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- Thread starter jshrager
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For an ideal relativistic Bose gas you have

$$f(\vec{p})=(2 \pi)^3 n_c \delta^{(3)}(\vec{p})+\frac{1}{\exp[(E(\vec{p})-m)/T]-1},$$

where ##n_c## is the particle-number density of the particles in the condensate, ##m## the mass of the particles and (in the thermodynamic limit) ##\mu=m##.

For the energy density, it follows

$$\epsilon=\int_{\mathbb{R}^3} \frac{\mathrm{d}^3 \vec{p}}{(2 \pi)^3} E(\vec{p}) f(\vec{p})=m n_c + \int_{\mathbb{R}^3} \frac{\mathrm{d}^3 \vec{p}}{(2 \pi)^3} E(\vec{p}) \frac{1}{\exp[(E(\vec{p})-m)/T]-1}.$$

Thus, for a relativistic gas, the energy density for the particles in the condensate is ##n_c m##.

The calculation is the same for a non-relativistic ideal Bose gas. The only difference is that one doesn't include the rest-mass energy in the energy but uses kinetic energy ##E(\vec{p})=\vec{p}^2/(2m)##, and then (in the thermodynamic limit) you have ##\mu=0## and thus

$$f(\vec{p})=(2 \pi)^3 n_c \delta^{(3)}(\vec{p}) + \frac{1}{\exp[E(\vec{p})/T]-1},$$

and the energy density of the condensate particles is 0.

Concerning the question in the OP, I don't know, what it means. Of course interchanging a particle within the BEC doesn't change anything, because all these particles are completely indistinguishable. I don't get the idea about a fermion exchanged with a boson at all. Of course, you can consider mixtures of fermions with bosons, where the bosons can be in a condensate, but I can't make sense about the question also in this context.

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