Meaning of the chemical potential for a boson gas

Question

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).

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

Answer 1

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: $\frac{\partial E}{\partial N}=\mu$.

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

Answer 2

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.

Answer 3

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 $i$.