Classical Mechanics Lec 23-26 (V Balakrishnan) Opus 5

`E' is some kind of thermodynamic potential

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

\[ \begin{aligned} T &= \left.\frac{\partial E}{\partial S}\right)_{V,N}\\ P &= \left.-\frac{\partial E}{\partial V}\right)_{S,N}\\ \mu &= \left.\frac{\partial E}{\partial N}\right)_{S,V} \end{aligned} \] \[ \Leftarrow \left( dE = TdS - Pdv + \mu dN \right) \]

What is significance of above eq\(^{\rm n}\) set ?

we know that \(E(S,V,N)\) `s' is not a control variable

So lets change that by legendre transformations.

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

\(\downarrow\)

(Internal Energy)

Assumption

"\(E\) is homogeneous function of thermodynamic variables
*\(S, V, N\)*".

for that we have to assume we are in thermodynamic limit of
thermodynamic system i.e. \(V \to \infty\), \(N \to \infty\) &
\(\frac{N}{V} = S\) (fixed).

then \(S, V, N\) which increase with system size (extensive quantities)
which leads to some \(E\).

then we can assume \(E\) is homogeneous fun of \(S, V, N\) of degree
1.

for any homogenuous function of degree `r'

\[ f = f(x, y, z \cdots) \]

i.e. \(f(\lambda x, \lambda y, \lambda z, \cdots) = \lambda^{r} f(x,y,z,\cdots)\).

\[ \left(x \frac{\partial f}{\partial x} + y \frac{\partial f}{\partial y} + z \frac{\partial f}{\partial z} + \cdots \right) = r f \]

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

\[ S \left.\frac{\partial E}{\partial S}\right|_{V,N} + V \left.\frac{\partial E}{\partial V}\right|_{S,N} + N \left.\frac{\partial E}{\partial N}\right|_{S,V} = E \] \[ S(T) + V(-P) + N(\mu) = E \qquad \text{(Euler's relations)} \]

So \(E\) must be equal to \(TS - PV + \mu N\).

\[ E - TS + PV = \mu N \]

we also know that \(G = E - TS + PV\)

so \(G = \mu N\)

\[ \text{So} \Rightarrow \mu = \frac{G}{N} = \text{Gibbs free energy per particle} \]

let us find \(dE\)

\[ dE = \left(TdS - Pdv + \mu dN\right) + \left(SdT - vdP + N d\mu\right) \]

But laws of thermodynamics tells us that,

\[ \left(dE = TdS - Pdv + \mu dN\right) \text{ only} \]

So it implies that,

\[ SdT - vdP + N d\mu = 0 \] \[ ([?]) \Rightarrow d\mu = \left(\frac{V}{N}\right) dP - \left(\frac{S}{N}\right) dT \] \[ = v\, dP - s\, dT \]

volume per particle \(\longleftarrow\) specific volume per particle

\(s = \) entropy per particle \(\to\) specific entropy (per particle)

So, \(d\mu = v\, dP - s\, dT\) \(\to\) Gibbs--Duhem relation

it implies, \(\boxed{\mu = \mu(P,T).}\)

So once we know \((P,T)\) we know the chemical potential \((\mu)\)

If we assume `T' is some kind of generalised force & \(ds\) as some
generalised flux

\[ \text{then,} \quad dE = \left(Tds\right) + \sum_{i} F_{i}\, dX_{i} \] \[ dE = \sum_{i} F_{i}\, dX_{i} \]

So homogeneity arguments tells

\[ E = \sum_{i} F_{i} X_{i} \Rightarrow \left(\sum X_{i}\, dF_{i} = 0\right). \]

\(\downarrow\)

(Generalised Duhem relation)

This comes from our assumption of \(E\) to be a homogeneous function of
degree 1.

If \(E\) is not homogeneous function.

\[ E(S,V,N) \longrightarrow H(S,P,N) \]

(Enthalpy)

\[ \begin{aligned} H &= E + PV\\ dH &= dE + Pdv + vdP\\ &= Tds - Pdv + \mu dN + Pdv + vdP\\ dH &= Tds + vdP + \mu dN \end{aligned} \]

So \(H = H(S,P,N)\)

\[ E(S,V,N) \longrightarrow F(T,V,N) \] \[ \begin{aligned} F &= E - TS\\ dF &= TdS + \mu dN - Pdv - TdS - SdT\\ dF &= -SdT - Pdv + \mu dN \qquad \Rightarrow F = F(T,V,N) \end{aligned} \] \[ E(S,V,N) \longrightarrow G(T,P,N) \] \[ G = E - TS + PV. \qquad \text{(Gibbs free energy)} \] \[ E(S,V,N) \longrightarrow \Phi_{1}(S,V,\mu) \] \[ \Phi_{1} = E - \mu N \] \[ E(S,V,N) \longrightarrow \Phi_{2}(S,P,\mu) \] \[ \Phi_{2} = E + PV - \mu N \] \[ E(S,V,N) \longrightarrow \Phi_{3}(T,V,\mu) \] \[ \Phi_{3} = E - TS - \mu N \]

But we know \(E = TS - PV + \mu N\)

\[ \Rightarrow E - TS - \mu N = -PV \] \[ \text{So} \quad \boxed{\Phi_{3} = -PV} \; \to \; \text{``grand potential''} \]

\(\downarrow\)

The reason is that this potential \(\Phi_{3}\) is connected to grand
canonical ensamble but others above mentioned potentials are only
linked to Canonical ensamble of stat. mech.

In,

\[ E = TS - PV + \mu N \]

\((T, P, \mu) \to\) Intensive variables

\((S, V, N) \to\) Extensive ".

