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Lecture 27: Phase Transitions

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 parameter in thermodynamics.

There is a place where thermodynamics fails (completely and no longer valid); it is hard problem of phase transitions of matter.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

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

We will use vander Waals gas to demonstrate real gases.

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

Phase diagram

\usetikzlibrary{arrows.meta,decorations.pathreplacing,calc,angles,quotes}
\begin{tikzpicture}[scale=1]
\draw[->] (0,0) -- (0,4) node[above] {$P$};
\draw[->] (0,0) -- (5,0) node[right] {$T$};
\draw[thick] (0.8,0) .. controls (1.0,2.0) .. (1.6,3.8); % solid-liquid (b)
\draw[thick] (0.3,0.3) .. controls (0.8,0.3) .. (0.8,0); % solid-gas curve (c) below triple point
\draw[thick] (0.8,0) .. controls (1.5,0.6) .. (3.2,1.8); % liquid-gas (a)
\filldraw (0.8,0) circle (1.5pt) node[below] {(Triple point)};
\filldraw (3.2,1.8) circle (1.5pt) node[right] {(a) $T_c$};
\node at (1.0,3.3) {solid};
\node at (0.4,1.0) {(crystalline)};
\node at (2.6,2.5) {liquid};
\node at (2.2,0.7) {gas};
\node at (4.0,1.2) {Homogeneous};
\node at (4.0,0.9) {fluid phase};
\node[right] at (5.2,3.5) {(a) $\to$ liquid-gas coexisty curve};
\node[right] at (5.2,3.1) {\quad or (Boiling curve)};
\node[right] at (5.2,2.6) {(b) $\to$ solid-liquid coexisty};
\node[right] at (5.2,2.2) {\quad curve or melting curve};
\node[right] at (5.2,1.7) {(c) $\to$ solid-gas coexistat};
\node[right] at (5.2,1.3) {\quad curve or sublimation};
\node[right] at (5.2,0.9) {\quad (curve).};
\end{tikzpicture}

Every point on this phase plane is assumed to be at thermal eqm state of the system.

(phase \(\to\) A homogeneous thermodynamic eqm system)

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

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

(Gibbs phase rule): For one kind of molecular species we can at most have 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.

Critical Point \((V_c, T_c, P_c)\)

The boiling curve/liquid-gas coexisty 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 & gas phase. (Homogeneous fluid phase).

To convert liquid to gas phase, we supply latent heat. This transition is discontinuous as density of system changes discontinuously.

\usetikzlibrary{arrows.meta,decorations.pathreplacing,calc,angles,quotes}
\begin{tikzpicture}
\draw[->] (0,0) -- (0,3) node[above] {$P$};
\draw[->] (0,0) -- (4,0) node[right] {$T$};
\draw[->] (0.5,0.5) -- (2.5,2.5) node[midway,above,sloped]{$L_2$};
\draw[->] (0.5,0.3) -- (2.5,1.2) node[midway,below,sloped]{$L_1$};
\node at (2.6,2.6) {$(P_c,T_c)$};
\node at (0,-0.4) {$(L_2 < L_1)$};
\end{tikzpicture}

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.

Let's draw melting curve in the \((V-T)\) plane

\usetikzlibrary{arrows.meta,decorations.pathreplacing,calc,angles,quotes}
\begin{tikzpicture}[scale=0.9]
\draw[->] (0,0) -- (0,4) node[above] {$V$};
\draw[->] (0,0) -- (5,0) node[right] {$T$};
\draw[thick] (0.5,3.6) .. controls (1.5,1.2) .. (2.2,1.0) -- (2.2,0.4);
\draw[thick] (2.2,0.4) .. controls (3.0,1.0) .. (3.8,3.6);
\filldraw (2.2,1.0) circle (1.5pt);
\node at (2.7,1.2) {$(V_c,T_c)$};
\node at (4.0,3.4) {gas};
\node at (0.7,0.6) {liquid};
\node at (2.2,-0.4) {$(T_c)$};
\node at (2.2,2.2) {region of coexistance};
\node at (2.2,1.9) {of liquid \& gas};
\end{tikzpicture}

This boiling curve becomes region in \(V\)-\(T\) diagram (homogeneous fluid phase).

\usetikzlibrary{arrows.meta,decorations.pathreplacing,calc,angles,quotes}
\begin{tikzpicture}[scale=0.9]
\draw[->] (0,0) -- (0,2.5) node[above] {$\rho$};
\draw[->] (0,0) -- (3,0) node[right] {$T$};
\draw[thick] (0.3,2.2) .. controls (1.2,1.6) .. (1.5,1.3);
\draw[thick] (1.5,0.7) .. controls (1.8,0.4) .. (2.6,0.1);
\draw[dashed] (1.5,0) -- (1.5,1.3);
\filldraw (1.5,1.3) circle (1.2pt);
\filldraw (1.5,0.7) circle (1.2pt);
\node at (1.5,-0.4) {$T_c$};
\node[right] at (3.1,1.4) {liquid};
\node[right] at (3.1,0.9) {\& gas};
\end{tikzpicture}

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

Critical point is 2\(^{\text{nd}}\) order or continuous phase transition, and coexisting curve is series of discontinuous / 1\(^{\text{st}}\) order phase transitions.

P-V diagram (only for boiling curve)

\usetikzlibrary{arrows.meta,decorations.pathreplacing,calc,angles,quotes}
\begin{tikzpicture}[scale=0.9]
\draw[->] (0,0) -- (0,4) node[above] {$P$};
\draw[->] (0,0) -- (6,0) node[right] {$V$};
\foreach \k in {0,1,2,3} {
  \draw[domain=0.6:5.5,smooth,variable=\x]
    plot ({\x},{1.6+\k*0.35 + 3.0/(\x+0.6-0.2*\k)});
}
\draw[thick] (1.0,0.6) .. controls (1.6,3.0) and (3.4,3.0) .. (4.0,0.6);
\filldraw (2.5,2.35) circle (1.5pt);
\node[right] at (2.6,2.4) {$(P_c,V_c)$};
\node[right] at (5.6,3.4) {$(T>T_c)$ (ideal gas approx).};
\node[right] at (5.6,2.9) {$T>T_c$};
\node[right] at (5.6,2.4) {$T=T_c$ (critical isotherm).};
\draw[thick] (1.0,0.6) -- (4.0,0.6);
\node at (2.5,0.35) {A};
\end{tikzpicture}

wrong curves as \(\dfrac{\partial P}{\partial V} > 0\) [?] (violates principle of stability).

