What Are the Conditions for Bose-Einstein Condensation to Occur?

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SUMMARY

Bose-Einstein condensation occurs when the density of particles exceeds a critical density, defined as ##\rho_C(T) = A(kT)^{3/2} \int_0^{\infty} \frac{x^{1/2}}{ e^x − 1} dx##. The density of particles is expressed as ##\rho = \rho_o + \rho_+##, where ##\rho_o## represents the density of the ground state and ##\rho_+## accounts for the density of excited states. The energy of particles on the lattice is modified to ##\epsilon_{j\chi} = E\chi + \frac{\hbar^2 k^2_j}{(2m)}##, with conditions ##\chi = 0, 1## and ##E > 0## influencing the overall density calculation. Understanding these parameters is crucial for deriving the conditions necessary for Bose-Einstein condensation.

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Homework Statement


Explain the conditions under which Bose-Einstein condensation occurs and show this happens for density ##\rho \equiv N/V > \rho_C(T)##, where $$\rho_C(T) = A(kT)^{3/2} \int_0^{\infty} \frac{x^{1/2}}{ e^x − 1} dx .$$

Suppose the energy of the particles on the lattice is now ## \epsilon_j → \epsilon_{j\chi} = E\chi+ \frac{\hbar^2 k^2_j}{(2m)}##, where ##\chi = 0, 1## and ##E > 0##. Obtain an expression for ##\rho \equiv N/V##.

Homework Equations


##\rho = \rho_o + \rho_+## where ##\rho_o## is density of ground state and ##\rho_+## density of all others.

The Attempt at a Solution


[/B]
The first part is ok, in the integral, ##x = \beta \epsilon##. In the second part, are we just shifting the ground state energy to ##E\chi##? and then evalaute the integral ##\rho_C(t)## using that expression for ##\epsilon_{j \chi}##? I'm not sure what the conditions ##\chi = 0,1## and ##E>0## imply yet either. Thanks!
 
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