• #Lecture 27
  • #(Phase transitions)
  • #Phase diagram
  • #Critical Point (V_c, T_c, P_c) :-
  • #P-V diagram :- (only for boiling curve).
  • #Maxwells equal Area rule :-
  • #Mechanical example of hysteresis (Double well potentials).
• #Lecture 28
  • [[#\(-\) Model of paramagnetism]]
  • #Isothermal susceptibility
  • #Weiss molecular field theory
• #Lecture 29
  • #Susceptibility at (H=0).

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(After) \((0:00)\)

Thermodynamics is a kind of limiting case of statistical mech. where fluctuations are neglected and so on; And these fluctuations can be taken into account if we use stat. mech. and it provides correction to thermodynamic cases, goes beyond thermostatics & calculate \((C_P, C_V)\) which are input parameters in thermodynamics.

There is a place where thermodynamics fails (completly and no longer valid); it is hard problem of phase transitions of matter. In principle stat. mech. (eq\(^m\) stat. mech.) should tell us everything about system at thermal eq\(^n\), including processes at phase transitions.

Thermodynamics fails at critical point \((P_c, V_c, T_c)\).

We will use vander vaals gas to demostrate real gases.

We have thermodynamic variables \((P, V, T)\) out of which we only use any two of them, as third is found of using equation of state (which is different for different phases of matter -- solid, liquid, gas).

\usetikzlibrary{arrows.meta,decorations.pathreplacing,calc,angles,quotes,positioning,patterns}
\begin{tikzpicture}[scale=0.95,>=Stealth]
  % axes
  \draw[->] (0,-0.2) -- (0,4.3) node[above left] {$P$};
  \draw[->] (-1.6,0) -- (7.2,0) node[below] {$T$};
  % sublimation curve (c): from lower left to triple point
  \draw[thick] (-1.4,-0.05) .. controls (0.2,0.15) and (1.1,0.5) .. (1.7,0.95);
  % melting curve (b): steep from triple point
  \draw[thick] (1.7,0.95) .. controls (2.1,2.4) and (2.4,3.4) .. (2.6,4.1);
  \node at (2.75,4.25) {\small\textcircled{b}};
  % boiling curve (a): triple point to critical point
  \draw[thick] (1.7,0.95) .. controls (3.0,1.15) and (4.2,1.5) .. (5.0,2.2);
  \fill (5.0,2.2) circle (1.6pt);
  \node[anchor=south west] at (5.0,2.25) {\small $(P_c,T_c)$};
  \node at (5.3,1.95) {\small\textcircled{a}};
  % triple point marker
  \node at (1.7,0.6) {\small\textcircled{c}};
  \draw[->] (-1.2,1.0) -- (1.45,1.0);
  \node[anchor=east] at (-1.25,0.75) {(Triple point)};
  % dashed lines at critical point
  \draw[dashed] (1.9,2.2) -- (5.0,2.2);
  \draw[dashed] (5.0,0) -- (5.0,3.6);
  % region labels
  \node at (1.15,3.0) {\small solid};
  \node at (1.15,2.7) {\footnotesize (crystalline)};
  \node at (3.5,3.4) {\small liquid};
  \node at (3.4,0.55) {\small gas};
  \node[anchor=west,align=left] at (5.2,1.0) {\small Homogeneous\\ \small fluid\\ \small phase};
\end{tikzpicture}

• (a) \(\to\) liquid-gas coexisting curve or (Boiling curve)
• (b) \(\to\) solid-liquid coexisting curve or melting curve
• (c) \(\to\) solid-gas coexistant curve or sublimation (curve).

every point on this phase plane is assumed to at thermal eq\(^m\) state of the system.

(phase -- A homogeneous thermodynamic eq\(^m\) system)

( Ice melting of glaciers is measured using rise of water level but Ice can diractly convert to vapour phase, so this is a challanging problem).

(Note :- It looks like the melting curve goes on with same slope, but it does not happen, As pressure increases solid have more densly packed atoms, at very high pressure (FCC) Face centered cubic crystals formes.

(Gibbs phase rule) :- For one kind of molecular species at most we can have only three coexisting phases. For more than one molecular species we can have more than 3 kind of phase coexisting at common point.

The boiling curve/liquid-gas coexisting curve stops at some point on phase plane called critical point where it becomes impossible to distinguish liquid phase from gas phase.

At this point the conventional thermodynamics fails. on right of \((P_c,T_c)\) or for \((T > T_c)\) we can't tell diff between liquid & phase. (Homogeneous fluid phase).

To convert form liquid to gas phase, we supply latent heat.

\usetikzlibrary{arrows.meta,decorations.pathreplacing,calc,angles,quotes,positioning,patterns}
\begin{tikzpicture}[scale=0.9,>=Stealth]
  \draw[->] (0,-0.2) -- (0,2.2) node[above] {};
  \node at (-0.35,1.9) {$P$};
  \draw[->] (-0.7,0) -- (3.6,0);
  \node at (1.5,-0.35) {$T$};
  \draw[thick] (0.7,0.55) .. controls (1.5,0.75) and (2.2,1.1) .. (2.7,1.55);
  \fill (0.7,0.55) circle (1.2pt);
  \fill (2.7,1.55) circle (1.2pt);
  \node[anchor=south west] at (2.6,1.6) {\small $(P_c,T_c)$};
  \draw[->] (1.35,1.35) -- (1.55,0.9);
  \node[anchor=south] at (1.35,1.35) {\small $L_2$};
  \draw[->] (0.75,1.15) -- (1.0,0.68);
  \node[anchor=south] at (0.75,1.15) {\small $L_1$};
  \node at (1.5,0.3) {\small gas};
  \node[anchor=west] at (2.3,-0.45) {$(L_2 < L_1)$};
\end{tikzpicture}

this transition is discontinuous as density of system changes discontinuously

As we go up this curve (i.e. increase pressure), latent heat becomes less and less, & latent heat vanishes at critical point (i.e. we require no heat to convert from liquid to gas phase).

Also the surface tension also vanishes at critical point i.e. the meniscus (the curved upper surface of liquid) also disappears. "we have mush at critical point."

Everything that distinguish liquid from gas disappears, even the density start matching.

Lets draw melting curve in \((V-T)\) plane

volume of gas is very high

\usetikzlibrary{arrows.meta,decorations.pathreplacing,calc,angles,quotes,positioning,patterns}
\begin{tikzpicture}[scale=0.95,>=Stealth]
  \draw[->] (0,-0.2) -- (0,4.0);
  \node at (-0.3,3.8) {$V$};
  \draw[->] (-0.6,0) -- (5.6,0);
  \node[anchor=west] at (5.65,0) {$T$};
  % coexistence dome (opening left, critical point at right tip)
  \draw[thick] (0.9,3.4) .. controls (2.4,3.2) and (3.1,2.6) .. (3.1,2.0)
        .. controls (3.1,1.4) and (2.2,0.9) .. (1.1,0.7);
  \fill (3.1,2.0) circle (1.5pt);
  \node[anchor=south east] at (3.05,2.05) {\small $(V_c,T_c)$};
  \draw[dashed] (3.1,-0.05) -- (3.1,3.2);
  \draw[dashed] (0,2.0) -- (3.1,2.0);
  \node[anchor=east] at (-0.05,2.0) {\small $(V_c)$};
  \node at (3.1,-0.4) {\small $(T_c)$};
  \node at (1.7,3.7) {\small gas};
  \node at (1.5,0.35) {\small liquid};
  \node[anchor=west,align=left,rotate=-30] at (3.3,1.1) {\footnotesize homogeneous\\ \footnotesize fluid phase};
  \draw[->] (3.4,2.5) -- (4.3,2.5);
  \node[anchor=west,align=left] at (4.3,2.5) {\small region of coexistance\\ \small of liquid \& gas.};
\end{tikzpicture}

\usetikzlibrary{arrows.meta,decorations.pathreplacing,calc,angles,quotes,positioning,patterns}
\begin{tikzpicture}[scale=0.9,>=Stealth]
  \draw[->] (0,-0.15) -- (0,1.3);
  \node at (-0.3,1.1) {$P$};
  \draw[->] (-0.5,0) -- (2.2,0);
  \node at (0.9,-0.35) {$T$};
  \draw[thick] (0.4,0.3) .. controls (0.9,0.45) and (1.3,0.6) .. (1.6,0.85);
  \draw[->] (1.75,0.9) -- (2.5,0.9);
  \node[anchor=west,align=left] at (2.5,0.55) {this boiling curve\\ becomes region\\ in V-T diagram};
\end{tikzpicture}

\usetikzlibrary{arrows.meta,decorations.pathreplacing,calc,angles,quotes,positioning,patterns}
\begin{tikzpicture}[scale=0.9,>=Stealth]
  \draw[->] (0,-0.2) -- (0,2.6);
  \node[anchor=east] at (-0.1,1.3) {$\rho \left(\rho \propto \frac{1}{V}\right)$};
  \draw[->] (-0.5,0) -- (4.0,0);
  \node[anchor=west] at (4.05,0) {$T$};
  % rho vs T coexistence curve: liquid branch upper, gas branch lower, meeting at Tc
  \draw[thick] (1.05,2.3) .. controls (1.6,2.25) and (2.05,2.0) .. (2.1,1.35)
        .. controls (2.15,0.7) and (1.6,0.4) .. (1.05,0.35);
  \fill (2.1,1.35) circle (1.5pt);
  \draw[dashed] (2.1,-0.05) -- (2.1,2.4);
  \node at (1.15,2.55) {\small liquid};
  \node at (1.15,0.1) {\small gas};
  \node at (2.1,-0.4) {\small $T_c$};
  \draw[<->] (1.45,0.55) -- (1.45,2.15);
  \draw[->] (2.3,1.9) -- (3.2,1.9);
\end{tikzpicture}

discontinuous phase transition jump in density \((\rho)\).

\usetikzlibrary{arrows.meta,decorations.pathreplacing,calc,angles,quotes,positioning,patterns}
\begin{tikzpicture}[scale=0.9]
  \draw (0,-0.6) -- (0,0.9);
  \draw (-0.6,0) -- (1.4,0);
  \node[anchor=west] at (0.15,0.65) {\small liquid};
  \node[anchor=west] at (0.35,0.3) {\small gas};
\end{tikzpicture}

Critical point is 2nd order or continuous phase transition, and coexisting curve is series of discontinuous / 1st order phase transitions.

\usetikzlibrary{arrows.meta,decorations.pathreplacing,calc,angles,quotes,positioning,patterns}
\begin{tikzpicture}[scale=0.72,>=Stealth]
  \useasboundingbox (-1.2,-2.6) rectangle (16.0,5.6);
  % axes
  \draw[->] (0,-0.2) -- (0,5.2);
  \node at (-0.35,4.9) {$P$};
  \draw[->] (-0.4,0) -- (6.8,0);
  \node at (3.4,-0.45) {$V$};
  % coexistence dome
  \draw[thick] (1.25,1.0) .. controls (1.55,2.6) and (2.05,3.3) .. (2.6,3.3)
        .. controls (3.3,3.3) and (3.95,2.4) .. (4.6,1.0);
  % critical isotherm
  \draw[very thick] (1.05,4.9) .. controls (1.6,3.9) and (2.1,3.4) .. (2.6,3.3)
        .. controls (3.4,3.15) and (4.3,1.8) .. (6.3,1.15);
  \fill (2.6,3.3) circle (2pt);
  \node[anchor=south] at (2.6,3.42) {\small $(P_c,V_c)$};
  % isotherms above Tc
  \draw (1.35,5.0) .. controls (2.2,4.4) and (3.5,3.0) .. (6.4,1.95);
  \draw (1.75,5.0) .. controls (2.8,4.6) and (4.0,3.5) .. (6.4,2.7);
  % isotherms below Tc (van der Waals loops)
  \draw (0.95,4.5) .. controls (1.3,2.6) and (1.5,1.9) .. (1.9,1.9)
        .. controls (2.4,1.9) and (2.6,2.6) .. (3.0,2.6)
        .. controls (3.8,2.6) and (4.5,1.1) .. (6.3,0.6);
  \draw (0.8,3.5) .. controls (1.1,1.6) and (1.3,0.9) .. (1.7,0.95)
        .. controls (2.25,1.0) and (2.55,1.8) .. (3.4,1.75)
        .. controls (4.2,1.7) and (4.8,0.6) .. (6.3,0.28);
  % Pc, Vc guide lines
  \draw (0,3.3) -- (2.6,3.3);
  \node[anchor=east] at (-0.05,3.3) {$P_c$};
  \draw[->] (2.6,3.3) -- (2.6,0.15);
  \node[anchor=north] at (2.6,0.1) {$V_c$};
  % right hand labels
  \draw[->] (5.4,3.6) -- (6.6,3.9);
  \node[anchor=west] at (6.6,3.9) {\small $(T>T_c)$ \ (Ided gas approx.).};
  \draw[->] (5.6,2.6) -- (6.6,3.1);
  \node[anchor=west] at (6.6,3.1) {\small $T>T_c$};
  \draw[->] (5.9,1.5) -- (6.6,2.2);
  \node[anchor=west] at (6.6,2.2) {\small $T=T_c$ (critical Isotherm).};
  % wrong-curve annotation
  \draw[->] (4.9,-1.3) -- (2.6,2.15);
  \node[anchor=north west,align=left] at (5.0,-1.1)
     {\small wrong curves as \underline{$P\uparrow$ $V\uparrow$.} (violets lessatelier's\\
      \small \hspace*{3.6cm} principle of stability).};
\end{tikzpicture}

\[ P = \frac{nRT}{V-nb} - \frac{an^2}{V^2} \qquad \text{or} \qquad \left(\left(P + \frac{an^2}{V^2}\right)(V-nb) = nRT\right) \]

