Huh? Condensates can exchange particles without effect?

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1. Apr 20, 2015

jshrager

I have now seen it repeated multiple times that a particle (a fermion, perhaps?) moving in a condensate can exchange particles (bosons, most probably) "without effect" -- the version of this that I run into usually goes something like that the energy of the condensate does not change AT ALL when, say, a virtual photon (or z-boson) is absorbed into a BEC. Now, first off, hunh? Does this really mean that the energy of the condensate doesn't change when something is absorbed by it? How can that be? And if so, what aspect of what equation(s) tell us this? One possible way that I can resolve this apparent impossibility is that to require that the emitting particle be a part of the condensate to begin with, so that the absorption is balanced by the emission, but this doesn't seem to be what is being said.

2. Apr 21, 2015

DrDu

One usually assumes that the condensate is infinitely large, so its total energy is infinite or at least undefined anyhow.

3. Apr 21, 2015

vanhees71

If you go into the thermodynamical limit $V \rightarrow \infty$, total energy, total particle number etc. get undefined, and only the corresponding densities make sense. You also must say, whether you look at relativistic particles, where energy is usually defined as including the rest-mass contribution, $E=\sqrt{m^2+p^2}$ (setting $c=\hbar=1$), or a non-relativistic gas.

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.