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

To find critical point (inflection point), from van der Waals equation of state,

\[ \frac{\partial P}{\partial V}\bigg|_{(T_c,V_c)} = 0 \]

also

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

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

There is no universal eq. of state as eq. of state depends upon interactions and interactions are hard to model universally.

Lecture 27 (continued)

The portion where \(\dfrac{\partial P}{\partial V}\) is positive makes compressibility negative and violates thermodynamic stability. So we have to correct this portion.

\[ k_B = -\frac{1}{V}\frac{\partial V}{\partial P} \] 
\usetikzlibrary{arrows.meta,decorations.pathreplacing,calc,angles,quotes}
\begin{tikzpicture}[scale=0.9]
\draw[->] (0,0) -- (0,3) node[above] {$P$};
\draw[->] (0,0) -- (5,0) node[right] {$V$};
\draw[thick] (0.5,2.5) .. controls (1.0,0.5) and (1.5,2.2) .. (2.2,0.8) .. controls (3.0,0.2) .. (4.5,1.5);
\node at (3,2.0) {hysteresis is the problem here};
\node at (3,1.7) {to avoid, it is observed that};
\end{tikzpicture}

tieline transition occurs such that Area are equal (flip).

\usetikzlibrary{arrows.meta,decorations.pathreplacing,calc,angles,quotes}
\begin{tikzpicture}[scale=0.9]
\draw[->] (0,0) -- (0,2.5) node[above] {$P$};
\draw[->] (0,0) -- (5,0) node[right] {$V$};
\draw[thick] (0.5,2.2) .. controls (1.0,0.5) and (1.5,1.8) .. (2.0,0.9);
\draw[thick] (2.0,0.9) -- (3.3,0.9);
\draw[thick] (3.3,0.9) .. controls (3.8,0.5) .. (4.4,0.3);
\node at (1.5,1.5) {$A_1$};
\node at (2.7,0.7) {$A_2$};
\node at (1.5,-0.4) {(tieline)};
\node at (0,-0.4) {$P = \dfrac{NRT}{V-Nb}-\dfrac{aN^2}{V^2}$};
\end{tikzpicture}

So some portions of allowed curve also gets omitted in observation. So vander Waals gas eqn. 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}
\begin{tikzpicture}[scale=0.9]
\draw[->] (0,0) -- (0,3.5) node[above] {$P$};
\draw[->] (0,0) -- (5.5,0) node[right] {$V$};
\draw[thick] (0.6,0.6) .. controls (1.4,2.6) and (2.8,2.6) .. (3.6,0.6);
\draw[thick,domain=0.5:5.2,smooth,variable=\x] plot ({\x},{1.6+2.4/(\x+0.5)});
\draw[thick,domain=0.7:5.2,smooth,variable=\x] plot ({\x},{1.0+1.6/(\x+0.5)});
\filldraw (2.1,2.6) circle (1.5pt) node[above] {$(P_c,V_c)$};
\node at (0.9,1.3) {liquid};
\node at (4.8,1.0) {gas};
\node at (3.0,3.1) {$T>T_c$};
\node at (2.4,0.5) {$T<T_c$};
\node at (1.4,1.6) {metastable};
\node at (2.1,1.4) {unstable};
\node at (2.1,1.1) {region};
\end{tikzpicture}

(system separates slowly into gas & liquid) (system separates to liquid & gas abruptly).

Any system on unstable region will instantaneously separate into some liquid & remaining to gas depending upon lever rule.

Note: In case of binary alloys (liquid alloys), The metastable portion can be made extremely slow, it may take hours, years to happen.

Maxwell's Equal Area Rule

\usetikzlibrary{arrows.meta,decorations.pathreplacing,calc,angles,quotes}
\begin{tikzpicture}[scale=0.9]
\draw[->] (0,0) -- (0,3) node[above] {$P$};
\draw[->] (0,0) -- (5,0) node[right] {$V$};
\draw[thick] (0.5,2.6) .. controls (1.2,0.6) and (1.6,1.6) .. (2.0,1.2)
  .. controls (2.4,0.9) and (2.6,1.5) .. (3.0,1.2) .. controls (3.6,0.7) .. (4.4,0.4);
\node at (0.9,2.2) {A};
\node at (1.4,1.9) {$A_1$};
\node at (1.9,1.4) {B};
\node at (2.4,1.0) {C};
\node at (1.4,0.8) {$A_2$};
\node at (1.2,0.6) {E};
\draw[dashed] (1.7,0.4) -- (1.7,2.0);
\node at (3.0,1.4) {D};
\end{tikzpicture}

(Maxwell's tieline construction) — "Isn't quite correct though as It goes through \((D\to B)\), i.e. from thermodynamic unstable region."

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

\[ dG = dU - TdS - SdT + PdV + VdP \]

Since \(dU = TdS - PdV + \mu dN\),

\[ dG = -SdT + VdP + \mu dN \]

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

\[ G_E - G_A = 0 = \int_A^E dG = \int_A^B VdP + \int_B^C VdP + \int_C^D VdP + \int_D^E VdP \]

\(G = \mu\) should be same in gas phase or liquid phase.

\[ \left(\int_A^B VdP - \int_C^B VdP\right) + \left(-\int_C^D VdP + \int_E^D VdP\right) = 0 \] \[ \underbrace{\phantom{xxxxxxxxxxx}}_{(A_1)} \;+\; \underbrace{\phantom{xxxxxxxxxxx}}_{(-A_2)} \] \[ \boxed{A_1 = A_2} \Rightarrow A \]

Lecture 27 (continued): Mechanical Example of Hysteresis

Mechanical example of hysteresis (Double well potentials).

\usetikzlibrary{arrows.meta,decorations.pathreplacing,calc,angles,quotes}
\begin{tikzpicture}[scale=0.55]
\foreach \i/\lab/\note in {0/a/{},1/b/{$(L\to R)$},2/c/{},3/d/{},4/e/{},5/f/{$(R\to L)$},6/g/{}}{
  \begin{scope}[xshift=\i*2.3cm]
    \draw[domain=-1:1,smooth,variable=\x] plot ({\x},{1.2*\x^4-1.0*\x^2+0.5});
    \node at (0,-0.6) {(\lab)};
    \node at (0,-1.0) {\note};
  \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 (a), 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 eqm.

Lecture 28

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).

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\begin{tikzpicture}[scale=0.8]
\draw[->] (0,0) -- (0,3) node[above] {$P$};
\draw[->] (0,0) -- (4,0) node[right] {$T$};
\draw[thick] (0.6,0.5) .. controls (1.4,1.5) .. (2.4,2.2);
\draw[->] (1.6,1.6) circle (0.5);
\node at (1.6,1.6) {(A$\to$A)};
\node at (0.8,0.3) {gas};
\node at (0.6,1.5) {liquid};
\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 a phase, less symmetric it is; in the sense that the set of transformations under which it remains unchanged is smaller and 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.