To find critical point (Inflection point).

\[ \left.\frac{\partial P}{\partial V}\right|_{(T_c,P_c,V_c)} = 0 \]

also

\[ \left.\frac{\partial^2 P}{\partial V^2}\right|_{(P_c,V_c,T_c)} = 0 \]

and from eq\(^n\) of state, we can find \((P_c, V_c, T_c)\).

There is no universal eq\(^n\) of state as eq\(^n\) of state depends upon interactions and interactions are hard to model universally.

The portion where \(\dfrac{\partial P}{\partial V}\) is positive makes compressibility

\[ K_B = -\frac{1}{V}\frac{\partial V}{\partial P} \]

negative and violets thermodynamic stability. So we have to correct this portion.

\usetikzlibrary{arrows.meta,decorations.pathreplacing,calc,angles,quotes,positioning,patterns}
\begin{tikzpicture}[scale=0.9,>=Stealth]
  \draw[->] (0,-0.2) -- (0,3.4);
  \node at (-0.35,3.0) {$P$};
  \draw[->] (-0.6,0) -- (5.6,0);
  \node at (2.8,-0.45) {$V$};
  % van der Waals isotherm with loop
  \draw[thick] (0.5,3.2) .. controls (0.85,2.0) and (1.0,1.35) .. (1.4,1.35)
        .. controls (1.9,1.35) and (2.1,2.4) .. (2.7,2.35)
        .. controls (3.4,2.3) and (4.0,1.1) .. (5.2,0.75);
  % direction arrows along the curve
  \draw[->] (2.4,2.42) -- (2.05,2.42);
  \draw[->] (1.8,1.5) -- (2.15,1.62);
  \draw[->] (3.3,2.05) -- (3.6,1.9);
  \draw[->] (0.9,2.7) -- (0.8,2.45);
\end{tikzpicture}

hysteresis is the problem here. So to avoid, it is observed that, (flip)

\usetikzlibrary{arrows.meta,decorations.pathreplacing,calc,angles,quotes,positioning,patterns}
\begin{tikzpicture}[scale=0.9,>=Stealth]
  \draw[->] (0,-0.2) -- (0,3.0);
  \node at (-0.35,2.6) {$P$};
  \draw[->] (-0.6,0) -- (5.2,0);
  \node at (2.6,-0.45) {$V$};
  \draw[thick] (0.45,2.8) .. controls (0.8,1.6) and (0.95,1.0) .. (1.35,1.0)
        .. controls (1.85,1.0) and (2.05,2.05) .. (2.65,2.0)
        .. controls (3.35,1.95) and (3.9,0.85) .. (5.0,0.55);
  % tie line
  \draw[very thick] (0.95,1.5) -- (3.15,1.5);
  \node at (1.5,1.15) {\small $A$};
  \node at (2.5,1.75) {\small $A'$};
  \draw[->] (-1.2,1.9) -- (0.75,1.6);
  \node[anchor=east] at (-1.25,1.95) {(Tieline)};
  \draw[->] (3.6,2.4) -- (3.05,1.6);
  \node[anchor=west,align=left] at (3.7,2.3)
    {Tieline\\ Transition occurs\\ such that Area are\\ equal};
\end{tikzpicture}

So some portions of allowed curve also gets omitted in observation. So vaander vaals gas eq\(^n\) of state is not accurate, we not only have to remove unstable region but slightly bigger region also.

So envelope goes like this --

\usetikzlibrary{arrows.meta,decorations.pathreplacing,calc,angles,quotes,positioning,patterns}
\begin{tikzpicture}[scale=0.85,>=Stealth]
  \useasboundingbox (-2.4,-4.6) rectangle (9.6,5.6);
  \draw[->] (0,-0.2) -- (0,5.1);
  \node at (-0.35,4.6) {$P$};
  \draw[->] (-0.6,0) -- (6.4,0);
  % binodal (outer envelope)
  \draw[thick] (1.35,0.15) .. controls (1.45,1.8) and (1.9,3.3) .. (2.75,3.35)
        .. controls (3.55,3.4) and (3.85,1.6) .. (3.9,0.15);
  % spinodal (inner, dashed)
  \draw[dashed] (1.85,0.6) .. controls (2.0,1.8) and (2.3,3.0) .. (2.75,3.35)
        .. controls (3.2,3.0) and (3.4,1.7) .. (3.45,0.6);
  \fill (2.75,3.35) circle (1.8pt);
  \node[anchor=south east] at (2.7,3.5) {\small $(P_c,V_c)$};
  \draw[->] (2.95,3.35) -- (4.15,3.45);
  \node[anchor=west] at (4.15,3.45) {\small critical point};
  % isotherms
  \draw (1.0,4.9) .. controls (1.7,4.2) and (3.0,2.7) .. (6.1,1.9);
  \node[anchor=west] at (3.3,4.75) {\small gas};
  \draw[->] (3.25,4.7) -- (2.35,4.55);
  \node[anchor=west] at (4.15,4.2) {\small $(T>T_c)$};
  \draw[very thick] (0.85,4.6) .. controls (1.6,3.7) and (2.2,3.4) .. (2.75,3.35)
        .. controls (3.6,3.25) and (4.4,1.7) .. (6.1,1.25);
  \node[anchor=west] at (6.2,1.25) {\small $T=T_c$};
  \draw (0.7,3.9) .. controls (1.1,2.0) and (1.4,1.15) .. (1.85,1.2)
        .. controls (2.4,1.25) and (2.55,2.2) .. (3.45,2.1)
        .. controls (4.3,2.0) and (4.8,1.0) .. (6.1,0.75);
  \node[anchor=west] at (6.2,0.75) {\small $T<T_c$};
  \draw (0.55,3.1) .. controls (0.95,1.2) and (1.25,0.5) .. (1.7,0.55)
        .. controls (2.3,0.6) and (2.5,1.4) .. (3.3,1.3)
        .. controls (4.2,1.2) and (4.7,0.5) .. (6.1,0.3);
  \node[anchor=west] at (6.2,0.3) {\small $T<T_c$};
  \node[anchor=east] at (1.25,4.3) {\small liquid};
  % labels below
  \draw[->] (-0.7,-1.5) -- (1.5,0.5);
  \node[anchor=north,align=left] at (-1.4,-1.5) {\small Meta\\ \small -stable\\ \small \ region};
  \node[anchor=north west,align=left] at (-2.3,-3.1) {\small (System seperates\\ \small slowly into gas \& liquid)};
  \draw[->] (2.8,-1.0) -- (2.8,0.4);
  \node[anchor=north,align=left] at (3.0,-1.1) {\small unstable region\\ \small (system seperates to\\ \small liquid \& gas abruptly).};
\end{tikzpicture}

Any system on unstable region will instantane-ously separate into some liquid & remainy to gas depending upon lever rule.

Note :- In case of binary alloys (liquid alloys), the metastable portion can be made extremly slow, It may take hours, yeares to happen.

\usetikzlibrary{arrows.meta,decorations.pathreplacing,calc,angles,quotes,positioning,patterns}
\begin{tikzpicture}[scale=0.95,>=Stealth]
  \draw[->] (0,-0.2) -- (0,3.4);
  \node at (-0.3,3.0) {};
  \draw[->] (-0.5,0) -- (5.4,0);
  % van der Waals loop
  \draw[thick] (0.5,3.2) .. controls (0.85,1.9) and (1.0,1.15) .. (1.45,1.2)
        .. controls (2.0,1.25) and (2.15,2.3) .. (2.85,2.2)
        .. controls (3.55,2.1) and (4.1,1.0) .. (5.1,0.7);
  % tie line
  \draw (0.75,1.7) -- (3.35,1.7);
  \draw[dashed] (3.35,0) -- (3.35,2.4);
  \node[anchor=south east] at (0.78,1.75) {$A$};
  \node[anchor=north] at (1.45,1.15) {$B$};
  \node[anchor=south west] at (2.0,1.72) {$C$};
  \node[anchor=south] at (2.6,2.25) {$D$};
  \node[anchor=north west] at (3.3,1.68) {$E$};
  \node at (2.6,1.95) {$A_1$};
  \node at (1.35,1.45) {$A_2$};
\end{tikzpicture}

(maxwell's tieline construction)

(Is not quite correct though as It goes through ((D) \(\to\) (B)) i.e. from thermodynamic unstable region.)

let us consider \(G(T,P,N) = U - TS + PV\)

\[ \begin{aligned} dG &= dU - TdS - SdT + PdV + VdP\\ &= -SdT + VdP + \mu dN \end{aligned} \]

for fixed \((N,P,T)\) \((dG = 0)\)

On an isotherm, *\(dG = VdP\)* (as \(dT = 0\)).

\[ G_E - G_A = 0 = \int_A^E dG = \int_A^B VdP + \int_B^C VdP + \int_C^D VdP + \int_D^E VdP \] \[ \left(G = \frac{\mu}{N} \ \text{should be same in gas phase \& liquid phase}\right) \] \[ = \left(\int_A^B VdP - \int_C^B VdP\right) + \left(-\int_D^C VdP + \int_D^E VdP\right) = 0 \] \[ \underbrace{\hphantom{XXXXXXXXXXXX}}_{(A_1)} \quad + \quad \underbrace{\hphantom{XXXXXXXXXXXX}}_{(-A_2)} \] \[ = \boxed{(A_1 = A_2) = A} \]

