I will answer your second question because it's the one with which I'm more familiar.

The question we're answering is: "Why does current in a superconductor move with no resistance?"

To understand this we should first understand why normal metals have nonzero resistivity. Imagine an electron in the metal and suppose it is traveling in some direction. If the electron never interacted with anything else then it would just go along merrily in that direction and in fact the resistance would be zero. However, in a normal metal the traveling electrons interact ("collide") with the ions of the metal via the Coulomb force because they're both charged. This transfers energy and momentum from the electron to the ion. Since that ion is tightly connected to the other metal ions, this energy and momentum transfer causes the lattice of ions to vibrate. An excitation of this kind of vibration of the lattice is called a "phonon". When you will hear people say that, in a normal metal, electrons scatter off of phonons, this is what they're talking about.

By the way, excitation of phonons is precisely heating of the metal, so we see that as the electron loses energy the metal heats up. This is Joule heating.

So, we can say that in a normal metal the resistivity is non-zero because the electrons scatter off of phonons. In a superconductor you'd think the situation would be the same. There are still vibrational modes in the metal, and the electrons and ions are still charged so of course they still interact via the Coulomb force. To understand why this doesn't happen we have to more carefully consider what's going on in a normal metal.

When an electron scatters off of a phonon the electron's state (ie. it's direction and speed of travel) changes. Electrons are in a class of particles called Fermions. Fermions are subject to the Pauli exclusion principle, which is that it is impossible to have two Fermions occupying the same state. That means, for example, that you can't have two electrons in the same momentum state. That means that in order for an electron to scatter off of a phonon, we have to have a situation where there is an unoccupied final state into which the electron can scatter. Otherwise, the scattering process simply cannot happen. In a normal metal at finite temperature, the electrons occupy a set of momentum states up to a certain level. Think of it as follows. When you put the first electron in the system it goes in the lowest energy state (which is at zero momentum). Because of the exclusion principle the next electron goes in a higher energy state, which also has more momentum. As you continue to add electrons you fill of a 3D ball of momentum states. This is called the "Fermi Sea" and is illustrated in part **a** of the attached diagram.

At zero temperature the electrons would fill the ball up to a cutoff energy. This is shown in dark red in the diagram. However, at finite temperature there's a bit of thermal energy that allows the electrons to jump around to slightly higher energy (and momentum) states. This little band is shown in pink in the diagram. In part **b** we zoom in on the pink band. The dark circles indicate filled electron states and the open circles indicate empty electron states. When an electron interacts with a phonon, it can only scatter and change state if there is a final state available. The green arrow indicates a possible scattering process, while the gray arrow indicates one forbidden by the exclusion principle.

In a superconductor, something really interesting happens. Because of interactions with the phonons, it turns out that there's an effective *attractive* interaction between the electrons! Basically what happens is that a traveling electron with its negative charge causes the positively charged ions of the metal to get sucked together a little bit. The electron leaves really fast, but the ions take longer to move around, so after the electron is gone there is still a region of increased density of positively charged ions. This causes other electrons to get attracted to that point, because of the higher positive charge. In this way, there is a weird time dependent attraction between electrons.

The attraction between electrons makes it turn out that there's a lower energy state than the filled red region shown in part **a** of the figure. This lower energy state is such that the usual single electron excitations are at an energy level that is a significant distance away from this new ground state. The gap in energy, shown in part **c** of the figure, is the reason that superconductivity happens. Now if electrons were to scatter off of a phonon, the only available states they have to scatter *into* are a large distance away in energy. That means that you need really high energy phonons (or something else) to disturb this ground state. So, as long as you don't whack your superconductor too hard, there's is NO scattering, and thus no resistance.

This argument also explains why superconductors have zero resistance at DC, but nonzero resistance at AC. If you put in high frequency perturbations, you can introduce enough energy to actually kick a superconducting electron out of the gap into those available states in the pink region. Remember, the energy of a perturbation is related to frequency by $E=h f$.

**Summary:** Superconducting metals have zero resistance because there aren't any available states for electrons to scatter into. No scattering means no resistance.

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 $\mu_\max<\epsilon_0$.

## Best Answer

Maybe it would be useful to clear up a little bit the all thing :

First of all, Bose-Einstein Condensation does not necessarily implies superfluidity:

Bose-Einstein condensation (BEC) is characterized by a macroscopic occupation of a many-particle state $\Phi$ which correspond to the product of individual particle groud-states with zero-momentum $|\textbf{k=0}\rangle$. Therefore, such system shows strong (quantum) phase coherence effects, said "particles are acting as one".

Superfluidity characterize a fluid that do not undergoes any dissipation by viscosity phenomena (i.e. its viscosity $\eta=0$). A superfluid has also a quasi-infinite thermal conductivity.

One can show that superfluidity only occurs in a BEC composed by interacting particles. In other words, an ideal gas of non-interacting boson can not be superfluid.

The reason is that interactions between particles allows to conserve the phase coherence for the all system. Since BEC generally occurs at pretty low temperatures, the thermal De Broglie wavelength associated to each particle is pretty large :

$$\lambda_{th}=\sqrt\frac{2\pi\hbar^2}{m\,k_B\,T}\underset{T\rightarrow 0}{\rightarrow}+\infty$$

so that there is an overlap between each particle wave-function in the condensate (hence the single "big" wave-function for the all condensate). This overlap is what I would call "a phase coherence effect".

And, the all game is that interactions between particles preserves such very special state from external perturbations, i.e. from phase decoherence phenomena. Before explaining this, lets introduce a key notion, which is the healing length $\xi$ of a condensate :

$$\xi=\sqrt\frac{\hbar^2}{m\,g\,n}$$

where $g$ is a constante characterizing interactions, and $n$ the density of the BEC. $\xi$ can be interpreted as the length scale in which density and $\Phi$ wave-function phase fluctuations are removed in the BEC.

Let us now considere a BEC flowing in a $D$ diameter pipe.

Without interaction $g=0$, leading $\xi\rightarrow +\infty$, more precisely $\xi\rightarrow D$. Thus, such condensate can not recover from any external perturbation in any point $\vec{r}$ of the BEC (for instance, lets say that a particle "hitting" the pipe is a phase perturbation). Therefore, the system has to dissipate its energy, leading to $\eta\neq 0$.

With interactions $g\neq 0$, $\xi$ is finite, eventually microscopic so that $D>>\xi$. In this case, phase coherence will be preserved in any point $\vec{r}$ of the BEC,

in a small $\xi$ width area near the contact surface between the pipe and the BEC. Then, the BEC is disspating energy only in this tiny area, and preserving the phase cohenrence everywhere else. This leads to $\eta\simeq 0$.exceptSo that was for qualitative explanation. For more theoretical insight, one can show that the Gross-Pitaevskii equation, which describes dynamical properties of a BEC, can be equivalently re-written (with some assumptions) like a Navier-Stokes equation for an irrotational inviscid flow.