Pressure of Bose-Einstein-Condensate in decreasing volume Consider a Bose-Gas where $T_0$ is the critical temperature at which the temperature dependant chemical potential is $\mu(T_0) = 0$. Looking at the interval $ 0 \leq T < T_0 $ one can show that the pressure of the system is:
$$ P = \dfrac{2}{3}\dfrac{E}{V} = \zeta \bigg(\dfrac{5}{2}\bigg) \dfrac{k_BT_0}{\lambda_{T_0}^3}\bigg(\dfrac{T}{T_0}\bigg)^{\dfrac{5}{2}}$$
where $\zeta$ is the Riemann zeta function and $\lambda_T$ the termal de-Broglie-wavelength of the wavepackages. As one can see
$$P \propto T^{\dfrac{5}{2}} $$
But the interesting fact is that the pressure of an ideal bose gas in the interval $ 0 \leq T < T_0 $ is independant of the particle density N/V, where N is the number of particles and V the volume. One can deduce that there is no resistence of the system when decreasing the volume in which the gas is placed.
Now, in my lecture notes it says this is due to more particles being pushed into ground state when the volume is reduced, but I do not get how this is related to the effect.
I would be very happy if someone could explain me this effect in more detail as I have not found much about it yet.
 A: I restate the question for my own clarity:


*

*The Expression for pressure in a Bose-Einstein condensate is independent of the Volume and Number of particles.  $$P(N,V,T)=P(\frac{N}{V},T)=P_{BEC}(T)$$ While the first equality is thermodynamic in origin(intensive nature of $P,T$), the second is  an expression that is exclusively true for Bose-Einstein condensates. 

*This expression suggests that reducing the volume, holding $N,T$ fixed, causes no change in pressure. This is in stark contrast to ideal gasses where a increase in pressure would occur. 

*Another observation from the expression for total number $$N=\frac{V}{\lambda^3} g_{\frac{3}{2}}(z)+n_0$$ is if we reduced the volume holding $N,T$ fixed, then $n_0$ would increase. Here $n_0$ is number of particles in the ground state, $V$ is the volume, $\lambda$ is the thermal wavelength and $g$ is a special function whose nature is unimportant.

*Your question is: Are these two phenomenon related and if so, how?


The Intuition: The hint to the relationship between the two phenomenon is in the equation of state 
$$P=\frac{3E}{2V}$$
In an ideal gas, holding $N,T$ fixed fixes $E$. Corrections to $E$ due to volume change in real gasses at high temperature are small and can be ignored. 
However, suppose there was a mechanism to reduce $E$ by the same proportion(or drastically)when $V$ is reduced. Then the pressure independence(or weak dependance) on volume would be explained.
In a BEC, this mechanism is provided for by the "pushing" of particles to the zero energy ground state. This reduces the average energy drastically enough for the Pressure to remain independent of the volume. 
The calculation: Divide up the pressure into two parts
$$P=\frac{3}{2V}\left(E_>+E_0\right)$$
Where $E_>$ is the energy contribution from higher states and $E_0$ is that due to the ground state. In the ground state, the BEC has zero energy(Unlike a particle in a box). All that remains to be shown is that $E_>$ scales with $V$.This is carried out by integrating the Bose distribution to get the energy:
$$E_>=\int_0^\infty dE g(E)E \frac{1}{z^{-1}e^{\beta E}-1}$$
Plugging in the expression for $g(E)$, one gets
$$E_>= \frac{V}{4\pi^2}\left(\frac{2m}{\hbar}\right)^2\int_0^\infty dE  \frac{E^{3/2}}{z^{-1}e^{\beta E}-1}$$
This shows that $E_>$ does scale linearly with V hence establishing the pressure (in)dependence on volume. Remember that at high temperature $z$ depends strongly on $\frac{N}{V}$, while at low temperatures it is almost fixed by the average number of particles.
$$N=\frac{V}{\lambda^3} g_{\frac{3}{2}}(z)+\frac{z}{1-z}$$
In the low temperature limit $\lambda \to 0$ and $z$ is fixed by $N$.
Edit: At the request of the OP, I add a few details to the number of particles formula.
The total number of particles is given by integrating the Bose Einstein distribution:
$$N=\int_0^\infty dE g(E) \frac{1}{z^{-1}e^{\beta E}-1}$$
However, this integral is exactly $0$, in the neighborhood of the region $E=0$ since the density of states vanishes near $E=0$. Therefore this formula does not account for the particles in the ground state. At high temperatures, this is okay since most particles are in fact in higher excited states. However, in anticipation of the importance on the ground state at low temperature, we will need to account for the ground state separately, by hand. So the corrected formula looks like:
$$N=\int_0^\infty dE g(E) \frac{1}{z^{-1}e^{\beta E}-1}+n_0$$
Where $n_0$ is the average number of particles in the ground state. Now, explicitly working out the density of states gives us:
$$g(E)=\frac{V\sqrt{E}}{4\pi^2}\left(\frac{2m}{\hbar}\right)^{\frac{3}{2}}$$
Plugging this back into the equation for $N$ gives you:
$$N= \frac{V}{4\pi^2}\left(\frac{2m}{\hbar}\right)^{\frac{3}{2}} \int_0^\infty dE \sqrt{E}  \frac{1}{z^{-1}e^{\beta E}-1}+n_0$$
This equation tells you exactly why reducing the volume increases the number of particles in the ground state. If we reduce the volume, the first term in the equation reduces(remember $z=1$ for a condensate). But the total number of particles is held fixed in the system. Therefore the second term must increase for the left hand hand side to remain constant. This means that the number of particles in the ground state increases. 
Intuitively, The density of states decreases when we reduce the volume. This makes the spacing between states larger, and the higher states become less accessible at very low temperature. Therefore, particles tend to get pushed into the ground state.