\usetikzlibrary{arrows.meta,decorations.pathreplacing,calc,angles,quotes,positioning,patterns}
\begin{tikzpicture}[scale=0.85,>=Stealth]
  % (a) single well
  \begin{scope}[xshift=0cm]
    \draw[thick] plot[smooth,tension=0.7] coordinates
      {(-0.75,2.2) (-0.6,1.2) (-0.35,0.45) (0,0.25) (0.35,0.45) (0.6,1.2) (0.75,2.2)};
    \fill (0,0.25) circle (1.4pt);
    \node at (0,-0.1) {\footnotesize L R};
    \node at (0,-0.65) {(a)};
  \end{scope}
  % (b)
  \begin{scope}[xshift=2.2cm]
    \draw[thick] plot[smooth,tension=0.7] coordinates
      {(-0.8,2.2) (-0.6,1.05) (-0.38,0.85) (-0.15,0.98) (0.05,0.6) (0.3,0.3) (0.55,0.9) (0.8,2.2)};
    \fill (0.3,0.3) circle (1.4pt);
    \node at (-0.55,1.25) {\footnotesize L};
    \node at (0.3,-0.1) {\footnotesize R};
    \node at (0,-0.65) {(b)};
    \draw[->] (0,-1.15) -- (0,-0.85);
    \node[anchor=west] at (0.1,-1.15) {$(L\to R)$};
  \end{scope}
  % (c)
  \begin{scope}[xshift=4.6cm]
    \draw[thick] plot[smooth,tension=0.7] coordinates
      {(-0.8,2.2) (-0.6,1.0) (-0.4,0.7) (-0.18,0.78) (0.02,1.0) (0.22,0.55) (0.42,0.28) (0.65,1.2) (0.8,2.2)};
    \fill (0.42,0.28) circle (1.4pt);
    \node at (-0.4,0.35) {\footnotesize L};
    \node at (0.42,-0.1) {\footnotesize R};
    \node at (0,-0.65) {(c)};
  \end{scope}
  % (d)
  \begin{scope}[xshift=7.0cm]
    \draw[thick] plot[smooth,tension=0.7] coordinates
      {(-0.85,2.2) (-0.62,0.9) (-0.4,0.3) (-0.2,0.6) (0,0.95) (0.2,0.6) (0.4,0.3) (0.62,0.9) (0.85,2.2)};
    \fill (-0.4,0.3) circle (1.4pt);
    \fill (0.4,0.3) circle (1.4pt);
    \node at (-0.4,-0.1) {\footnotesize L};
    \node at (0.4,-0.1) {\footnotesize R};
    \node at (0,-0.65) {(d)};
  \end{scope}
  % (e)
  \begin{scope}[xshift=9.4cm]
    \draw[thick] plot[smooth,tension=0.7] coordinates
      {(-0.8,2.2) (-0.65,1.2) (-0.42,0.28) (-0.22,0.55) (-0.02,1.0) (0.18,0.78) (0.4,0.7) (0.6,1.0) (0.8,2.2)};
    \fill (-0.42,0.28) circle (1.4pt);
    \node at (-0.42,-0.1) {\footnotesize L};
    \node at (0.4,0.35) {\footnotesize R};
    \node at (0,-0.65) {(e)};
  \end{scope}
  % (f)
  \begin{scope}[xshift=11.8cm]
    \draw[thick] plot[smooth,tension=0.7] coordinates
      {(-0.8,2.2) (-0.55,0.9) (-0.3,0.3) (-0.05,0.6) (0.15,0.98) (0.38,0.85) (0.6,1.05) (0.8,2.2)};
    \fill (-0.3,0.3) circle (1.4pt);
    \node at (-0.3,-0.1) {\footnotesize L};
    \node at (0.55,1.25) {\footnotesize R};
    \node at (0,-0.65) {(f)};
    \draw[->] (0,-1.15) -- (0,-0.85);
    \node[anchor=west] at (0.1,-1.15) {$(R\to L)$};
  \end{scope}
  % (g)
  \begin{scope}[xshift=14.2cm]
    \draw[thick] plot[smooth,tension=0.7] coordinates
      {(-0.75,2.2) (-0.6,1.2) (-0.35,0.45) (0,0.25) (0.35,0.45) (0.6,1.2) (0.75,2.2)};
    \fill (0,0.25) circle (1.4pt);
    \node at (0,-0.1) {\footnotesize L};
    \node at (0,-0.65) {(g)};
  \end{scope}
\end{tikzpicture}

If we start at (R) i.e. right well, Transition occurs to `L' at (f).

If we start at `L' i.e. left wall (g) then transitions occur at (b). This is called hysteresis in mechanical case.

There is a branch of mathematics called catastrophe theory where people talk about change in generic potentials. (BTW) we can't have such minimum in thermal eq\(^n\).

Near critical point, many systems behave (independent of their interactions/details) identically.

We can convert from liquid to gas or gas to liquid without encountering a sharp transitions. (This is only possible because boiling curve ends at critical point).

\usetikzlibrary{arrows.meta,decorations.pathreplacing,calc,angles,quotes,positioning,patterns}
\begin{tikzpicture}[scale=0.95,>=Stealth]
  \draw[->] (0,-0.2) -- (0,3.0);
  \node at (-0.35,2.6) {$P$};
  \draw[->] (-0.6,0) -- (4.8,0);
  \node at (2.1,-0.4) {$T$};
  % boiling curve ending at critical point
  \draw[thick] (0.6,0.35) .. controls (1.4,0.7) and (2.0,1.1) .. (2.5,1.55);
  \fill (2.5,1.55) circle (1.6pt);
  \node[anchor=south west] at (2.5,1.6) {$(P_c,T_c)$};
  % path looping around critical point
  \draw[->] (1.5,1.5) .. controls (2.2,2.5) and (3.6,2.3) .. (3.5,1.5)
        .. controls (3.4,0.7) and (2.2,0.5) .. (1.6,0.75);
  \node[anchor=east] at (1.45,1.35) {\small liquid};
  \node[anchor=north] at (1.7,0.6) {\small gas};
\end{tikzpicture}

we can't have continuous transitions from liquid to solid or vice versa. (Because this curve (melting curve) does not end. It does not end because solid has crystalline symmetry

It is symmetric under group of specific transformations like space rotations, translations. On the other hand liquid is homogeneous and isotropic, therefore liquid has much greater degree of symmetry. Solid has long range order.

order and symmetry are kind of opposites. More ordered is phase less symmetric it is; in the sense that the set of transformations under which it remains unchanged is smaller and st smaller.

The most disordered state is in fact most symmetric because it looks exactly the same in every direction.

on the other hand,

liquid and gas phases are both symmetric (in every dir.).

So \(\exists\) a critical point

we can understand behaviour at critical point via fluid--magnet analogy.

Assumption -- system consists of elementary magnetic dipole moments (atomic (Spin + Angular)). oriented randomly in presence of ext. \(\vec{H}_{ext}\) \(\left(H = \dfrac{\vec{B}}{\mu_0}\right)\).

magnetic dipole moment will try to align in direction of \(\vec{H}_{ext}\). So system will have some kind of symmetry.

\usetikzlibrary{arrows.meta,decorations.pathreplacing,calc,angles,quotes,positioning,patterns}
\begin{tikzpicture}[scale=0.9,>=Stealth]
  \node[anchor=east,align=center] at (-1.4,1.0) {(auxilary\\ field)};
  \node at (-0.95,1.15) {$\vec{H}$};
  \draw[->] (-0.55,0.3) -- (-0.55,1.9);
  \draw (0,0) rectangle (3.6,2.4);
  \foreach \px/\py/\a in {0.45/1.95/45, 0.95/1.95/270, 1.55/1.95/0, 2.35/1.95/90,
                     0.5/1.5/225, 1.15/1.5/0, 1.9/1.5/0, 2.6/1.5/60, 2.9/1.35/300,
                     0.4/1.0/0, 1.2/1.0/0, 2.0/0.95/40, 2.55/1.0/30,
                     0.4/0.45/225, 0.9/0.45/300, 1.35/0.45/90, 2.1/0.45/280, 2.8/0.45/90}
    \draw[->] (\px,\py) -- ++(\a:0.42);
  \node[anchor=west,align=left] at (3.75,1.6)
     {N dipole moments, $\vec{\mu}$\\ placed in heat bath at temp $T$.};
\end{tikzpicture}

P.E of a dipole \(\vec{\mu}\) in field \(\vec{H}\)

\[ U = -\vec{\mu}\cdot\vec{H} \] \[ U_{min} = -\mu H \quad (\theta = 0) \] \[ U_{max} = +\mu H \quad (\theta = \pi). \]

lets look at very simple model, such that

\[ U = +\mu H \ \underline{\text{or}} \ -\mu H \qquad \underline{\text{(one--dimension case)}} \]

(magnetic dipole moment of \(e^-\) can only take two values).

\usetikzlibrary{arrows.meta,decorations.pathreplacing,calc,angles,quotes,positioning,patterns}
\begin{tikzpicture}[scale=0.9,>=Stealth]
  \draw[thick] (0,1.6) -- (2.6,1.6);
  \node[anchor=east] at (-0.1,1.6) {$E_1$};
  \node[anchor=west] at (2.75,1.6) {$(\mu H)$};
  \draw[->] (5.4,1.3) -- (5.4,1.95); \node[anchor=west] at (5.45,1.95) {$\vec{\mu}$};
  \draw[->] (6.4,1.95) -- (6.4,1.3); \node[anchor=west] at (6.45,1.95) {$\vec{H}$};
  \draw[thick] (0,0) -- (2.6,0);
  \node[anchor=east] at (-0.1,0) {$E_0$};
  \node[anchor=west] at (2.75,0) {$(-\mu H)$};
  \draw[->] (5.4,-0.3) -- (5.4,0.35); \node[anchor=west] at (5.45,0.35) {$\vec{\mu}$};
  \draw[->] (6.4,-0.3) -- (6.4,0.35); \node[anchor=west] at (6.45,0.35) {$\vec{H}$};
\end{tikzpicture}

lets find magnetisation (or better average Magnetisation). in *\((1-D)\)*

\[ \vec{M} = N \ (\text{Avg dipolemoments of each dipole moment}). \]

(\(\hookrightarrow N\langle \mu\rangle\) is only true iff all \(\vec{\mu}\) are independent assuming no--dipole--dipole interactions).

(since we only want to see \(\vec{M}\) along dir. of \(\vec{H}\). lets drop vector)

\[ M = N\left(\frac{\mu\,P(\mu) + (-\mu)\,P(-\mu)}{P(\mu) + P(-\mu)}\right) \] \[ = \frac{N\left(\mu e^{\beta\mu H} - \mu e^{-\beta\mu H}\right)}{e^{\beta\mu H} + e^{-\beta\mu H}} \qquad \left(\tanh x = \frac{e^x - e^{-x}}{e^x + e^{-x}}\right). \]

Av. magnetisation \((M)\) \(= N\mu\tanh(\beta\mu H)\)

So, \(\boxed{M = N\mu \tanh\left(\dfrac{\mu H}{k_BT}\right)}\) is magnetic eq\(^n\) of state.

fluid \((P, V, T)\) of [?] \((H, M)\) system

}cc}
\(\uparrow\) \(\uparrow\) \(\uparrow\) \(\uparrow\)
field responce field responce
(Intensive) extensive (Intensive) (extensive)

\(\left(\dfrac{M}{N\mu}\right)\) is magnetisation per atom per dipole moment.

\usetikzlibrary{arrows.meta,decorations.pathreplacing,calc,angles,quotes,positioning,patterns}
\begin{tikzpicture}[scale=1.0,>=Stealth]
  \useasboundingbox (-4.2,-1.9) rectangle (6.4,2.0);
  \clip (-4.2,-1.9) rectangle (6.4,2.0);
  \draw[->] (0,-1.7) -- (0,1.7);
  \node[anchor=south east] at (0,1.5) {$\left(\frac{M}{N\mu}\right)$};
  \draw[->] (-3.6,0) -- (3.6,0);
  \node[anchor=west] at (3.6,0) {$H$};
  \draw (-3.6,1) -- (3.6,1);
  \draw (-3.6,-1) -- (3.6,-1);
  \node[anchor=south west] at (0.05,1.0) {$+1$};
  \node[anchor=north west] at (0.05,-1.0) {$-1$};
  \draw[thick,domain=-3:3,samples=80,smooth] plot (\x,{tanh(1.5*\x)});
  \draw[thick] (3,0.99978) -- (3.5,1);
  \draw[thick] (-3,-0.99978) -- (-3.5,-1);
  \draw (0,0) ellipse (1.1 and 0.6);
  \draw[->] (1.15,-0.35) -- (1.9,-0.45);
  \node[anchor=west,align=left] at (1.9,-0.6) {linear\\ region};
\end{tikzpicture}

our old def\(^n\) of susceptibility \((\chi)\)

\[ \begin{aligned} \vec{M} &\propto \vec{H}\\ \vec{M} &= \chi\vec{H}\\ \frac{M}{H} &= \text{const} \end{aligned} \]

is not quite right as, this graph is non linear in most range of \(\vec{H}\).

\[ \chi_T = \left.\left(\frac{\partial M}{\partial H}\right)_{T,N}\right|_{H=0} \]

or

\[ \left(\chi_T = \text{slope at origin}\right) \] \[ \chi_T = \left.\left(\frac{\partial M}{\partial H}\right)_T\right|_{H=0} \]

\(\tanh\theta \approx \theta\) for very small \(\theta\).