Field variables (Intesive) State variables. (extensive)
\(T\) \(S\)
\(P\) \(V\)
\(\mu\) \(N\)
Stess \(\sigma_{ij}\) Strain \(\epsilon_{ij} (= q_{i})\)
\(\vec{E}\) \(P \to\) polarisation per unit volume
\(\vec{H}\) \(M\) ; magnetic dipole moment per unit volume

\[ dF = -s\,dT - P\,dv + \mu\,dN \qquad \text{and} \qquad ; \qquad dG = -s\,dT + v\,dP + \mu\,dN \]

lets look at these 2 relations,

\[ S = -\left(\frac{\partial F}{\partial T}\right)_{V,N} \]

So,

\[ \left(\frac{\partial F}{\partial T}\right)_{V,N} = \left(\frac{\partial G}{\partial T}\right)_{P,N} \]

Similarly,

\[ P = -\left(\frac{\partial F}{\partial V}\right)_{T,N} \] \[ \mu = \left(\frac{\partial F}{\partial N}\right)_{T,V} \]

It says these thermodynamic potentials \((U, G, F, H \cdots)\), their \(1^{st}\) partial derivative with their individual variables on which they depend are other thermodynamic variables.

\[ \begin{aligned} dQ &= dU + dW\\ TdS &= dE + P\,dv - \mu\,dN \end{aligned} \] \[ \left(\frac{dQ}{dT}\right)_{V,N} \text{ or } \left(\frac{\partial Q}{\partial T}\right)_{V,N} = C_V = T\left(\frac{\partial S}{\partial T}\right)_{V,N} = \left(\frac{\partial E}{\partial T}\right)_{V,N} \quad {}^{+0-0} \] \[ C_V = T\left(\frac{\partial S}{\partial T}\right)_{V,N} \qquad \text{but,} \qquad S = -\left(\frac{\partial F}{\partial T}\right)_{V,N} \]

\[ \boxed{\;C_V = -T\left(\frac{\partial^2 F}{\partial T^2}\right)_{V,N}\;} \]

response functions are \(2^{nd}\) derivative of thermodynamic potentials

But we know \(C_V\) can not be negative \(\Big\}\) lesea teliers principle.

So,

\[ {}_{V}\left(\frac{\partial^2 F}{\partial T^2} < 0\right). \] \[ C_P = T\left(\frac{\partial S}{\partial T}\right)_{P,N} = -T\left(\frac{\partial^2 G}{\partial T^2}\right)_{P,N} \qquad \text{as,} \left(S = -\left(\frac{\partial G}{\partial T}\right)_{P,N}\right). \]

also

\[ (C_P > C_V) \qquad \text{so,} \qquad \left|\frac{\partial^2 G}{\partial T^2}\right| > \left|\frac{\partial^2 F}{\partial T^2}\right|. \]

\[ K_T = \frac{1}{\text{bulk modulus}} = \frac{-\Delta v}{V\,\Delta P} = -\frac{1}{v}\left(\frac{\partial v}{\partial P}\right)_{T,N} \qquad (\text{as } P\uparrow\ v\downarrow) \]

Since \(\left(V = \left(\dfrac{\partial G}{\partial P}\right)_{T,N}\right)\)

\[ \left(\frac{\partial v}{\partial P}\right)_{T,N} = \left(\frac{\partial^2 G}{\partial P^2}\right)_{T,N} \] \[ \boxed{\;K_T = -\frac{1}{v}\left(\frac{\partial^2 G}{\partial P^2}\right)_{T,N}\;} \]

Poisson ratio is measure of Poissons effect, that describes the expansion or contraction of a material in direction perpendicular to the direction of loading

\[ \left(\nu = \frac{\Delta x}{\Delta y} = \frac{\text{contraction strain}}{\text{extension strain}}\right. \]

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It's possible that \((\nu<0)\) such that upon extending in one direction object/material also expand in \(\perp\) dir.

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It turns out \(\boxed{-1 < \nu \le \tfrac{1}{2}}\)

For ideal gas

\[ \begin{aligned} PV &= nRT\\ P\,dv + v\,dP &= nR\,dT\\ v\,dP &= -P\,dv\\ \frac{dv}{v} &= -\frac{dP}{P}\\ \Rightarrow\quad K_T &= -\frac{1}{v}\frac{\partial v}{\partial P} = -\frac{1}{V}\cdot\left(\frac{-v}{P}\right) = \frac{1}{P} \end{aligned} \]

If \(K_T\) is high pressure is low.

for adiabatic process

\[ \begin{aligned} P v^{\gamma} &= \text{const}\\ dP\,v^{\gamma} + P\,\gamma\,P^{\gamma-1}\,dV &= 0\\ v^{\gamma-1}\left(v\,dP + \gamma P\,dv\right) &= 0\\ \Rightarrow\quad K_T &= -\frac{1}{v}\frac{\partial v}{\partial P} = +\frac{1}{v}\cdot\frac{v}{\gamma P} = \frac{1}{\gamma P} \end{aligned} \] \[ \left(K_{S,N} = \frac{1}{\gamma P}\right). \]

What happens if we have van der vaals gas :-

\[ P = \frac{nRT}{V - bn} - \frac{an^2}{V^2} \]

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\(\left\{\text{This repulsion arose from pauli exclusion principle}\right.\)

\[ V(r) = V_0\left[\left(\frac{a}{r}\right)^{12} - \left(\frac{a}{r}\right)^{6}\right] \qquad \text{``6--12'' potential} \qquad \left(\begin{aligned}&\text{lennard--Jonnes}\\ &\text{potential}\end{aligned}\right) \]

\(\left(\frac{a}{r}\right)^{6} \to\) attractive force potential

\(\left(\frac{a}{r}\right)^{12} \to\) need quantum mech effect. (repulsive term).

Attractive force term arise from dipole moment attraction.

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Random fluctuations can cause dipole moment.

interaction energy of any of dioples goes like

\[ (U) \sim \vec{p}\cdot\vec{E} \]

greater the field greater the seperation

\[ \vec{p} = \alpha \vec{E} \] \[ U \sim \alpha |E|^2 \qquad\qquad \vec{E} \text{ of dipole} = \frac{2kp}{r^3} \] \[ \boxed{\;U \propto E^2 \propto \frac{1}{r^6}\;} \]

This formula holds true only for molecules with temperory dipole moment.

