Contents

overview | playlist | << prev | next >>

Lecture 17

(Q) Does liapunov exponent depends on the initial condition or not?

\usetikzlibrary{arrows.meta,decorations.pathreplacing,calc,angles,quotes,positioning,patterns}
\begin{tikzpicture}[scale=0.9,>=Stealth]
  % outer phase space blob
  \draw[thick] (0,0) .. controls (0.3,2.2) and (1.6,3.4) .. (3.2,3.3)
        .. controls (4.8,3.2) and (5.4,1.8) .. (5.0,0.3)
        .. controls (4.6,-1.2) and (2.6,-1.8) .. (1.2,-1.3)
        .. controls (0.3,-1.0) and (-0.2,-0.7) .. (0,0);
  % attractor 1 (right inner blob)
  \draw (3.1,2.4) .. controls (3.6,2.2) and (3.6,1.4) .. (3.2,1.0)
        .. controls (2.9,0.7) and (2.8,1.6) .. (2.9,2.1)
        .. controls (2.95,2.35) and (3.0,2.44) .. (3.1,2.4);
  % attractor 2 (left inner blob)
  \draw (1.5,2.0) .. controls (1.9,1.8) and (1.9,1.0) .. (1.6,0.5)
        .. controls (1.3,0.1) and (1.2,1.1) .. (1.3,1.6)
        .. controls (1.35,1.85) and (1.4,2.03) .. (1.5,2.0);
  % x0 and arrows
  \node at (2.35,2.85) {$x_0$};
  \draw[->] (2.4,2.65) -- (2.5,1.4);
  \draw[->] (2.0,2.6) -- (1.75,2.15);
  \draw[->] (3.05,2.75) -- (3.05,2.5);
  % labels
  \node[right] at (6.2,2.9) {attractor$_1$};
  \draw (6.1,2.9) .. controls (5.0,2.7) and (4.2,2.6) .. (3.45,2.2);
  \node[left] at (-1.4,-0.9) {attractor$_2$};
  \draw (-1.3,-0.85) .. controls (0.2,-0.2) and (1.0,0.4) .. (1.42,0.6);
\end{tikzpicture}

If the initial condition is such that the trajectory comes to arbitary point in phase space.

Since liapunov exponent is calculated by taking time steps to infinity, and the point has explored whole phase space, therefore in that sense liapunov exponent should not depend on initial condition.

\usetikzlibrary{arrows.meta,decorations.pathreplacing,calc,angles,quotes,positioning,patterns}
\begin{tikzpicture}[scale=0.9]
  \draw (0,0) ellipse (2.7 and 2.3);
  \node[align=left,text width=4.6cm] at (0,0) {Though there could be multiple attractors which suggests it depends on $x_0$, but generally otherwise where ever we start, $x$ is going to fill up whole phase space.};
  \draw (-1.9,1.6) -- (-1.5,3.2) -- (-1.0,1.95);
\end{tikzpicture}

So liapunov exponent now becomes the property of trejectory rather than of a point.

Baker's map :--

( measure (area, volume) is preserved, still system is chaotic ), conservative system.

It is like bernoulli map where we have to cut the graph above \(x_{n+1} = 1\) and put it back. We could have also folded the peice back rather than cutting to gain non-linearity.

\[ x_{n+1} = 2x_n \bmod 1 \qquad \longrightarrow \ \text{(fold just)} \quad \text{(analogy)} \] 
\usetikzlibrary{arrows.meta,decorations.pathreplacing,calc,angles,quotes,positioning,patterns}
\begin{tikzpicture}[scale=3.0,>=Stealth]
  % axis
  \draw (-0.1,0) -- (1.25,0);
  % box
  \draw (0,0) rectangle (1,1);
  % bernoulli branches
  \draw[thick] (0,0) -- (0.5,1);
  \draw[dashed] (0.5,0) -- (1,1);
  % folded (tent) branch
  \draw[thick] (0.5,1) -- (1,0);
  % dashed vertical
  \draw[dashed] (0.5,1) -- (0.5,0);
  % fold arrow
  \draw[->] (0.62,1.15) .. controls (0.85,0.95) and (0.9,0.6) .. (0.78,0.45);
  % labels
  \node[left] at (-0.05,0.45) {$x_{n+1}$};
  \node[below left] at (0,0) {$0$};
  \node[below] at (1,-0.02) {$1$};
  \node[below] at (0.5,-0.16) {$x_n$};
\end{tikzpicture}

Tent map (non-linear, contineous)

\(\lambda(x_0) = \ln 2\), all rational no are periodic unstable orbits.

So the idea is to stretch the interval then put the second half part on top of first half map.

\usetikzlibrary{arrows.meta,decorations.pathreplacing,calc,angles,quotes,positioning,patterns}
\begin{tikzpicture}[scale=0.9,>=Stealth]
  % interval 0 to 1
  \draw[thick] (0,3) -- (3,3);
  \draw (0,2.88) -- (0,3.12);
  \draw (3,2.88) -- (3,3.15);
  \node[below] at (0,2.85) {$0$};
  \node[below] at (3,2.85) {$1$};
  % double arrow down
  \draw[->,double] (1.5,2.6) -- (1.5,2.0);
  % stretched interval 0 to 2
  \draw[thick] (0,1.6) -- (6,1.6);
  \draw (0,1.48) -- (0,1.72);
  \draw (3,1.45) -- (3,1.75);
  \draw (6,1.48) -- (6,1.75);
  \node[below] at (6,1.45) {$2$};
  % folding arrow
  \draw[->] (4.3,1.75) .. controls (4.5,2.5) and (3.2,2.6) .. (3.0,1.95);
  % arrow down
  \draw[->] (1.5,1.2) -- (1.5,0.55);
  % flattened slab
  \draw[thick] (0,0.0) -- (3,0.05) -- (3,0.3) -- (0,0.3) -- cycle;
  \node[below] at (0,-0.02) {$0$};
  \node[below] at (3,-0.02) {$1$};
\end{tikzpicture}

Backer's map takes one more diamencion (increasing diamension by one) in such a way that measure is preserved.
\(x_{n+1} = 2x_n \bmod 1\) [?]

\[ y_{n+1} = \begin{cases} \dfrac{1}{2}\,y_n & \left(x_n < \tfrac{1}{2}\right)\\ \dfrac{1}{2} + \dfrac{1}{2}\,y_n & \left(x_n > \tfrac{1}{2}\right)\end{cases} \]

lets draw, \(x_{n+1}\) vs \(y_{n+1}\); and \(x_n\) vs \(y_n\).

\usetikzlibrary{arrows.meta,decorations.pathreplacing,calc,angles,quotes,positioning,patterns}
\begin{tikzpicture}[scale=0.85,>=Stealth]
  % first square
  \draw[thick] (0,0) rectangle (3,3);
  \draw[pattern=north east lines] (1.5,0) rectangle (3,3);
  \filldraw (0.4,0.7) circle (0.05);
  \filldraw (0.9,0.35) circle (0.05);
  \node at (2.2,1.0) {$\surd$};
  \node[left] at (0,3) {$1$};
  \node[left] at (0,1.5) {$\frac{1}{2}$};
  \node[left] at (-0.5,1.6) {$y$};
  \node[below] at (0,-0.05) {$0$};
  \node[below] at (1.5,-0.05) {$\frac{1}{2}$};
  \node[below] at (1.5,-0.75) {$x$};
  \draw[dashed] (1.5,0) -- (1.5,3);
  \draw[->] (3.3,3.6) -- (3.05,3.1);
  \node[right] at (3.3,3.7) {$A = |x| = 1$};
  % arrow with labels
  \draw[->,thick] (4.0,1.5) -- (7.0,1.5);
  \node[above] at (5.5,1.6) {expanding $x$ by 2};
  \node[below] at (5.5,1.4) {contracting $y$ by $\frac{1}{2}$};
  % second rectangle
  \draw[thick] (7.6,0.8) rectangle (13.6,2.3);
  \draw[dashed] (10.6,0.8) -- (10.6,2.3);
  \filldraw (8.2,1.0) circle (0.05);
  \filldraw (8.8,1.05) circle (0.05);
  \node at (12.4,1.1) {$\surd$};
  \node[left] at (7.6,2.3) {$\frac{1}{2}$};
  \node[below] at (13.6,0.75) {$2$};
  \node[below] at (10.6,0.7) {$x$};
  % fold arrow
  \draw[->] (12.6,3.3) .. controls (11.3,3.9) and (10.0,3.6) .. (9.6,3.0);
\end{tikzpicture}

\(\Rightarrow\) Jacobian of transformation is 1

As long as measure is preserved sum of liapunov exponents is zero

All liapunov exponents are zero for integrable hamiltonian syst[?]{} since it is integrable none of \(\lambda(x_0)\) should be \(+\)ve, \(-\)ve (expand[?]{} volume space/contracting), sum of these also will be zero, \((0+0+\cdots)\)

For non-integrable hamilton systems liapunov exponents occure in pairs such that total liapunov exponent is zero. (\(-\)ve for each \(+\)ve).