Model of Paramagnetism

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}
\begin{tikzpicture}[scale=0.9]
\draw (0,0) rectangle (3,3);
\foreach \p in {(0.5,0.5),(1.2,0.6),(2.0,0.4),(2.6,1.0),(0.4,1.6),(1.4,1.6),(2.3,1.8),(0.6,2.4),(1.6,2.5),(2.4,2.5)}{
  \draw[->] \p -- ++(0.2,0.3);
}
\draw[->] (3.4,0.2) -- (3.4,2.8) node[above] {$\vec H$};
\node[right] at (3.6,1.5) {(auxiliary field)};
\node[right] at (3.3,3.1) {$N$ dipole moments, $\vec \mu_i$};
\node[right] at (3.3,-0.3) {placed in Heat bath at temp $T$.};
\end{tikzpicture}

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

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

Let's look at very simple model, such that

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

(magnetic dipole moment of \(e^-\) can only take two values), one along \(\vec H\).

\[ \begin{array}{ll} E_1 & (\mu H) \\ E_0 & (-\mu H) \end{array} \]

Let's find magnetisation (or better average magnetisation), in \((1\)-\(D)\).

\[ M = \sum_{i=1}^N \mu_i \quad \text{(Avg dipole moment of each dipole moment)}. \]

\(\big(N\langle\mu\rangle\) is only true iff all \(\mu\) are independent, since we only want to see \(\bar M\) along dir. of \(\vec H\), let's drop vector, assuming no dipole-dipole interactions\(\big)\).

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

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

So,

\[ \boxed{M = N\mu\tanh\left(\frac{\mu H}{k_BT}\right)} \quad \text{is eq.\ of state.} \]
fluid \((P,V,T)\) \(\longleftrightarrow\) magnetic \((H,M,T)\)
field \(\longleftrightarrow\) response
(extensive) \(\longleftrightarrow\) (Intensive) field \(\longleftrightarrow\) response (extensive)

\((M/N\mu)\) vs \(H\)

\usetikzlibrary{arrows.meta,decorations.pathreplacing,calc,angles,quotes}
\begin{tikzpicture}[scale=1]
\draw[->] (-3,0) -- (3,0) node[right] {$H$};
\draw[->] (0,-1.5) -- (0,1.5) node[above] {$M/N\mu$};
\draw[thick,domain=-3:3,smooth,variable=\x] plot ({\x},{tanh(1.2*\x)});
\draw[dashed] (-3,1) -- (3,1) node[right]{$+1$};
\draw[dashed] (-3,-1) -- (3,-1) node[right]{$-1$};
\node at (1.3,0.4) {linear};
\node at (1.3,0.1) {region};
\end{tikzpicture}

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

Our old defn. of susceptibility \((\chi)\): \(\vec M \propto \vec H\), \(\vec M = \chi \vec H\), \(\vec M = \) const \(\vec H\) ... is not quite right as, this graph is non linear in most range of \(H\).

Isothermal susceptibility

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

or \((\chi_T = \) slope at origin\()\)

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

\(\tanh(\beta\mu H) \simeq \beta\mu H\) for very small \(H\); \(\tanh(ax) \simeq ax\) for very small \(x\).

\[ \chi_T = \frac{N\mu^2}{k_BT} \;\to\; \underline{\text{Curie's law}} \quad \left(\chi_T \propto \frac{1}{T}\right) \quad \text{(Pierre Curie)}. \]

So slope at origin increases with decrease in temp.

\usetikzlibrary{arrows.meta,decorations.pathreplacing,calc,angles,quotes}
\begin{tikzpicture}[scale=1]
\draw[->] (-3,0) -- (3,0) node[right] {$H$};
\draw[->] (0,-1.5) -- (0,1.5) node[above] {$M/N\mu$};
\draw[thick,domain=-3:3,smooth,variable=\x] plot ({\x},{tanh(0.5*\x)}) node[right]{$T_1$};
\draw[thick,domain=-3:3,smooth,variable=\x] plot ({\x},{tanh(1.2*\x)}) node[right]{$T_2$};
\draw[thick,domain=-1.3:1.3,smooth,variable=\x] plot ({\x},{tanh(4*\x)});
\draw[thick] (1.3,0.9999) -- (3,0.9999) node[right]{$(T=0)$};
\draw[thick] (-3,-0.9999) -- (-1.3,-0.9999);
\end{tikzpicture}

at \(T=0\) (absolute zero): \(\chi_T \to \infty\), or slope \(\to \infty\).

Lecture 28 (continued): Interactions and Ferromagnetism

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

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

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 \(\big(S = k_B\ln\Omega(E)\big)\) \((\Omega(E)\) 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.

"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 & relation \(\left(M = N\mu\tanh\left(\dfrac{\mu H}{k_BT}\right)\right)\) fails.

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

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\begin{tikzpicture}[scale=1]
\draw[->] (0,0.3) -- (1,0.3);
\draw[->] (0,0) -- (1,0);
\node[right] at (1.1,0.15) {$U_{d_1}$ \quad more stable};
\begin{scope}[xshift=4.5cm]
\draw[->] (0,0.3) -- (1,0.3);
\draw[<-] (0,0) -- (1,0);
\node[right] at (1.1,0.15) {$U_{d_2}$};
\end{scope}
\end{tikzpicture}

\((U_{d_1} < U_{d_2})\)

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

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  \begin{scope}[xshift=\i cm]
  \draw[->] (0,0) -- (0.9,0);
  \draw[->] (0,0.5) -- (0.9,0.5);
  \node at (0.45,-0.4) {\lab};
  \end{scope}
}
\end{tikzpicture}

(plane) — these configurations do not have same energy.

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 direction \((\downarrow)\mu\), so that the net dipole moment cancels out & magnetisation becomes zero.

\[ (\uparrow\downarrow\uparrow\downarrow\uparrow\downarrow\uparrow\downarrow\uparrow\downarrow\cdots) \]

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. (Whereas dipole-dipole interaction is long range

\(\left(\dfrac{\textbf{[?]}}{r^3}\right)\) & weak).