Let us try to understand the distribution of energy in a subsystem of a (large isolated system.

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cond\(^n\) of eq\(^n\) implies,

\[ \frac{\partial \ln \Omega}{\partial E} = \frac{\partial \ln \Omega'}{\partial E'} = \frac{1}{k_B T} \] \[ (T_A = T_B). \]

Probability that subsystem A has energy E

\[ P(E) = \frac{\text{No of microstates of total system such that A has energy `E'}}{\text{Total no of accessible microstates of system with energy } E_{tot}.} \] \[ P(E) = \frac{\Omega(E)\cdot\Omega'(E')}{\Omega_{tot}(E_{tot})} \]

where \(\left(0 \le E \le E_{tot}\right)\).

We know that

\[ \Omega'(E') = \Omega'(E_{tot}-E) \propto \left(E_{tot}-E\right)^{N} \]

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\[ \ln P(E) = \ln \Omega(E) + \ln \Omega'(E_{tot}-E) - \ln \Omega_{tot}(E_{tot}) \]

let us expand \(\ln P(E)\) about some point \(\bar{E}\) or \(\langle E\rangle\).

\[ \ln P(E) \approx \ln P(\bar{E}) + (E-\bar{E})\left.\frac{\partial \ln P(E)}{\partial E}\right|_{(E=\bar{E})}^{\;=0} + \frac{(E-\bar{E})^2}{2!}\left.\frac{\partial^2 \ln P(E)}{\partial E^2}\right|_{E=\bar{E}} \]

Since \(P(E)\) is max at \(E=\bar{E}\), \(\left(\left.\dfrac{\partial \ln P(E)}{\partial E}\right|_{\bar{E}} = 0\right)\)

\[ \ln P(E) = \ln P(\bar{E}) + \frac{(E-\bar{E})^2}{2!}\left.\frac{\partial^2 \ln P(E)}{\partial E^2}\right|_{\bar{E}} \]

Since slope of slope at \(\bar{E}\) is negative.

\[ P(E) = P(\bar{E})\cdot e^{-K\frac{(E-\bar{E})^2}{2!}} \] \[ P(E) = (\text{Normalisation const})\cdot e^{-\alpha\frac{(E-\bar{E})^2}{2\bar{E}}} \qquad \text{`}\alpha\text{' to make dimensionless.} \]

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So relative fluctuations. goes like \(\dfrac{\sqrt{\bar{E}}}{\bar{E}} = \dfrac{1}{\sqrt{\bar{E}}}\)

\[ \left(\bar{E} = N\frac{f}{2}k_B T\right) \]

relative fluctuations about mean \(\propto \dfrac{1}{\sqrt{N}}\)

So for \((N \approx 10^{24})\) fluctuations in eq\(^m\) about mean value in avg Energy \(\propto 10^{-12}\). which is negligible. That is why in thermodynamics we only talk about \((\bar{E})\) or \(\langle E\rangle\) rather than \((E_{tot}.)\)

\[ \begin{aligned} \ln P(\varepsilon) &= \ln \Omega'(E') - \ln \Omega_{tot}(E_{tot})\\ &= \ln \Omega'(E_{tot}-E) - \ln \Omega_{tot}(E_{tot}) \end{aligned} \]

Since \((E_{tot} \gg \varepsilon)\) let us expand \(\ln \Omega'(E')\) about \((E_{tot})\).

\[ \ln P(\varepsilon) = \left(\ln \Omega'(E_{tot}) - \varepsilon\left.\frac{\partial \ln \Omega'(E')}{\partial E'}\right|_{E'=E_{tot}} + \cdots\right) - \ln \Omega_{tot}(E_{tot}) \] \[ P(\varepsilon) = (\text{const})\,e^{-\beta\varepsilon} \qquad\Bigg|\qquad \beta = \frac{\partial \ln \Omega'(E')}{\partial E'} = \text{inverse temp} = \frac{1}{k_B T} \text{ of heat bath} \]

to find (const) we need to know \(\Omega'(E_{tot})\) (info about heat bath). but we don't need that too, we can normalize \(P(\varepsilon)\) and find normalisation const.

\[ P(\varepsilon) = \frac{e^{-\beta\varepsilon}}{\displaystyle\sum_{\varepsilon} e^{-\beta\varepsilon}} \]

So Normalisation const. is fixed by properties of subsystem alone and we eliminate whatever happens outside"

for a moment, let us assume \((A)\) has discrete energy levels.

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  \node[right] at (4.6,0.0) {$\varepsilon_0$};
\end{tikzpicture}

\[ P(\varepsilon_i) = \frac{e^{-\beta\varepsilon_i}}{\displaystyle\sum_{j} e^{-\beta\varepsilon_j}} \qquad \to \text{ summed over all states not \underline{levels}}. \]

The only information about heat bath to be known is its temp. unlike (case of microcanonical ensamble where \(E = \bar{E}\) at eq\(^m\)) The total energy of system *in thermal eq\(^n\) with heat bath* is not fixed. It is fluctuating with in diff energy with \(P(\varepsilon_i) = e^{-\beta\varepsilon_i}\big/\sum_i e^{-\beta\varepsilon_j}\) but the avg. value of \(\langle \varepsilon_i\rangle\) is independent of time, because \(P(\varepsilon_i)\) is independent of time

The factor \(\left(\displaystyle\sum_{\text{states } j} e^{-\beta\varepsilon_j}\right)\) is called `Z' the canonical partition function.

