Contents

<< previous | overview | playlist | next >>

Lecture 23

`\(E\)' is some kind of thermodynamic potential

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

What is significance of above eqn set?

We know that in \(E(S,V,N)\), `\(S\)' is not a control variable. So let's change that by Legendre transformations.

\[ E = E(S,V,N) \ \longrightarrow\ \text{(Internal Energy)} \]

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

For that we have to assume we are in the thermodynamic limit of the thermodynamic system, i.e. \(V\to\infty\) and \(N\to\infty\) with \(N/V=s\) (fixed).

Then \(S,V,N\), which increase with system size (extensive quantities), lead to some \(E\). Then we can assume \(E\) is a homogeneous function of \(S,V,N\) of degree \(1\).

Euler's theorem

For any homogeneous function of degree `\(r\)'

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

i.e. \(f(\lambda x,\lambda y,\lambda z,\ldots) = \lambda^r f(x,y,z,\ldots)\).

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

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

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

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

\[ \text{or}\quad E - TS + PV = \mu N \]

we also know that \(\big(G(T,P,N) = E-TS+PV\big)\)

so \(\mu N = G\)

so \(\Rightarrow \mu = \dfrac{G}{N} = \) Gibbs free energy per particle.

Let us find \(dE\):

\[ dE = (TdS-PdV+\mu dN) + (Sd T - VdP+Nd\mu) \]

But laws of thermodynamics tell us that

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

so it implies that

\[ SdT - VdP + N\,d\mu = 0 \] \[ d\mu = -\frac{V}{N}dP+\frac{S}{N}dT \quad \Longrightarrow\quad d\mu = v\,dP - s\,dT \]

specific volume per particle specific entropy per particle

So,

\[ d\mu = v\,dP - s\,dT \qquad \text{(Gibbs--Duhem relation)} \]

it implies,

\[ \boxed{\mu = \mu(P,T)} \]

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

If we assume \(T\)' is some kind of generalized flux force and \(dS\)' as some generalized flux, then

\[ dE = T\,dS + \sum F_i\,dX_i \] \[ dE = \sum F_i\,dX_i \qquad \text{(homogeneity argument)} \]

so:

\[ \mu N + PV - E = \sum F_i X_i \ \Rightarrow\ \left(\sum X_i\,dF_i = 0\right) \qquad \text{(Generalized Duhem relation)} \]

This confirms (for our assumption of \(E\) to be a homogeneous function of degree \(1\).
If \(E\) is not homogeneous function, so \(\mu = G/N\), degree \(\neq 1\).

Legendre transforms

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

(Enthalpy)

\[ H = E+PV \] \[ \begin{aligned} dH &= dE + PdV + VdP \notag \\ &= (TdS-PdV+\mu dN) + PdV+VdP \notag \\ dH &= TdS + VdP + \mu dN \end{aligned} \]

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

\[ E(S,V,N) \longrightarrow F(T,V,N) \] \[ F = E-TS \] \[ \begin{aligned} dF &= TdS+\mu dN - PdV - TdS - SdT \notag\\ dF &= -SdT - PdV+\mu dN \ \Rightarrow\ F=F(T,V,N) \end{aligned} \] \[ E(S,V,N) \longrightarrow G(T,P,N) \qquad G = E-TS+PV \qquad \text{(Gibbs free energy)} \] \[ E(S,V,N) \longrightarrow \Phi_1(S,V,\mu) \qquad \Phi_1 = E-\mu N \] \[ E(S,V,N) \longrightarrow \Phi_2(S,P,\mu) \qquad \Phi_2 = E+PV-\mu N \] \[ E(S,V,N) \longrightarrow \Phi_3(T,V,\mu) \qquad \Phi_3 = E-TS-\mu N \]

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

So

\[ \boxed{\Phi_3 = -PV} \ \to\ \text{``grand potential''} \]

The reason is that this potential \(\Phi_3\) is connected to the grand canonical ensemble, but the other potentials above mentioned are only linked to the canonical ensemble of stat. mech.

