# Heat capacity of Bose-Einstein Condensation

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

Based on this expression for the internal energy:

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:

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

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 $$\frac{\lambda^3}{v} = g_{3/2}(z)$$ :

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

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