and our system

\usetikzlibrary{arrows.meta,decorations.pathreplacing,calc,angles,quotes,positioning,patterns}
\begin{tikzpicture}[scale=0.95,>=Stealth]
  \draw[thick] (0,0.6) .. controls (0.1,1.7) and (0.9,2.6) .. (2.1,2.6)
       .. controls (3.4,2.6) and (4.2,1.7) .. (4.1,0.8)
       .. controls (4.0,-0.15) and (3.1,-0.75) .. (2.0,-0.7)
       .. controls (1.0,-0.65) and (0.35,-0.35) .. (0.55,0.15)
       .. controls (0.7,0.5) and (0.15,0.35) .. (0,0.6);
  \draw (0.95,2.0) circle (0.26);
  \node at (0.95,2.0) {\footnotesize $T$};
  \draw (0.9,1.35) circle (0.28);
  \node at (0.9,1.35) {\footnotesize $A$};
  \draw[->] (1.05,1.7) -- (0.5,1.75);
  \node at (2.2,2.15) {$T$};
  \node at (2.3,1.1) {$B$};
  \node[align=center] at (2.0,0.35) {\underline{Heat}\\ \underline{Bath}};
\end{tikzpicture}

in Thermal-eq\(^m\) at temp \(T\) is called "canonical ensamble".

\[ \left(Z = Z(T,V,N) = \sum_{\text{states } j} e^{-\beta\varepsilon_j} = \sum_{\text{levels } i} g_i\, e^{-\beta\varepsilon_i}\right) \]

The subsystem "A" in thermal eq\(^m\) of this kind is not in fixed energy state. It has got probability distributions for certain energy values. For a quantum system "A" is not even in pure energy eigen state. It will be in superposition of all energy eigen states. \(\left\{\text{mind blowing fact.}\right.\)

\(\left(P(\varepsilon) \propto e^{-\beta\varepsilon_i}\right)\).

\usetikzlibrary{arrows.meta,decorations.pathreplacing,calc,angles,quotes,positioning,patterns}
\begin{tikzpicture}[scale=0.95,>=Stealth]
  \useasboundingbox (-1.6,-0.9) rectangle (6.4,2.6);
  \clip (-1.6,-0.9) rectangle (6.4,2.6);
  \draw[->] (0,-0.55) -- (0,2.2);
  \node[left] at (-0.15,1.4) {$(P(\varepsilon))$};
  \draw[->] (-0.55,0) -- (5.0,0);
  \draw[domain=0:4.6,samples=140,smooth] plot (\x,{1.75*exp(-0.62*\x)});
  \node[below] at (2.6,-0.15) {$\varepsilon$};
\end{tikzpicture}

distribution of energy goes like exponential decay rather than steep gaussian distribution.

\(\beta\) = property of bath.

In terms of energy levels,

\[ Z = \sum_{\text{levels } j} \left(g_j\right) e^{-\beta\varepsilon_j} \]

\(\to\) degeneracy of level \(\underline{\varepsilon_j}\).

for continuous energy levels,

\[ P(\varepsilon) = \frac{e^{-\beta\varepsilon}}{\displaystyle\int_{0}^{\infty} e^{-\beta\varepsilon}\, g(\varepsilon)\,d\varepsilon}. \]

now instead of counting degeneracy \((g_j)\) we ask

\usetikzlibrary{arrows.meta,decorations.pathreplacing,calc,angles,quotes,positioning,patterns}
\begin{tikzpicture}[scale=0.95,>=Stealth]
  \node[left] at (-0.2,0.75) {$d\varepsilon$};
  \draw (0,0.6) -- (2.6,0.6);
  \draw (0,0.9) -- (2.6,0.9);
  \foreach \xx in {0.25,0.6,0.95,1.3,1.65,2.0,2.35}{
     \draw (\xx,0.6) -- (\xx+0.22,0.9);
  }
  \node[right] at (2.7,0.75) {$\varepsilon$};
  \draw (0,-0.35) -- (2.6,-0.35);
  \node[right] at (2.7,-0.35) {$0$};
  \draw[->] (3.5,-0.45) -- (3.5,1.0);
  \node[right] at (3.6,0.5) {$\varepsilon$};
\end{tikzpicture}

how many states with energy \(\varepsilon\) & \(\varepsilon+d\varepsilon\)

\[ = g(\varepsilon)\,d\varepsilon \]

\(\hookrightarrow\) density of states

\(g(\varepsilon)\) = no of states in \(d\varepsilon\) range per unit energy \(\varepsilon\)

\[ Z(T,V,N) \equiv Z = \int_0^\infty e^{-\beta\varepsilon}\, g(\varepsilon)\, d\varepsilon \qquad (\text{general canonical partition fun.}) \]

So,

\[ p(\varepsilon) = \frac{e^{-\beta\varepsilon}}{\displaystyle\int_0^\infty e^{-\beta\varepsilon} g(\varepsilon)\, d\varepsilon}. \]

So \(Z\) is Laplace transform of \(g(\varepsilon)\), in which transform variable is \(\beta\), instead of `s'. (Laplace transform or Moment generating functions for continuous variables)

\usetikzlibrary{arrows.meta,decorations.pathreplacing,calc,angles,quotes,positioning,patterns}
\begin{tikzpicture}[scale=0.8]
  % doodle of a face
  \draw (0,0) circle [x radius=0.45, y radius=0.55];
  \draw (-0.18,0.18) circle [radius=0.07];
  \draw (0.10,0.20) circle [radius=0.07];
  \draw (-0.20,-0.18) .. controls (-0.05,-0.32) and (0.10,-0.30) .. (0.20,-0.14);
  \foreach \a in {100,112,124,136,148,160}{
    \draw (\a:0.5) -- ++(\a:0.28);
  }
  \draw (-0.44,0.35) -- (0.05,0.62);
\end{tikzpicture}

\[ \left. \begin{aligned} g(\varepsilon) &\longrightarrow \text{system properties} \quad (\text{mech. aspect classical/quantum}).\\ e^{-\beta\varepsilon} &\longrightarrow \text{statistical aspect}. \end{aligned} \right\} \] \[ e^{-\beta\varepsilon} g(\varepsilon) \longrightarrow \underline{\text{stat. mech.}} \]

(Q) What is average energy?