In,

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

\((T,P,\mu)\to\) Intensive variables; \((S,V,N)\to\) Extensive variables.

Field variables (Intensive) State variables (extensive)
\(T\) \(S\)
\(P\) \(V\)
\(\mu\) \(N\)
Stress \(\sigma_{ij}\) Strain \(\epsilon_{ij}\)
\(\vec E\) \(\vec P \to\) polarisation per unit volume
\(\vec H\) \(\vec M \to\) magnetic dipole moment per unit volume
\[ dF = -SdT-PdV+\mu dN \ ; \qquad dG = -SdT+VdP+\mu dN \]

Let's look at these 2 relations,

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

So,

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

Similarly,

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

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

\[ dQ = dU+dW \qquad TdS = dE+PdV-\mu dN \] \[ \left(\frac{dQ}{dT}\right)_{V,N} = C_V = T\left(\frac{\partial S}{\partial T}\right)_{V,N} = \left(\frac{\partial E}{\partial T}\right)_{V,N} + 0 - 0 \] \[ C_V = T\left(\frac{\partial S}{\partial T}\right)_{V,N}\ ,\qquad S = -\left(\frac{\partial F}{\partial T}\right)_{V,N} \]

So

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

the response functions are 2nd derivatives of thermodynamic potentials.

But we know \(C_V\) cannot be negative \(\to\) Le Chatelier's principle.

So,

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

also

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

Isothermal Compressibility (fixed \(T,N\))

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

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

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

Poisson's ratio (\(\nu\))

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

\[ \nu = \frac{\Delta x}{\Delta y} = \frac{\text{contraction strain}}{\text{extension strain}} \] 
\usetikzlibrary{arrows.meta,decorations.pathreplacing,calc,angles,quotes}
\begin{tikzpicture}[scale=0.8]
\draw (0,0) ellipse (0.6 and 0.2);
\draw (0,2) ellipse (0.6 and 0.2);
\draw (-0.6,0) -- (-0.6,2);
\draw (0.6,0) -- (0.6,2);
\draw[->] (0,2.2) -- (0,2.8) node[right] {extend};
\draw[->] (0,-0.2) -- (0,-0.8);
\draw[<->] (-0.9,1) -- (-0.6,1) node[midway,above] {contract};
\node at (2,1) {$(\nu>0)$};
\end{tikzpicture}

It is possible that \((\nu<0)\) such that upon extending in one direction, the object/material also expands in the \(\perp\) direction.

\usetikzlibrary{arrows.meta,decorations.pathreplacing,calc,angles,quotes}
\begin{tikzpicture}[scale=0.8]
\draw (0,0) ellipse (0.6 and 0.2);
\draw (0,2) ellipse (0.6 and 0.2);
\draw (-0.6,0) -- (-0.6,2);
\draw (0.6,0) -- (0.6,2);
\draw[->] (-1.4,1) -- (-0.7,1) node[midway,above] {stress};
\draw[->] (1.4,1) -- (0.7,1) node[midway,above] {stress};
\node at (0,-1) {extending in $\perp$ direction};
\end{tikzpicture}

It turns out

\[ \boxed{-1 < \nu < \tfrac{1}{2}} \]

For ideal gas

\[ PV = nRT \qquad PdV+VdP = nRdT \]

at constant \(T\): \(\ d(PV) = -PdV\), i.e. \(VdP = -PdV\)

\[ \frac{dV}{V} = -\frac{dP}{P} \ \Rightarrow\ K_T = -\frac{1}{V}\frac{\partial V}{\partial P} = -\frac{1}{V}\left(\frac{-V}{P}\right) = \frac{1}{P} \]

