I am trying to prove that the specific heat is related to the fluctuations in the energy:

$$c_V = \frac{\langle E^2 \rangle - \langle E \rangle^2}{k_\mathrm BT^2}$$


$$\beta = \frac{1}{k_\mathrm BT}$$

$$\langle E \rangle = -\frac{ \partial \log(Z)}{\partial \beta}$$

I did:

$$\frac{\partial^2}{\partial \beta^2}\ln Z = \frac{\partial}{\partial \beta}\frac{1}{Z}\frac{\partial Z}{\partial \beta} = -\frac{\partial \langle E \rangle}{\partial \beta} = -\frac{\partial \langle E \rangle}{\partial T} \frac{\partial T}{\partial \beta}$$

My issue is that I do not understand why:

$$-\frac{\partial T}{\partial \beta} = k_\mathrm BT^2$$


1 Answer 1


You know that:

$$\beta = \frac{1}{k_\mathrm BT}$$

Rearranging yields: $$\color{red}{\beta} = \frac{1}{k_\mathrm B\color{blue}{T}}\implies \color{blue}{T} = \frac{1}{k_\mathrm B \color{red}{\beta}}$$


$$\frac{\partial \color{blue}{T}}{\partial\beta} = \frac{\partial}{\partial\beta}\left(\frac{1}{k_\mathrm B \beta}\right) = \color{red}{-}\frac{1}{k_\mathrm B \beta^\color{red}{2}}$$

Substituting $\beta = \color{red}{\frac{1}{k_\mathrm BT}}$ yields:

$$\frac{\partial T}{\partial\beta} = -\frac{1}{k_\mathrm B \color{red}{\left(\frac{1}{k_\mathrm BT}\right)}^2} = -\frac{k_\mathrm B^2T^2}{k_\mathrm B} = -k_\mathrm BT^2$$

$$-\frac{\partial T}{\partial\beta} = k_\mathrm BT^2$$ Q.E.D.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.