Equations of State

Consider a system with kinetic energy $K=\frac{1}{2}\frac{{\mathbb{P}}^2}{m}$, $U(\mathbb{Q})$. So, $\frac{\partial H}{\partial \mathbb{P}}=\frac{\mathbb{P}}{m}=\dot{\mathbb{Q}}$ and $\frac{\partial H}{\partial \mathbb{Q}}=-\dot{\mathbb{P}}$

$\frac{dH(\mathbb{P}(t),\mathbb{Q}(t),V(t))}{dt}=\frac{dH}{d\mathbb{P}}\dot{\mathbb{P}}+\frac{\partial H}{\partial \mathbb{Q}}\dot{\mathbb{Q}}+\frac{\partial H}{\partial V}\frac{\partial V}{\partial t}=\frac{\partial H}{\partial V}\frac{\partial V}{\partial t}$.

Time change of ensemble average of $E$: $\frac{d⟨H⟩}{dt}=⟨\frac{\partial H}{\partial V}⟩\frac{\partial V}{\partial t}$. $⟨\frac{\partial H}{\partial V}⟩$ is the initial average of volume changes. The steps must be so we move along equilibrium states (quasistatic), called: slow or adiabatic.

$-p_m=⟨\frac{\partial H}{\partial V}⟩=\frac{1}{\mathrm{\Omega }(E^{\prime })}\int d\mathbb{P}d\mathbb{Q}\delta (E-H(\mathbb{P},\mathbb{Q},V))\frac{\partial H}{\partial V}$ [1].

In general, \(\left = \frac{1}{\Omega(E)}\int d\mathbb{P}d\mathbb{Q}\delta(E-H(\mathbb{P},\mathbb{Q},V))O\).

${\left(\frac{\partial S}{\partial V}\right)}_{EN}={\left(\frac{\partial }{\partial V}\right)}_{EN}(k_B\ln \mathrm{\Omega })=\frac{k_B}{\mathrm{\Omega }}{\left(\frac{\partial \mathrm{\Omega }}{\partial V}\right)}_{EN}$.

${\left(\frac{\partial \mathrm{\Omega }}{\partial V}\right)}_{EN}={\left(\frac{\partial }{\partial V}\right)}_{EN}\int d\mathbb{P}\mathbb{Q}\delta (E-H)={\left(\frac{\partial }{\partial V}\right)}_{EN}{\left(\frac{\partial }{\partial E}\right)}_{VN}\int d\mathbb{P}\mathbb{Q}\mathrm{\Theta }(E-H)={\left(\frac{\partial }{\partial E}\right)}_{VN}{\left(\frac{\partial }{\partial V}\right)}_{EN}\int d\mathbb{P}\mathbb{Q}\mathrm{\Theta }(E-H)={\left(\frac{\partial }{\partial E}\right)}_{VN}\int d\mathbb{P}\mathbb{Q}(-1)\delta (E-H)\frac{\partial H}{\partial V}$ [2].

Comparing [1] and [2], ${\left(\frac{\partial \mathrm{\Omega }}{\partial V}\right)}_{EN}=-{\left(\frac{\partial }{\partial E}\right)}_{VN}(-p_m\mathrm{\Omega }(E))={\left(\frac{\partial }{\partial E}\right)}_{VN}(p_m\mathrm{\Omega }(E))={\left(\frac{\partial \mathrm{\Omega }}{\partial E}\right)}_{VN}{p}_m+\mathrm{\Omega }(E){\left(\frac{\partial p_m}{\partial E}\right)}_{VN}$.

${\left(\frac{\partial S}{\partial V}\right)}_{EN}=\frac{k_B}{\mathrm{\Omega }}{\left(\frac{\partial \mathrm{\Omega }}{\partial V}\right)}_{EN}=k_B{p}_m\frac{1}{\mathrm{\Omega }(E)}{\left(\frac{\partial \mathrm{\Omega }}{\partial E}\right)}_{VN}+k_B{\left(\frac{\partial p_m}{\partial E}\right)}_{VN}={\left(\frac{\partial S}{\partial E}\right)}_{VN}{p}_m+k_B{\left(\frac{\partial p_m}{\partial E}\right)}_{VN}=\frac{p_m}{T}+k_B{\left(\frac{\partial p_m}{\partial E}\right)}_{VN}$ = intrinsic/ intrinsic + intrinsic/extrinsic $=O(N^0)+O(N^{-}1)\approx O(N^0)$. So, in the limit, ${\left(\frac{\partial S}{\partial V}\right)}_{EN}{=}_{\text{limit }N\to \mathrm{\infty }}\frac{p_m}{T}$.

Takeaway: ${\frac{\partial S}{\partial V}|}_{EN}=\frac{p_m}{T}+k_B{\frac{\partial p_m}{\partial E}|}_{VN}$.

Probability density in phase space: $\rho (\mathbb{P},\mathbb{Q})=\rho (p_1,\cdots ,p_{3N},q_1,\cdots ,q_{3N})$.

$\frac{d\rho }{dt}=\frac{\partial \rho }{\partial t}+\sum _{\alpha =1}^{3N}\frac{\partial \rho }{\partial q_{\alpha }}{\dot{q}}_{\alpha }+\frac{\partial \rho }{\partial p_{\alpha }}{\dot{p}}_{\alpha }=0$. In other words, this is incompressible. There are no fixed points it doesn’t flow. It doesn’t contract to particular solutions.