Where there are magnets where dipole-dipole interaction becomes more important/dominant and it leads to Anti-ferromagnetism:

\[ (\uparrow\downarrow\uparrow\downarrow\downarrow\uparrow\downarrow\uparrow\downarrow). \]

Magnetism and Phase Transitions

If we take exchange interactions in account we see that the curve b/w \(\left(\dfrac{M}{N\mu}\right)\) vs \(H\), 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}
\begin{tikzpicture}[scale=1]
\draw[->] (-0.3,0) -- (4,0) node[right] {$H$};
\draw[->] (0,-1.6) -- (0,1.8) node[above] {$\left(\dfrac{M}{N\mu}\right)$};
% T < Tc : steep (near-vertical) rise at origin, saturating
\draw[thick] (0,-1.3) .. controls (0.05,-0.2) and (0.05,0.2) .. (0,1.3);
\draw[thick] (0,1.3) .. controls (0.3,1.5) and (1.5,1.6) .. (3.7,1.6);
\draw[thick] (0,-1.3) .. controls (0.3,-1.5) and (1.5,-1.6) .. (3.7,-1.6);
\node at (1.2,1.9) {$(T \le T_c)$};
\end{tikzpicture}

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

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

Magnetization Calculation (General 3-D Case)

For \(\to\) general case \(\vec{\mu}\) at angles \((\theta,\varphi)\) (3-D case) --- let's compute the magnetisation.

Also assuming no dipole-dipole interaction.

[?]

\usetikzlibrary{arrows.meta,decorations.pathreplacing,calc,angles,quotes}
\begin{tikzpicture}[scale=1]
\draw[->] (0,0) -- (0,2.2) node[above] {$z$};
\draw[->] (0,0) -- (-1.3,-1) node[left] {$x$};
\draw[->] (0,0) -- (2.2,-0.5) node[right] {$y$};
\draw[thick,->] (0,0) -- (1,1.6) node[above] {$\vec{\mu}$};
\draw[dashed] (0,0) arc (90:58:1);
\node at (0.35,1.15) {$\theta$};
\draw[dashed] (0,0) arc (-13:0:1.5);
\node at (1.1,-0.15) {$\varphi$};
\end{tikzpicture}

Average magnetisation: \(M = N\langle \mu \rangle\) (over solid angle)

Exchange interaction (short range), energy:

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

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

\[ \int d\Omega = \int_{-1}^{1} d(\cos\theta) \int_0^{2\pi} d\varphi \qquad \text{or} \qquad \int d\Omega = \int_0^{\pi} \sin\theta\, d\theta \int_0^{2\pi} d\varphi \] 
\usetikzlibrary{arrows.meta,decorations.pathreplacing,calc,angles,quotes}
\begin{tikzpicture}[scale=1]
\draw (0,0) circle (1.4);
\draw[->] (0,0) -- (0.9,1.05);
\draw[->] (0,0) -- (1.1,0.6);
\node at (1.2,0.95) {$d\Omega$};
\node at (1.6,0.4) {$r$};
\end{tikzpicture}
\[ d\Omega = \frac{dA}{r^2} = \frac{(r\,d\theta)(r\sin\theta\,d\varphi)}{r^2} = \sin\theta\, d\theta\, d\varphi \] \[ \int d\Omega = \int\!\!\int \sin\theta \, d\theta\, d\varphi \] \[ M = N\langle \mu \rangle = N\mu \cdot \frac{\displaystyle\int_0^{\pi}\cos\theta\, e^{\beta H\mu\cos\theta}\sin\theta\, d\theta \int_0^{2\pi} d\varphi}{\displaystyle\int_0^\pi \sin\theta\, d\theta \int_0^{2\pi} d\varphi} \]

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

\[ = \frac{N\mu (2\pi)\displaystyle\int_{-1}^{1} t\, e^{\beta H \mu t}\, dt}{2\pi \displaystyle\int_0^\pi e^{\beta H \mu \cos\theta}\sin\theta\, d\theta} = \frac{2\pi N \mu \displaystyle\int_{-1}^{1} t\, e^{\beta H \mu t}\, dt}{2\pi \displaystyle\int_{-1}^{1} e^{\beta H \mu t}\, dt} \] \[ = N\mu \left(\frac{t\, e^{\beta H \mu t}}{\beta H \mu} - \frac{e^{\beta H \mu t}}{\beta^2 H^2 \mu^2}\right)\Bigg|_{-1}^{1} \Bigg/ \left(\frac{1}{\beta H \mu}\right)\left(e^{\beta H \mu t}\right)\Big|_{-1}^{1} \] \[ = \frac{N\mu^2 H \beta \left(e^{\beta H\mu}\left(\frac{1}{\beta H\mu} - \frac{1}{\beta^2\mu^2 H^2}\right) - e^{-\beta H\mu}\left(\frac{-1}{\beta H\mu} - \frac{1}{\beta^2\mu^2H^2}\right)\right)}{\left(e^{\beta H\mu} - e^{-\beta H \mu}\right)} \] \[ = N\mu \left[\frac{e^{\beta H\mu} + e^{-\beta H\mu}}{e^{\beta H \mu} - e^{-\beta H \mu}} - \frac{1}{\beta H \mu}\right] \] \[ \boxed{M = N\mu\left[\coth(\beta H \mu) - \frac{1}{\beta H \mu}\right]} \]

Langevin Function and Susceptibility

Previously when we only assumed \(\mathcal{E} = \pm \mu H\):

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

Now when we used all possible orientations

\[ M = N\mu\left(\coth x - \frac{1}{x}\right), \qquad x = \beta \mu H \qquad \text{(Langevin function)} \]

Let us plot \(\left(\dfrac{M}{N\mu}\right)\) vs \(x\), and \(\left(\coth x - \dfrac{1}{x}\right)\) vs \((x)\):

\usetikzlibrary{arrows.meta,decorations.pathreplacing,calc,angles,quotes}
\begin{tikzpicture}[scale=1]
\draw[->] (-2.2,0) -- (2.2,0) node[right] {$(x)$};
\draw[->] (0,-1.6) -- (0,1.8) node[above] {$\left(\coth x - \dfrac{1}{x}\right)$};
\draw[thick] (-2,-1.2) .. controls (-0.6,-1.2) and (-0.2,-0.3) .. (0,0);
\draw[thick] (0,0) .. controls (0.2,0.3) and (0.6,1.2) .. (2,1.2);
\draw[dashed] (-2,1.2) -- (2,1.2);
\draw[dashed] (-2,-1.2) -- (2,-1.2);
\end{tikzpicture}
\[ \coth x = \frac{\cosh x}{\sinh x} = \frac{1+\dfrac{x^2}{2}+\cdots}{x\left(1+\dfrac{x^2}{6}+\cdots\right)} = \frac{1}{x}\left(1+\frac{x^2}{2}+\cdots\right)\left(1-\frac{x^2}{6}+\cdots\right) \] \[ = \frac{1}{x}\left(1 + x^2\left(\frac{1}{2}-\frac{1}{6}\right)+\cdots\right) = \frac{1}{x}\left(1+\frac{x^2}{3}\right) \] \[ \coth x \simeq \frac{1}{x} + \frac{x}{3} \] \[ \left(\coth x - \frac{1}{x}\right) \simeq \frac{x}{3} \]

