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I am puzzled about the calculation of saddle point in canonical ensemble partition function in Kerson Huang's book. $$ Z=\int_0^{+\infty} dE \exp[\beta (TS-E)] $$ The maximum of integrand occurs at $\bar{E}$ satisfying $$ T \left(\frac{\partial S}{\partial E}\right)_{E= \bar{E}} =1$$ $$ \left(\frac{\partial^2 S}{\partial E^2} \right)_{E= \bar{E}} < 0 $$ and $$ \left(\frac{\partial^2 S}{\partial E^2} \right)_{E= \bar{E}}=\left(\frac{\partial\frac{1}{T}}{\partial E}\right)_{E= \bar{E}} =-\frac{1}{T^2C_V} $$

I do not know why sometimes we treat $T$ as contant and sometimes as thermodunamics function.

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UPDATE:

As Alexey suggested, thinking of $T$ as a funtion of $E$, there seems to be some problem. All energy $E$ in the integral are energies of microstates in canonical ensemble, and correspondingly $T$ is the temperature of canonical ensemble which is fixed. So $T$ is still a constant.

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Assume we have some dependence of entropy on system energy $S(E)$. Then the saddle point occurs when $$ TS(E)-E\rightarrow\max\quad\mbox{at}\quad E=\bar{E}. $$ Taking derivative, we get $TS'(\bar{E})-1=0$ and $$ \qquad\qquad\qquad\qquad\qquad\quad S'(\bar{E})=\frac1T,\qquad\qquad\qquad\qquad\qquad\qquad(1) $$ so the saddle-point energy $\bar{E}$ depends on $T$ from this equation: $\bar{E}=\bar{E}(T)$.

Taking the first differential of (1), we get $S''(\bar{E})d\bar{E}=-dT/T^2$, so $$ S''(\bar{E})=-\frac1{T^2}\frac{dT}{d\bar{E}}. $$ Defining something like "saddle-point heat capacity" $C_V=\bar{E\,}'\!(T)=d\bar{E}/dT$, we get $$ S''(\bar{E})=-\frac1{T^2C_V}. $$

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  • $\begingroup$ Generally, when we judge the sign of second derivative term, we first calculate double prime and then take values as $E=\bar{E}$ $\endgroup$ – thone Nov 24 '16 at 0:35

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