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- TL;DR Summary
- Flaws in Bose-Einstein statistics for μ>ε
The Gibbs sum is given by
$$Z=\sum[\lambda \exp(-\varepsilon/\tau)]^N$$
where ##\lambda\equiv\exp(\mu/\tau)##. Since we are assuming ##\mu>\varepsilon##, we take only the last term of the sum because all others can be neglected.
thus
$$Z\approx[\lambda \exp(-\varepsilon/\tau)]^N$$
Now
$$\langle N\rangle =\lambda\frac{\partial}{\partial\lambda}\ln Z=\lambda\frac{\partial}{\partial\lambda}(N\ln\lambda-N\varepsilon/\tau)=N$$
But in general, we see that for any parameter ##X##
$$\langle X\rangle=\sum X_N[\lambda \exp(-\varepsilon/\tau)]^N/Z\approx X_N$$
where ##X_N## is the value at ##N\rightarrow\infty##, thus making the whole model useless.
But is this the only flaw in taking ##\mu>\varepsilon##?
That is, for the usual Bose-Einstein distribution
$$f(\varepsilon)=\frac1{\exp[(\varepsilon-\mu)/\tau]-1}$$
we get ##f(\varepsilon)<0## for ##\mu>\varepsilon##, which is physically wrong.
Are any such "physical" conditions present for the above model?
$$Z=\sum[\lambda \exp(-\varepsilon/\tau)]^N$$
where ##\lambda\equiv\exp(\mu/\tau)##. Since we are assuming ##\mu>\varepsilon##, we take only the last term of the sum because all others can be neglected.
thus
$$Z\approx[\lambda \exp(-\varepsilon/\tau)]^N$$
Now
$$\langle N\rangle =\lambda\frac{\partial}{\partial\lambda}\ln Z=\lambda\frac{\partial}{\partial\lambda}(N\ln\lambda-N\varepsilon/\tau)=N$$
But in general, we see that for any parameter ##X##
$$\langle X\rangle=\sum X_N[\lambda \exp(-\varepsilon/\tau)]^N/Z\approx X_N$$
where ##X_N## is the value at ##N\rightarrow\infty##, thus making the whole model useless.
But is this the only flaw in taking ##\mu>\varepsilon##?
That is, for the usual Bose-Einstein distribution
$$f(\varepsilon)=\frac1{\exp[(\varepsilon-\mu)/\tau]-1}$$
we get ##f(\varepsilon)<0## for ##\mu>\varepsilon##, which is physically wrong.
Are any such "physical" conditions present for the above model?