Note :--

unlike differential dynamics, in descrete time dynam[?]{} we can write explicit solution to a map, but stil[?]{} it will be chaotic.

\[ \begin{aligned} x_{n+1} &= 2x_n \bmod 1\\ x_n &= 2^n x_0 \bmod 1 \quad \text{---- explicit fun. of time}\\ &\qquad\qquad\qquad\qquad \hookrightarrow \text{still chaotic.} \end{aligned} \]

we can see how we do not see loss of information in Backers map, if we represent \(x_n\), \(y_n\), \(x_{n+1}\), \(y_{n+1}\) in binary

\[ \begin{aligned} x_n &= 0.\,a_0\,a_1\,a_2\,a_3 \cdots\\ y_n &= 0.\,b_0\,b_1\,b_2\,b_3 \cdots\\ &\quad \downarrow\\ x_{n+1} &= 0.\,a_1\,a_2\,a_3 \cdots\\ y_{n+1} &= 0.\,(a_0 b_0)\,b_1\,b_2\,b_3 \cdots \end{aligned} \]

so we do not loss `\(a_0\)'.

Chaotic but (measure preserving and invertible.)

(wow!)

Note :-- The general rule to produce chaos in 1-d map is that it has to be non-invertible, for 2 and higher diamen- ncions invertible maps can also produce chaos.

Arnold's cat map :--

\[ \begin{aligned} x_{n+1} &= (x_n + y_n) \bmod 1\\ y_{n+1} &= (x_n + 2y_n) \bmod 1 \end{aligned} \] \[ \begin{pmatrix} x_{n+1}\\ y_{n+1}\end{pmatrix} = \begin{pmatrix} 1 & 1\\ 1 & 2\end{pmatrix} \begin{pmatrix} x_n\\ y_n\end{pmatrix} \]

determinent of Transformation matrix \(= 1\) (orientation preserving)

\usetikzlibrary{arrows.meta,decorations.pathreplacing,calc,angles,quotes,positioning,patterns}
\begin{tikzpicture}[scale=2.6,>=Stealth]
  % slightly tilted unit square
  \draw[thick] (0,0) -- (1.02,0.02) -- (1.05,1.03) -- (0.03,1.0) -- cycle;
  % stretched sliver going out of the top right
  \draw (0,0) .. controls (0.35,0.28) and (0.6,0.62) .. (0.86,1.28);
  \draw (0,0) .. controls (0.28,0.32) and (0.5,0.62) .. (0.86,1.28);
  \draw (0.86,1.28) -- (1.05,1.03);
  % small sliver near bottom right
  \draw (0.62,0.02) -- (0.72,0.42) -- (0.78,0.03);
  % stretching / contracting arrows
  \draw[<->] (0.34,0.42) -- (0.52,0.62);
  \draw[<->] (0.30,0.58) -- (0.48,0.46);
  % markers
  \filldraw (0.22,0.62) circle (0.008);
  \filldraw (0.42,0.35) circle (0.008);
  \filldraw (0.68,0.78) circle (0.008);
  \filldraw (1.02,0.02) circle (0.012);
  % labels
  \node[left] at (-0.03,1.0) {$1$};
  \node[left] at (-0.16,0.5) {$y$};
  \node[below left] at (0,0) {$0$};
  \node[below] at (0.5,-0.06) {$x$};
  \node[below right] at (1.05,0.0) {$1$};
\end{tikzpicture}

The reason it is named as arnold's cat is because he put face of cat in this graph and after few iteration we can't recognize the face of cat.

\(\to\) every point is having stretching direction and contracting direction.

*. Gauss (continued fraction) map :--

\[ x_0 = \cfrac{1}{a_0 + \cfrac{1}{a_1 + \cfrac{1}{a_2 + \cfrac{1}{a_3 + \cfrac{1}{a_4 + \cdots}}}}} \quad \xrightarrow[\text{writing}]{\text{eas of}} \quad \frac{1}{a_0} \oplus \frac{1}{a_1} + \frac{1}{a_2} + \frac{1}{a_3} \cdots \]

\(\hookrightarrow\) notice `\(+\)' sign is written down

If continued fraction terminates \(x_0 = \) rational else \(x_0\) is irrational number

\[ x_1 = \cfrac{1}{a_1 + \cfrac{1}{a_2 + \cfrac{1}{a_3 + \cfrac{1}{a_4 + \cdots}}}} \]

So to get \(x_1\); take reciprocal of \(x_0\), throw away integer part `\(a_0\)'. So,

\[ x_{n+1} = \frac{1}{x_n} - \left[\frac{1}{x_n}\right] \] \[ \hspace{2.2cm}\hookrightarrow \text{integer part of } (1/x_n) \] \[ \text{for,} \quad \left(\begin{array}{c} \tfrac{1}{2} < x_n < 1\\ 1 < \tfrac{1}{x_n} < 2\end{array}\right) \] 
\usetikzlibrary{arrows.meta,decorations.pathreplacing,calc,angles,quotes,positioning,patterns}
\begin{tikzpicture}[scale=3.4]
  \useasboundingbox (-0.45,-0.4) rectangle (1.2,1.15);
  \clip (-0.45,-0.4) rectangle (1.2,1.15);
  % box
  \draw (0,0) rectangle (1,1);
  % diagonal
  \draw (0,0) -- (1,1);
  % branches of the Gauss map
  \draw[thick,domain=0.5:1,samples=80] plot (\x,{1/\x - 1});
  \draw[thick,domain=0.3334:0.5,samples=60] plot (\x,{1/\x - 2});
  \draw[thick,domain=0.25:0.3333,samples=60] plot (\x,{1/\x - 3});
  \draw[thick,domain=0.2:0.25,samples=60] plot (\x,{1/\x - 4});
  \draw[thick,domain=0.1667:0.2,samples=60] plot (\x,{1/\x - 5});
  % markers at tops of branches
  \filldraw (0.5,1) circle (0.008);
  \filldraw (0.3334,1) circle (0.008);
  \filldraw (0.25,1) circle (0.008);
  \filldraw (0.2,1) circle (0.008);
  \filldraw (1,0) circle (0.008);
  \filldraw (0.5,0) circle (0.008);
  \filldraw (0.3334,0) circle (0.008);
  % dashed guides
  \draw[dashed] (0,0.5) -- (1,0.5);
  \draw[dotted] (0.5,0) -- (0.5,1);
  \draw[dotted] (0.618,0) -- (0.618,0.618);
  % labels
  \node[left] at (0,1) {$1$};
  \node[left] at (-0.02,0.55) {$x_{n+1}$};
  \node[below] at (0,-0.03) {$0$};
  \node[below] at (0.25,-0.03) {$\frac{1}{4}$};
  \node[below] at (0.3334,-0.03) {$\frac{1}{3}$};
  \node[below] at (0.5,-0.03) {$\frac{1}{2}$};
  \node[below] at (0.66,-0.03) {f.p.};
  \node[below] at (1,-0.03) {$1$};
  \node[below] at (0.35,-0.17) {$x_n$};
\end{tikzpicture}

for any given \(x_{n+1}\) there lies infinite pre images.

\(|\)slope at each point\(| > 1\)

\(\to\) unstable at each point.

Q.1) lets take any arbitrary no from 0 to 1. take its reciprocal throw away integer, and repeat the process for long time, what is the probability that final number is less than \(\alpha\). \((0,\alpha)\) \(\left(\text{Ans: } \dfrac{\ln(1+\alpha)}{\ln 2}\right)\)

Before we calculate let's find fixed point of gauss map.

\[ \begin{aligned} \text{f.p.} \Rightarrow \quad x &= \frac{1}{x} - 1\\ x^2 + x - 1 &= 0 \qquad x = \frac{-1 \pm \sqrt{1+4}}{2}\\ &\qquad\qquad\quad\ = \frac{-1 \pm \sqrt{5}}{2} \end{aligned} \] \[ \text{So} \quad \text{f.p.} = \frac{\sqrt{5}-1}{2} \qquad \text{as} \qquad 0 < x < 1 \] \[ \begin{aligned} \to \quad x + 1 &= \frac{1}{x}\\ \to \qquad x &= \frac{1}{1+x} \end{aligned} \]

continued fraction can be written as

\[ x = \cfrac{1}{1 + \cfrac{1}{1 + \cfrac{1}{1 + \cdots}}} \]

For f.point\(_2\) , \(x = \dfrac{1}{x} - 2\)

\[ \begin{aligned} \Rightarrow \quad x^2 + 2x - 1 &= 0\\ x = \frac{-2 \pm \sqrt{4+4}}{2} \quad &\quad \frac{-2 \pm 2\sqrt{2}}{2} \quad \Rightarrow \quad -1 \pm \sqrt{2} \end{aligned} \] \[ \left(x = \sqrt{2} - 1\right) \] \[ \begin{aligned} \text{continued fraction} \Rightarrow \quad \frac{1}{x} &= x + 2\\ \Rightarrow \quad x &= \frac{1}{x+2}\\ \Rightarrow \quad x &= \cfrac{1}{2 + \cfrac{1}{2 + \cfrac{1}{2 + }}} \end{aligned} \]

Q.) what short of this no is?

\[ x = \cfrac{1}{1 + \cfrac{1}{2 + \cfrac{1}{1 + \cfrac{1}{2 + \cdots}}}} \]

Ans. It would be fixed point of first iterate of map or period 2 cycle.