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

For adiabatic process

\[ PV^\gamma = \text{const.} \] \[ dP\,V^\gamma + P\gamma V^{\gamma-1}dV = 0 \] \[ V^{\gamma-1}\left(VdP+\gamma P dV\right) = 0 \] \[ \Rightarrow\ K_T = -\frac{1}{V}\frac{dV}{dP} = \frac{1}{\gamma}\cdot\frac{V}{VP} = \frac{1}{\gamma P} \] \[ \left(K_{S,N} = \frac{1}{\gamma P}\right) \]

Van der Waals gas

What happens if we have a van der Waals gas:

\[ P = \frac{nRT}{V-bn} - \frac{an^2}{V^2} \] 
\usetikzlibrary{arrows.meta,decorations.pathreplacing,calc,angles,quotes}
\begin{tikzpicture}[scale=0.9]
\useasboundingbox (-0.3,-1.7) rectangle (4.2,2.6);
\clip (-0.3,-1.7) rectangle (4.2,2.6);
\draw[->] (0,0) -- (4,0) node[right] {$r$};
\draw[->] (0,-1.5) -- (0,2) node[above] {$V(r)$};
\draw[thick,domain=0.62:3.5,samples=100] plot (\x, {2*( (0.7/\x)^12 - (0.7/\x)^6 )});
\node at (0.9,1.5) {repulsive};
\node at (2.5,-0.6) {attractive force};
\draw[dashed] (1.2,-1.2) -- (1.2,0.3);
\end{tikzpicture}

This repulsion arose from the Pauli exclusion principle \((r>0)\).

Attractive force potential: "6--12" potential (Lennard-Jones potential)

\[ V(r) = V_0\left[\left(\frac{a}{r}\right)^{12} - \left(\frac{a}{r}\right)^{6}\right] \qquad \text{need quantum mech.\ effect (repulsive term).} \]

Real gas approximates to van der Waals gas.

Attractive force term arises from dipole moment attraction.

\usetikzlibrary{arrows.meta,decorations.pathreplacing,calc,angles,quotes}
\begin{tikzpicture}[scale=0.7]
\draw (0,0) circle (0.6);
\draw[dashed] (0,0) circle (0.4);
\node at (0,0) {$\oplus$};
\draw[->] (0.7,0) -- (1.4,0) node[right] {$p$};
\draw (4,0) circle (0.6);
\draw[dashed] (4,0) circle (0.4);
\node at (4,0) {$\oplus$};
\draw[->] (4.7,0) -- (5.4,0) node[right] {$p$};
\end{tikzpicture}

Random fluctuations can cause dipole moment.

Interaction energy of any dipole goes like

\[ U(r) \sim \vec p\cdot \vec E \]

the greater the field, the greater the separation.

So \(\vec p = \alpha \vec E\)

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

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

Lecture 24

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

\usetikzlibrary{arrows.meta,decorations.pathreplacing,calc,angles,quotes}
\begin{tikzpicture}[scale=1]
\draw (0,0) ellipse (2 and 1.3);
\draw (0,-1.3) -- (0,1.3);
\node at (-1,0.3) {$A$};
\node at (1,0.3) {$B$};
\node at (-1,-0.5) {$E,V,N$};
\node at (1,-0.5) {$E',V',N'$};
\node at (0,1.8) {Isolated system $(E_{\text{tot}},V_{\text{tot}},N_{\text{tot}})$};
\end{tikzpicture}

cond. of eqm implies,

\[ \frac{\partial \ln\Omega}{\partial E} = \frac{\partial \ln\Omega'}{\partial E'} = \frac{1}{k_BT} \qquad (T_A=T_B) \]

Probability that subsystem \(A\) has energy \(E\):

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

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

We know that

\[ \Omega(E)\,\Omega'(E') \propto \Omega'(E_{\text{tot}}-E) \propto (E_{\text{tot}}-E)^N \] 
\usetikzlibrary{arrows.meta,decorations.pathreplacing,calc,angles,quotes}
\begin{tikzpicture}[scale=0.9]
\draw[->] (0,0) -- (5,0) node[right] {$E$};
\draw[->] (0,0) -- (0,2.3);
\draw[thick,domain=0.2:4.7,samples=100] plot (\x, {2.1*exp(-4*(\x-2.3)*(\x-2.3))});
\draw[thick,dashed,domain=0.2:4.7,samples=100] plot (\x, {0.4+0.25*\x});
\node at (2.3,-0.3) {avg.\ energy (typical energy)};
\end{tikzpicture}
\[ \ln P(E) = \ln \Omega(E) + \ln \Omega'(E_{\text{tot}}-E) - \ln \Omega_{\text{tot}}(E_{\text{tot}}) \]