*\(\tanh ax \approx ax\)*

\[ = N\mu\left(\frac{\mu H}{k_BT}\right) \] \[ \left(\chi_T = \frac{N\mu^2 H}{k_BT}\right) \ \to \ \underline{\text{Curie's law}} \ \text{(pierre curie)} \qquad \left(\chi_T \propto \frac{1}{T}\right) \]

So slope at origin increases with decrease in temp.

\usetikzlibrary{arrows.meta,decorations.pathreplacing,calc,angles,quotes,positioning,patterns}
\begin{tikzpicture}[scale=1.0,>=Stealth]
  \useasboundingbox (-4.4,-2.0) rectangle (7.6,2.7);
  \draw[->] (0,-1.7) -- (0,1.8);
  \node[anchor=east] at (-0.15,1.6) {$\left(\frac{M}{N\mu}\right)$};
  \draw[->] (-3.6,0) -- (4.0,0);
  \node[anchor=west] at (4.0,0) {$H$};
  \draw (-3.6,1) -- (3.4,1);
  \draw (-3.6,-1) -- (3.4,-1);
  \node[anchor=south west] at (-3.55,1.02) {$+1$};
  \node[anchor=north west] at (-3.55,-1.02) {$-1$};
  \node[anchor=west] at (2.2,2.0) {$(T_3 < T_2 < T_1)$};
  % T1 (shallowest)
  \draw[domain=-3.2:3.2,samples=80,smooth] plot (\x,{tanh(0.7*\x)});
  \node[anchor=south] at (1.9,1.28) {$T_1$};
  % T2
  \draw[domain=-3:3,samples=80,smooth] plot (\x,{tanh(1.5*\x)});
  \node[anchor=south] at (1.0,1.28) {$T_2$};
  % T3 (steepest)
  \draw[domain=-1.4:1.4,samples=80,smooth] plot (\x,{tanh(3.5*\x)});
  \draw (1.4,0.99926) -- (3.2,1);
  \draw (-1.4,-0.99926) -- (-3.2,-1);
  \node[anchor=south] at (0.3,1.28) {$T_3$};
  % T = 0 step
  \draw[very thick] (-3.2,-1) -- (0,-1) -- (0,1) -- (3.2,1);
  \draw[->] (1.4,-0.75) -- (0.55,-0.75);
  \node[anchor=west] at (1.4,-0.8) {$(T=0)$};
  \draw (0,0) ellipse (0.75 and 0.75);
\end{tikzpicture}

at \(T=0\) (absolute zero)

\[ \chi \to \infty \]

or slope \(\to \underline{\infty}\)

We have neglected interaction between magnetic dipoles.

At high temperature, we can neglect dipole--dipole interaction but at very low temp, dipole-dipole interactions become large as compared to random fluctuation driven by heat bath.

All phase transitions of this kind is dependent on minimization of free energy

\[ F = U - TS \] \[ U = \langle E\rangle \ \text{or} \ \bar{E} \ \text{or} \ \langle \mathcal{H}\rangle \qquad \text{(Hamiltonian)} \]

It so happens that at sufficiently low temperatures the ordering tendency of internal energy \((U)\) overcomes effect of entropy.

So Free energy \((F)\) is mostly governed by \(U\) as \(T\)' is very small and \(S\)' (entropy) also becomes small at low temperatures \((S = k_B \ln \Omega(E))\) \((\Omega(E)\) decreases with *\(T\)*\()\).

So at low temp, \(F\)' is governed by \(U\) and it is ordered phase, and at high temp \(F\)' is governed by `\(TS\)' which is very large compared to \(U\), and this is disordered phase.

\(\bullet\) "This is the main reason low temp is ordered state and high temp is disordered state."

If dipole--dipole interaction is taken in account our plot or relation \(\left(M = N\mu\tanh\left(\dfrac{\mu H}{k_BT}\right)\right)\) fails.

For two dipoles, if \(U_d\) represents interaction energy

\usetikzlibrary{arrows.meta,decorations.pathreplacing,calc,angles,quotes,positioning,patterns}
\begin{tikzpicture}[scale=0.9,>=Stealth]
  % first pair
  \draw (1.1,0.35) ellipse (1.15 and 0.75);
  \draw[->] (0.55,0.65) -- (1.35,0.65); \node[anchor=west] at (1.4,0.65) {$\mu$};
  \draw[<-] (0.55,0.05) -- (1.35,0.05); \node[anchor=west] at (1.4,0.05) {$\mu$};
  \node[anchor=west] at (2.4,0.35) {$U_{d_1}$};
  \node[anchor=east,align=center] at (-0.3,0.15) {More\\ stable};
  % second pair
  \draw (6.1,0.35) ellipse (1.15 and 0.75);
  \draw[->] (5.55,0.65) -- (6.35,0.65); \node[anchor=west] at (6.4,0.65) {$\mu$};
  \draw[->] (5.55,0.05) -- (6.35,0.05); \node[anchor=west] at (6.4,0.05) {$\mu$};
  \node[anchor=west] at (7.4,0.35) {$U_{d_2}$};
  \node at (6.1,-0.95) {$(U_{d_1} < U_{d_2})$};
\end{tikzpicture}

dipole--dipole interaction is not isotropic, below configuration does not have same energy.

\usetikzlibrary{arrows.meta,decorations.pathreplacing,calc,angles,quotes,positioning,patterns}
\begin{tikzpicture}[scale=0.9,>=Stealth]
  % (1) staggered collinear
  \begin{scope}[xshift=0cm]
    \draw[->] (0,0.9) -- (0.85,0.9); \node[anchor=west] at (0.9,0.9) {$\mu$};
    \draw[->] (0.45,0.25) -- (1.3,0.45); \node[anchor=west] at (1.32,0.4) {$\mu$};
    \node at (0.6,-0.5) {\textcircled{1}};
  \end{scope}
  % (2) parallel stacked
  \begin{scope}[xshift=3.6cm]
    \draw[->] (0,0.9) -- (0.85,0.9); \node[anchor=west] at (0.9,0.9) {$\mu$};
    \draw[->] (0,0.4) -- (0.85,0.4); \node[anchor=west] at (0.9,0.4) {$\mu$};
    \node at (0.6,-0.5) {(2)};
  \end{scope}
  % (3) head to tail
  \begin{scope}[xshift=7.2cm]
    \draw[->] (0,0.7) -- (0.85,0.7); \node[anchor=south] at (0.7,0.75) {$\mu$};
    \draw[->] (1.1,0.85) -- (1.85,0.95); \node[anchor=west] at (1.9,0.95) {$\mu$};
    \node at (0.9,-0.5) {(3)};
  \end{scope}
  % (4) out of plane
  \begin{scope}[xshift=11.2cm]
    \draw[->] (0,0.95) -- (0.85,0.95); \node[anchor=west] at (0.9,0.95) {$\mu$};
    \draw[->] (0.1,0.7) -- (0.8,0.35); \node[anchor=west] at (0.85,0.3) {$\mu$};
    \node[anchor=north west,align=left] at (0.75,0.05) {\footnotesize out\\ \footnotesize of\\ \footnotesize (plane[?]};
    \node at (0.05,-0.5) {(4)};
  \end{scope}
\end{tikzpicture}

If we rely on dipole--dipole interactions alone, we won't have bar magnet at all. Because one \((\uparrow \mu)\) will flip another in reverse diraction \((\downarrow)\mu\), so that the net dipole moment cancels out & magnatisation becomes zero.

\[ \left(\uparrow\downarrow\ \uparrow\downarrow\ \uparrow\downarrow\ \uparrow\downarrow\ \uparrow\downarrow \ \cdots\ \right). \] \[ \text{net } (M=0). \]

But we do see permanent magnets, so the phenomena of permanent magnetism is more subtle than this and is not connected to dipole--dipole interactions, It is connected to a quantum mechanical effect called the exchange Interaction which actually favours \((\uparrow\uparrow)\) over \((\uparrow\downarrow)\). and it in fact overcomes dipole--dipole interactions. \((\uparrow\downarrow)\).

Exchange Interaction is very strong short range interaction It decays exponentially. (where as dipole--dipole inter-action is long range \(\left(\dfrac{2K(\vec{p}\cdot\hat{r})}{r^3}\right)\) & weak).

There are magnets where dipole--dipole interaction becomes more important/dominent and it leads to Anti-ferromagnetism : \((\ \uparrow\downarrow\uparrow\downarrow\uparrow\downarrow\uparrow\downarrow\uparrow\downarrow)\).

If we take exchange interactions in account we see that the curve b/w \(\left(\frac{M}{NH}\right)\) vs \(\vec{H}\), attains infinite slope at finite temp. \((T>0K)\) instead of \((T=0K)\).

Below curie temp \((T_c)\) the curve attains infinite slope at origin.

\usetikzlibrary{arrows.meta,decorations.pathreplacing,calc,angles,quotes,positioning,patterns}
\begin{tikzpicture}[scale=0.9,>=Stealth]
  \draw[->] (-3,0) -- (3.2,0) node[right] {$\vec{H}$};
  \draw[->] (0,-2) -- (0,2.3) node[above] {$(M/NH)$};
  % steep curve (T < Tc)
  \draw[thick] (-2.6,-1.5) .. controls (-1.2,-1.45) and (-0.25,-1.35) .. (0,0)
        .. controls (0.25,1.35) and (1.2,1.45) .. (2.6,1.5);
  % shallower curve (T > Tc)
  \draw[thick] (-2.6,-1.0) .. controls (-1.0,-0.9) and (-0.5,-0.75) .. (0,0)
        .. controls (0.5,0.75) and (1.0,0.9) .. (2.6,1.0);
  \node at (1.55,1.95) {\small $(T<T_c)$};
  \node[anchor=west] at (1.9,0.75) {\small $T>T_c$};
  \draw[->] (1.85,0.75) -- (1.3,0.62);
  \node at (-1.6,-1.95) {\small $(T<T_c)$};
\end{tikzpicture}

So this is discontinuous phase transitions at \(\left(T_c\right)\) curie temp.

This phase transition is from paramagnetic phase to ferromagnetic phase (\(M\) is very high even at \(H=0\)).

For general case \(\left(E = -\mu H\cos\theta\right)\) (3-D case)

\usetikzlibrary{arrows.meta,decorations.pathreplacing,calc,angles,quotes,positioning,patterns}
\begin{tikzpicture}[scale=0.8,>=Stealth]
  \draw[->] (0,0) -- (0,1.5) node[above] {$\hat{n}$};
  \draw[->] (0,0) -- (0.9,1.2);
  \draw[->] (0,-0.2) -- (0,-1.2) node[below] {$\vec{H}$};
\end{tikzpicture}

lets compute the magnetisation.