So,

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

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

\[ \frac{M}{N\mu}\bigg|_{H\to 0} \simeq \frac{\mu H}{3 k_BT} \]

So

\[ \chi_T = \frac{N\mu^2}{3k_BT} \qquad (\chi_T = \text{Isothermal susceptibility}), \qquad \text{No. of dimensions is `3' in one case.} \]

Two-Dimensional Case

(For 2-D) :- \(\mu\) is allowed to have any direction in a plane.

\[ \frac{M}{N\mu} = \frac{\displaystyle N\int_0^{2\pi} \mu\cos\theta \, e^{\beta H \mu \cos\theta}\, d\theta}{\displaystyle\int_0^{2\pi} e^{\beta H \mu \cos\theta}\, d\theta} \]

(This integral is related to the modified Bessel function \(I_0\); not elementary.)

Since we are only interested in finding \(\chi_T\), i.e. slope at origin, let's expand \(e^{\beta H \mu \cos\theta}\) to first order:

\[ \frac{M}{N\mu} = \frac{\displaystyle\int_0^{2\pi}\cos\theta\left(1+\beta H \mu \cos\theta + \frac{(\beta H \mu \cos\theta)^2}{2!}+\cdots\right)d\theta}{\displaystyle\int_0^{2\pi}\left(1+\beta H\mu\cos\theta + \frac{(\beta H \mu \cos\theta)^2}{2!}+\cdots\right)d\theta} \]

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

\[ \frac{M}{N\mu} = \frac{0 + \beta H \mu \displaystyle\int_0^{2\pi}\cos^2\theta\, d\theta}{2\pi + \beta H \mu \left(\displaystyle\int_0^{2\pi}\cos\theta\, d\theta\right)^{\to 0}} = \frac{\beta H \mu \, \pi}{2\pi} \] \[ = \frac{\beta H \mu}{2} \]

So slope

\[ \frac{\partial M}{\partial H}\bigg|_{H=0} = \frac{N\beta \mu^2}{2} = \frac{\mu^2 N}{2k_BT} \qquad \text{(dimension is 2)} \] \[ \boxed{\chi_T = \frac{N\mu^2}{2k_BT}} \]

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

Weiss Molecular Field Theory

Just like to go from ideal gas to real gas (use interactions) we used the van der Waals gas eq\(^n\), similarly here, we use Weiss molecular field theory.

*Weiss molecular field theory*

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

replace \(H \to H_{\text{eff}} = H + \dfrac{\lambda M}{N}\) (to make intensive quantity to add to \(H\))

\usetikzlibrary{arrows.meta,decorations.pathreplacing,calc,angles,quotes}
\begin{tikzpicture}[scale=0.9]
\draw[->] (0,0) -- (0,2) node[above] {$\vec{H}$};
\draw[->] (0,0) -- (1.5,1.5) node[right] {$\dfrac{\vec{M}}{N}$};
\draw (3,1) circle (0.9);
\foreach \x/\y in {2.7/1.4,3/1.5,3.3/1.4,2.7/0.7,3/0.6,3.3/0.7} {\draw[->] (\x,\y-0.3) -- (\x,\y+0.3);}
\node at (4.6,1.5) {$H_{\text{int}} \propto M$};
\node at (4.6,1) {$H_{\text{int}} = \lambda M$};
\end{tikzpicture}
\[ M = N\mu \tanh\left(\frac{\mu\left(H+\dfrac{\lambda M}{N}\right)}{k_BT}\right) \qquad (\text{for } T\to 0) \]

This is transcendental equation, we can only solve it by numerical methods.

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

\[ M_0 = N\mu \tanh\left(\frac{\mu \lambda M_0}{N k_B T}\right) \qquad (M_0 = \text{magnetisation per unit dipole moment}) \]

(1) \(M_0=0\) is always a solution (Paramagnetic solution). \((H=0, M=0)\)

(2) Real ferromagnets have hysteresis.

\usetikzlibrary{arrows.meta,decorations.pathreplacing,calc,angles,quotes}
\begin{tikzpicture}[scale=1.1]
\draw[->] (-2.2,0) -- (2.2,0) node[right] {$H$};
\draw[->] (0,-1.8) -- (0,1.8) node[above] {$M$};
\draw[thick] (-1.8,-1) .. controls (-1,1.3) and (-0.3,1) .. (0,1) .. controls (0.6,1) and (1.6,1) .. (1.8,1);
\draw[thick] (1.8,1) .. controls (1,-1.3) and (0.3,-1) .. (0,-1) .. controls (-0.6,-1) and (-1.6,-1) .. (-1.8,-1);
\node at (0.15,1.2) {$+1$};
\node at (0.15,-1.2) {$-1$};
\node[left] at (-0.05,0.55) {$M_0$};
\node[left] at (-0.05,-0.55) {$-M_0$};
\end{tikzpicture}

[?] (slope/intercept annotation near origin of the hysteresis loop is unclear)

*Near \(M \ge 0\)*:

\[ M_0 = N\mu \tanh\left(\frac{\mu\lambda M_0}{Nk_BT}\right) \simeq \frac{N\mu^2\lambda M_0}{Nk_BT} \simeq \frac{\mu^2\lambda M_0}{k_BT} \]

Let's plot \(\left(\dfrac{\mu^2\lambda M_0}{k_BT}\right)\) vs \((M_0)\), i.e. (at high temp \(T'>T_c\), slope is less), (for smaller temp \(T''\) slope is even higher).

\usetikzlibrary{arrows.meta,decorations.pathreplacing,calc,angles,quotes}
\begin{tikzpicture}[scale=1]
\draw[->] (-2.2,0) -- (2.2,0) node[right] {$M_0$};
\draw[->] (0,-1.8) -- (0,1.8) node[above] {$\dfrac{\mu^2\lambda M_0}{k_BT}$};
\draw[thick] (-1.8,-1.2) .. controls (-0.6,-1.2) and (-0.2,-0.3) .. (0,0) .. controls (0.2,0.3) and (0.6,1.2) .. (1.8,1.2);
\draw[thick,domain=-1.8:1.8] plot (\x,{0.4*\x}) node[right] {$T'$};
\draw[thick,domain=-1.3:1.3] plot (\x,{0.9*\x}) node[right] {$T''$};
\end{tikzpicture}

Intersection happens if (slope of line) \(> 1\):

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

So \(T=\dfrac{\mu^2\lambda}{k_B}\) 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.

