8
$\begingroup$

How do we justify taking the chemical potential, $\mu$ as $0$ when calculating the critical temperature of Bose-Einstein Condensates (BECs)?

I apologise as I do not how to use LaTeX, for if I did the elegance of mathematics would’ve allowed me to construct my question with ease...

I understand to calculate the total number of particles in a system comprised of non-relativistic bosons of mass m at thermal equilibrium at temperature $T$. One must simply some over the occupancies for each energy state, the occupancies is given by the bose-einstein distribution...

For some reason during the derivation setting chemical potential to zero within the bose-einstein distribution gives us the largest possible number of particles for a given temperature, can someone explain why this is true?

Edit: Also I know the within the bose-einstein distribution, the energy of the states must always be greater than the chemical potential, this confines the distribution to a range of $$ 0<\text{bose-einstein distribution}<+\infty$$ I can say that the lowest energy state (ground state) has an energy of 0 and thus chemical potential < 0, but if my ground state has an arbitrary non zero energy would the chemical potential = 0?

$\endgroup$
1
  • $\begingroup$ Look into MathJax; it's pretty easy to use. $\endgroup$ Jan 11, 2016 at 14:05

2 Answers 2

8
+100
$\begingroup$

To determine the upper limit on chemical potential for a gas of $\mathcal N$ bosons, look at the form of the Bose distribution in the grand canonical ensemble with $\langle N \rangle = \mathcal N$. When using the GCE, it's easiest to work at chemical potential $\mu$ and to then choose $\mu(\mathcal N)$ so that $\langle N\rangle(\mu)=\mathcal N$. Each state $s$ has average occupancy $$ \langle n_s\rangle=\frac{\sum_{n\geq 0} ne^{-\beta n(\epsilon_s-\mu)}}{\sum_{n\geq 0} e^{-\beta n(\epsilon_s-\mu)}}=\frac{1}{\Xi_s}\frac{\partial}{\partial(\beta\mu)}\Xi_s,\quad \Xi_s=\frac{1}{1-e^{-\beta(\epsilon_s-\mu)}},\\ =-\partial_{(\beta\mu)}\log(1-e^{-\beta\epsilon_s+(\beta\mu)})=\frac{e^{\beta\mu}}{1-e^{-\beta(\epsilon_s-\mu)}}. $$ This is finite as long as $\mu<\epsilon_s$. In order for $\langle N\rangle=\sum_s \langle n_s\rangle$ to be finite we need $\mu<\min_s \epsilon_s=\epsilon_0$. Hence, for any system of bosons where $N$ is conserved we have $\mu<\epsilon_0$. It is conventional to set $\epsilon_0=0$ for simplicity, but you can have systems with $\epsilon_0\neq 0$. As you implied by your final question, in these systems the critical value of $\mu$ is $\epsilon_0$ in the thermodynamic limit, with $N, V, E\rightarrow \infty$ and $\mu, p, T$ held constant. Of course, if the system has no BEC phase then as $T\rightarrow 0$, the chemical potential $\mu$ never exceeds some value $\mu_\max<\epsilon_0$.

$\endgroup$
3
$\begingroup$

You can think of the chemical potential as the amount of free energy needed to add one additional particle to the system. Because the ground state of a BEC is degenerate and can hold an infinite number of particles, there's no energy cost to add another particle to that state. So, $\mu = 0$.

$\endgroup$
2
  • $\begingroup$ @Couchyam Yes, thanks for catching that. I've edited my answer. $\endgroup$ Jan 19, 2016 at 1:29
  • $\begingroup$ One additional thing: that there is no energy cost to adding another particle to the ground state only implies that $\mu\leq 0$. For example, the ground state of a classical ideal gas has zero energy, but there is no BEC phase. The reason is that the entropy associated with excited states of a classical gas overcomes the energy cost of these states at any (arbitrarily small) positive temperature, so $\mu$ is strictly negative in this case. For a gas of bosons, the situation is different. $\endgroup$
    – TLDR
    Jan 19, 2016 at 5:04

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.