\usetikzlibrary{arrows.meta,decorations.pathreplacing,calc,angles,quotes,positioning,patterns}
\begin{tikzpicture}[scale=0.9,>=Stealth]
  \draw[->] (0,0) -- (0,2.2) node[left] {$z$};
  \draw[->] (0,1.6) -- (0,2.4);
  \node[right] at (0.05,2.25) {$H$};
  \draw[->] (0,0) -- (3,0) node[right] {$Y$};
  \draw[->] (0,0) -- (-1.2,-1.2) node[below left] {$X$};
  \draw[->] (0,0) -- (1.15,1.6) node[above right] {$\vec{\mu}$};
  \node at (0.42,0.95) {$\theta$};
  \draw (0,0.9) arc (90:54:0.9);
\end{tikzpicture}

-- Also assuming no dipole-dipole interaction.

\[ M = N\langle \mu\rangle \]

(our magnetisation)

\[ \langle \mu\rangle \;=\; \frac{\displaystyle\int \mu\cos\theta\, P(\theta)\, d\Omega}{\displaystyle\int P(\theta)\, d\Omega} \;=\; \frac{\displaystyle\int \mu\cos\theta\; e^{+\beta \mu H\cos\theta}\, d\Omega}{\displaystyle\int e^{+\beta \mu H\cos\theta}\, d\Omega} \]

(\(d\Omega\) = solid angle)

Here we have ignored K.E, Rotational energy of these dipole moments, we have only used magnetic energy.

\[ \int d\Omega \;=\; \int_{-1}^{1} d(\cos\theta) \int_{0}^{2\pi} d\varphi \]

or

\[ \int d\Omega \;=\; \int_{0}^{\pi} \sin\theta\, d\theta \int_{0}^{2\pi} d\varphi \]

\usetikzlibrary{arrows.meta,decorations.pathreplacing,calc,angles,quotes,positioning,patterns}
\begin{tikzpicture}[scale=0.8,>=Stealth]
  \draw (0,0) -- (2.2,0.55);
  \draw (0,0) -- (2.2,-0.15);
  \draw[thick] (2.2,0.55) -- (2.2,-0.15);
  \draw[->] (3.1,0.6) -- (2.35,0.3);
  \node[right] at (3.1,0.6) {$dA$};
\end{tikzpicture}

\[ \begin{aligned} d\Omega &= \frac{dA}{r^2}\\ &= \frac{(r\,d\theta)(r\sin\theta\, d\varphi)}{r^2}\\ d\Omega &= \sin\theta\, d\theta\, d\varphi\\ \int d\Omega &= \iint \sin\theta\, d\theta\, d\varphi \end{aligned} \] \[ \begin{aligned} M &= N\langle \mu\rangle\\ &= N\mu\; \frac{\displaystyle\int \cos\theta\; e^{\beta\mu H\cos\theta}\,\sin\theta\, d\theta \int_{0}^{2\pi} d\varphi} {\displaystyle\int_{0}^{\pi}\sin\theta\, d\theta \int_{0}^{2\pi} d\varphi} \end{aligned} \]

(\(\cos\theta = t\), \(-\sin\theta\, d\theta = dt\))

\[ = \; \frac{\displaystyle N\mu\,(2\pi)\,(-)\int_{1}^{-1} t\, e^{\beta\mu H t}\, dt} {\displaystyle 2\pi \int_{0}^{\pi} e^{\beta\mu H\cos\theta}\,\sin\theta\, d\theta} \] \[ = \; \frac{\displaystyle 2\pi N\mu \int_{-1}^{1} t\, e^{\beta\mu H t}\, dt} {\displaystyle 2\pi \int_{-1}^{1} e^{\beta\mu H t}\, dt} \] \[ = \; \frac{\displaystyle N\mu\left(\frac{t\, e^{\beta\mu H t}}{(\beta\mu H)} - \frac{e^{\beta\mu H t}}{\beta^2\mu^2 H^2}\right)_{-1}^{1}} {\displaystyle \left(\frac{1}{\beta\mu H}\right)\left(e^{\beta\mu H}\right)_{-1}^{1}} \] \[ = \; \frac{\displaystyle N\mu^2 H\beta\left(e^{\beta\mu H}\left(\frac{1}{\beta\mu H} - \frac{1}{\beta^2\mu^2H^2}\right) - e^{-\beta\mu H}\left(\left(\frac{-1}{\beta\mu H}\right) - \frac{1}{\beta^2\mu^2H^2}\right)\right)} {\left(e^{\beta\mu H} - e^{-\beta\mu H}\right)} \] \[ = N\mu\left[\frac{e^{\beta\mu H} + e^{-\beta\mu H}}{e^{\beta\mu H} - e^{-\beta\mu H}} \;-\; \frac{1}{\beta\mu H}\right] \] \[ \boxed{\,M \;=\; N\mu\left[\coth(\beta\mu H) \;-\; \frac{1}{\beta\mu H}\right]\,} \]

Previously when we only assumed \(\left(E = \pm\mu H\right)\)

\[ M = N\mu\tanh x \qquad\qquad x = \beta\mu H \]

\(\downarrow\) Now when we used all possible orientations

\[ M = N\mu\left(\coth x - \tfrac{1}{x}\right) \qquad\qquad = \text{Langevin functions.} \]

Let us plot \(\left(\frac{M}{N\mu}\right)\) vs \(\left(\coth x - \frac{1}{x}\right)\) vs \((x)\)

\usetikzlibrary{arrows.meta,decorations.pathreplacing,calc,angles,quotes,positioning,patterns}
\begin{tikzpicture}[scale=0.8,>=Stealth]
  \useasboundingbox (-3.6,-3.2) rectangle (3.8,3.4);
  \clip (-3.6,-3.2) rectangle (3.8,3.4);
  \draw[->] (-3.4,0) -- (3.5,0) node[right] {$(x)$};
  \draw[->] (0,-3.0) -- (0,3.2);
  \node[anchor=south east] at (-0.1,2.6) {$\dfrac{M}{N\mu}$};
  \node[anchor=south east] at (-0.1,3.0) {$(\coth x)$};
  \draw (-3.4,1) -- (3.4,1);
  \draw (-3.4,-1) -- (3.4,-1);
  \draw[thick] plot coordinates {(0.33,3.05) (0.4,2.6) (0.5,2.16) (0.7,1.65)
      (1.0,1.31) (1.5,1.10) (2.0,1.04) (2.6,1.01) (3.3,1.00)};
  \draw[thick] plot coordinates {(-3.3,-1.00) (-2.6,-1.01) (-2.0,-1.04) (-1.5,-1.10)
      (-1.0,-1.31) (-0.7,-1.65) (-0.5,-2.16) (-0.4,-2.6) (-0.33,-3.05)};
\end{tikzpicture}

\usetikzlibrary{arrows.meta,decorations.pathreplacing,calc,angles,quotes,positioning,patterns}
\begin{tikzpicture}[scale=0.8,>=Stealth]
  \useasboundingbox (-3.6,-2.4) rectangle (3.8,3.0);
  \clip (-3.6,-2.4) rectangle (3.8,3.0);
  \node[anchor=south] at (0,2.3) {$\left(\coth x - \frac{1}{x}\right)$};
  \draw[->] (-3.4,0) -- (3.5,0) node[right] {$(x)$};
  \draw[->] (0,-2.2) -- (0,2.2);
  \draw (-3.4,1) -- (3.4,1);
  \draw (-3.4,-1) -- (3.4,-1);
  \draw[thick] plot coordinates {(-3.4,-0.95) (-2.5,-0.87) (-1.8,-0.78)
      (-1.2,-0.62) (-0.6,-0.34) (0,0) (0.6,0.34) (1.2,0.62) (1.8,0.78)
      (2.5,0.87) (3.4,0.95)};
\end{tikzpicture}

\[ \begin{aligned} \coth x = \frac{\cosh x}{\sinh x} &= \frac{\left(1 + \frac{x^2}{2} + \cdots\right)}{x\left(1 + \frac{x^2}{6} + \cdots\right)}\\ &= \left(\frac{1}{x}\right)\cdot\left(1 + \frac{x^2}{2} + \cdots\right)\left(1 - \frac{x^2}{6} + \cdots\right)\\ &= \frac{1}{x}\left(1 + x^2\left(\frac{1}{2} - \frac{1}{6}\right) + \cdots\right)\\ &= \frac{1}{x}\left(1 + \frac{2x^2}{6}\right) \end{aligned} \] \[ \coth x \approx \frac{1}{x} + \frac{x}{3} \]

So \(\left(\coth x - \frac{1}{x}\right) \approx \left(\frac{x}{3}\right)\)

So,

\[ \frac{M}{N\mu} \;=\; \left(\coth\frac{\mu H}{k_BT} \;-\; \frac{k_BT}{\mu H}\right) \]

So, near origin \((H=0)\)

\[ \left.\frac{M}{N\mu}\right|_{H\to 0} \;\approx\; \frac{\mu H}{3k_BT} \]

So \(\left(\chi_T = \dfrac{N\mu^2}{3k_BT}\right)\) (\(\chi_T\) = Isothermal susceptibility)

\(\longrightarrow\) No of dimensions is `3' in our case

\(\left(\text{For 2-D}\right)\) :-- \(\mu\) is allowed to have any dirn. in a plane

\usetikzlibrary{arrows.meta,decorations.pathreplacing,calc,angles,quotes,positioning,patterns}
\begin{tikzpicture}[scale=0.8,>=Stealth]
  \draw[->] (0,0) -- (0.2,1.4);
  \node[left] at (0.05,1.2) {$\vec{H}$};
  \draw[->] (0,0) -- (1.5,1.0) node[right] {$\vec{\mu}$};
  \node at (0.55,0.85) {$\theta$};
  \draw (0.12,0.85) arc (81:34:0.9);
\end{tikzpicture}

\[ \begin{aligned} M &= N\langle \mu\rangle\\ M &= N\;\frac{\displaystyle\int_{0}^{2\pi} \mu\cos\theta\; e^{+\beta\mu H\cos\theta}\, d\theta} {\displaystyle\int_{0}^{2\pi} e^{\beta\mu H\cos\theta}\, d\theta} \end{aligned} \] \[ \frac{M}{N\mu} \;=\; \frac{\displaystyle\int_{0}^{2\pi}\cos\theta\; e^{\beta\mu H\cos\theta}\, d\theta} {\displaystyle\int_{0}^{2\pi} e^{\beta\mu H\cos\theta}\, d\theta} \qquad \longleftarrow \text{(Bessels function }J_0\text{)} \]

\usetikzlibrary{arrows.meta,decorations.pathreplacing,calc,angles,quotes,positioning,patterns}
\begin{tikzpicture}[scale=0.8,>=Stealth]
  \draw (0,0) .. controls (0.2,1.2) and (1.6,1.5) .. (1.9,0.6)
        .. controls (2.2,-0.4) and (1.0,-1.1) .. (0.4,-0.8)
        .. controls (-0.1,-0.55) and (-0.15,-0.6) .. (0,0);
  \draw[->] (0.35,-0.55) -- (0.35,0.05);
  \draw[->] (0.9,-0.6) -- (1.0,0.0);
  \draw[->] (1.45,-0.35) -- (1.35,0.25);
  \draw[->] (0.45,0.3) -- (0.55,0.9);
  \draw[->] (1.05,0.35) -- (0.95,0.95);
  \draw[->] (1.55,0.45) -- (1.65,1.0);
\end{tikzpicture}

\[ \frac{M}{N\mu} \;=\; \frac{\displaystyle\int \cos\theta\left(1 + \frac{\beta\mu H\cos\theta}{1} + \frac{(\beta\mu H\cos\theta)^2}{2!} + \cdots\right) d\theta} {\displaystyle\int_{0}^{2\pi}\left(1 + \beta\mu H\cos\theta + \frac{(\beta\mu H\cos\theta)^2}{2!} + \cdots\right) d\theta} \]

We are only interested in finding \(\left(\chi_T\right)\) i.e. slope at origin. So, lets expand \(\left(e^{\beta\mu H\cos\theta}\right)\) to first [?] order.

\[ \frac{M}{N\mu} \;=\; \frac{\displaystyle 0 + \beta\mu H\int_{0}^{2\pi}\cos^2\theta\, d\theta} {\displaystyle 2\pi + \beta\mu H\left(\underbrace{\int_{0}^{\pi}\cos\theta\, d\theta}_{\to 0}\right)} \;=\; \frac{\beta\mu H\,\pi}{2\pi} \] \[ = \frac{\beta\mu H}{2} \]

So slope

\[ \left.\frac{\partial M}{\partial H}\right|_{H\to 0} \;=\; \frac{N\beta\mu^2}{2} \;=\; \frac{\mu^2 N}{\left(2k_BT\right)} \qquad \longrightarrow \underline{\text{dimension is 2}}. \] \[ \boxed{\;\chi_T \;=\; \frac{N\mu^2}{2k_BT}\;} \]

what we saw that \(\left(\chi_T \propto \frac{1}{T}\right)\) but it is not valid at small temp \(\left(T\to 0 \text{ or } T=0\right)\). So we need to fix it.

Just like to go from ideal gas to real gas (use interactions) we used the van der vaals gas eqn, similarly here we use Weiss moleculer field theory.

The idea is that in the ferromagnetic medium each dipole moment experience not only the ext. field \(\vec{H}\) but also the internal mag. field due to other dipole moments.

\[ H \;\xrightarrow{\ \text{replace}\ }\; H_{\text{eff}} \;=\; H + \lambda\frac{M}{\boxed{N}} \]

\(\downarrow\) to make intensive quantity to add to H.