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

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.

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

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

Lecture 29: Ferroelectric Transitions and the Critical Region

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 eq\(^n\) of state

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

would be very similar to that of fluid near its critical region.

[?] (a boxed condition, likely of the form \(\lambda N/k_BT \gg 1\), is only partially legible)

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: Hysteresis and Domains

Also our model \(M = N\mu\tanh\left(\dfrac{\mu(H+\lambda M/N)}{k_BT}\right)\) 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 \(H\). For a single domain our model works perfectly; a single domain can't explain/show hysteresis.

Family of Isotherms

\usetikzlibrary{arrows.meta,decorations.pathreplacing,calc,angles,quotes}
\begin{tikzpicture}[scale=1]
\draw[->] (-0.3,0) -- (3.6,0) node[right] {$H$};
\draw[->] (0,-0.3) -- (0,2.2) node[above] {$\dfrac{M}{N\mu}$};
\draw[thick] (0,0) .. controls (0.2,1.7) and (1,1.9) .. (3.4,1.95);
\node at (3.4,1.6) {$T<T_c$};
\draw[thick] (0,0) .. controls (0.5,1.2) and (1.5,1.7) .. (3.4,1.85);
\node at (3.4,1.3) {$T=T_c$};
\draw[thick] (0,0) .. controls (1,0.6) and (2,1.2) .. (3.4,1.7);
\node at (3.4,1) {$T>T_c$};
\node at (0.6,-0.2) {$H'$};
\end{tikzpicture}

Critical Isotherm: \(\left.\dfrac{\partial M}{\partial H}\right|_{T_c} = 0\), \(\left.\dfrac{\partial^2 M}{\partial H^2}\right|_{T_c}=0\).

\usetikzlibrary{arrows.meta,decorations.pathreplacing,calc,angles,quotes}
\begin{tikzpicture}[scale=1]
\draw[->] (-1.8,0) -- (1.8,0) node[right] {$H$};
\draw[->] (0,-1.8) -- (0,1.8) node[above] {$\dfrac{M}{N\mu}$};
\draw[thick] (-1.5,-1.3) .. controls (-0.3,-1) and (0.3,1) .. (1.5,1.3);
\draw[thick,dashed] (-1.5,1.3) .. controls (-0.3,1) and (0.3,-1) .. (1.5,-1.3);
\node at (1.1,1.5) {$A$};
\node at (1.6,0.4) {$C$};
\node[right] at (0.2,0.6) {Jump};
\end{tikzpicture}

Curve \(A\to C\) is unphysical, as (\(H\uparrow\) in \(-\)ve direction, \(\lambda M \uparrow\) in \(+\)ve direction).

Isotherms in case of fluid:

\usetikzlibrary{arrows.meta,decorations.pathreplacing,calc,angles,quotes}
\begin{tikzpicture}[scale=1]
\draw[->] (0,0) -- (0,2.2) node[above] {$P$};
\draw[->] (0,0) -- (3.6,0) node[right] {$V$};
\foreach \k in {0,1,2,3} {\draw[thick] (0.4+0.2*\k,2-0.2*\k) .. controls (1.2+0.2*\k,0.9-0.15*\k) and (1.8+0.2*\k,0.9-0.15*\k) .. (3.4,0.5-0.05*\k);}
\node[left] at (0.4,2) {$P_c$};
\end{tikzpicture}

Critical Isotherm: \(\left.\dfrac{\partial P}{\partial V}\right|_{(P_c,V_c)}=0\), \(\left.\dfrac{\partial^2 P}{\partial V^2}\right|_{(P_c,V_c)}=0\).

Let us plot \((M\) vs \(T)\):

\usetikzlibrary{arrows.meta,decorations.pathreplacing,calc,angles,quotes}
\begin{tikzpicture}[scale=1]
\draw[->] (0,0) -- (4,0) node[right] {$T$};
\draw[->] (0,-1.8) -- (0,1.8) node[above] {$M$};
\draw[thick] (0,1.4) .. controls (1.5,1.3) and (2,0.3) .. (2.3,0);
\node[left] at (0,1.4) {$M_0(H>0)$ (ferro $\uparrow$)};
\draw[thick] (0,-1.4) .. controls (1.5,-1.3) and (2,-0.3) .. (2.3,0);
\node[left] at (0,-1.4) {$M_0(H<0)$ (ferro $\downarrow$)};
\draw[thick,dashed] (2.3,1.6) -- (2.3,-1.6);
\node[above] at (2.3,1.6) {$T_c$ (Curie temp)};
\draw[thick] (2.3,0) -- (3.8,0);
\node[right] at (3.4,0.3) {$M(H=0)$ (paramagnet)};
\end{tikzpicture}

In case of (magnet): discontinuous transitions of \(M_0\) at \(T_c\).

In case of fluid:

\usetikzlibrary{arrows.meta,decorations.pathreplacing,calc,angles,quotes}
\begin{tikzpicture}[scale=1]
\draw[->] (0,0) -- (4,0) node[right] {$T$};
\draw[->] (0,-0.3) -- (0,2) node[above] {$\rho$};
\draw[thick] (0,1.6) .. controls (1.5,1.5) and (2,0.9) .. (2.3,0.7);
\node[left] at (0,1.6) {gas};
\draw[thick] (0,-0.1) .. controls (1.5,0.1) and (2,0.5) .. (2.3,0.7);
\node[left] at (0,-0.1) {liq.};
\filldraw (2.3,0.7) circle (1.5pt);
\node[right] at (2.4,0.9) {$(\rho_c,V_c)$};
\end{tikzpicture}

Discontinuous transitions, parabolic region near critical point.