Lecture 20: (classical statistical mechanics)

let us start with hamiltonian of a system. system contains many particles and is isolated from the rest of the universe.

Isolated system (exchange of matter & energy are prohibited).

\usetikzlibrary{arrows.meta,decorations.pathreplacing,calc,angles,quotes,positioning,patterns}
\begin{tikzpicture}[scale=1.0,>=Stealth]
  % outer blob
  \draw[thick] (0,0) .. controls (-0.5,1.2) and (0.1,2.6) .. (1.2,2.9)
        .. controls (2.3,3.2) and (3.1,2.3) .. (3.0,1.3)
        .. controls (2.9,0.3) and (2.0,-0.6) .. (1.0,-0.5)
        .. controls (0.4,-0.45) and (0.15,-0.4) .. (0,0);
  % inner squiggles
  \draw (1.4,2.1) .. controls (0.5,2.0) and (0.4,1.0) .. (1.0,0.6)
        .. controls (1.6,0.25) and (1.9,0.6) .. (1.6,0.9)
        .. controls (1.3,1.2) and (0.7,0.6) .. (1.1,0.1)
        .. controls (1.4,-0.25) and (2.0,-0.2) .. (2.1,0.15);
  \draw (1.5,2.2) .. controls (2.1,2.2) and (2.2,1.6) .. (1.8,1.3);
  % small tilted boxes
  \draw (1.05,2.45) -- (1.35,2.6) -- (1.2,2.3) -- (0.92,2.2) -- cycle;
  \draw (2.1,1.75) -- (2.4,1.9) -- (2.3,1.55) -- (2.0,1.45) -- cycle;
  \draw (1.2,1.75) -- (1.45,1.85) -- (1.35,1.55) -- (1.1,1.5) -- cycle;
  % particles
  \foreach \p in {(0.35,1.6),(0.55,0.9),(0.45,0.35),(0.8,-0.15),(1.6,-0.3),
                  (2.3,0.7),(2.6,1.5),(2.4,2.2),(1.8,2.6),(0.9,2.75),
                  (2.1,0.2),(0.25,2.0),(2.75,1.05),(1.9,2.05),(0.7,2.35)}
    \filldraw \p circle (0.035);
\end{tikzpicture}
\[ H(q,p) \]

If rules are time-translation invariant then we know that Hamiltonian is constant of motion. \(\left(\dfrac{dH}{dt} = 0\right)\)

\[ H(q,p) = E \quad \text{(total energy of system).} \] \[ \{\text{with uncertainty } \delta E\} \]

phase space becomes \(6n\)-diamentional for \(n\)-particles. So we need to know \(3n\) -- constants of motion which are in involution with each other. \(\{c_i\,c_j\} = 0\) \(i \neq j\).

constants of motion will be

  1. \(H\) 1
  2. total linear momentum \((P)\) 3
  3. Total angular momentum \((L)\) 3
  4. C.o.M position. \((\vec{R})\) 3

Total 10 COM

To specify state of system we need a point in phase space \((q_1\,q_2 \cdots q_n,\ p_1\,p_2 \cdots p_n)\). And that point will move on some \(6n\)-diamentional hypersurface.

Any way exponential seperation \(\Rightarrow\) mixing (implies.)

\[ \Downarrow \] \[ \text{ergodicity} \]

Ergodicity :--

Fundamental Assumption/postulate of equilibrium statistical mechanics.

"This single statement is enough to derive all equations of stat. mech, whereas thermal eq\(^n\) turns out to be a special case of it".

Thermal equilibrium :--

Two systems are said to be in thermal equilibrium if there is no heat flows between them when they are connected to each other.

A system is said to be in thermodynamic eq\(^m\) if it is in mechanical, physical and thermal equilibrium.

"All physical macroscopic quantities are time independent in system of thermal eq\(^m\) state."

Note :--

micro state :--

specifying the state of every constituent of the system. The set \((q_1 \cdots q_n,\ p_1 \cdots p_n)\) is microstate of the system.

or we can say that position in phase space is microstate where as the total (avg) momentum of all particles is macrostate.

By accessible microstate we restrict the velocity a particle can have, i.e.

\[ H(q,p) = E_{\text{total}} \]

so a particle of such isolated system can not have \(K.E > E\). So some micro states are not accessible anymore.

"All those micro states which are compatible with all restrictions applied on system are termed as accessible micro-state."

"microstate tells what each and every particle is doing and macrostate is giving some gross (overall) information."

If microstate energy is the only parameter of microstate \(-\) (In quantum stat mech)

"If a isolated system has 10J of total energy then we can have a single particle which has energy close to 10J and rest. of all are close to zero energy. This (microstate) is equally probable than other state; followed from postulate of eq\(^m\) stat. mech."

In phase-space probability density obeys

\[ \frac{\partial \rho}{\partial t} = \{H, \rho\} \]

\(\rho = \) const. of motion

\[ \frac{d\rho}{dt} = \{\rho, H\} + \frac{\partial \rho}{\partial t} = 0 \] \[ \frac{\partial \rho_{eq}}{\partial t} = \{H, \rho_{eq}\} \]

if probability distribution is independent of time

\[ \frac{\partial \rho_{eq}}{\partial t} = 0 = \{H, \rho_{eq}\} \Rightarrow \text{implies } \underline{\rho_{eq} = f(H)}. \]

remember \(\rho_{eq}\) is not physical observable.

\[ \rho_{eq} = \rho_{eq}(H) = \delta(H(q,p) - E) \]

since such isolated system uses microstates, such ensamble is termed as microcanonical ensamble. A stat. ensambles with \(\rho_{eq} = \delta(H(q,p)-E)\) in phase space with which corresponds to isolated system in thermal eq\(^n\) is called microcanonical ensamble

\(>\) If we assume we can not have a point in phase space but a volume shell of finite resolution then only we can define probability of occupying certain state (volume element in phase space (shelle).

\[ \Delta q \, \Delta p \geq \frac{\hbar}{2} \]

for 1 particle volume element

\[ (\Delta q_1 \Delta p_1)(\Delta q_2 \Delta p_2)(\Delta q_3 \Delta p_3) \] \[ \sim \hbar^3 \text{ or } h^3 \]

for \(N\) particle the volume element would be

\[ \sim h^{3N}. \]

If volume of accessible phase space is `\(M\)',

\[ \# \text{ of micro states} = \frac{M}{h^{3N}} = \Omega(E) \text{ (finite \& large)} \]

let probability of any microstate \(= P\) \(\sum_i P_i = 1\)

\[ P \cdot (\text{no. of microstates}) = 1 \] \[ \Rightarrow \quad \left\{ \; P = \frac{1}{\Omega} = \frac{h^{3N}}{M} \right. \]

Toy Model :--

set of \(N\) distinguishable (identifiable) coins. where each coin has \(\{H, T\}\) two states.

\usetikzlibrary{arrows.meta,decorations.pathreplacing,calc,angles,quotes,positioning,patterns}
\begin{tikzpicture}
\node[anchor=west] (s) at (0,0) {htt \; -- \; hhh \; --{}--\,tth \; ---\, hhh};
\node[anchor=east] at ([xshift=-0.25cm]s.west) {$\{2^N \text{ microstates}\}$};
\draw[-{Latex[length=2mm]}] ([xshift=0.15cm]s.east) -- ++(0.75,0);
\node[anchor=west] (ms) at ([xshift=1.05cm]s.east) {microstate};
\draw[decorate,decoration={brace,amplitude=6pt,mirror}]
  ([yshift=-2pt]s.south west) -- ([yshift=-2pt]s.south east);
\node[anchor=north] at ([yshift=-0.4cm]s.south) {`$N$' letters};
\node[anchor=north west,align=left] at ([yshift=-0.15cm]ms.south west)
  {(still a state\\ of system as\\ a whole)\\ but describes\\ what each object is\\ doing};
\end{tikzpicture}

Let,

Total no. of heads \(= H\)

of Tails \(= T\)

\[ H + T = N \]

Let,

\[ H - T = M \]

\((N,M)\) or \((N,H)\) or \((H,T)\) are macrostate of the system.