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

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

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

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

Since the slope of the slope at \(\bar E\) is negative,

\[ = -k\,\frac{(E-\bar E)^2}{2} \] \[ P(E) = P(\bar E)\cdot e^{-k(E-\bar E)^2/2} \qquad \text{`$k$' to make dimensionless} \] \[ P(E) = (\text{Normalisation const.})\cdot e^{-(E-\bar E)^2/(2\bar E)} \] 
\usetikzlibrary{arrows.meta,decorations.pathreplacing,calc,angles,quotes}
\begin{tikzpicture}[scale=1]
\draw[->] (0,0) -- (5,0) node[right] {$E$};
\draw[->] (0,0) -- (0,2.3) node[above] {$P(E)$};
\draw[thick,domain=0.2:4.7,samples=100] plot (\x, {2*exp(-6*(\x-2.5)*(\x-2.5))});
\draw[dashed] (2.5,0) -- (2.5,2);
\node at (2.5,-0.3) {$\bar E$};
\node at (3.6,1) {$\alpha \sqrt{\bar E}$};
\node at (1,2.1) {$\sigma^2 = \bar E$};
\end{tikzpicture}

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

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

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

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

So internal energy used in thermodynamics is actually the average energy, but when we study statistical mech., we have to become precise, and talk about \(E\) and \(\bar E\).

(B) What if `\(A\)' is a very small ensemble/subsystem sitting in a huge heat bath (B)?

\(A\)' might contain a few atoms or a single atom/molecule, so energy of \(A\)' will have very large fluctuations from its mean value \(\langle E\rangle\).

\usetikzlibrary{arrows.meta,decorations.pathreplacing,calc,angles,quotes}
\begin{tikzpicture}[scale=1]
\draw (0,0) ellipse (2.2 and 1.5);
\draw (-1,0.3) circle (0.5);
\node at (-1,0.3) {$A$};
\node at (-0.7,0.9) {$E$};
\node at (0.8,0) {$B$};
\node at (0,1.9) {heat bath};
\end{tikzpicture}

Let us assume total energy of \(A\) is \(E\), and that of \(B\) is \(E'\): \((E+E'=E_{\text{tot}})\), \((E'=E_{\text{tot}}-E)\).

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

Since \((E \ll E_{\text{tot}})\), \(A\) is strongly affected by \(B\), but \(B\) is not affected by \(A\) at all (as \(N_A\ll N_B\)).

So,

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

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

\[ \ln P(E) = \left[\ln \Omega'(E_{\text{tot}}) - E\underbrace{\left.\frac{\partial \ln \Omega'(E')}{\partial E'}\right|_{E'=E_{\text{tot}}}}_{\beta} + \cdots\right] - \ln \Omega_{\text{tot}}(E_{\text{tot}}) \] \[ P(E) = (\text{const})\,e^{-\beta E} \qquad \beta = \left.\frac{\partial \ln \Omega'(E')}{\partial E'}\right. = \text{inverse temp} = \frac{1}{k_BT} \text{ of heat bath} \]

To find (const) we need to know \(\Omega'(E_{\text{tot}})\) (info about heat bath). But we don't need that either — we can normalize \(P(E)\) and find the normalisation const.

\[ P(E) = \frac{e^{-\beta E}}{\sum_E e^{-\beta E}} \]

So, normalisation const. is fixed by properties of the subsystem alone, and we eliminate whatever happens outside.

For a moment, let us assume `\(A\)' has discrete energy levels.