Analogy with the Fluid Critical Point

\usetikzlibrary{arrows.meta,decorations.pathreplacing,calc,angles,quotes}
\begin{tikzpicture}[scale=1]
\draw[->] (0,0) -- (0,2.2) node[above] {$P$};
\draw[->] (0,0) -- (3,0) node[right] {$T$};
\node at (1.3,0.8) {liq};
\node at (2.2,0.3) {gas};
\draw[thick,dashed] (0.3,0.2) -- (2,1.4);
\filldraw (2,1.4) circle (1.5pt);
\node[right] at (2,1.4) {$cp$};
\draw[thick,->] (2,1.4) -- (2.7,1.8) node[right] {Homogeneous fluid};
\end{tikzpicture}

\usetikzlibrary{arrows.meta,decorations.pathreplacing,calc,angles,quotes}
\begin{tikzpicture}[scale=1]
\draw[->] (0,1) -- (3,1) node[right] {$T$};
\draw[->] (1.5,0) -- (1.5,2.2) node[above] {$H$};
\node at (0.8,1.4) {ferro $\uparrow$};
\node at (0.8,0.6) {ferro $\downarrow$};
\draw[thick,dashed] (1.5,1) -- (3,1);
\node[above] at (2.5,1) {paramagnet};
\draw[thick,dashed] (1.5,0) -- (1.5,2.2);
\node[above] at (1.6,2.1) {$T_c$};
\end{tikzpicture}

In case of fluid as well as ferromagnet, we have one line ending at critical point.

\[ \frac{dP}{dT} = \frac{L}{T\Delta v} = \frac{T\Delta S}{T\Delta v} = \frac{\Delta S}{\Delta v} = \frac{S_{\text{gas}}-S_{\text{liq}}}{v_{\text{gas}}-v_{\text{liq}}} \qquad (\text{specific volume}) \] \[ \frac{dH}{dT} = \frac{\Delta S}{\Delta M} \]

\(\uparrow\uparrow\uparrow\) has same entropy as \(\downarrow\downarrow\downarrow\)

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

Curie Temperature from the Mean-Field Equation

Let's find \(M_0 = N\mu\tanh\left(\dfrac{\mu\lambda M_0}{Nk_BT}\right)\).

Let \(\left(\dfrac{M_0}{N\mu}\equiv m_0\right)\):

\[ m_0 = \tanh\left(\frac{\mu^2\lambda m_0}{k_BT}\right) \] 
\usetikzlibrary{arrows.meta,decorations.pathreplacing,calc,angles,quotes}
\begin{tikzpicture}[scale=1]
\draw[->] (-1.8,0) -- (1.8,0) node[right] {$m_0$};
\draw[->] (0,-1.6) -- (0,1.6) node[above] {$m_0$};
\draw[thick] (-1.6,-1.3) .. controls (-0.5,-1.2) and (-0.2,-0.4) .. (0,0) .. controls (0.2,0.4) and (0.5,1.2) .. (1.6,1.3);
\draw[thick,domain=-1.5:1.5] plot (\x,{1.6*\x}) node[right] {$T<T_c$};
\draw[thick,domain=-1.5:1.5] plot (\x,{0.5*\x}) node[right] {$T=T_c$};
\filldraw (0,0) circle (1.5pt);
\filldraw (0.85,0.85) circle (1.5pt);
\end{tikzpicture}

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

\[ \left(\frac{\mu^2\lambda}{k_BT}=1\right) \qquad \Rightarrow \qquad \boxed{T_c = \frac{\mu^2\lambda}{k_B}} \]

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

Susceptibility at \(H=0\)

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

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

\usetikzlibrary{arrows.meta,decorations.pathreplacing,calc,angles,quotes}
\begin{tikzpicture}[scale=1.1]
\draw[->] (-2.2,0) -- (2.2,0) node[right] {$H,\, T$};
\draw[->] (0,-1.8) -- (0,1.8) node[above] {$M_0$};
\draw[thick] (-1.8,-1) .. controls (-1,1.3) and (-0.3,1) .. (0,1) .. controls (0.6,1) and (1.6,1) .. (1.8,1);
\draw[thick] (1.8,1) .. controls (1,-1.3) and (0.3,-1) .. (0,-1) .. controls (-0.6,-1) and (-1.6,-1) .. (-1.8,-1);
\end{tikzpicture}

Let's plot \((\chi_T\) vs \(T)\):

\usetikzlibrary{arrows.meta,decorations.pathreplacing,calc,angles,quotes}
\begin{tikzpicture}[scale=1]
\draw[->] (0,0) -- (4,0) node[right] {$T$};
\draw[->] (0,-0.2) -- (0,2.2) node[above] {$\chi_T$};
\draw[thick] (0.3,0.7) .. controls (1.3,1.3) and (1.8,2.2) .. (2,2.2);
\draw[thick] (2,2.2) .. controls (2.2,2.2) and (2.7,1.3) .. (3.6,0.5);
\draw[dashed] (2,0) -- (2,2.2);
\node[below] at (2,0) {$T_c$};
\end{tikzpicture}
\[ m_0 = \tanh\left(\frac{\mu^2\lambda(m_0)}{k_BT}\right) = \tanh\left(\frac{T_c m_0}{T}\right) \]

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 [origin]:

\[ \tanh x = \frac{\sinh x}{\cosh x} = \frac{e^x-e^{-x}}{e^x+e^{-x}} = \frac{\left(1+x+\dfrac{x^2}{2!}+\cdots\right)-\left(1-x+\dfrac{x^2}{2!}-\dfrac{x^3}{3!}+\cdots\right)}{\left(1+x+\dfrac{x^2}{2!}+\dfrac{x^3}{3!}\right)+\left(1-x+\dfrac{x^2}{2!}-\dfrac{x^3}{3!}\right)} \] \[ = \frac{2\left(x+\dfrac{x^3}{3!}\right)}{2\left(1+\dfrac{x^2}{2!}\right)} = \left(x+\frac{x^3}{3!}+\cdots\right)\left(1-\frac{x^2}{2!}+\cdots\right) \] \[ \tanh x \simeq x + x^3\left(\frac{1}{6}-\frac{1}{2}\right) = x - \frac{x^3}{3}+\cdots \]

Critical Exponent \(\beta\)

\[ m_0 = \tanh\left(\frac{T_c}{T}m_0\right) \] \[ m_0 \simeq \frac{T_c}{T}m_0 - \frac{1}{3}\left(\frac{T_c}{T}m_0\right)^3 \]

for \(m_0\) other than zero,

\[ m_0\left(1-\frac{T_c}{T}\right) \simeq -\frac{1}{3}\left(\frac{T_c}{T}\right)^3 m_0^2 \] \[ 1-\frac{T_c}{T} \simeq -\frac{1}{3}\left(\frac{T_c}{T}\right)^3 m_0^2 \] \[ \left(\frac{T_c-T}{T}\right) \simeq \frac{1}{3}\left(\frac{T_c}{T}\right)^3 m_0^2 \]

(mean field critical exponent \((\beta)\); here \(\beta\ne \dfrac{1}{k_BT}\), just a symbol)

\((T\lesssim T_c)\): if \(T\) is very near to \(T_c\) \(\left(\dfrac{T_c}{T}\approx 1\right)\)

\[ \Rightarrow \qquad m_0 \sim \pm\sqrt{T_c-T} \qquad \text{i.e.} \quad m_0 \simeq \pm(T_c-T)^{1/2} \] \[ \left.\frac{dm_0}{dT}\right|_{T=T_c} = \frac{\pm(-1)}{2\sqrt{T_c-T}} \to \infty \]

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

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\begin{tikzpicture}[scale=1]
\draw[->] (0,0) -- (3.6,0) node[right] {$T$};
\draw[->] (0,-0.2) -- (0,2) node[above] {$m_0$};
\draw[thick] (0,1.7) .. controls (1.3,1.6) and (1.9,0.3) .. (2,0);
\draw[dashed] (2,0) -- (2,1.9);
\node[below] at (2,0) {$T_c$};
\end{tikzpicture}

Exactly same relation arose in case of fluid: \((v-v_c) \sim \pm\sqrt{T_c-T}\).