For given \(N\),

\[ \# \text{ of macrostates} = N+1. \]

ex for \(N = 3\) let \((N,H)\) represents macrostate

we have: \((3,0)\), \((3,1)\), \((3,2)\), \((3,3)\) 4 macrostates.

no. of microstates are much larger than no. of macrostates.

\[ 2^N > N+1 \] \[ N = d.o.f. \]

\(>\) probability of any possible/specific microstate would be \(\left(\frac{1}{2^N}\right)\).

\(>\) Probability of attaining a given macrostate :--

we have \(0 \leq H \leq N\)

\[ P(H) = {}^N C_H \cdot \left(\frac{1}{2}\right)^H \left(\frac{1}{2}\right)^{(N-H)} \]

Suppose coin is biased then

\[ \begin{aligned} P(h) &= p\\ P(t) &= 1-p = q \end{aligned} \]

then,

\[ P(H) = {}^N C_H \; p^H \, q^{N-H} \] \[ \sum_{H=0}^{N} P(H) = (p+q)^N = 1^N = 1 \qquad \text{binomial expansion.} \]

Generating function for the binomial distribution :--

\[ \begin{aligned} f(x) &= \sum_{H=0}^{N} P(H)\, x^H\\ &= (px+q)^N = \sum_{H=0}^{N} \left({}^N C_H \, p^H q^{N-H}\right) x^H \end{aligned} \] \[ P(H) = \text{coefficient of } x^H \text{ in expanded generating function} \]

Property of G.F :-

\[ f(1) = 1 \; ; \; (p+q)^N = 1 \] \[ \langle H \rangle = ? \qquad (\text{avg. no. of Heads}) \] \[ \langle H \rangle = \frac{\sum_{H=0}^{N} P(H)\, H}{\sum_{H=0}^{N} P(H)} = \frac{\sum_{H=0}^{N} P(H)\, H}{1} \]

to find \(\langle H \rangle = \sum_{H=0}^{N} P(H) H\) we differentiate G.F

\[ \begin{aligned} f'(x) &= \sum_{H=0}^{N} P(H) \cdot H\, x^{H-1}\\ f'(1) &= \sum_{H=0}^{N} P(H) \cdot H = \left(N (px+q)^{N-1} \cdot p\right)_{x=1} \end{aligned} \]

so,

\[ \langle H \rangle = \sum_{H=0}^{N} P(H) H = N (p+q)^{N-1} \cdot p = Np \] \[ \boxed{\langle H \rangle = Np} \] \[ \langle H^2 \rangle = ? \]

find

\[ f''(1) = \left(N p^2 (N-1) (px+q)^{N-2}\right)_{x=1} = p^2 N (N-1) \qquad \text{(1)} \] \[ f''(x)\Big|_{x=1} = \left(\sum_{H=0}^{N} P(H)\, H(H-1)\, x^{H-2}\right)_{x=1} = \begin{aligned}[t] &\langle H(H-1)\rangle\\ &= \langle H^2 - H \rangle\\ &= \langle H^2 \rangle - \langle H \rangle \qquad \text{(2)} \end{aligned} \]

Equating (1) and (2)

\[ p^2 N (N-1) = \langle H^2 \rangle - \langle H \rangle \] \[ \Rightarrow \quad \langle H^2 \rangle = p^2 N^2 - p^2 N + \langle H \rangle \] \[ \Rightarrow \quad \langle H^2 \rangle = p^2 N^2 - p^2 N + Np \]

Also, Now we can find variance,

variance \(=\) mean square \(-\) square of mean

\[ \begin{aligned} \mathrm{var}(H) &= \langle H^2 \rangle - \langle H \rangle^2 \qquad \left(= \langle (H - \langle H \rangle)^2 \rangle\right)\\ &= N^2 p^2 - p^2 N + Np - N^2 p^2\\ \mathrm{var}(H) &= Np(1-p)\\ \sigma^2 = \mathrm{var}(H) &= Npq \qquad , \text{ as, } (q = 1-p) \end{aligned} \]

Therefore \(\Delta H = \) std. deviation \(=\) uncertainty \(= \sqrt{Npq}\)

\[ \boxed{\Delta H = \sigma = \sqrt{Npq}} \]

Therefore relative fluctuation will be :--

\[ \text{fluctuation about mean} = \frac{\Delta H}{\langle H \rangle} = \frac{\sqrt{Npq}}{Np} = \sqrt{\frac{q}{Np}} \] \[ \text{Relative scatter} = \frac{1}{\sqrt{N}} \sqrt{\frac{q}{p}} \]

(dispersion about mean)

\(>\) It says relative fluctuation becomes smaller as no. of coin toss increases / no. of coins increases.

In thermodynamics we deal with \(\sim 10^{23}\) d.o.f so relative fluctuations becomes very small. This is the main reason "why thermodynamics works".

"In statistical mechanics generating function is called as partition function ( laplace transformation / green fn.)"

Note :--

out of 100 coin toss \(P(50H)\) or \(P(50T)\) is most probable. Because of \({}^{100}C_{50}\) is biggest for all \(H\) in \(\left({}^{100}C_H\right)\).

each microstate is equally probable, but each macrostate is made up of many microstates. so attaining 1H and 99T is having lesser no of micro-states then that of 50H, 50T So our example that any one molecule can have nearly all total energy and

Rest of all are nearly at rest; Indeed such micro-states are equally probable but the macrostate has contribution from very less no. of microstates so such macrostate is less likely to exist then other possible macrostates.

Lecture 21

\(N\) coins, \(H\) heads, \(T\) tails

\[ P(H) = {}^N C_H \; p^H q^{N-H} \]

\(p = \) prob of heads

\(q = \) prob of tails

\[ \begin{aligned} f(x) &= (px+q)^N\\ \langle H \rangle &= Np\\ \Delta H &= \sqrt{Npq} \end{aligned} \] \[ \begin{aligned} f(x) &= \sum_{n=0}^{N} P(n)\, x^n\\ &= \sum {}^N C_n \, p^n q^{N-n} x^n\\ &= (px+q)^N \end{aligned} \]

What does this distribution looks like.

\usetikzlibrary{arrows.meta,decorations.pathreplacing,calc,angles,quotes,positioning,patterns}
\begin{tikzpicture}[scale=0.9]
\draw[-{Latex[length=2mm]}] (0,0) -- (9.2,0);
\draw[-{Latex[length=2mm]}] (0.9,-0.6) -- (0.9,4.3);
\node[anchor=east] at (0.75,2.4) {$P(H)$};
\node[anchor=north] at (9.2,-0.1) {$H$};
\foreach \x/\h in {1.35/0.35, 1.75/0.8, 2.35/1.9, 2.7/2.2, 3.0/2.5, 3.3/2.7, 3.65/3.9, 4.1/2.6, 4.45/3.2, 4.85/2.4, 5.3/2.0, 5.9/1.5, 6.35/1.7, 6.75/1.05, 7.2/0.75, 7.7/0.6}{
  \draw (\x,0) -- (\x,\h);
  \fill (\x,\h) circle (1.1pt);
}
\draw (3.9,-0.12) -- (3.9,0.12);
\node[anchor=north] at (3.9,-0.15) {$N/2$};
\draw (7.0,-0.12) -- (7.0,0.12);
\node[anchor=north] at (7.0,-0.15) {$N$};
\end{tikzpicture}
\[ P(H) = \frac{N!}{H! \, (N-H)!} \; p^H q^{N-H} \]

we would like to find out what does \({}^N C_H\) do at large no. of `\(N\)'

What is \(N!\) (for large \(N\)) :--

Stirling formula :--

\[ \begin{aligned} n! &= n(n-1)(n-2) \; -- \; 3.2.1\\ &= n^n \left(1 - \tfrac{1}{n}\right)\left(1 - \tfrac{2}{n}\right) \; ---\\ &= n^n e^{-n} \sqrt{2\pi n} \cdot \left\{ 1 + \tfrac{1}{12n} + \cdots \right\} \end{aligned} \] \[ 1! = 1^1 e^{-1}\sqrt{2\pi} \qquad \text{so \quad is } e \simeq \sqrt{2\pi} \; ? \]