\[ \begin{aligned} \langle E\rangle &= \sum_{(\text{states } j)} P(E_j)\, E_j\\ &= \frac{\displaystyle\sum_{(\text{states } i)} E_i\, e^{-\beta E_i}}{Z(T,V,N)}\\ &= \frac{\displaystyle\sum_{(\text{states } i)} E_i\, e^{-\beta E_i}}{\left(\displaystyle\sum_{(\text{states } j)} e^{-\beta E_j}\right)} \;=\; \frac{1}{Z}\left(\sum_{\text{states } i} E_i\, e^{-\beta E_i}\right). \end{aligned} \] \[ \boxed{\;\langle E\rangle = -\frac{\partial(\ln Z)}{\partial\beta}\;} \qquad \left(\frac{-1}{Z}\cdot\frac{\partial Z}{\partial\beta} = +\frac{1}{Z}\sum E_i\, e^{-\beta E_i}\right) \] \[ = -\frac{1}{Z}\frac{\partial Z}{\partial\beta} \]

So, \((C_v \geq 0.)\) \(\therefore\)

In thermodynamics we only deal with \(\langle E\rangle\). So we can't calculate \((C_v)\) or \(\underline{C_p}\). So \(C_p, C_v\) are input in Thermo-dynamics.

Recall that the canonical partition function of small subsystem in heat bath of temp. \(T = \frac{1}{k_B\beta}\)

\[ Z = \sum_{\text{states } j} e^{-\beta E_j} \;\longrightarrow\; \int_0^\infty dE\; e^{-\beta E}\, g(E) \] \[ g(E) \propto E^{1/2} \] \[ g(E) \propto E^{\frac{N}{2}} \ \text{or} \ \propto E^{N} \qquad \text{for } \underline{N = 10^{24}}. \]

Lets plot \((Z\) vs \(E)\)

\usetikzlibrary{arrows.meta,decorations.pathreplacing,calc,angles,quotes,positioning,patterns}
\begin{tikzpicture}[scale=0.95,>=Stealth]
  \useasboundingbox (-0.9,-1.0) rectangle (7.0,3.6);
  \clip (-0.9,-1.0) rectangle (7.0,3.6);
  % axes
  \draw[->] (0.45,-0.05) -- (0.30,3.3);
  \draw[->] (-0.15,0) -- (6.6,0);
  \node at (-0.55,1.6) {$Z$};
  \node at (6.5,-0.4) {$E$};
  % decaying exponential e^{-beta E}
  \draw[thick] plot[smooth] coordinates
    {(0.30,2.55) (0.9,2.05) (1.6,1.65) (2.4,1.30) (3.2,1.05)
     (4.0,0.83) (4.8,0.66) (5.6,0.52) (6.3,0.42)};
  \node at (1.15,2.45) {$e^{-\beta E}$};
  % rising density of states g(E)
  \draw[thick] plot[smooth] coordinates
    {(1.05,0.03) (1.6,0.15) (2.1,0.40) (2.5,0.75) (2.9,1.30)
     (3.25,1.95) (3.55,2.70) (3.70,3.15)};
  \node at (4.35,3.20) {$g(E)$};
  % product: peaked curve
  \draw[thick] plot[smooth] coordinates
    {(1.35,0.03) (1.75,0.35) (2.05,1.05) (2.30,2.05) (2.55,2.75)
     (2.85,2.90) (3.15,2.55) (3.45,1.75) (3.75,1.05) (4.15,0.68)
     (4.7,0.48) (5.3,0.38) (5.8,0.35)};
  % dashed line at Ebar
  \draw[dashed] (2.80,2.85) -- (2.80,0);
  \node at (2.80,-0.45) {$\bar{E}$};
\end{tikzpicture}

\[ Z = \int_0^\infty dE\; e^{-\beta E}\, g(E) \;=\; e^{-\beta\bar{E}}\, \Omega(\bar{E}) \]

(?) maybe \(\sim\) (by me) \(\left(\Omega(E) \approx \Omega(\bar{E}) = \int_0^\infty g(E)\,dE\right)\)

\(\Omega(\bar{E}) = \) total no of microstates

So,

\[ \ln Z = \ln \Omega(\bar{E}) - \beta\bar{E} \] \[ -k_B T \ln Z = -k_B T \ln\Omega(\bar{E}) + \bar{E} \]

But we know that entropy (S) is defined as

\[ \boxed{\,S = k_B \ln\Omega\,} \] \[ -k_B T\ln Z = -TS + \bar{E} \qquad (\bar{E} = \langle E\rangle) = U \ (\text{Internal energy}) \] \[ = F/A \ (\text{Hemholtz free energy}). \] \[ \Rightarrow \boxed{\,Z = e^{-\beta F}\,} \]

So, the whole idea was to write

\[ Z = \sum_{\text{states } j} e^{-\beta E_j} \quad \text{in terms of} \quad e^{-\beta E_{\text{eff}}} \]

\(\hookrightarrow\) effective energy

\[ \left(Z = e^{-\beta F}\right). \qquad (F = \text{Helmholtz free energy}). \]

It needs coherent punch from bath molecules to provide maximum energy to subsystem `A'. So it is less likely to get high energy from a heat bath; that is the reason the energy distribution goes like

\[ P(E_i) = \frac{e^{-\beta E_i}}{\displaystyle\sum_{\text{states } j} g_j\, e^{-\beta E_j}} \]

We know that

\[ \begin{aligned} F &= U - TS\\ dF &= dU - TdS - SdT\\ &= TdS - PdV + \mu dN - TdS - SdT\\ dF &= -SdT - PdV + \mu dN\\ P &= -\left(\frac{\partial F}{\partial V}\right)_{T,N} \end{aligned} \]

(Partition function for classical ideal gas):