\usetikzlibrary{arrows.meta,decorations.pathreplacing,calc,angles,quotes}
\begin{tikzpicture}[scale=1]
\draw (0,0) -- (2,0) node[right] {$\varepsilon_0$};
\draw (0,0.6) -- (2,0.6) node[right] {$\varepsilon_1$};
\draw (0,1.4) -- (2,1.4) node[right] {$\varepsilon_2$};
\draw (0,2.4) -- (2,2.4) node[right] {$\varepsilon_3$};
\node at (-1,1.5) {degenerate states};
\end{tikzpicture}
\[ P(\varepsilon_i) = \frac{e^{-\beta \varepsilon_i}}{\sum_j e^{-\beta \varepsilon_j}} \qquad \text{(summed over all states $j$, states not levels.)} \]

The only information about the heat bath to be known is its temperature. Unlike the case of the microcanonical ensemble where \(E=\bar E\) at eqm, the total energy of the system in thermal eqm with a heat bath is not fixed. It is fluctuating with different energy with \(P(\varepsilon_i) = e^{-\beta \varepsilon_i}/\sum_j e^{-\beta \varepsilon_j}\), but the average value of \(\langle \varepsilon_i\rangle\) is independent of time, because \(P(\varepsilon_i)\) is independent of time.

The factor \(\left(\displaystyle\sum_{\text{states } j} e^{-\beta \varepsilon_j}\right)\) is called \(Z\)', the <span id="Lecture 24-(B) What if \(A\)' is a very small ensemble/subsystem sitting in a huge heat bath (B)?-canonical partition function">canonical partition function.

And our system in thermal eqm at temp \(T\) is called a "canonical ensemble."

\usetikzlibrary{arrows.meta,decorations.pathreplacing,calc,angles,quotes}
\begin{tikzpicture}[scale=1]
\draw (0,0) ellipse (2 and 1.3);
\node at (-1,0.3) {$A$};
\node at (0.5,0.3) {$B$};
\node at (0,1.7) {$T$};
\node at (0.5,-0.5) {heat bath};
\end{tikzpicture}
\[ \boxed{Z = Z(T,V,N) = \sum_{\text{states } j} e^{-\beta \varepsilon_j} = \sum_{\text{levels } i} g_i\, e^{-\beta \varepsilon_i}} \]

The subsystem "A" in thermal eqm of this kind is not in a fixed energy state. It has got probability distributions for certain energy values. For a quantum system, "A" is not even in a pure energy eigenstate. It will be in a superposition of all energy eigenstates. (mind blowing fact)

\[ \big(P(E) \propto e^{-\beta E}\big). \]

Distribution of energy goes like exponential decay rather than a steep Gaussian distribution.

\usetikzlibrary{arrows.meta,decorations.pathreplacing,calc,angles,quotes}
\begin{tikzpicture}[scale=1]
\draw[->] (0,0) -- (4,0) node[right] {$\varepsilon$};
\draw[->] (0,0) -- (0,2) node[above] {$P(\varepsilon)$};
\draw[thick,domain=0.1:3.7,samples=100] plot (\x,{1.8*exp(-1.1*\x)});
\end{tikzpicture}

\(\beta = \) property of the bath.

In terms of energy levels,

\[ Z = \sum_{\text{levels } i} g_i\,e^{-\beta \varepsilon_i} \qquad g_i \to \text{degeneracy of level } \varepsilon_i. \]

For continuous energy levels,

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

Now, instead of counting degeneracy (\(g_i\)), we ask how many states with energy \(\varepsilon \le \varepsilon+d\varepsilon\):

\[ d\varepsilon\ \text{(band at }\varepsilon\text{)} = f(\varepsilon)\,d\varepsilon \qquad \text{density of states} \]

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

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

So,

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

So \(Z\) is the Laplace transform of \(f(\varepsilon)\), in which the transform variable is \(\beta\), instead of `\(s\)'. (Laplace transform, or moment generating function, for a continuous variable.)