\(n!\) is defined as

\[ \int_0^{\infty} dx \; e^{-x} x^n = \Gamma(n+1) \quad (n = 1,2,3 \cdots) \] 
\usetikzlibrary{arrows.meta,decorations.pathreplacing,calc,angles,quotes,positioning,patterns}
\begin{tikzpicture}[scale=0.9]
\useasboundingbox (-0.6,-1.0) rectangle (7.2,3.9);
\clip (-0.6,-1.0) rectangle (7.2,3.9);
\draw (-0.4,0) -- (6.6,0);
\draw (0.55,-0.9) -- (0.55,3.4);
\node[anchor=east] at (0.5,2.55) {$1$};
\draw (0.42,2.55) -- (0.68,2.55);
% e^{-x} decaying curve
\draw[thick] (0.55,2.55) .. controls (1.6,1.2) and (2.6,0.45) .. (4.4,0.12);
\node[anchor=south west] at (1.05,2.15) {$e^{-x}$};
% x^n rising curve
\draw[thick] (0.55,0.02) .. controls (2.4,0.12) and (3.6,0.9) .. (4.6,3.3);
\node[anchor=west] at (4.35,3.5) {$x^n$};
% product: peaked bell curve
\draw[thick] (1.1,0.03) .. controls (2.0,0.1) and (2.2,2.6) .. (2.75,2.6)
  .. controls (3.3,2.6) and (3.5,0.35) .. (4.6,0.08) -- (5.6,0.03);
\node[anchor=north] at (3.3,-0.1) {$x$};
\end{tikzpicture}
\[ g(x) = e^{-x} x^n = e^{-x} e^{\ln x^n} = e^{-x + n \log_e x} = e^{-(x - n \ln x)} \] \[ g'(x) = e^{-(x - n\ln x)} \cdot (-1)\left(1 - \frac{n}{x}\right) = 0 \Rightarrow \boxed{g'(n) = 0} \]

let, \(f(x) = x - n\ln x\)

\[ f(n) = n - n\ln n \] \[ \begin{aligned} f'(x) &= 1 - \frac{n}{x}\\ f''(x) &= 0 + \frac{n}{x^2} \end{aligned} \] \[ f(x) = f(n) + f'(n)(x-n) + \frac{f''(n)(x-n)^2}{2!} + \cdots \] \[ f(x) = (n - n\ln n) + 0 + \frac{(x-n)^2}{2n} + \cdots \] \[ n! = \int_0^{\infty} e^{-x} x^n \, dx = \int_0^{\infty} e^{-(x - n\ln x)} \] \[ \Rightarrow \quad n! = \int_0^{\infty} e^{-\left((n - n\ln n) + \frac{(x-n)^2}{2n} + \cdots\right)} dx \] \[ = \int_0^{\infty} \left(e^{-n} \cdot n^n\right) \cdot e^{-\frac{(x-n)^2}{2n}} \left\{ 1 + \cdots \right) dx \]

for very large `\(n\)'

\[ n! = e^{-n} n^n \int_{-\infty}^{\infty} dx \; e^{-\frac{(x-n)^2}{2n}} \, dx \] \[ = e^{-n} n^n \sqrt{2\pi n} \cdot \left\{ 1 + \text{Order}\left(\tfrac{1}{n}\right) \right\} \]

\(\displaystyle \int_{-\infty}^{\infty} e^{-\frac{t^2}{a}} dt = \sqrt{\pi a}\)

Gaussian integral

\[ I = \int_{-\infty}^{\infty} e^{-ax^2} dx = \sqrt{\pi/a} \]

also

\[ I = \int_{-\infty}^{\infty} e^{-ay^2} dy \] \[ I^2 = \int_{-\infty}^{\infty}\int_{-\infty}^{\infty} e^{-a(x^2+y^2)} \, dx\, dy \qquad ; \left\{ \begin{aligned} &x^2 + y^2 = r^2\\ &x = r\cos\theta\\ &y = r\sin\theta\\ &dA = (r\, d\theta)\, dr \end{aligned} \right. \] \[ = \int\int e^{-ar^2} \; r\, dr \, d\theta \] \[ = \int_{-\infty}^{\infty} e^{-ar^2} r\, dr \int_0^{\pi} d\theta = \left(\sqrt{\pi/a}\right)^2 \] \[ \left( I = \sqrt{\pi/a} \right) \]

so for large \(N\)

\usetikzlibrary{arrows.meta,decorations.pathreplacing,calc,angles,quotes,positioning,patterns}
\begin{tikzpicture}[scale=0.9]
\useasboundingbox (-0.4,-1.2) rectangle (8.0,3.9);
\clip (-0.4,-1.2) rectangle (8.0,3.9);
\draw[-{Latex[length=2mm]}] (-0.3,0) -- (7.4,0);
\draw[-{Latex[length=2mm]}] (1.35,-1.1) -- (1.35,3.5);
\node[anchor=east] at (1.2,2.3) {$P(H)$};
\node[anchor=north] at (5.3,-0.15) {$H$};
\draw[thick] (0.1,0.02) .. controls (1.2,0.05) and (1.6,0.7) .. (2.3,1.6)
  .. controls (2.75,2.3) and (2.9,3.35) .. (3.35,3.35)
  .. controls (3.8,3.35) and (3.95,1.6) .. (4.5,0.9)
  .. controls (5.1,0.2) and (5.8,0.1) .. (6.9,0.05);
\end{tikzpicture}

Take ideal gas contained in volume \(V\).

\usetikzlibrary{arrows.meta,decorations.pathreplacing,calc,angles,quotes,positioning,patterns}
\begin{tikzpicture}[scale=0.9]
\draw[thick] (2.0,3.6) .. controls (3.4,3.8) and (4.2,2.9) .. (4.15,1.9)
  .. controls (4.1,0.7) and (3.4,-0.2) .. (2.3,-0.2)
  .. controls (1.1,-0.2) and (0.35,0.7) .. (0.45,1.8)
  .. controls (0.55,2.9) and (1.1,3.5) .. (2.0,3.6);
\draw[thick] (2.0,3.6) .. controls (1.75,3.35) and (1.6,3.25) .. (1.55,3.15);
\node at (2.25,2.55) {$N, V$};
\draw (1.15,0.50) -- (1.78,0.68) -- (1.92,0.16) -- (1.29,-0.02) -- cycle;
\draw (1.62,0.62) -- (1.88,1.18);
\node[anchor=south west] at (1.78,1.12) {$n, v$};
\end{tikzpicture}

\(N = \) total no. of particles.

What is probability that \(v\) contains \(n\) particles?

\[ P(n) = \text{Probability that `}v\text{' contains } n \text{ particle} \]

Probability that a given particle is in volume `\(v\)' \(= \dfrac{v}{V}\)

assume that all the particles are moving independent of each other, therefore the probability that there are \(n\)' of them inside \(v\)' is

\[ = {}^N C_n \left(\frac{v}{V}\right)^n \left(1 - \frac{v}{V}\right)^{N-n} \]

Here `\(n\)' is random variable whose no. can vary from \((0 \leq n \leq N)\).

\[ \text{here } p = \frac{v}{V}, \qquad q = 1 - \frac{v}{V} \]

this is the binomial distribution.

lets put \(\rho = \dfrac{N}{V}\)

\[ \Rightarrow \quad \frac{1}{V} = \frac{\rho}{N} \]

so,

\[ P(n) = {}^N C_n \left(\frac{v\rho}{N}\right)^n \left(1 - \frac{v\rho}{N}\right)^{N-n} \qquad \left(\begin{array}{l}\text{binomial}\\ \text{distribution}\end{array}\right) \]

What happens to \(P(n)\), when \(N \to \infty\) and \(V \to \infty\) but keeping `\(\rho\)' fixed.

\[ P(n) = \xrightarrow{\;N \to \infty\;} \frac{N!}{n! \, (N-n)!} \left(\frac{v\rho}{N}\right)^n \left(1 - \frac{v\rho}{N}\right)^{N-n} \] \[ = \frac{N^N e^{-N} \sqrt{2\pi N}}{\left(n^n e^{-n}\sqrt{2\pi n}\right) (N-n)^{N-n} e^{-(N-n)} \sqrt{2\pi (N-n)}} \left(\frac{\rho v}{N}\right)^n \left(1 - \frac{\rho v}{N}\right)^{N-n} \]

$\left(\begin{array}{l}\text{we can't}\\ \text{use } n! = e{-n} nn \sqrt{2\pi n}\\ \text{as `}n\text{' can take 0 too } \infty\end{array}\right)\( \)(n!)$

\[ \boxed{\rho v = \bar{n}} \] \[ P(n) \longrightarrow \; = \frac{e^{-\bar{n}} \, (\bar{n})^n}{(n!)} \qquad \text{where } n = 0,1,2,3 \cdots \infty \]

and \(\bar{n} = \rho v \; = \) avg. no of particles in sub volume `\(v\)'.

\[ \left(\bar{n} = Np = N\frac{v}{V}\right) = \left(\bar{n} = \rho v\right) \]

we have assumed that particles behave like newtonian particle mech. and is distinguishable from one another.