\usetikzlibrary{arrows.meta,decorations.pathreplacing,calc,angles,quotes,positioning,patterns}
\begin{tikzpicture}[scale=1.0,>=Stealth]
  % outer blob = heat bath
  \draw plot[smooth cycle] coordinates
    {(0,1.7) (1.6,2.5) (3.3,2.2) (4.2,0.9) (3.6,-0.7) (2.0,-1.4)
     (0.2,-1.1) (-0.9,0.2)};
  % inner blob = ideal gas
  \draw plot[smooth cycle] coordinates
    {(1.15,1.55) (2.15,1.75) (2.85,1.15) (2.75,0.25) (1.85,-0.25)
     (0.95,0.05) (0.75,0.55) (1.05,1.0)};
  \node at (2.05,0.85) {$V,N$};
  \foreach \p in {(1.35,1.2),(1.75,1.35),(1.55,0.55),(2.25,0.35),(1.15,0.6),(2.45,1.35),(1.95,0.15)}
    {\fill \p circle (0.03);}
  \draw[->] (3.9,2.3) -- (3.0,1.9);
  \node[anchor=west] at (3.9,2.35) {mass of each particle \underline{`m'}};
  \draw[->] (3.15,0.05) .. controls (3.6,-0.35) .. (4.05,-0.2);
  \node[anchor=west,align=left] at (4.05,-0.25) {ideal\\ gas};
  \draw[->] (0.2,-2.0) .. controls (1.6,-2.5) and (3.2,-2.2) .. (3.9,-1.0);
  \node[anchor=east] at (0.2,-2.05) {Heat bath \ $T$};
\end{tikzpicture}

Assume, All particles are independent of each other and there is no interaction b/w the particles.

\[ Z = \sum_{\substack{\text{states}\\ \text{of}\\ \text{gas}}} e^{-\beta E} = \sum e^{-\beta(\varepsilon(1)+\varepsilon(2)+\varepsilon(3)+\cdots)} \]

where \(\varepsilon(i)\) is energy of \(i\)th particle of whole system.

If all particles are independent of each other, then partition function factorize into --

\[ Z = \sum e^{-(\beta\varepsilon+\beta\varepsilon+\beta\varepsilon+\cdots)} = \sum_{\substack{\text{states of}\\ \text{gas}}}\left(e^{-\beta\varepsilon}\cdot e^{-\beta\varepsilon}\cdot e^{-\beta\varepsilon}\cdots\right) \] \[ Z = \left(\sum_{\substack{\text{states of}\\ \text{any particle}}} e^{-\beta\varepsilon}\right)^{N} \]

\(\longrightarrow\) Only true if particles are distinguish*-able*.

\[ \begin{aligned} Z &= \left(\int_0^\infty e^{-\beta\varepsilon}\, \underbrace{g(\varepsilon)d\varepsilon}_{\Omega(E)}\right)^{N} = \left(\frac{1}{h^3}\left(\int_V d^3r \int d^3p\; e^{-\frac{\beta p^2}{2m}}\right)\right)^{N}\\ Z &= \left(\int_0^\infty e^{-\beta E}\,\Omega(E)\right)^{N}\\ &= \left(\frac{V}{h^3}\left(\int e^{-\frac{\beta p_x^2}{2m}}dp_x \int e^{-\frac{\beta p_y^2}{2m}}dp_y \int e^{-\frac{\beta p_z^2}{2m}}dp_z\right)\right)^{N}\\ &= \left(\frac{V}{h^3}\left((2\pi m k_BT)^{1/2}\cdot(2\pi m k_BT)^{1/2}(2\pi m k_BT)^{1/2}\right)\right)^{N}\\ Z &= \left(\frac{V}{h^3}\left(2\pi m k_BT\right)^{3/2}\right)^{N} \end{aligned} \] \[ \Omega(\varepsilon) = \frac{1}{h^3}\int_V d^3r \int_{-\infty}^{\infty} d^3p \qquad \left(\int_{-\infty}^{\infty} e^{-ax^2}dx = \sqrt{\frac{\pi}{a}}\right) \] \[ Z = e^{-\beta F} \] \[ \ln Z = -\beta F \;\Rightarrow\; \left(F = \frac{-1}{\beta}\ln Z\right) = -k_B T N \ln\left(\frac{V(2\pi m k_BT)^{3/2}}{h^3}\right) \] \[ \begin{aligned} P &= -\left(\frac{\partial F}{\partial V}\right)_{T,N} = \frac{1}{\beta}\cdot\frac{1}{Z}\cdot\frac{\partial Z}{\partial V}\bigg|_{T,N}\\ &= \frac{1}{\beta Z}\cdot\frac{1}{h^3}\\ &= Nk_BT\cdot\frac{1}{V\left(\dfrac{(2\pi m k_BT)^{3/2}}{h^3}\right)}\cdot\frac{(2\pi m k_BT)^{3/2}}{h^3} \end{aligned} \] \[ P = \frac{Nk_BT}{V} \qquad \Rightarrow \qquad \boxed{\,PV = Nk_BT\,} \]

When we will do Q-M ; \(E = pc\) for photons in black body cavity at temp \(T\). then

\[ \bar{E} \neq \frac{3}{2}k_BT \quad (\text{even in 3-D}). \]

\(\hookrightarrow E = p^1 c\) \(\underline{\text{So}}\)

\[ \underline{\bar{E} \propto 3k_B T^4 f(\ )} \]

\(\hookrightarrow\) Bose statistics.

So,

\[ PV = Nk_BT \qquad ; \qquad U = \frac{3}{2}Nk_BT \] \[ U = \frac{3}{2}\cdot PV \] \[ \Rightarrow \boxed{\,PV = \frac{2U}{3}\,} \]

\(\longrightarrow\) avg energy density.

or, \(\left(P = \dfrac{2u}{3}\right)\) \(\left(u = \dfrac{U}{V} = \text{avg Energy per unit volume}\right)\).

This relation is even true for quantum gases. (quantum ideal gas).

(Q) Is it correct that

\[ Z = \sum e^{-\beta E} = \left(\sum e^{-\beta\varepsilon}\right)^{N} \ ? \]

summed over all states of gas summed over states of any particle

No, its not true (always) here we have overcounted the number of states, since the particles are indistinguishable. (physical fact).