\usetikzlibrary{arrows.meta,decorations.pathreplacing,calc,angles,quotes,positioning,patterns}
\begin{tikzpicture}[scale=0.8,>=Stealth]
  \draw[->] (-1.6,0.2) -- (-1.6,1.2);
  \node[left] at (-1.65,0.9) {$\vec{H}$};
  \draw (0,0) .. controls (0.1,1.3) and (1.8,1.6) .. (2.1,0.5)
        .. controls (2.35,-0.6) and (1.0,-1.2) .. (0.4,-0.9)
        .. controls (-0.15,-0.6) and (-0.1,-0.55) .. (0,0);
  \draw[->] (0.45,-0.55) -- (0.45,0.05);
  \draw[->] (1.0,-0.6) -- (1.1,0.0);
  \draw[->] (1.55,-0.4) -- (1.45,0.2);
  \draw[->] (0.4,0.35) -- (0.5,0.95);
  \draw[->] (1.5,0.4) -- (1.6,1.0);
  \draw[->] (1.05,0.25) -- (1.05,0.85);
  \draw (1.05,0.05) circle (0.16);
  \draw[->] (1.05,-0.05) -- (1.05,0.2);
\end{tikzpicture}

\[ H_{int} \propto M \qquad\qquad H_{int} = \lambda M \] \[ M \;=\; N\mu\tanh\left(\frac{\mu\left(H + \frac{\lambda M}{N}\right)}{k_BT}\right) \qquad \text{(for 1-D)}. \]

This is transedental equation, we can only solve it by Numerical methods.

Let us see its behaviour near *\(H=0\)*.

\[ M_0 \;=\; N\mu\tanh\left(\frac{\mu\lambda M_0}{N k_B T}\right) \qquad \left(\begin{aligned} M_0 &= \text{magnetisation}\\ &\ \ \text{per unit}\\ &\ \ \text{dipole moments}\end{aligned}\right) \]

1. \(M_0 = 0\) is always a solution (Paramagnetic solution). \((H=0,\ M=0)\)
2. But ferromagnet have hysteresis.

   \usetikzlibrary{arrows.meta,decorations.pathreplacing,calc,angles,quotes,positioning,patterns}
   \begin{tikzpicture}[scale=0.9,>=Stealth]
     \draw[->] (-3.4,0) -- (3.5,0) node[right] {$H$};
     \draw[->] (0,-2.2) -- (0,2.3) node[above right] {$M$};
     \draw (-3.2,1.5) -- (3.2,1.5);
     \node[anchor=south east] at (-0.05,1.5) {$+1$};
     \draw (-3.2,-1.5) -- (3.2,-1.5);
     \node[anchor=north east] at (-0.05,-1.5) {$-1$};
     \draw[thick] (2.6,1.45) .. controls (0.6,1.3) and (-0.6,0.9) .. (-1.0,0.45)
           .. controls (-1.6,-0.2) and (-2.0,-1.4) .. (-2.6,-1.45)
           -- (-2.6,-1.45);
     \draw[thick] (-2.6,-1.45) .. controls (-0.6,-1.3) and (0.6,-0.9) .. (1.0,-0.45)
           .. controls (1.6,0.2) and (2.0,1.4) .. (2.6,1.45);
     \node[anchor=east] at (-0.1,0.5) {$M_0$};
     \draw[fill] (0,0.5) circle (0.04);
     \node[anchor=west] at (0.1,-0.5) {$-M_0$};
     \draw[fill] (0,-0.5) circle (0.04);
   \end{tikzpicture}
   

*Near \(M\approx 0\)*

\[ \begin{aligned} M_0 &= N\mu\left(\frac{\mu\lambda M_0}{N k_B T}\right)\\ &\approx \frac{N\mu^2\lambda M_0}{N k_B T} \;\approx\; \frac{\mu^2\lambda M_0}{k_B T} \end{aligned} \]

Lets plot \(\left(\dfrac{\mu^2\lambda M_0}{k_BT}\right)\) vs \(\left(M_0\right)\)

\(\left(\text{at high temp } T' > T_0,\ \text{slope is less}\right)\),
\(\left(\text{for smaller temp } T'' \text{ slope is even higher}\right)\).

\usetikzlibrary{arrows.meta,decorations.pathreplacing,calc,angles,quotes,positioning,patterns}
\begin{tikzpicture}[scale=0.9,>=Stealth]
  \draw[->] (-3.2,0) -- (3.4,0) node[right] {$M_0$};
  \draw[->] (0,-2.0) -- (0,2.2);
  \node[anchor=south west] at (0.05,1.8) {$\dfrac{\mu^2\lambda M_0}{k_BT}$};
  % tanh-like saturating curve
  \draw[thick] plot coordinates {(-2.9,-1.35) (-2.0,-1.3) (-1.2,-1.15)
      (-0.6,-0.85) (-0.25,-0.45) (0,0) (0.25,0.45) (0.6,0.85) (1.2,1.15)
      (2.0,1.3) (2.9,1.35)};
  % steeper straight line T''
  \draw (-1.6,-2.0) -- (1.6,2.0) node[above right] {$T''$};
  % shallower straight line T'
  \draw (-3.0,-1.5) -- (3.0,1.5) node[right] {$T'$};
  \draw (1.15,1.15) circle (0.11);
  \draw (-1.15,-1.15) circle (0.11);
  \draw[dashed] (1.15,1.15) -- (1.15,0);
  \draw[dashed] (-1.15,-1.15) -- (-1.15,0);
\end{tikzpicture}

Intersection happens if slope \(>1\)

\[ \frac{\mu^2\lambda}{k_BT} > 1 \qquad \text{or} \qquad \boxed{\,T < \frac{\mu^2\lambda}{k_B}\,} \]

So \(\left(T = \dfrac{\mu^2\lambda}{k_B}\right)\) is critical point.

The negative solution for \(M_0\) is unstable; so we only have one positive root of \(M_0\). Which leads to ferromagnetism.

So below a certain critical temp, the cooperative tendency of all these magnetic moments to align in same dir. will dominate over disrupting tendency of entropy \((S)\) and we have order.

\[ F = U - TS \qquad\qquad (U >> TS) \]

at low temp \((T)\) \(F \approx U\) (ordered phase) (ferromagnetic)

But above \((T_c)\) disorder wins and interaction is not strong enough to maintain order.

\[ F = U - TS \] \[ (TS >> U) \]

This is how phase transition occurs b/w paramagnet and ferromagnet. Now we will see how it is similer to case of fluids.

We can also have ferroelectric transitions of permanent electric dipole moments, which are accompanied by structural phase transitions (crystalline phase transitions which would lead to shapes of unit cells which have permanent electric dipole moments.

The behaviour near the critical region of eqn of state

\[ M = N\mu\tanh\left(\frac{\mu\left(H + \frac{\lambda M}{N}\right)}{k_BT}\right) \]

would be very similer to that of fluid near its

critical region. We would like to find out the non zero magnetisation \((M)\) for \(H=0\).

\[ M_0 = N\mu\tanh\left(\frac{\mu\lambda M_0}{N k_B T}\right) \]

(Digression) also our model \(M = N\mu\tanh\left(\frac{\mu\left(H + \frac{\lambda M}{N}\right)}{k_BT}\right)\) does not include hysteresis phenomena. Hysteresis actually arose from presence of multiple domains in magnetic material. These domains align in right direction in presence of strong \(\vec{H}\). For a single domain our model works perfectly as a single domain can't show hysteresis.

Family of Isotherms :--

\usetikzlibrary{arrows.meta,decorations.pathreplacing,calc,angles,quotes,positioning,patterns}
\begin{tikzpicture}[scale=0.9,>=Stealth]
  \node at (0,2.9) {(family of isotherms)};
  \draw[->] (-3.4,0) -- (3.6,0) node[right] {$H$};
  \draw[->] (0,-2.4) -- (0,2.6) node[above left] {$\frac{M}{N\mu}$};
  \draw (-3.2,1.6) -- (3.2,1.6);
  \draw (-3.2,-1.6) -- (3.2,-1.6);
  % T < Tc : steepest
  \draw[thick] plot coordinates {(-3.0,-1.55) (-1.6,-1.5) (-0.7,-1.35)
     (-0.2,-0.9) (0,0) (0.2,0.9) (0.7,1.35) (1.6,1.5) (3.0,1.55)};
  % T = Tc
  \draw[thick] plot coordinates {(-3.0,-1.5) (-1.8,-1.4) (-0.9,-1.15)
     (-0.4,-0.75) (0,0) (0.4,0.75) (0.9,1.15) (1.8,1.4) (3.0,1.5)};
  % T > Tc : shallowest
  \draw[thick] plot coordinates {(-3.0,-1.4) (-2.0,-1.25) (-1.2,-1.0)
     (-0.6,-0.6) (0,0) (0.6,0.6) (1.2,1.0) (2.0,1.25) (3.0,1.4)};
  \node[anchor=south] at (0.9,1.75) {\small $T<T_c$};
  \node[anchor=south] at (1.9,1.75) {\small $T=T_c$};
  \node[anchor=south] at (2.9,1.75) {\small $T>T_c$};
  \node[anchor=north] at (0.45,-0.1) {\small $H'$};
\end{tikzpicture}

\[ \left(\text{Critical Isotherm}\quad \frac{\partial M}{\partial H} = 0 \ \&\ \frac{\partial^2 M}{\partial H^2} = 0\right) \]

\usetikzlibrary{arrows.meta,decorations.pathreplacing,calc,angles,quotes,positioning,patterns}
\begin{tikzpicture}[scale=0.9,>=Stealth]
  \draw[->] (-2.8,0) -- (3.0,0) node[right] {$H$};
  \draw[->] (0,-2.2) -- (0,2.4) node[above right] {$\frac{M}{N\mu}$};
  \draw (-2.6,1.4) -- (2.8,1.4);
  \draw (-2.6,-1.4) -- (2.8,-1.4);
  % multivalued (unphysical) curve
  \draw[thick] plot coordinates {(-2.4,-1.35) (-1.4,-1.2) (-0.7,-0.95)
     (-0.9,-0.3) (-0.55,0.4) (0.55,-0.4) (0.9,0.3) (0.7,0.95)
     (1.4,1.2) (2.4,1.35)};
  % ordinary sigmoid
  \draw[thick] plot coordinates {(-2.4,-1.3) (-1.5,-1.15) (-0.8,-0.85)
     (-0.3,-0.5) (0,0) (0.3,0.5) (0.8,0.85) (1.5,1.15) (2.4,1.3)};
  \node[anchor=east] at (-0.95,0.6) {$A$};
  \node[anchor=west] at (0.95,-0.5) {$C$};
  \node[anchor=east] at (-2.0,-0.1) {Jump.};
\end{tikzpicture}

curve \(A\to C\) is unphysical as \(\left(H\uparrow \text{ in }-\text{ve direction } \&\ M\uparrow \text{ in +ve direction}\right)\).