In \(\left(P(n) = \dfrac{e^{-\rho v} (\rho v)^n}{n!}\right)\) formula, \((N,V)\) don't appear anymore, as \((N, V \to \infty)\) Such that their ratio \(\rho = \frac{N}{V}\) is finite (this is called thermodynamic limit of system when \(N \to \infty\), \(V \to \infty\) such that \(\rho = \frac{N}{V}\) is finite). Statistical mech. Reduce to thermodynamics in thermodynamic limit.

Poisson Distribution :-

for random var. \(n = 0, 1, 2 \cdots \infty\)

\[ P(n) = \frac{e^{-\lambda}\,\lambda^{n}}{n!} \qquad (\lambda = \langle n\rangle) \]

\(\lambda\) is avg value of random variable

Generating function

\[ f(x) = \sum_{n=0}^{\infty} P(n)\, x^{n} = e^{\lambda(x-1)} \qquad (\text{How ?}) \] \[ f(1) = 1 \quad \text{for} \quad \sum_{n=0}^{\infty} P(n) = \underline{1} \]

varience \(= \langle (n - \langle n\rangle)^{2}\rangle = \lambda = \langle n\rangle = \langle n^{2}\rangle - \langle n\rangle^{2}\)

In poisson distribution varience = mean

\[ \text{Relative fluctuation} = \frac{\sigma}{\langle n\rangle} = \frac{\Delta n}{\langle n\rangle} = \frac{\sqrt{\langle n\rangle}}{\langle n\rangle} = \frac{1}{\sqrt{\langle n\rangle}} \]

for poisson's distribution, \(\langle n\rangle = \lambda\), \((\Delta n)^{2} = \lambda\).
& all higher moments \(= \lambda\)., so its a single parameter
distribution;

In gaussian distribution except for \(\langle n\rangle\) & \((\Delta n)^{2}\)
all higher moments vanish.

\(\bar{n} = \) avg. no of particles in subvolume `v'.

\usetikzlibrary{arrows.meta,decorations.pathreplacing,calc,angles,quotes,positioning,patterns}
\begin{tikzpicture}[scale=0.9]
  \draw[->] (0,0) -- (0,3.4) node[left] {$P(n)$};
  \draw[->] (0,0) -- (7.2,0) node[below right] {$n$};
  % dashed envelope
  \draw[dashed] (0.35,2.55) .. controls (0.7,0.35) and (1.3,0.6) .. (2.0,1.6)
     .. controls (2.6,2.5) and (3.0,3.0) .. (3.6,3.0)
     .. controls (4.4,3.0) and (5.0,1.2) .. (6.3,0.35);
  % spikes
  \draw (0.35,0) -- (0.35,2.55);
  \draw (0.75,0) -- (0.75,0.45);
  \draw (1.15,0) -- (1.15,0.75);
  \draw (1.55,0) -- (1.55,1.15);
  \draw (2.0,0) -- (2.0,1.6);
  \draw (2.45,0) -- (2.45,2.25);
  \draw (2.9,0) -- (2.9,2.75);
  \draw (3.6,0) -- (3.6,3.0);
  \draw (4.2,0) -- (4.2,2.6);
  \draw (4.8,0) -- (4.8,1.85);
  \draw (5.9,0) -- (5.9,0.6);
  \node[below] at (2.9,-0.05) {$\bar{n}$};
\end{tikzpicture}

if \(\bar{n}\) itself become very large then deviation from mean
\((n - \bar{n})\) takes gaussian curve. and call that a continuous
variable if \(\bar{n}\) is very large.

from

\[ P(n) = \frac{e^{-\bar{n}}\,(\bar{n})^{n}}{n!} \]

using sterling formula again, and replacing variable to

\[ x = n - \bar{n} \] \[ P(x) \alpha \; \frac{e^{-\frac{x^{2}}{2\sigma^{2}}}}{\sqrt{[?]}} \qquad (\text{Gaussian distribution}) \]

" The whole point of above disscussion is that if we start with a
binomial distribution of burnoulli trials and then from that we tke
no. of particles / trials to be very large such that probability of
succes in trial is vanishingly small. Then Binomial distribution goes
over into poisson distribution. Then if mean value of poisson
distribution is very large compared to unity, The deviation from the
mean \((n - \bar{n})\) is approximatly a contineous variable that has a
Gaussian shape".

So A probability distribution can shifts over to another probability
distribution.

Generating function :- \(f(x) = \sum\limits_{n=0}^{\infty} P(n) x^{n} = P(0)x^{0} + P(1)x^{1} + P(2)x^{2} + \cdots P(n)x^{n}\)

\[ f(x) = \sum_{n=0}^{\infty} P(n)\, x^{n} = \sum_{n=0}^{\infty} \frac{e^{-\lambda}\lambda^{n}}{n!}\, x^{n} \Rightarrow e^{-\lambda} \sum \frac{(\lambda x)^{n}}{n!} = e^{-\lambda} e^{\lambda x} \] \[ \boxed{f(x) = e^{\lambda(x-1)}} \]

lets go back to our problem. of isolated system in thermal equilibrium.

\usetikzlibrary{arrows.meta,decorations.pathreplacing,calc,angles,quotes,positioning,patterns}
\begin{tikzpicture}[scale=0.95,>=Stealth]
  % blob
  \draw (0.6,3.0) .. controls (0.5,4.1) and (1.6,4.6) .. (2.4,4.45)
        .. controls (3.2,4.3) and (3.9,3.7) .. (3.9,3.0)
        .. controls (3.9,2.3) and (3.0,2.1) .. (3.1,1.4)
        .. controls (3.2,0.7) and (2.4,0.2) .. (1.6,0.4)
        .. controls (0.9,0.6) and (0.7,1.4) .. (0.7,2.0)
        .. controls (0.7,2.5) and (0.65,2.7) .. (0.6,3.0);
  % dashed partition
  \draw[dashed] (1.75,4.65) -- (1.6,3.6) -- (1.75,2.4) -- (1.5,1.3) -- (1.65,0.15);
  \node at (1.15,3.55) {$A$};
  \node at (2.35,3.35) {$B$};
  \node at (1.05,2.35) {$N,V,E$};
  \node at (2.55,2.35) {$N',V',E'$};
  % dots
  \foreach \p in {(1.0,3.0),(1.3,1.6),(0.95,1.2),(2.1,4.0),(2.6,3.9),(2.9,2.9),
                  (2.3,1.8),(2.6,1.1),(2.0,0.8),(1.2,0.9),(3.3,3.3),(2.0,2.9)}
     \fill \p circle (0.03);
  % arrows to label
  \draw[->] (1.9,4.75) -- (3.9,4.95);
  \draw[->] (4.3,4.85) -- (2.6,4.35);
  \node[anchor=west] at (4.0,5.0) {$N_{tot},\; V_{tot},\; E_{Tot.},\; \Omega_{tot}$};
  \draw[->] (7.9,5.35) -- (7.6,5.12);
  \node[anchor=west] at (7.6,5.5) {no of accessible microstate.};
\end{tikzpicture}

Imagine the system is made of two sub-systems, in thermal eq\(^{\rm m}\)
with each other, such that

\[ \begin{aligned} N + N' &= N_{Tot}\\ V + V' &= V_{Tot}\\ E + E' &\simeq E_{tot} \end{aligned} \]

(Isolated, in thermal eq\(^{\rm m}\))

not necessarily

\((E + E' \neq E_{Tot}\) always if pot. energy plays role\()\)

\(E + E' \simeq E_{Tot}\) to a very good approximation.

"as no. of d.o.f which are interecting near partition is
\(\left(\frac{1}{10^{8}}\right)^{th}\) of the total d.o.f".
(assuming short range forces").

Then at any instant of time, what is the probability that Energy of
\(A\) is \(E\).; \(P(E)\)

\(P(E) = \) no. of microstate of this entire system such that \(A\) has
energy \(E\). \((\Omega(E))\) (also (no. of microstates such that \(B\) has
energy \(E'\).)

\[ = \frac{\Omega(E)\,\Omega'(E')}{\Omega_{tot}(E_{tot})} \qquad \text{(normalization)} \]

\(\Omega(E)\) & \(\Omega'(E')\) can have totally different kind of functions.

As \(A\) can be a large container of oil and \(B\) can be our atmosphere.
so \(A, B\) have different d.o.f.

Lecture 22

Note that Energy of \(A\) and \(B\) are not fixed. Because there are
fluctuations.