\[ \left\{ \begin{aligned} f(\varepsilon) &\to \text{system property (mech.\ aspect: classical/quantum)}\\ e^{-\beta \varepsilon} &\to \text{statistical aspect.} \end{aligned} \right\} \qquad e^{-\beta \varepsilon}f(\varepsilon) \to \text{stat.\ mech.} \]

(Q) What is average energy?

\[ \begin{aligned} \langle E\rangle &= \sum_{\text{states } j} P(\varepsilon_j)\,\varepsilon_j \notag\\ &= \frac{\sum_i \varepsilon_i\, e^{-\beta \varepsilon_i}}{Z(T,V,N)} = \frac{\sum_i \varepsilon_i\,e^{-\beta \varepsilon_i}}{\sum_j e^{-\beta \varepsilon_j}} = \frac{1}{Z}\sum_i \varepsilon_i\,e^{-\beta \varepsilon_i} \end{aligned} \] \[ \boxed{\langle E\rangle = -\frac{\partial(\ln Z)}{\partial \beta} = -\frac{1}{Z}\frac{\partial Z}{\partial \beta}} \]

\(\langle E\rangle = U = \) internal energy of the system (or) \(\langle H\rangle\) expectation of Hamiltonian.

(Q) Var\((E) = ?\)

\[ \begin{aligned} \sigma^2 &= \big\langle (E-\langle E\rangle)^2\big\rangle \notag\\ &= \big\langle E^2 + \langle E\rangle^2 - 2E\langle E\rangle\big\rangle \notag\\ &= \langle E^2\rangle + \langle E\rangle^2 - 2\langle E\rangle^2 \notag\\ \sigma^2 &= \langle E^2\rangle - \langle E\rangle^2 \end{aligned} \] \[ \langle E^2\rangle = \frac{\sum_{\text{states } i} E_i^2\, e^{-\beta E_i}}{Z} = \frac{1}{Z}\frac{\partial^2 Z}{\partial \beta^2} \] \[ \sigma^2 = \frac{1}{Z}\frac{\partial^2 Z}{\partial \beta^2} - \frac{1}{Z^2}\left(\frac{\partial Z}{\partial \beta}\right)^2 = \frac{\partial^2 (\ln Z)}{\partial \beta^2} \] \[ \begin{aligned} \sigma^2 &= -\frac{\partial \langle E\rangle}{\partial \beta} \notag\\ &= -\frac{\partial \langle E\rangle}{\partial T}\cdot\frac{\partial T}{\partial \beta} \notag\\ &= \left(\frac{\partial \langle E\rangle}{\partial T}\right)k_BT^2 \end{aligned} \] \[ \boxed{\sigma^2 = C_V\,k_BT^2} \] \[ \left(\frac{\partial \langle E\rangle}{\partial T} = \frac{\partial U}{\partial T} = C_V\right) \]

So variance of energy is directly related to \(C_V\).

\[ \boxed{C_V = \frac{\sigma^2}{k_BT^2}} \]

So, \((C_V \geq 0)\).

In thermodynamics we only deal with \(\langle E \rangle\). So we can't calculate \(C_V\) or \(C_P\). So \(C_P, C_V\) are input in thermodynamics.

Lecture 25: Connection with Thermodynamics

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

\[ Z = \sum_{\text{states } j} e^{-\beta E_j} \;\longrightarrow\; \int_0^{\infty} dE\, e^{-\beta E}\, g(E) \]

where \(g(E) \propto E^{1/2}\), \(g(E) \propto E^{N/2}\) or \(\propto \alpha E^N\) for \(N = 10^{24}\).