\usetikzlibrary{arrows.meta,decorations.pathreplacing,calc,angles,quotes,positioning,patterns}
\begin{tikzpicture}[scale=0.9,>=Stealth]
  \draw[->] (0,0) -- (0,3.2) node[left] {$P$};
  \draw[->] (-0.2,0) -- (4.6,0) node[right] {$V$};
  \draw[thick] (0.35,3.0) .. controls (1.2,1.3) and (2.2,0.95) .. (4.2,0.85);
  \draw[thick] (0.35,2.5) .. controls (1.1,1.05) and (2.1,0.72) .. (4.2,0.62);
  \draw[thick] (0.35,2.0) .. controls (1.0,0.85) and (2.0,0.52) .. (4.2,0.42);
  \draw[thick] (0.35,1.5) .. controls (0.95,0.62) and (1.9,0.32) .. (4.2,0.25);
  \draw[fill] (1.35,1.35) circle (0.06);
  \node[anchor=south] at (1.15,1.55) {$(P_c,V_c)$};
\end{tikzpicture}

\(\longleftarrow\) Isotherms in case of fluid.

\[ \longrightarrow \left(\text{Critical Isotherm}\ \left.\frac{\partial P}{\partial V}\right|_{(P_c,V_c)} = 0,\ \&\ \left.\frac{\partial^2 P}{\partial V^2}\right|_{(P_c,V_c)} = 0\right) \]

let us plot \((M\) vs \(T)\)

\usetikzlibrary{arrows.meta,decorations.pathreplacing,calc,angles,quotes,positioning,patterns}
\begin{tikzpicture}[scale=0.9,>=Stealth]
  \draw[->] (0,-2.4) -- (0,2.6) node[left] {$M$};
  \draw[->] (-0.3,0) -- (4.4,0) node[right] {$T$};
  \draw[thick] plot coordinates {(0.15,2.1) (0.6,2.05) (1.2,1.9)
     (1.8,1.65) (2.3,1.3) (2.65,0.85) (2.85,0.4) (2.95,0)};
  \draw[thick] plot coordinates {(0.15,-2.1) (0.6,-2.05) (1.2,-1.9)
     (1.8,-1.65) (2.3,-1.3) (2.65,-0.85) (2.85,-0.4) (2.95,0)};
  \draw (2.95,0) circle (0.22);
  \node[anchor=north west] at (3.0,-0.05) {$T_c$};
  \node[anchor=south west] at (0.6,2.15) {$M_0(H=0)$ \ (ferro $\uparrow$)};
  \node[anchor=north west] at (0.6,-2.15) {$M_0$ \ $(M(H=0))$ \ (ferro $\downarrow$)};
  \draw[->] (3.2,0.3) -- (4.1,0.55);
  \node[anchor=west] at (4.1,0.55) {Critical temp or curic temp};
  \node[anchor=north] at (1.4,-2.6) {(ferromagnet)};
  \node[anchor=north] at (3.4,-2.6) {(paramagnet)};
\end{tikzpicture}

In case of (magnet)

In case of fluid

\usetikzlibrary{arrows.meta,decorations.pathreplacing,calc,angles,quotes,positioning,patterns}
\begin{tikzpicture}[scale=0.9,>=Stealth]
  \draw[->] (0,-0.3) -- (0,2.9);
  \draw[->] (-0.3,0) -- (4.2,0) node[right] {$T$};
  \draw[thick] plot coordinates {(0.3,0.5) (0.8,0.75) (1.35,1.05)
     (1.8,1.3) (2.0,1.45)};
  \draw[thick] plot coordinates {(0.3,2.4) (0.8,2.2) (1.35,1.9)
     (1.8,1.62) (2.0,1.45)};
  \draw (2.0,1.45) circle (0.22);
  \node[anchor=west] at (2.25,1.45) {$(P_c,V_c)$};
  \draw[dashed] (0.05,1.45) -- (1.78,1.45);
  \node[anchor=south] at (1.1,2.35) {gas};
  \node[anchor=north] at (1.0,0.55) {Liq};
\end{tikzpicture}

\(\longrightarrow\) discontinuous transitions

\(\longrightarrow\) parabolic region

\usetikzlibrary{arrows.meta,decorations.pathreplacing,calc,angles,quotes,positioning,patterns}
\begin{tikzpicture}[scale=0.9,>=Stealth]
  \draw[->] (0,0) -- (0,3.0) node[left] {$P$};
  \draw[->] (0,0) -- (4.4,0) node[right] {$T$};
  \draw[thick] (0.7,1.2) .. controls (1.2,1.7) and (1.5,1.9) .. (1.9,2.15);
  \draw[fill] (1.9,2.15) circle (0.06);
  \node[anchor=south west] at (1.9,2.15) {CP};
  \node[anchor=south east] at (1.15,1.7) {liq};
  \node[anchor=north west] at (0.85,1.15) {gas};
  \draw[dashed] (1.9,-0.35) -- (1.9,2.9);
  \draw[dashed] (2.35,0.35) -- (2.9,2.5);
  \node[anchor=west] at (2.55,2.45) {Homogeneous};
  \node[anchor=west] at (2.55,2.05) {fluid $\longrightarrow$};
\end{tikzpicture}

\usetikzlibrary{arrows.meta,decorations.pathreplacing,calc,angles,quotes,positioning,patterns}
\begin{tikzpicture}[scale=0.9,>=Stealth]
  \draw[->] (0.6,-2.2) -- (0.6,3.0) node[above] {$H$};
  \draw[->] (-0.4,0) -- (4.4,0) node[right] {$T$};
  \draw[thick] (-0.3,0) -- (2.3,0);
  \node[anchor=east] at (-0.35,0) {$H_c\to$};
  \node[anchor=south] at (1.3,0.08) {ferro $\uparrow$};
  \node[anchor=north] at (1.3,-0.08) {ferro $\downarrow$};
  \node[anchor=north west] at (2.3,-0.08) {$T_c$};
  \draw[dashed] (2.3,-0.6) -- (2.3,2.7);
  \draw[dashed] (2.75,0.4) -- (3.3,2.3);
  \node[anchor=west] at (3.0,2.3) {paramagnet};
\end{tikzpicture}

\[ \frac{dP}{dT} \;=\; \frac{L}{T\Delta V} \;=\; \frac{T\Delta S}{T\Delta V} \;=\; \frac{\Delta S}{\Delta V} \;=\; \frac{S_{gas} - S_{liq}}{v_{gas} - v_{liq}} \]

\(\downarrow\) specific volume

\[ \frac{dH}{dT} \;=\; \frac{\Delta S}{\Delta M} \]

In case of fluid as well as f. magnet, we have one line ending at critical pt. [?] from ferro \(\uparrow\) to ferro \(\downarrow\) [?] with \(\Delta S\) [?]

\usetikzlibrary{arrows.meta,decorations.pathreplacing,calc,angles,quotes,positioning,patterns}
\begin{tikzpicture}[scale=0.9,>=Stealth]
  \draw (0,0) ellipse (0.9 and 0.6);
  \foreach \x in {-0.45,0,0.45} {
    \draw[->] (\x,0.05) -- (\x,0.4);
    \draw[->] (\x,-0.4) -- (\x,-0.05);
  }
  \node[anchor=west] at (1.2,0.15) {has same};
  \node[anchor=west] at (1.2,-0.25) {entropy as};
  \begin{scope}[xshift=4.6cm]
    \draw (0,0) ellipse (0.9 and 0.6);
    \foreach \x in {-0.45,0,0.45} {
      \draw[->] (\x,0.4) -- (\x,0.05);
      \draw[->] (\x,-0.05) -- (\x,-0.4);
    }
  \end{scope}
\end{tikzpicture}

\[ \left(\frac{dH}{dT} \;=\; \frac{\Delta S}{\Delta M} \;=\; \frac{0}{\text{nonzero}} \;=\; 0\right) \quad\checkmark \]

let's find

\[ M_0 = N\mu\tanh\left(\frac{\mu\lambda M_0}{N k_B T}\right) \]

let \(\left(\dfrac{M_0}{N\mu} = m_0\right)\)

\[ m_0 = \tanh\left(\frac{\mu^2\lambda m_0}{k_B T}\right) \]

\usetikzlibrary{arrows.meta,decorations.pathreplacing,calc,angles,quotes,positioning,patterns}
\begin{tikzpicture}[scale=0.9,>=Stealth]
  \draw[->] (-2.8,0) -- (3.4,0) node[right] {$m_0$};
  \draw[->] (0,-1.9) -- (0,2.3) node[above left] {$m_0$};
  \draw[thick] plot coordinates {(-2.6,-1.35) (-1.8,-1.3) (-1.1,-1.15)
     (-0.6,-0.9) (-0.25,-0.5) (0,0) (0.25,0.5) (0.6,0.9) (1.1,1.15)
     (1.8,1.3) (2.6,1.35)};
  \draw (-1.35,-1.9) -- (1.35,1.9);
  \draw (-1.75,-1.75) -- (1.75,1.75);
  \draw (-2.6,-1.3) -- (2.9,1.45);
  \node[anchor=south west] at (0.85,1.25) {\small $T<T_c$};
  \node[anchor=west] at (1.8,1.75) {\small $T>T_c$};
  \node[anchor=north west] at (0.1,0.55) {\small $(T=T_c)$};
  \draw (1.02,1.13) circle (0.11);
  \draw (-1.02,-1.13) circle (0.11);
\end{tikzpicture}

\(T_c\) is given by when slope is *\(45^\circ\)*

\[ \left(\frac{\mu^2\lambda}{k_BT} = \underline{1}\right) \] \[ \left(T_c = \frac{\mu^2\lambda}{k_B}\right). \]

\((M_0=0)\) root is not stable for \((T<T_c)\) as magnet becomes ferromagnet instead of paramagnet. We can show it using \(F = U-TS\) \((T<T_c)\) \((F\approx U)\to\) ordered phase

\[ \chi_T \;=\; \left.\left(\frac{\partial M}{\partial H}\right)_T\right|_{H=0} \]

even though \((M)\) is discontinuous we can define \(\chi_T\) as slope are continuous.

\usetikzlibrary{arrows.meta,decorations.pathreplacing,calc,angles,quotes,positioning,patterns}
\begin{tikzpicture}[scale=0.9,>=Stealth]
  \draw[->] (-3.2,0) -- (3.4,0) node[right] {$H$};
  \draw[->] (0,-1.9) -- (0,2.3) node[above right] {$M_0$};
  \draw (-3.0,1.5) -- (3.0,1.5);
  \draw (-3.0,-1.5) -- (3.0,-1.5);
  \draw[thick] (2.6,1.45) .. controls (0.7,1.3) and (-0.35,0.95) .. (-0.55,0.5)
     .. controls (-0.9,-0.3) and (-1.6,-1.35) .. (-2.6,-1.45);
  \draw[thick] (-2.6,-1.45) .. controls (-0.7,-1.3) and (0.35,-0.95) .. (0.55,-0.5)
     .. controls (0.9,0.3) and (1.6,1.35) .. (2.6,1.45);
  \draw (-0.95,0.35) -- (-0.25,0.75);
  \draw (0.25,-0.75) -- (0.95,-0.35);
\end{tikzpicture}

*Lets plot \(\left(\chi_T \text{ vs } T\right)\)*

\usetikzlibrary{arrows.meta,decorations.pathreplacing,calc,angles,quotes,positioning,patterns}
\begin{tikzpicture}[scale=0.9,>=Stealth]
  \useasboundingbox (-0.8,-0.8) rectangle (5.2,3.4);
  \clip (-0.8,-0.8) rectangle (5.2,3.4);
  \draw[->] (0,0) -- (0,3.2) node[left] {$\chi_T$};
  \draw[->] (0,0) -- (5.0,0) node[right] {$T$};
  \draw[thick] plot coordinates {(0.5,0) (0.6,0.5) (0.8,1.0) (1.1,1.5)
     (1.5,2.1) (1.75,2.7) (1.85,3.1)};
  \draw[thick] plot coordinates {(2.05,3.1) (2.15,2.5) (2.35,1.9)
     (2.7,1.35) (3.2,0.95) (3.9,0.65) (4.7,0.5)};
  \draw[dashed] (1.95,0) -- (1.95,3.1);
  \node[anchor=north] at (1.95,-0.05) {$T_c$};
\end{tikzpicture}

\[ m_0 = \tanh\left(\frac{\mu^2\lambda\, m_0}{k_B T}\right) \;=\; \tanh\left(\frac{T_c m_0}{T}\right) \]

at \(T=T_c\) \(m_0 = \tanh m_0\) \(\Rightarrow\) \((m_0 = 0)\) is only solution (paramagnet).

but below \((T<T_c)\) we have other solutions.

let us expand \(\tanh x\), near [?]

\[ \tanh x = \frac{\sinh x}{\cosh x} = \frac{e^x - e^{-x}}{e^x + e^{-x}} = \frac{\left(1+x+\frac{x^2}{2!}+\cdots\right) - \left(1-x+\frac{x^2}{2!}-\frac{x^3}{3!}+\cdots\right)} {\left(1+x+\frac{x^2}{2!}+\frac{x^3}{3!}\right) + \left(1-x+\frac{x^2}{2!}-\frac{x^3}{3!}+\cdots\right)} \]