\usetikzlibrary{arrows.meta,decorations.pathreplacing,calc,angles,quotes}
\begin{tikzpicture}[scale=1]
\draw[->] (0,0) -- (3.6,0) node[right] {$T$};
\draw[->] (0,-0.2) -- (0,2) node[above] {$V$};
\draw[thick] (2,1) .. controls (2.3,1.4) and (3,1.6) .. (3.4,1.6);
\draw[thick] (2,1) .. controls (2.3,0.6) and (3,0.4) .. (3.4,0.4);
\node[left] at (0,1) {$V_c$};
\draw[dashed] (0,1) -- (2,1);
\draw[dashed] (2,0) -- (2,1);
\node[below] at (2,0) {$T_c$};
\end{tikzpicture}

(H.W.) Show that \((v-v_c) \sim \pm\sqrt{T_c-T}\).

We can see that in both cases critical exponent is \(\left(\dfrac{1}{2}\right)\), which arose because both are examples of mean field theory models, i.e. in van der Waals case we assume \(P \ne \dfrac{RT}{v-b}\) but \(P=\dfrac{RT}{v-b}-\dfrac{a}{v^2}\) because of attraction of each molecule. In same way here we assume every magnetic moment [feels] another magnetic moment and internal \(H\) is proportional to magnetisation itself. We have exactly the same philosophy, and that is what led to critical exponent.

(Q) Does experiment agree with critical exponent to be \((1/2)\), or how accurately we can get closer to \(T_c\) and control 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 things to control; best we can control 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 \(10^{19}\).

Critical Isotherm and Exponent \(\delta\)

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

\usetikzlibrary{arrows.meta,decorations.pathreplacing,calc,angles,quotes}
\begin{tikzpicture}[scale=1]
\draw[->] (0,0) -- (0,2.2) node[above] {$P$};
\draw[->] (0,0) -- (3.6,0) node[right] {$V$};
\draw[thick] (0.5,0.3) .. controls (1.2,1.4) and (1.6,1.5) .. (2,1.5) .. controls (2.4,1.5) and (2.8,1.4) .. (3.4,0.4);
\filldraw (2,1.5) circle (1.5pt);
\node[above] at (2,1.5) {$(P_c,V_c)$};
\node[left] at (0,1.5) {$P_c$};
\draw[dashed] (0,1.5) -- (2,1.5);
\draw[dashed] (2,0) -- (2,1.5);
\node[below] at (2,0) {$V_c$};
\node[right] at (3.4,0.4) {$T=T_c$};
\end{tikzpicture}
\[ \left.\frac{\partial P}{\partial V}\right|_{T_c} = \left.\frac{\partial^2 P}{\partial V^2}\right|_{T_c} = 0 \]

Shifting origin to see behaviour near \((V_c,P_c)\):

\[ \tilde{P} = \frac{P-P_c}{P_c}, \qquad \tilde{V} = \frac{V-V_c}{V_c} \]

The curve near \((V_c,P_c)\) is like: (mean field critical exponent \(\delta\))

\[ \left(\tilde{P} \sim |\tilde{V}|^\delta\right) \qquad \text{or} \qquad \left(|\tilde{V}| \sim \tilde{P}^{1/3}\right) \]

Let's look at critical isotherm for ferromagnetic transitions:

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

Let \(m = \dfrac{M}{N\mu}\):

\[ m = \tanh\left(\frac{\mu}{k_BT}(H+\lambda\mu m)\right) = \tanh\left(\frac{\lambda\mu^2}{k_BT}\left(\frac{H}{\lambda\mu}+m\right)\right) \]

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

\[ = \tanh\left(\frac{T_c}{T}(h+m)\right) \]

For critical isotherm \((T=T_c)\):

\[ \boxed{m = \tanh(h+m)} \] 
\usetikzlibrary{arrows.meta,decorations.pathreplacing,calc,angles,quotes}
\begin{tikzpicture}[scale=1]
\draw[->] (-1.8,0) -- (1.8,0) node[right] {$h$};
\draw[->] (0,-1.6) -- (0,1.6) node[above] {$m$};
\draw[thick] (-1.6,-1.2) .. controls (-0.5,-1.2) and (-0.1,-0.3) .. (0,0) .. controls (0.1,0.3) and (0.5,1.2) .. (1.6,1.2);
\node[right] at (1.2,1.3) {$T<T_c$};
\draw[thick] (-1.6,-0.9) .. controls (-0.5,-0.6) and (-0.2,-0.2) .. (0,0) .. controls (0.2,0.2) and (0.5,0.6) .. (1.6,0.9);
\node[right] at (1.2,0.7) {$T=T_c$ (critical isotherm)};
\draw[thick] (-1.6,-0.5) -- (1.6,0.5);
\node[right] at (1.2,0.35) {$T>T_c$};
\end{tikzpicture}

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

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

\(h\) is much smaller (near origin), so \(\underline{m^3 \sim h}\).

Susceptibility Exponent \(\gamma\) and Landau Theory

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

Just like \(\left(|\tilde{V}| \sim \tilde{P}^{1/3}\right)\) \((\tilde{P}\) is field variable\()\).

\((m,\tilde{V})\) are response variables.

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

\[ \chi_T = \left.\frac{\partial M}{\partial H}\right|_T\Bigg|_{H=0} \, ; \qquad \kappa_T = -\frac{1}{v}\frac{\partial v}{\partial P}, \qquad \left.\frac{\partial P}{\partial v}\right|_{\text{crit. pt}} = 0 \]

(Curie Weiss law) \(\longleftrightarrow\)

\[ \chi_T = \frac{1}{|T-T_c|} \] \[ \left(\chi_T\big|_{T=T_c}=\infty\right) \quad \longleftrightarrow \quad (\kappa_T=\infty) \] \[ \chi_T = \frac{1}{|T-T_c|^\gamma} \quad \to \gamma \text{ critical point} \qquad (\gamma \ne C_P/C_V, \text{ just a symbol}) \]

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

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

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