make that happen. You may have to push that fermion in there, in which case the system will not be happy about it; you'd have to do some work ($\mu>0$).

In the case of photons, the system will take any energy you give it, but it won't reward you for it; it just doesn't care. $\mu=0$. It would be weird if $\mu$ were negative, because that would make it suck in all the photons (energy) it could get its hands on.


Edit in response to question in comment:

Why is $\mu$ the energy needed to stick in another particle? Let's work with a Maxwell-Boltzmann distribution because it's simpler. (Truthfully, I'm not sure how to do it with Bose-Einstein or Fermi-Dirac, but I'm not going to lose sleep over it; you can have fun with that.) Say you have states of energy $\epsilon_i$, $N$ particles, and $E$ total energy. You then have two normalization conditions:

$$N=\sum_in\left(\epsilon_i\right)=\sum_ie^{\alpha+\beta\epsilon_i}$$ $$E=\sum_i\epsilon_in\left(\epsilon_i\right)=\sum_i\epsilon_ie^{\alpha+\beta\epsilon_i}$$

(I like this notation better; $\beta=\left(k_bT\right)^{-1}$, and $\alpha=-\beta\mu$)

We want to show that the chemical potential is the change in system energy when increasing the number of particles: $\frac{\partial E}{\partial N}=-\mu$. (Why the minus sign? That's just how it's defined. There are lots of odd definitions in stat mech.)

Starting off:

$$N=e^\alpha\sum_ie^{\beta\epsilon_i}$$ $$e^\alpha=\frac{N}{Z}$$

where $Z=\sum_ie^{\beta\epsilon_i}$. Note that $\frac{\partial Z}{\partial\beta}=\sum_i\epsilon_ie^{\beta\epsilon_i}$

Then

$$E=\frac{N}{Z}\sum_i\epsilon_ie^{\beta\epsilon_i}$$ $$E=\frac{N}{Z}\frac{\partial Z}{\partial\beta}$$ $$E=-N\frac{\partial}{\partial\beta}\ln{Z}$$ $$\frac{\partial E}{\partial N}=-\frac{\partial}{\partial\beta}\ln{Z}$$

Putting it all together in terms of $\mu$

$$e^{-\beta\mu}=\frac{N}{Z}$$ $$-\beta\mu=\ln{N}-\ln{Z}$$ $$-\ln{Z}=\ln{N}-\beta\mu$$ $$-\frac{\partial}{\partial\beta}\ln{Z}=-\mu$$ $$\frac{\partial E}{\partial N}=-\mu$$ QED

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  • $\begingroup$ Ya, I see what you mean, but can you tell me why chemical potential can be thought of as the work you have to do to stick a new particle in the system Mathematically? $\endgroup$ – el psy Congroo Jan 5 '14 at 3:22
  • $\begingroup$ Doesn't it follow directly from the first law: $dE= -P dV + T dS - \mu dN$ or was this modification done after motivation from statistical mechanics ? $\endgroup$ – user7757 Apr 25 '14 at 13:53
  • $\begingroup$ But wasn't it the other way around? i.e. $dE=−PdV+TdS+\mu dN$ and $\mu= +\frac{\partial E}{\partial N}$... $\endgroup$ – Ralph Oct 1 '14 at 13:23
  • $\begingroup$ I guess I'd have to check a textbook; the definitions include random plus and minus signs. Either way, the $dE$ expression reeks of empirical thermodynamics, not statistical mechanics. You need to derive that $dE$ expression if you really want to see where that $\mu$ comes from; IIRC, that $dE$ experssion is specifically for an ideal gas -- not for every system. $\endgroup$ – lnmaurer Oct 6 '14 at 15:43
  • $\begingroup$ This is a little nitpicky, but the third line from the bottom should read $\ln Z = \ln N + \beta \mu$. $\endgroup$ – Joel Feb 16 '15 at 21:43
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I think you can consider the zero chemical potential to be an effect of popping up of photon. Both chemical potential and the temperature appears as a undetermined multiplier due to conservation of energy and number of particle respectively. For photon, there is no conservation of particle number and hence the corresponding undetermined multiplier, i.e. the chemical potential ($\mu/kT$ to be precise) is also zero for all energy states.

In other way you can say that since the number of particle is indefinite for each energy level, the equilibrium configuration is described by the minimisation of the free energy, $\partial A / \partial N_i =0$. When all the $N_i$ are independent of each other (as a result of spontaneous production and destruction of photon), this implies $\mu_i=0$ for all $i$.

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  • $\begingroup$ I see, I think I get what you mean for your first point. If the number of particle does not conserve, we don't have to worry about the undetermined multiplier ( the constraint of total number). But I'm not sure if I get what you mean for minimisation of free energy,I don't really get why people defined the term free energy and what that was used for. Anyway, your answer is helpful. Thanks! $\endgroup$ – el psy Congroo Jan 4 '14 at 14:55
  • $\begingroup$ thanks @elpsyCongroo. Equilibrium is defined as a state corresponding to minimum energy or maximum entropy. So you can obtain the equilibrium by either minimising the free energy or maximising the entropy. Depending on the constants of your process, you may find different free energy (Gibbs ($G$) or Helmholtz ($A$)) to be more useful for practical purpose. They are connected by Legendre transformations. So Equilibrium can be defined as a vanishing first derivative of free energy which corresponds no change of instantaneous configuration. $\endgroup$ – Sumit Jan 5 '14 at 7:30
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To be a little more precise: the chemical potential of a non-interacting Bose gas must be exceed the energy of the ground state single particle energy of that gas. If there are (as say, in $^4\rm He$) repulsive interactions between the particles, the chemical potential can be anything.

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