Invert expression to find the Chemical potential of an Ideal Bose Gas

**thelostchild** · February 27th 2011, 11:26 AM

Hi I'm having trouble with this step of the derivation of the chemical potential for an ideal fermi gas in the low temperature limit

I've got as far as to find
$$ E_F = \mu \left(1+\frac{\pi^2}{12} \left(\frac{kT}{\mu}\right)^2 + ...\right) $

and need to invert this expression to get the chemical potential as a function of fermi energy

I've tried writing this as (which is ok as this is the same as above to first order)
$$ E_F = \frac{\mu}{1-\frac{\pi^2}{12} \left(\frac{kT}{\mu}\right)^2} $

but I cant seem to get it into a form which is easily invertable any chance of a hand here please?

NB the answer I'm trying to get to is
$$ \mu = E_F \left(1-\frac{\pi^2}{12} \left(\frac{kT}{E_F}\right)^2 + ...\right) $

**zzzoak** · February 27th 2011, 03:11 PM

$ \displaystyle E=\mu(1+\frac{a}{\mu^2}) $

$ \displaystyle \mu^2-E\mu+a=0 $

$ \displaystyle \mu=\frac{E \pm \sqrt{E^2-4a}}{2}=\frac{E \pm E \sqrt{1-4a/E^2}}{2}= $

$ \displaystyle =\frac{E \pm E (1-2a/E^2)}{2}=E(1-a/E^2) $

**thelostchild** · February 27th 2011, 03:33 PM

ok that makes me feel stupid for not spotting it

Cheers!

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