\(E, V, N, N', V', E'\). are variable quantities.

what we were trying is to find, is

"Given the postulate of equal apriori probabilities of all the
accessible microstates of the total system, what can we say about
probability distribution of (for instance say) Energy in system \(A\)."

\(P(E) = \)

\[ \frac{\text{\# of microstates of tot. system such that }A\text{ has energy }E\text{. (\& }B\text{ has }E'\text{.)}} {\text{Total no. of microstates of total system such that T.E }= E_{tot}\text{.}} \]

If \(A\) & \(B\) are assumed as independent system (they actually are not)
then \(\Omega_{tot}(E) = \Omega(E).\Omega'(E')\)

\[ = \frac{\Omega(E) \cdot \Omega'(E')}{\Omega_{tot}(E_{tot}).} \]

\(P(A \& B) = P(A).P(B)\) only iff \(A, B\) are independent events

In above discussion we have assumed that interection Energy is fairly
negligible.

\(\Omega(E)\) is actually a fun. of \(V, N\) also

Remember always

\[ \begin{aligned} \Omega(E) &= \Omega(E,V,N)\\ \Omega'(E') &= \Omega'(E',V',N') \end{aligned} \]

lets see how big these no. are :

Suppose we take a single particle & put that into volume \(V\) and the
particle is moving around with some energy \(\leq E\). lets calculete
how many microstates it has.

\usetikzlibrary{arrows.meta,decorations.pathreplacing,calc,angles,quotes,positioning,patterns}
\begin{tikzpicture}[scale=0.9,>=Stealth]
  \draw[thick] (0,0) rectangle (2.2,1.4);
  \fill (0.15,0.05) circle (0.05);
  \fill (2.05,0.05) circle (0.05);
  \fill (0.15,1.35) circle (0.05);
  \fill (2.05,1.35) circle (0.05);
  \node at (1.75,1.0) {$V$};
  \draw (0.5,0.35) .. controls (0.3,0.9) and (1.0,1.1) .. (1.15,0.75)
        .. controls (1.25,0.45) and (0.55,0.5) .. (0.6,0.25);
  \draw[->] (0.6,0.25) -- (0.95,0.15);
  \fill (0.95,0.75) circle (0.04);
\end{tikzpicture}

No. of microstates is the no. of shells in phase-space

true in any dimensions \(\longleftarrow\) \(\left(E = \dfrac{p^{2}}{2m}\right)\)
\(|p| \leq \sqrt{2mE}\)

let no. of such microstates be \(\phi(E) = \dfrac{\text{phase space volume}}{\text{min volume}(h^{3})}\)

\[ \phi(E) = \frac{1}{h^{3}} \int_{V} d^{3}q \int_{|p| \leq \sqrt{2mE}} d^{3}p \qquad \to \text{phase space volume} \]

\(d^{3}p = 4\pi p^{2} dp\) \; volume element

\[ = \frac{V}{h^{3}} \int_{0}^{\sqrt{2mE}} 4\pi\, p^{2}\, dp \; \alpha \; \frac{V}{h^{3}}\, E^{3/2} \qquad \to \text{no of dimensions} \]

\((p\, \alpha\, E^{1/2})\).

\[ \begin{aligned} &= \frac{V}{h^{3}} \frac{4\pi}{3} \left. \left(p^{3}\right)\right|_{0}^{\sqrt{2mE}}\\ &= \frac{V}{h^{3}} \frac{4\pi}{3} \cdot E^{3/2}\, 2m^{3/2} \end{aligned} \]

relation b/w \(E\) & \(p\) for a free particle is

\[ E^{2} = c^{2}p^{2} + m^{2}c^{4} \]

\(m = \) rest mass always

mass does not increase with speed, its just \(p = \gamma m v\) instead
of \(mv\). \(\left(\gamma = \frac{1}{\sqrt{1 - v^{2}/c^{2}}}\right)\)

\[ E = \left(c^{2}p^{2} + m^{2}c^{4}\right)^{1/2} = c\left(p^{2} + m^{2}c^{2}\right)^{1/2} \]

If \(p \ll mc\) \(\{ mc = \) compton momentum

In `n' dimentions, \((n-1)\) angles runs from (\(0-\pi\) or \(0-2\pi\)).

In \(n\)-spacial dimen.

\[ \phi(\varepsilon) \; \alpha \int_{0}^{\sqrt{2mE}} p^{n-1}\, dp \;\; \alpha \;\; \varepsilon^{n/2} \] \[ \begin{aligned} E &= mc^{2}\left(1 + \frac{p^{2}}{m^{2}c^{2}}\right)^{1/2}\\ &= mc^{2}\left(1 + \frac{p^{2}}{2m^{2}c^{2}} - \frac{1}{8}\frac{p^{4}}{m^{4}c^{4}} + \cdots \right)\\ &= mc^{2} + \frac{p^{2}}{2m} + \cdots \qquad = \text{origin of } \frac{p^{2}}{2m} \end{aligned} \]

No of microstates with energy in \((\varepsilon, \varepsilon + d\varepsilon) \sim \varepsilon^{1/2} d\varepsilon = d\Phi(\varepsilon)\)
(given energy shell) \(p = \sqrt{2m\varepsilon}\)

\usetikzlibrary{arrows.meta,decorations.pathreplacing,calc,angles,quotes,positioning,patterns}
\begin{tikzpicture}[scale=0.9,>=Stealth]
  \draw (0,0) circle (1.1);
  \draw[dashed] (0,0) circle (1.28);
  \draw[->] (0,0) -- (-0.78,0.78);
  \node[anchor=east] at (-0.72,0.62) {$p \sim \varepsilon^{1/2}$};
  \draw[<->] (1.1,0.35) -- (1.28,0.42);
  \node[anchor=west] at (1.3,0.45) {$d\varepsilon$};
  \draw (0,0) -- (1.1,0);
\end{tikzpicture}
\[ \begin{aligned} dV &= 4\pi p^{2}\, dp\\ &= 4\pi (2m)\varepsilon \frac{(2m)}{2\sqrt{2m\varepsilon}}\\ dv &\simeq 8\pi \sqrt{2}\, m^{3/2} \varepsilon^{1/2}\, d\varepsilon \end{aligned} \] \[ \begin{aligned} \Omega(\varepsilon)\, d\varepsilon &= \frac{\int d^{3}q \cdot \int d^{3}p}{h^{3}}\\ &= \frac{V \cdot 4\pi p^{2}\, dp}{h^{3}}\\ &= \frac{V}{h^{3}} \cdot 4\pi \cdot 2m\varepsilon \cdot \sqrt{2m}\, \frac{1}{2\sqrt{\varepsilon}}\, d\varepsilon \end{aligned} \] \[ \omega(\varepsilon) \sim \varepsilon^{1/2} d\varepsilon = g(\varepsilon)\, d\varepsilon \]

Density of states \((g(\varepsilon) = \varepsilon^{1/2}) = g_{i}\)
(degeneracy at Energy \(\varepsilon\)).

{no. of microstates per unit energy interval at energy \(\varepsilon\)}

*for \(n\)-D* : \(V \; \alpha \int p^{n-1} dp\)
\(\left(\Omega\, \alpha\, \varepsilon^{n/2}\right)\) for \((n \geq 2)\).

case

for 2-D container \(dv = 2\pi p\, dp = 2\pi \sqrt{2m\varepsilon}\, \dfrac{2m}{2\sqrt{\varepsilon}}\)

\[ = \text{Konstant} \]

density of states becomes constant. in 2-D

case for 1-D

\[ \phi(\varepsilon) = \frac{d}{d\varepsilon} \left(\frac{V}{h^{3}}\, \varepsilon^{1/2}\right) = \frac{1}{2}\, \varepsilon^{-1/2}\, \frac{V}{h^{3}} \]

Density of states \(\uparrow\) as \(\varepsilon \downarrow\)

\usetikzlibrary{arrows.meta,decorations.pathreplacing,calc,angles,quotes,positioning,patterns}
\begin{tikzpicture}[scale=0.95,>=Stealth]
  \draw[->] (0,-0.3) -- (0,4.6);
  \draw[->] (-0.9,0) -- (7.4,0) node[below] {$\varepsilon$};
  \node[anchor=south] at (-0.4,4.2) {$\phi(\varepsilon)$};
  \node[anchor=north] at (-0.4,3.9) {No of microstates};
  \draw (-0.15,0.65) -- (0.15,0.65) node[left,xshift=-6mm] {$1$};
  \draw (-0.15,1.2) -- (0.15,1.2) node[left,xshift=-6mm] {$2$};
  % staircase
  \draw (0,0.12) -- (1.6,0.12) -- (1.6,0.65) -- (2.5,0.65) -- (2.5,1.2)
        -- (3.4,1.2) -- (3.4,1.9) -- (4.3,1.9) -- (4.3,2.75)
        -- (5.0,2.75) -- (5.0,3.5) -- (5.5,3.5) -- (5.5,4.05)
        -- (5.85,4.05) -- (5.85,4.45);
  % smooth classical curve
  \draw[domain=0:6.1,samples=60,smooth] plot (\x,{0.28*pow(\x,1.5)});
  \draw (6.3,4.45) -- (5.95,4.35);
  \node[anchor=west] at (6.35,4.4) {classicl};
  \draw (6.3,3.7) -- (5.6,3.6);
  \node[anchor=west] at (6.35,3.7) {quantum mech. states};
  \draw (6.55,2.6) circle (0.55) node {$\varepsilon^{3/2}$};
  \draw (6.0,2.55) -- (4.3,2.2);
  \node[anchor=north] at (1.6,-0.05) {$\varepsilon_{qm}$};
  \node[anchor=north,align=center] at (2.6,-0.35) {1st\\excited\\state};
\end{tikzpicture}