\usetikzlibrary{arrows.meta,decorations.pathreplacing,calc,angles,quotes,positioning,patterns}
\begin{tikzpicture}[scale=0.9,>=Stealth]
  \draw[->] (0,-0.8) -- (0,2.2);
  \node[anchor=east] at (-0.05,1.6) {$m_0$};
  \draw[->] (-0.3,0.9) -- (2.6,0.9) node[right] {$T$};
  \draw[thick] (0.1,1.9) .. controls (0.6,1.85) and (1.0,1.5) .. (1.15,0.9);
  \draw (1.15,0.9) circle (0.16);
  \node[anchor=north] at (1.15,0.7) {$T_c$};
\end{tikzpicture}

\[ \begin{aligned} &= \frac{2\left(x + \frac{x^3}{3!}\right)}{2\left(1 + \frac{x^2}{2!} + \frac{x^4}{4!} + \cdots\right)} \;=\; \left(x + \frac{x^3}{3!} + \cdots\right)\left(1 - \frac{x^2}{2!} - \cdots\right)\\ &= \left(x - \frac{x^3}{2!} + \frac{x^3}{3!} - \frac{x^5}{3!\,2!} + \cdots\right) \end{aligned} \] \[ \tanh x \;\approx\; x + x^3\left(\frac{1}{6} - \frac{1}{2}\right) \;=\; x - \frac{x^3}{3} + \cdots \] \[ m_0 = \tanh\left(\frac{T_c}{T} m_0\right) \] \[ m_0 \approx \frac{T_c}{T} m_0 - \frac{1}{3}\left(\frac{T_c}{T} m_0\right)^3 \]

for \(m_0\) other than zero,

\[ \begin{aligned} m_0\left(1 - \frac{T_c}{T}\right) &\approx -\frac{1}{3}\left(\frac{T_c}{T} m_0\right)^3\\ 1 - \frac{T_c}{T} &\approx -\frac{1}{3}\left(\frac{T_c}{T}\right)^3 m_0^2\\ \left(\frac{T_c - T}{T}\right) &\approx \frac{1}{3}\left(\frac{T_c}{T}\right)^3 m_0^2 \end{aligned} \]

\((T<T_c)\) if \(T\) is very near to \(T_c\) \(\left(\frac{T_c}{T}\approx 1\right)\)

\[ \Rightarrow \boxed{\;m_0 \sim \pm\sqrt{T_c - T}\;} \] \[ m_0 \approx \pm\left(T_c - T\right)^{1/2} \longrightarrow \text{mean field critical exponent } (\beta) \]

here \(\left(\beta \neq \frac{1}{k_BT}\right)\) (Just a symbol)

\[ \left.\frac{dm_0}{dT}\right|_{T=T_c} \;=\; \frac{(-1)}{\pm 2\sqrt{T_c - T}} \;\longrightarrow\; \infty. \]

So behaviour of \(m_0\) near \(T_c\) is

\usetikzlibrary{arrows.meta,decorations.pathreplacing,calc,angles,quotes,positioning,patterns}
\begin{tikzpicture}[scale=0.9,>=Stealth]
  \draw[->] (0,-0.5) -- (0,2.4);
  \node[anchor=east] at (-0.05,1.9) {$m_0$};
  \draw[->] (-0.3,0.7) -- (3.2,0.7) node[right] {$T$};
  \draw[thick] (0.1,2.0) .. controls (0.7,1.95) and (1.25,1.7) .. (1.5,0.7);
  \node[anchor=north] at (1.5,0.6) {$T_c$};
  \node[anchor=west] at (1.75,1.7) {$\left(\left.\dfrac{dm_0}{dT}\right|_{T_c} = \infty\right)$};
\end{tikzpicture}

Exactly same relation arose in case of fluid.

\[ \left(\left(v - v_c\right) \sim \pm\sqrt{T_c - T}\right) \]

\usetikzlibrary{arrows.meta,decorations.pathreplacing,calc,angles,quotes,positioning,patterns}
\begin{tikzpicture}[scale=0.9,>=Stealth]
  \draw[->] (0,-0.3) -- (0,2.8) node[left] {$V$};
  \draw[->] (-0.3,0) -- (3.4,0) node[right] {$T$};
  \node[anchor=east] at (-0.05,1.4) {$v_c$};
  \draw[dashed] (0.05,1.4) -- (2.0,1.4);
  \draw[thick] (0.35,0.15) .. controls (1.3,0.25) and (2.0,0.6) .. (2.05,1.4)
     .. controls (2.0,2.2) and (1.3,2.55) .. (0.35,2.65);
  \draw[fill] (2.05,1.4) circle (0.06);
  \node[anchor=north] at (2.05,-0.05) {$T_c$};
\end{tikzpicture}

\(\left(\underline{\text{H.W}}\right)\) show that \(\left(v - v_c\right) \sim \pm\sqrt{T_c - T}\) ).

we can see that (in both cases critical exponent is \(\left(\frac{1}{2}\right)\)) which arose because (both models are examples of mean field theory models. i.e. In van der vaals case we assume \(P \neq \frac{RT}{v-b}\) but \(P = \frac{RT}{v-b} - \frac{a}{v^2}\) because of attraction of each molecules. In same way

here we assume every magnetic moment another magnetic moment and internal \(\vec{H}\) is proportional to magnetisation itself, we have exactly the same philosopy and that is what led to critical exponent.

\((\underline{Q})\) Does experiments agree with critical exponent to be \((1/2)\) or how accuratly we can get closer to \(T_c\) and controll temp. to see it.

Ans \(\sim\) In most cases we essentially see exponent closer to \(1/2\) but, but, if we do careful experiments near \(T_c\) we do not see half at all. we see \((1/3\) or \(0.3)\) for it we have to get very--very close to \(T_c\) to see it.

Temp is one of hardest thing to controlt, best we can controll is of (milli-degree) of accuracy.

Frequency is most accurately measurable quantity, we can measure frequency in one part in \(10^{15}\). There are astronomical objects/clocks (pulsars) whose rate of change of time period can be computed to one part in \(10^{19}\).

\((\underline{Q})\) How does critical Isotherm behave? (near \((V_c,P_c)\)).

\usetikzlibrary{arrows.meta,decorations.pathreplacing,calc,angles,quotes,positioning,patterns}
\begin{tikzpicture}[scale=0.9,>=Stealth]
  \draw[->] (0,0) -- (0,3.4) node[left] {$P$};
  \draw[->] (0,0) -- (4.6,0) node[right] {$V$};
  \draw[thick] (0.55,3.2) .. controls (1.1,1.9) and (1.9,1.7) .. (4.2,1.55);
  \draw[thick] (0.55,2.7) .. controls (1.15,1.55) and (1.9,1.25) .. (4.2,1.15)
     node[right] {\small $T=T_c$};
  \draw[thick] (0.55,2.2) .. controls (1.2,1.15) and (1.9,0.85) .. (4.2,0.78);
  \draw[fill] (1.35,1.75) circle (0.06);
  \node[anchor=south east] at (1.3,1.8) {\small $(V_c,P_c)$};
  \draw[dashed] (0.05,1.75) -- (1.33,1.75);
  \node[anchor=east] at (-0.05,1.75) {$P_c$};
  \draw[dashed] (1.35,0.05) -- (1.35,1.73);
  \node[anchor=north] at (1.35,-0.05) {$V_c$};
  \draw[->] (1.35,1.75) -- (1.35,3.1) node[above] {\small $(p)$};
  \draw[->] (1.35,1.75) -- (2.9,1.75) node[right] {\small $(v)$};
\end{tikzpicture}

shifting origin to see behaviour near \((V_c,P_c)\)

\[ \left(\frac{\partial P}{\partial V}\right)_{T_c} \;=\; \left(\frac{\partial^2 P}{\partial V^2}\right)_{T_c} \;=\; 0 \] \[ p = \frac{P - P_c}{P_c} \qquad\qquad v = \frac{V - V_c}{V_c} \]

The curve near \((V_c,P_c)\) is like

\[ \left(p \sim |v|^{\delta}\right) \longrightarrow \text{mean field critical exponent } (\delta) \]

or \(\left(|v| \sim p^{1/3}\right)\)

lets look at critical isotherm for ferro magnetic transitions

\[ M = N\mu\tanh\left(\frac{\mu\left(H + \frac{\lambda M}{N}\right)}{k_BT}\right) \]

let \(m = \dfrac{M}{N\mu}\)

\[ \begin{aligned} m &= \tanh\left(\frac{\mu}{k_BT}\left(H + \lambda m\mu\right)\right)\\ &= \tanh\left(\frac{\lambda\mu^2}{k_BT}\left(\frac{H}{\lambda\mu} + m\right)\right)\\ &= \tanh\left(\frac{T_c}{T}\left(h+m\right)\right) \end{aligned} \]

let \(\left(\dfrac{H}{\lambda\mu} = h\right)\)

for critical isotherm \((T=T_c)\)

\[ \boxed{\;m = \tanh(h+m)\;} \]

\usetikzlibrary{arrows.meta,decorations.pathreplacing,calc,angles,quotes,positioning,patterns}
\begin{tikzpicture}[scale=0.9,>=Stealth]
  \draw[->] (-2.6,0) -- (3.6,0) node[right] {$h$};
  \draw[->] (0,-2.2) -- (0,2.4) node[above right] {$m$};
  \draw[thick] plot coordinates {(-1.8,-2.1) (-1.1,-1.7) (-0.55,-1.25)
     (-0.15,-0.6) (0,0) (0.15,0.6) (0.55,1.25) (1.1,1.7) (1.8,2.1)};
  \draw[thick] plot coordinates {(-2.4,-2.05) (-1.5,-1.6) (-0.8,-1.15)
     (-0.3,-0.6) (0,0) (0.3,0.6) (0.8,1.15) (1.5,1.6) (2.4,2.05)};
  \draw[thick] plot coordinates {(-2.4,-1.6) (-1.6,-1.35) (-0.9,-1.0)
     (-0.4,-0.55) (0,0) (0.4,0.55) (0.9,1.0) (1.6,1.35) (2.9,1.55)};
  \draw (0,0) circle (0.16);
  \node[anchor=south east] at (0.15,1.55) {\small $T<T_c$};
  \node[anchor=south west] at (0.55,1.55) {\small $(T=T_c)$};
  \node[anchor=west] at (1.7,1.3) {\small $T>T_c$};
\end{tikzpicture}

\(\longrightarrow\) critical isotherm

at \((h=0)\), \((m=0)\) \(m\) is zero. (for \(T>T_c\))

\[ m = \tanh(h+m) \;\approx\; (h+m) - \frac{(h+m)^3}{3} \] \[ \Rightarrow \qquad (m+h)^3 \;\approx\; h \]

\(h\) is much smaller (near origin) so \(m^3 \sim h\)

\[ \boxed{\;m \sim h^{1/3}\;} \qquad\qquad (h \text{ is field variable}) \]

Just like \(\left(|v| \sim p^{1/3}\right)\) (\(p\) is field variable)

\((m,v)\) are responce variables.

Similerly, \(\chi_T\) and compressibility behave exactly same near critical point.

\[ \chi_T \;=\; \left.\left(\left.\frac{\partial M}{\partial H}\right|_T\right)\right|_{\underline{H=0}} \qquad ; \] \[ k_T \;=\; \frac{-1}{v}\frac{\partial v}{\partial P} \] \[ \frac{\partial P}{\partial v} = 0 \ \text{at critical point} \]

\usetikzlibrary{arrows.meta,decorations.pathreplacing,calc,angles,quotes,positioning,patterns}
\begin{tikzpicture}[scale=0.8,>=Stealth]
  \draw[->] (0,0) -- (0,1.5) node[left] {$P$};
  \draw[->] (0,0) -- (2.4,0) node[right] {$v$};
  \draw[thick] (0.25,1.35) .. controls (0.7,0.55) and (1.2,0.5) .. (2.1,0.4);
\end{tikzpicture}

\(\left(\begin{aligned}&\text{curie}\\ &\text{weiss law}\end{aligned}\right) \Longleftarrow\)

\[ \chi_T \;=\; \frac{1}{|T-T_c|} \] \[ \left(\left.\chi_T\right|_{T=T_c} = \infty\right) \qquad \longleftrightarrow \qquad \left(k_T = \infty\right) \] \[ \chi_T \;=\; \frac{1}{|T-T_c|^{\gamma}} \longrightarrow \gamma \ \text{critical point} \]

\(\left(\gamma \neq C_P/C_V\right)\) just a symbol.

Experimental values of \((\gamma \sim 1.3)\).

All these phase transitions can be explained by Landau's theory ((1937). (Idea of brocken symmetry leads to such phase transitions of all kinds of systems).