\usetikzlibrary{arrows.meta,decorations.pathreplacing,calc,angles,quotes,positioning,patterns}
\begin{tikzpicture}[scale=0.9,>=Stealth]
  % ---- left set ----
  \draw[->] (-0.35,0) -- (-0.35,2.1);
  \node[anchor=east] at (-0.4,1.1) {level};
  \foreach \x/\d in {0.4/1, 1.7/0, 3.0/2}{
    \foreach \y in {0,1,2}{ \draw (\x,\y) -- (\x+0.8,\y); }
    \fill (\x+0.4,\d+0.07) circle (0.07);
  }
  \node[anchor=north,align=center,font=\scriptsize] at (0.8,-0.15) {state\\ of 1st\\ particle};
  \node[anchor=north,align=center,font=\scriptsize] at (2.1,-0.15) {states\\ of\\ 2nd particle};
  \node[anchor=north,align=center,font=\scriptsize] at (3.4,-0.15) {states\\ 3rd\\ particle};
  % arrow between
  \draw[->] (4.2,1.1) -- (5.5,1.1);
  \node[anchor=south,align=center,font=\scriptsize] at (4.85,1.15) {same\\ configuration};
  % ---- right set ----
  \draw[->] (5.9,0) -- (5.9,2.1);
  \node[anchor=east] at (5.85,1.1) {level};
  \foreach \x in {6.3,7.6,8.9}{
    \foreach \y in {0,1,2}{ \draw (\x,\y) -- (\x+0.8,\y); }
  }
  \node at (6.7,2.2) {$\times$};
  \node at (8.0,0.2) {$\times$};
  \node at (9.3,1.2) {$\times$};
  \node at (9.3,0.2) {$\times$};
  \draw (9.1,0.2) -- (9.5,0.28);
  \node[anchor=north,align=center,font=\scriptsize] at (6.7,-0.15) {state\\ 1st};
  \node[anchor=north,align=center,font=\scriptsize] at (8.0,-0.15) {2nd\\ particle\\ states};
  \node[anchor=north,align=center,font=\scriptsize] at (9.3,-0.15) {3rd\\ particle\\ states};
\end{tikzpicture}

since particles are indistinguishable, we have overcounted the number of states.

In quantum statistics we will corect this mistake.

Instead of asking in which level particle 1 is.. and 2 is.. we ask how many particles are in ground state, 1st excited state. and so on....

So in nutshell in quantum statistics we do row wise counting instead of column wise counting performed in classical statistics. From here idea of occupation no. arose in quantum statistics.

\[ dF = -SdT - PdV + \mu dN \] \[ S = -\left(\frac{\partial F}{\partial T}\right)_{V,N} \qquad ; \qquad e^{-\beta F} = Z \] \[ \left(F = \frac{-1}{\beta}\ln Z\right) \]

and,

\[ Z = \left(\frac{V}{h^3}\left(2\pi m k_BT\right)^{3/2}\right)^{N} \] \[ \Rightarrow \quad F = \frac{-N}{\beta}\ln\left(\frac{V}{h^3}\left(\frac{2\pi m}{\beta}\right)^{3/2}\right) = -Nk_BT\ln\left(\frac{V}{h^3}\left(2\pi m k_BT\right)^{3/2}\right) \]

So,

\[ \begin{aligned} S &= -\left(\frac{\partial F}{\partial T}\right)_{V,N} = +Nk_B \ln\left(\frac{V}{h^3}\left(2\pi m k_BT\right)^{3/2}\right)\\ &\qquad + \frac{Nk_BT}{\dfrac{V}{h^3}\left(2\pi m k_BT\right)^{3/2}}\cdot\frac{V}{h^3}\cdot\frac{3}{2}\left(2\pi m k_BT\right)^{1/2}\left(2\pi m k_B\right)\\ &= -Nk_B \ln\left(\frac{V}{h^3}\left(2\pi m k_BT\right)^{3/2}\right) + \frac{3}{2}Nk_B \end{aligned} \]

at \(T=0\) \(\ln\left(\frac{V}{h^3}(0)\right) \longrightarrow \infty\) , so ideal gas is no more a good assumption for our real gas, there are interactions

and physical real gas condenses to liquid & solid which also can't be explained by assuming gas to be ideal.

Also this formula of entropy is more flawed and it is due to incorect partition function arose from overcounting of states (also known as Gibbs paradox).

Let us find

\[ \mu = +\frac{\partial F}{\partial N}\bigg|_{T,V} \qquad ; \qquad F = -Nk_BT\ln\left(\frac{V}{h^3}\left(2\pi m k_BT\right)^{3/2}\right) \] \[ \mu = -k_BT \ln\left(\frac{V}{h^3}\left(2\pi m k_BT\right)^{3/2}\right) \]

this formula is also not sensible

Even the formula for Free energy \(F = -Nk_BT\ln\left(\frac{V}{h^3}(\ )^{3/2}\right)\) as, we expect free energy to be extensive quantity.

The `V' in \(\ln\left(\frac{V}{h^3}(\ )^{3/2}\right)\) is trouble here.

(\(F \propto\) any one extensive variable

\(F = N f(\text{intensive variable})\).)

There are \((n!\) no of ways\()\) particles can be arranged

ex 3 particles.