As \(\varepsilon \uparrow\) no of states increases [3-D case]

(microscopic)

Now, we want to find \(\Omega(E)\) so we will like to assume again that
particles are free and

\[ E = E_{1} + E_{2} + E_{3} + \cdots E_{N} \]

Sum of energy of indivisual particles = total energy

For any two particles with `3' states, together they occupy \(3^{2} = 9\)
states.

\[ \Omega(E) \; \alpha \; \varepsilon^{\alpha N} \; \alpha \; E^{\alpha N} \]

{ if one particle can take \(\omega(\varepsilon)\) states \(= k\varepsilon^{1/2}\)
with energy \(\varepsilon\) & \(\varepsilon + d\varepsilon\)

indepandant \(N\) particles will take ---
\(k \varepsilon^{(1/2)N}\) }

if all \(\varepsilon_{i}\) are of same order of megnitude \(\varepsilon\)
} approximation

\[ E = N\varepsilon \]

So, \(\Omega(E) \; \alpha \; \varepsilon^{\alpha N} \; \alpha \; \left(\dfrac{E}{N}\right)^{\alpha N}\).

for very large \(N\), \(\Omega(E)\) is astronomically large no.

\[ \alpha = \tfrac{3}{2} \; \text{or} \; \tfrac{1}{2} \; \text{or} \; \tfrac{5}{2} \cdots \]

Lecture 22 (time 30:15)

Note

So the system which has largest no of microstates, that macrostate is
the most probable state.

For Most probable macrostate it implies;

\[ \frac{\partial P(E)}{\partial E} = 0 \qquad\qquad P(E) = \frac{\Omega(E)\, \Omega'(E')}{\Omega_{Tot}(E_{Tot})} \]

as \(P(E)\) is very large, lets take \(\log P(E)\) & differentiate.

\[ \frac{\partial}{\partial E}\left(\ln P(E)\right) = 0 \]

Probability

\[ dP = \frac{\partial P}{\partial E} dE + \frac{\partial P}{\partial V} dv + \frac{\partial P}{\partial N} dN \]

for eq\(^{\rm m}\)

\[ \left(\frac{\partial P(E)}{\partial E} = \frac{\partial P}{\partial V} = \frac{\partial P}{\partial N} = 0\right) \] \[ \frac{\partial \left(\ln \Omega(E)\right)}{\partial E} + \frac{\partial}{\partial E}\left(\ln \Omega'(E')\right) = 0 \] \[ E + E' = E_{tot} \] \[ E' = E_{tot} - E \qquad \Rightarrow \qquad dE' = -dE \]

there fore

\[ \boxed{\frac{\partial \ln\left(\Omega(E)\right)}{\partial E} = \frac{\partial \ln \Omega'(E')}{\partial E'}} \]

cond\(^{\rm n}\) for thermal eq\(^{\rm m}\) / most probable macrostate.

\(A, B\) two different kind of matter fluid, (gas - liquid).

\[ \frac{\partial}{\partial E}\left(\ln \Omega(E)\right) = \frac{\partial}{\partial E'}\left(\ln \Omega'(E')\right) = [E^{-1}] \; \to \text{dimensions} \]

or,

some property of \(A\) = some property of \(B\) = \(\dfrac{1}{k_{B}\,\text{Temp}}\)
} Imp.

Temperature

\(\left(dE = TdS - Pdv + \mu dN \right.\)

\(\left. T = \dfrac{\partial E}{\partial S}\bigg|_{V,N} \; ; \; S = k_{B} \ln \Omega \right)\).

lets define fun. \(\beta(E)\) so for this property of \(A/B\).

\[ \boxed{\beta(E) = \frac{\partial \ln \Omega(E)}{\partial E}} = \frac{1}{k_{B}\, T(E)} \]

By def\(^{\rm n}\) in equilibrium

\[ \boxed{T_{A} = T_{B}} \]

" Any two parts have same temp in large isolated system under thermal
equilibrium". (large d.o.f).

" ofcourse each part must be large enough to talk about temp, as
temp. is not defined for a single particle".

If \(\Omega(E)\) is not strictly increasing function and if so

\[ \frac{\partial}{\partial E} \ln\left(\Omega(E)\right) < 0 \Rightarrow \text{negative temperature.} \] 
\usetikzlibrary{arrows.meta,decorations.pathreplacing,calc,angles,quotes,positioning,patterns}
\begin{tikzpicture}[scale=0.9,>=Stealth]
  \node[anchor=east] at (-0.15,1.3) {for $\ln \Omega(E)$};
  \draw[->] (0,0) -- (4.6,0);
  \node[anchor=north] at (1.6,-0.1) {$E$};
  \draw (0.15,0.05) .. controls (0.9,0.15) and (1.1,1.4) .. (1.75,1.45)
        .. controls (2.4,1.5) and (2.6,0.35) .. (3.3,0.1);
  \draw (1.15,0) -- (1.15,1.15);
  \draw (1.75,0) -- (1.75,1.45);
  \node at (1.45,0.35) {$(T>0)$};
  \draw[->] (2.4,0.55) -- (3.4,0.15);
  \node[anchor=west] at (3.4,0.1) {$(T<0)$};
  \draw[->] (2.4,1.6) -- (3.4,1.75);
  \node[anchor=west] at (3.4,1.8) {$(-ve)$ slope implies negative temp.};
\end{tikzpicture}

So possibility of \((-ve)\) temp is built in our def\(^{\rm n}\) of temp.

since \(\Omega = \Omega(N,V,E)\) from
\(\left(\dfrac{\partial \ln \Omega}{\partial V} = 0\right)\) &
\(\left(\dfrac{\partial \ln \Omega}{\partial N} = 0\right)\)

\[ \frac{\partial \ln \Omega}{\partial V} = \frac{\partial \ln \Omega'}{\partial V'} \equiv \left(\frac{P_{A}}{T_{A}} = \frac{P_{B}}{T_{B}}\right) \Rightarrow \left(P_{A} = P_{B}\right) \] \[ \frac{\partial \ln \Omega}{\partial N} = \frac{\partial \ln \Omega'}{\partial N'} \equiv \Rightarrow \left(\frac{\mu_{A}}{T_{A}} = \frac{\mu_{B}}{T_{B}}\right) \Rightarrow \left(\mu_{A} = \mu_{B}\right) \]

(in thermodynamic terms)

\[ \begin{aligned} E &= TS - PV + \mu N\\ S &= \frac{E}{T} + \frac{PV}{T} - \frac{\mu N}{T} \end{aligned} \] \[ \left(\frac{\partial S}{\partial N} = \frac{-\mu}{T}\right) \; ; \; \left(\frac{\partial S}{\partial V} = \frac{P}{T}\right) \]

Entropy

\[ S(E,V,N) = k_{B} \ln\left(\Omega(E,V,N)\right) \]

Thermodynamics in Nutshell :-

*1\(^{st}\) law of thermodynamics :-*

\[ dQ = dU + dW \]

\(dQ = \) not a perfect differential

\[ \boxed{du = dQ - dW} \]

chemical pot.

\[ = dU + Pdv - \mu\, dN \qquad \to \text{Generalized fluxes.} \] \[ dQ = dU - \sum F_{i}\, dx_{i} \]

\(\hookrightarrow\) generalized forces

\(E \leftrightarrow U\)

Entropy representation of microcanonical ensamble \(S = S(E,V,N)\)

\[ \boxed{dS = \frac{dE}{T} + \frac{Pdv}{T} - \frac{\mu dN}{T}} \]

tells us

\[ \left(\frac{\partial S}{\partial E}\right)_{V,N} = \frac{1}{T} \]

(Energy representation) \(\Rightarrow\) \(dE = TdS - Pdv + \mu dN\)
\(\Rightarrow\) \(E = E(S,V,N)\)

Translate this page