statistical mechanics - Meaning of the chemical potential for a boson gas

My lecturer told me that the mu is the Chemical potential is zero or negative, in the following example, mathematically it acts as a Normalisation 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 crop up from nowhere (i.e. Vacuum fluctuation).

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

Answer

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.

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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