# Equations of State

\(S = S(E,V,N)\)

\(dS = \left(\frac{\partial S}{\partial E}\right)_{VN}dE + \left(\frac{\partial S}{\partial V}\right)_{EN}dV + \left(\frac{\partial S}{\partial N}\right)_{EV}dN = \frac{1}{T}dE + \frac{P}{T}dV - \frac{\mu}{T}dN\). \(dE = TdS - pdV + \mu dN\).

Is this pressure the same as mechanical pressure? \(p = \frac{\langle F\rangle}{A} \leftrightarrow p = -\frac{\Delta E}{\Delta V}\). We get to the relation from work. \(dE = W = \int \vec{F}\cdot ds = \int Fdx\), \(-\frac{dE}{dV} = -F\frac{dx}{dV} = -\frac{F}{A}\frac{dx}{dx} = +p\), since \(F = -p\hat{n}dA\).

## Example

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 \(E\): \(\frac{d\langle H\rangle}{dt} = \left<\frac{\partial H}{\partial V}\right>\frac{\partial V}{\partial t}\). \(\left<\frac{\partial H}{\partial V}\right>\) is the initial average of volume changes. The steps must be so we move along equilibrium states (quasistatic), called: slow or adiabatic.

\(-p_m = \left<\frac{\partial H}{\partial V}\right> = \frac{1}{\Omega(E')}\int d\mathbb{P}d\mathbb{Q}\delta(E-H(\mathbb{P},\mathbb{Q},V))\frac{\partial H}{\partial V}\) [1].

In general, \(\left

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

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

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

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

**Takeaway**: \(\left.\frac{\partial S}{\partial V}\right|_{EN} = \frac{p_m}{T} + k_B \left.\frac{\partial p_m}{\partial E}\right|_{VN}\).

## Louiville Equation

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. \(\frac{d\rho}{dt} = \frac{\partial\rho}{\partial t} + \vec{V}\nabla\rho\), the second term is the current.

## Ergodic Hypothesis

(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 \(S\).

\(\overline{O}(t) = \lim_{T\to\infty}\frac{1}{T}\int_{t}^{t+T}O(\mathbb{P}(t),\mathbb{Q}(t))dt\).

## Irreversibility and Carnot Engines

- 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
- Kelvin Statement. No process is possible whose sole result is the complete conversion of heat into work. I.e. No perfect engine.
- Carnot’s Theorem. No engine operating between two reservoirs (\(T_H,T_C\)) is more efficient than a Carnot engine.
- 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: \(dS \geq \frac{dQ}{T}\). \(S(B) - S(A) \geq \int_A^B \frac{dQ}{T}\).

Consider \(\overline{c}\) 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 \(Q_H-Q_H'\) going to the engine from the hot reservoir and \(Q_C-Q_C'\) flows from the engine to the cold reservoir. But \(Q_H-Q_H'\geq 0\) thus, \(Q_H\geq Q_H'\). For \(\eta_S = \frac{W}{Q_H}\), \(\eta_C = \frac{W}{Q_H'}\). Then, \(\eta_C\geq \eta_S\).

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, \(1-\eta(T_a,T_b) = \frac{f(T_A)}{f(T_B)} := \frac{\alpha}{\alpha}\frac{T_A}{T_B} = \frac{T_A}{T_B} \Rightarrow \eta = 1 - \frac{T_C}{T_H}\).

## Examples

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 \(dQ_R\) from reservoir \(T_0\) and does some work \(dW_C\) on the environment. Along any step \(\frac{dW_C}{dQ_R} = \eta_C = 1-\frac{T}{T_0} = \frac{dQ_R-dQ}{dQ_R} = 1 - \frac{dQ}{dQ_R}\). So, \(\frac{dQ}{dQ_R} = \frac{T}{T_0} \Leftrightarrow dQ_R = T_0\frac{dQ}{T}\). After a cycle, \(Q_R = \oint dQ_R\). For our closed system, \(W = Q_R\). But by Kelvin statement, \(W \leq 0\). Thus, \(Q_R\leq 0\). \(Q_R = \oint dQ_R = T_0\oint \frac{dQ}{T}\leq 0\). Therefore, \(\oint\frac{dQ}{T}\leq 0\). Consequences:

- If this process is reversible then \(\oint\frac{dQ_rev}{T}=0\). \(S(B) - S(A) = \int_A^B\frac{dQ_{rev}}{T} = \int_A^B dS \geq \int_A^B \frac{dQ}{T}\).