This is the same as the continuity equation in phase space. $\frac{d\rho }{dt}=\frac{\partial \rho }{\partial t}+\overrightarrow{V}\mathrm{\nabla }\rho$, the second term is the current.

(Boltzmann)

Ensemble averages will be time averages if it is ergodic. It is hard to prove a system is ergodic.

Energy surface in phase space is stirred by time evolution, i.e. it is not divided into components that do not intermingle.

Definition: In an ergodic system, the trajectory of almost energy point in phase space eventually passes arbitrarily close to every other point.

Definition: A time evolution in a set S is ergodic iff all the ergodic components $R$ in $S$ have either zero volume or have a volume equal to $S$.

$\bar{O}(t)=\underset{T\to \mathrm{\infty }}{lim}\frac{1}{T}{\int }_t^{t+T}O(\mathbb{P}(t),\mathbb{Q}(t))dt$.

1. Clausius Statement. No process is possible whose sole result is the transfer of energy ’by heating’ from a cooler body to a hot body. I.e. no perfect refridgerator
2. Kelvin Statement. No process is possible whose sole result is the complete conversion of heat into work. I.e. No perfect engine.
3. Carnot’s Theorem. No engine operating between two reservoirs ($T_H,T_C$) is more efficient than a Carnot engine.
4. Entropy always increases (or stays the same).

Definition: Carnot engine. A Carnot engine is any engine that is reversible. Runs in a cycle with all heat exchanges taking place at a source temperature $T_H$ and a sink temperature $T_C$.

Definition: $dS\ge \frac{dQ}{T}$. $S(B)-S(A)\ge {\int }_A^B\frac{dQ}{T}$.

Consider $\bar{c}$ a cooler such that $Q_H=Q_C=Q$, that extracts heat from heat from the cold reservoir and puts it into a hot reservoir. Consider $E$ be an engine with $Q_C=Q$ and extracts work. Composing these two machines we get a machine that extracts heat from a hot reservoir and does work and does not output heat to a cold reservoir. Thus, the statements are equivalent.

For statement 3, assume a super engine, ${\eta }_S>{\eta }_C$. Then, have the engine drive a Carnot refrigerator. The net effect is $Q_H-Q_H^{\prime }$ going to the engine from the hot reservoir and $Q_C-Q_C^{\prime }$ flows from the engine to the cold reservoir. But $Q_H-Q_H^{\prime }\ge 0$ thus, $Q_H\ge Q_H^{\prime }$. For ${\eta }_S=\frac{W}{Q_H}$, ${\eta }_C=\frac{W}{Q_H^{\prime }}$. Then, ${\eta }_C\ge {\eta }_S$.

Corallary: All carnot engines have the same efficiency, ${\eta }_C={\eta }_C(T_C,T_H)$.

Thermodynamic Temprature: Let a Carnot engine 1 have $T_H=T_1$, $T_C=T_2$ and Carnot engine 2 have $T_H=T_2$, $T_C=T_3$. So, We get $Q_1$ from $T_1$, $Q_2$ from and to $T_2$, and $Q_3$ to $T_3$. The work extracted is labeled $W_{12}$ and $W_{23}$. Then, Composing the engines, we should get $Q_1$ to the composed engine and $Q_3$ out of the composed engine with work $W_{13}$.

$Q_2=Q_1-W_{12}=Q_1(1-{\eta }_{12})$. $Q_3=Q_2-W_{23}=Q_2(1-{\eta }_{23})=Q_1(1-{\eta }_{12})(1-{\eta }_{23})=Q_1-W_{23}=Q_1(1-{\eta }_{13})$. So, $(1-\eta (T_1,T_3))=(1-\eta (T_1,T_2))(1-\eta (T_2,T_3))$. So, $1-\eta (T_a,T_b)=\frac{f(T_A)}{f(T_B)}:=\frac{\alpha }{\alpha }\frac{T_A}{T_B}=\frac{T_A}{T_B}\Rightarrow \eta =1-\frac{T_C}{T_H}$.

Consider a closed environment with our (non-equilibrium) system and a Carnot engine that dumps $dQ$ heat into the system and the system does some $dW$ on the environment. The carnot engine takes in some heat $d{Q}_R$ from reservoir $T_0$ and does some work $d{W}_C$ on the environment. Along any step $\frac{d{W}_C}{d{Q}_R}={\eta }_C=1-\frac{T}{T_0}=\frac{d{Q}_R-dQ}{d{Q}_R}=1-\frac{dQ}{d{Q}_R}$. So, $\frac{dQ}{d{Q}_R}=\frac{T}{T_0}\iff d{Q}_R=T_0\frac{dQ}{T}$. After a cycle, $Q_R=\oint d{Q}_R$. For our closed system, $W=Q_R$. But by Kelvin statement, $W\le 0$. Thus, $Q_R\le 0$. $Q_R=\oint d{Q}_R=T_0\oint \frac{dQ}{T}\le 0$. Therefore, $\oint \frac{dQ}{T}\le 0$. Consequences:

• If this process is reversible then $\oint \frac{d{Q}_{r}ev}{T}=0$. $S(B)-S(A)={\int }_A^B\frac{d{Q}_{rev}}{T}={\int }_A^{B}dS\ge {\int }_A^B\frac{dQ}{T}$.