\usetikzlibrary{arrows.meta,decorations.pathreplacing,calc,angles,quotes,positioning,patterns}
\begin{tikzpicture}[scale=1.0]
  \draw (0,0) rectangle (3.0,0.7);
  \draw (1.0,0) -- (1.0,0.7);
  \draw (2.0,0) -- (2.0,0.7);
  \node at (0.5,0.35) {3};
  \node at (1.5,0.35) {2};
  \node at (2.5,0.35) {1};
  \draw (0.5,0) -- (0.45,-0.6);
  \draw (1.5,0) -- (1.45,-0.6);
  \draw (2.5,0) -- (2.45,-0.6);
  \node[anchor=north,align=center,font=\scriptsize] at (0.45,-0.6) {1st\\ particle};
  \node[anchor=north,align=center,font=\scriptsize] at (1.45,-0.6) {2nd\\ particle};
  \node[anchor=north,align=center,font=\scriptsize] at (2.45,-0.6) {3rd\\ particle};
  \node[anchor=west] at (3.5,0.35) {$= 6$ \underline{ways}.};
\end{tikzpicture}

So A rough way to corect \(Z\) is to divide by \((n!)\) but that is only true if no two particles attains same (state)

\usetikzlibrary{arrows.meta,decorations.pathreplacing,calc,angles,quotes,positioning,patterns}
\begin{tikzpicture}[scale=0.9]
  \foreach \x/\d in {0/0, 1.3/1, 2.6/2}{
    \foreach \y in {0,1,2}{ \draw (\x,\y) -- (\x+0.8,\y); }
    \fill (\x+0.4,\d+0.07) circle (0.07);
  }
\end{tikzpicture}

But that is only very likely if no. of states available

Since particles are quantum object, we have to move to quantum statistics for more precise understanding.

(Criterion for classical statistics to be valid) :-

In Q.M. particles are indistinguishable and \(x\) & \(p\) of particle can't be precisely defined instantaneously.

\usetikzlibrary{arrows.meta,decorations.pathreplacing,calc,angles,quotes,positioning,patterns}
\begin{tikzpicture}[scale=0.9]
  % particle 1 wave packet (broader)
  \draw[thick] plot[smooth,domain=-1.5:1.5,samples=60] (\x,{1.2*exp(-\x*\x/0.5)});
  \draw[->] (1.55,1.25) -- (0.35,1.25);
  \node[anchor=west] at (1.6,1.25) {wave packet};
  \node at (0,-0.5) {(Particle 1)};
  % particle 2 wave packet (narrower)
  \begin{scope}[xshift=9cm]
    \draw[thick] plot[smooth,domain=-1.2:1.2,samples=60] (\x,{1.4*exp(-\x*\x/0.22)});
    \draw[->] (-1.5,1.55) -- (-0.55,1.45);
    \node[anchor=east] at (-1.55,1.55) {wave packet};
    \node at (0,-0.5) {Particle 2};
  \end{scope}
\end{tikzpicture}

We can say that particle 1 is at this position \((x_1 \pm \Delta x_1)\) & particle 2 is at some other position \((x_2 \pm \Delta x_2)\), iff the avg distance b/w them is much larger than spread of wave packets. (i.e. non-interacting identical particles can be distinguishable).

When particles 1 & 2 interact, we can't identify them.

\usetikzlibrary{arrows.meta,decorations.pathreplacing,calc,angles,quotes,positioning,patterns}
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  \node at (1.3,-0.35) {2};
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  \node at (-1.3,-1.3) {$1'$};
  \node at (1.05,1.25) {$2'$};

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    \draw[->] (0.9,-0.5) -- (0.3,-0.5);
    \node at (0.7,-0.25) {2};
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    \node at (0.65,0.75) {$2'$};
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  \node at (6.2,0) {or};
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(particles are exchanged).

So, for classical statistics to work

\[ \text{mean interparticle seperation} \ggg \lambda_{\text{debroglie}} \]

we still have to divide by \((N!)\) to count states correctly.

If we have \(N\) particles in volume \(V\) then volume available for each particle is \(\left(\dfrac{V}{N}\right)\)

So linear dimension available is \(\left(\dfrac{V}{N}\right)^{1/3}\)

\[ \left(\frac{V}{N}\right)^{1/3} \gg \frac{h}{p_{rms}} \qquad \left(\lambda_{deB} = \frac{h}{p}\right). \] \[ \left\langle \frac{p^2}{2m} \right\rangle = \frac{3}{2} k_B T \qquad\qquad p_{rms} = \sqrt{3mk_BT} \]

for one particle

\[ \sqrt{\langle p^2\rangle} = p_{rms} \Longleftarrow \;=\; (3mk_BT)^{1/2} \ \text{ or } \ (mk_BT)^{1/2} \] \[ \left(\frac{V}{N}\right)^{1/3} \gg \frac{h}{(mk_BT)^{1/2}} \] \[ \left(\frac{V}{N}\right) \gg \frac{h^3}{(mk_BT)^{3/2}} \qquad \text{or} \qquad \left(\frac{1}{n}\right) \gg \frac{h^3}{(mk_BT)^{3/2}} \] \[ \boxed{1 \ggg \frac{nh^3}{(mk_BT)^{3/2}}} \qquad\qquad \frac{nh^3}{(mk_BT)^{3/2}} = \text{degeneracy factor} \]

\[ \boxed{\frac{nh^3}{(mk_BT)^{3/2}} \ll 1} \]

is criteria for statistical to be valid. (\(n\) = number density of particles)

Nitrogen \(m \approx 10^{-26}\) kg \[ h = 10^{-34} \] \[ n = 10^{24} \text{ per cubic meter} \] \[ T = 300\,\text{K} , \qquad k_B = 1.38\times 10^{-23} \] \[ \frac{10^{24}\cdot 10^{-34\times 3}}{\left(10^{-26}\cdot 10^{-23}\cdot 300\right)^{3/2}} \;=\; \frac{10^{-102+24}}{\left(10^{-51}\right)^{3/2}} \] \[ = \frac{10^{-78}}{10^{-75}} \;\approx\; 10^{-3} \ll 1 . \]

So classical statistics works.

If \((n\uparrow)\) this approximation fails. (like in case of liquids/neutron stars. the degeneracy factor is of order of \(10^6\)).

Also at low temp \((T\to 0)\) this limit/condition fails and we use quantum statistics.

(Lec -26 -- is of probability distributions so, It will be written in mathametical notebook of V. balki sir).

Contents

Lecture 23

Euler's theorem

Legendre transforms

Isothermal compressibility (fixed T,N)

Poisson's ratio :-

Lecture 24

Lecture 25: Connection with thermodynamics

Let us find Pressure:

Correction to partition function :-