I'm studying BEC and came across this result (book "Huang K Statistical Mechanics 2 edition", page 292) enter image description here

Based on this expression for the internal energy: enter image description here

The upper value is when v>vc and the lower one is when v<vc.

$g_{n}(z)$ are polylogarithms.

I started to derivate my internal energy, but I don't know how to proceed from: $\frac{d g_{5/2}(z)}{dz}\frac{dz}{dT}$, I can solve $\frac{d g_{5/2}(z)}{dz}$, how can I solve $\frac{dz}{dT}$?

z is obtained graphically: enter image description here

  • 1
    $\begingroup$ Please use MathJax for all math. $\endgroup$
    – G. Smith
    Jan 4, 2021 at 6:59
  • 1
    $\begingroup$ This question would be a lot clearer if you explained how $z$ is defined. $\endgroup$
    – G. Smith
    Jan 4, 2021 at 7:29

1 Answer 1


For $T > T_{\text{c}}$, you are looking at the bottom expression in all three sets of equations.

So: $$ \frac{U}{N} = \frac{3}{2}\frac{k T v}{\lambda^3}g_{5/2}(z),$$ but $$ \frac{\lambda^3}{v} = g_{3/2}(z),$$ so: $$ \frac{U}{N} = \frac{3}{2}k T \frac{g_{5/2}(z)}{g_{3/2}(z)},$$

The trick is to work this out: $$ \left ( \frac{\partial z}{\partial T} \right)_V = \left [ \frac{\partial z}{\partial z g_{3/2}(z)} \frac{\partial z g_{3/2}(z)}{\partial T} \right]_V = -\frac{3 z}{2 T} \frac{g_{3/2}(z)}{g_{1/2}(z)}.$$

Because then: $$ \frac{C_v}{N k} = \frac{1}{k} \left (\frac{\partial (U/V)}{\partial T} \right )_V = \frac{1}{k} \frac{\partial}{\partial T} \left (\frac{3}{2}k T \frac{g_{5/2}(z)}{g_{3/2}(z)} \right )_V.$$

Then, with symbolic integration on Mathematica: $$ -\frac{3 T g_{\frac{1}{2}}(z(T)) g_{\frac{5}{2}}(z(T)) z'(T)}{2 z(T) g_{\frac{3}{2}}(z(T)){}^2}+\frac{3 g_{\frac{5}{2}}(z(T))}{2 g_{\frac{3}{2}}(z(T))}+\frac{3 T z'(T)}{2 z(T)},$$

and using the derivative of $z$ that I have derived earlier:

$$ \frac{C_v}{N k} = \frac{15}{4}\frac{g_{5/2}(z)}{g_{3/2}(z)} - \frac{9}{4}\frac{g_{3/2}(z)}{g_{1/2}(z)},$$

and if you plug in the expression of the root $\frac{\lambda^3}{v} = g_{3/2}(z)$ :

$$ \frac{C_v}{N k} = \frac{15}{4}\frac{v}{\lambda^3}g_{5/2}(z) - \frac{9}{4}\frac{g_{3/2}(z)}{g_{1/2}(z)}.$$

  • $\begingroup$ thank you, I eventually arrived at that result with what is in page 259 of Kerson Huang, Introduction to Statistical Physics $\endgroup$
    – cmmigl
    Jan 4, 2021 at 9:25
  • $\begingroup$ I never forget who helped me. Once in a while I would be happy to have your answers to my questions as well. :-) $\endgroup$
    – Sebastiano
    Jan 19, 2021 at 22:45
  • $\begingroup$ Ahah I guess I have to see then and get there first :) $\endgroup$
    – SuperCiocia
    Jan 19, 2021 at 22:52

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