Let's plot \((Z \text{ vs } E)\):

\usetikzlibrary{arrows.meta,decorations.pathreplacing,calc,angles,quotes}
\begin{tikzpicture}[scale=1]
\draw[->] (0,0) -- (6,0) node[right] {$E$};
\draw[->] (0,0) -- (0,4) node[above] {$Z$};
% decaying exponential e^{-beta E}
\draw[thick, domain=0.1:5.8, smooth, variable=\x, black]
  plot ({\x}, {3.5*exp(-0.6*\x)}) node[right] {};
\node at (0.6,3.3) {$e^{-\beta E}$};
% g(E) rising curve, peaked
\draw[thick, domain=0.001:5.8, smooth, variable=\x, blue]
  plot ({\x}, {2.6*pow(\x,1.4)*exp(-0.9*\x)});
\node[blue] at (3.3,2.6) {$g(E)$};
\draw[dashed] (2.0,0) -- (2.0,2.0);
\node at (2.0,-0.3) {$\bar{E}$};
\end{tikzpicture}
\[ Z = \int_0^\infty dE\, e^{-\beta E} g(E)\,dE \;=\; e^{-\beta \bar E}\,\Omega(\bar E) \] \[ \text{(?) maybe} \sim \Big(\Omega(E) \approx \Omega(\bar E) = \int g(E)\,dE\Big) \quad \text{(by me)} \] \[ \Omega(\bar E) = \text{total no.\ of microstates} \]

So,

\[ \ln Z = \ln \Omega(\bar E) - \beta \bar E \] \[ -k_BT \ln Z = -k_BT \ln \Omega(\bar E) + \bar E \]

But we know that entropy \((S)\) is defined as

\[ S = k_B \ln \Omega \] \[ = \bar E - TS \;=\; F \quad \text{(Helmholtz free energy)} \] \[ \Rightarrow \boxed{Z = e^{-\beta F}} \]

So, the whole idea was to write

\[ Z = \sum_{\text{status } j} e^{-\beta E_j} \;\; \equiv \;\; \text{in terms of } e^{-\beta E_{\text{eff}}}, \text{ effective energy.} \] \[ (Z = e^{-\beta F}). \qquad (F = \text{Helmholtz free energy}). \]

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

\[ P(E_i) = \frac{e^{-\beta E_i}}{\sum_j e^{-\beta E_j}} \qquad \text{(sum over states $j$ of gas)} \]

Let us find Pressure

We know that \(F = U - TS\)

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

Partition function for classical ideal gas

\usetikzlibrary{arrows.meta,decorations.pathreplacing,calc,angles,quotes}
\begin{tikzpicture}
\draw[thick] (0,0) ellipse (3 and 1.6);
\draw[thick] (0.3,0.1) circle (1.1);
\node at (0.3,0.1) {$V,N$};
\node at (0.3,-1.0) {ideal gas};
\node at (-2.3,0) {Heat bath $T$};
\end{tikzpicture}

mass of each particle \(m\).

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

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

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

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

\[ Z = \sum e^{-\beta\varepsilon} = \sum\left(e^{-\beta\varepsilon}\cdot e^{-\beta\varepsilon}\cdot e^{-\beta\varepsilon}\cdots\right) \] \[ Z = \left(\sum_{\substack{\text{state of} \\ \text{any particle}}} e^{-\beta\varepsilon}\right)^N \longrightarrow \text{only true if particles are distinguishable.} \] \[ Z = \left(\int_0^\infty e^{-\beta E}\,\Omega(E)\,dE\right)^N = \left(\frac{1}{h^3}\int_V d^3x \int d^3p\, e^{-\beta \frac{p^2}{2m}}\right)^N \] \[ = \left(\frac{V}{h^3}\int e^{-\beta \frac{p_x^2}{2m}}dp_x \int e^{-\beta \frac{p_y^2}{2m}}dp_y \int e^{-\beta \frac{p_z^2}{2m}}dp_z\right)^N \] \[ = \left(\frac{V}{h^3}\Big((2\pi m k_BT)^{1/2}(2\pi m k_BT)^{1/2}(2\pi m k_BT)^{1/2}\Big)\right)^N \] \[ \boxed{Z = \left(\frac{V}{h^3}(2\pi m k_BT)^{3/2}\right)^N} \qquad \left(\int_{-\infty}^{\infty} e^{-ax^2}dx = \sqrt{\frac{\pi}{a}}\right) \] \[ Z = e^{-\beta F}, \qquad \ln Z = -\beta F \;\Rightarrow\; F = -\frac{1}{\beta}\ln Z = -k_BT\, N \ln\left(\frac{V(2\pi m k_BT)^{3/2}}{h^3}\right) \] \[ P = -\left(\frac{\partial F}{\partial V}\right)_{T,N} = \frac{1}{\beta}\cdot\frac{1}{Z}\frac{\partial Z}{\partial V}\bigg|_{T,N} = \frac{1}{\beta Z}\cdot\frac{Z}{h^3}\cdot N\left(\frac{V}{h^3}(2\pi m k_BT)^{3/2}\right)^{N-1}\!\!(2\pi mk_BT)^{3/2} \] \[ = Nk_BT\cdot\frac{1}{V}\cdot\frac{(2\pi m k_BT)^{3/2}/h^3}{(2\pi m k_BT)^{3/2}/h^3} \] \[ P = \frac{Nk_BT}{V} \;\Rightarrow\; \boxed{PV = Nk_BT} \]

