PX284 - C2 - temperature

the statistical definition

considering two systems that can exchange energy, and that the combined system is isolated
equilibrium is reached when $Ω_{t o t}$ is maximized:

\begin{matrix} \frac{d Ω}{d E_{1}} = \frac{d Ω_{1} Ω_{2}}{d E_{1}} = 0 \\ Ω_{2} \frac{d Ω_{1}}{d E_{1}} + Ω_{1} \frac{d Ω_{2}}{d E_{2}} \frac{d E_{2}}{d E_{1}} = 0 \\ \frac{1}{Ω_{1}} \frac{d Ω_{1}}{d E_{1}} + \frac{1}{Ω_{2}} \frac{d Ω_{2}}{d E_{2}} \frac{d E_{2}}{d E_{1}} = 0 \end{matrix}

the total energy is fixed, $E = E_{1} + E_{2}$
so, $d E_{1} = - d E_{2} ⟹ \frac{d E_{2}}{d E_{1}} = - 1 :$

\begin{matrix} \frac{1}{Ω_{1}} \frac{d Ω_{1}}{d E_{1}} - \frac{1}{Ω_{2}} \frac{d Ω_{2}}{d E_{2}} = 0 \\ \frac{d \ln Ω_{1}}{d E_{1}} = \frac{d \ln Ω_{2}}{d E_{2}} \end{matrix}

statistically defining the temperature using the expression:

\frac{1}{k_{B} T} = \frac{d \ln Ω}{d E}

it is found that $T_{1} = T_{2}$ at equilibrium

thermodynamics

for an isolated system, the first law states:

\begin{matrix} d U = T d S - p d V \\ d S = \frac{1}{T} d U + \frac{p}{T} d V = {(\frac{\partial S}{\partial V})}_{V} d U + {(\frac{\partial S}{\partial V})}_{T} d V \\ ⟹ \frac{1}{T} = {(\frac{\partial S}{\partial V})}_{V} \end{matrix}

statistical mechanics

for an isolated system:

\frac{1}{T} = \frac{d}{d E} (k_{B} \ln Ω) = \frac{d S}{d E}

$k_{B} T$ has the same units as energy
defining:

β = \frac{1}{k_{B} T}