Understanding the partial derivative of temperature with respect to thermodynamic beta (coldness)

Question

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

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

Where:

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

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

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:

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

Answer 1

You know that:

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

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

Thus:

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

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.