\(N=\) no. of particles \(T=\) temp. of gas & Heat bath.

(Q) What is average energy?

\[ U = \bar E = -\frac{\partial}{\partial\beta}\ln Z = -\frac{\partial}{\partial\beta}\left(N\ln\left(\frac{V}{h^3}\left(\frac{2\pi m}{\beta}\right)^{3/2}\right)\right) \] \[ = -N\cdot\frac{1}{\frac{V}{h^3}\left(\frac{2\pi m}{\beta}\right)^{3/2}}\cdot\frac{V}{h^3}\cdot\frac{3}{2}\left(\frac{2\pi m}{\beta}\right)^{1/2}\cdot 2\pi m\cdot\left(-\frac{1}{\beta^2}\right) \] \[ U = N\cdot\frac{3}{2}\cdot\frac{1}{\beta} = \frac{3}{2}Nk_BT \] \[ \bar E = U = \frac{3}{2}Nk_BT \] \[ \left(\frac{U}{N} = \frac{3}{2}k_BT\right) \rightarrow \text{equipartition theorem.} \]

\(\downarrow\)

Tells us that every quadratic term in `H' (Hamiltonian) gives a contribution of \(\left(\dfrac{1}{2}k_BT\right)\).

Note: not applicable for vibrational d.o.f. where the possible states [?] the particles are indistinguishable.

For \(E = H = \dfrac{1}{2}kx^2 + \dfrac{p^2}{2m}\)

\[ \bar E = \frac{1}{2}k_BT + \frac{1}{2}k_BT \] \[ (\bar E = k_BT) \]

for \(V = \dfrac{1}{2}kx^4\)

\[ H = \frac{1}{2}kx^4 + \frac{p^2}{2m} \] \[ \left(\bar E = \frac{k_BT}{4} + \frac{k_BT}{2} = \frac{3}{4}k_BT\right) \]

Lecture 25 (continued)

When we will do Q.M.; \(E = pc\) for photons in black body cavity at temp \(T\). Then \(\bar E \neq \dfrac{3}{2}k_BT\) (even in 3-D).

\[ \rightarrow E = pc, \text{ so} \] \[ \bar E \propto T^4 \qquad \text{(Bose statistics)} \]

So,

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

or, \(P = \dfrac{2U}{3}\) \(\rightarrow\) avg energy density.

\[ \left(u = \frac{U}{V} = \text{avg energy per unit volume}\right) \]

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

(Q) Is it correct that

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

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

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

\usetikzlibrary{arrows.meta,decorations.pathreplacing,calc,angles,quotes}
\begin{tikzpicture}[scale=0.9]
\foreach \x/\lab in {0/1st,1.6/2nd,3.2/3rd} {
  \draw (\x,0) -- (\x+0.8,0);
  \draw (\x,0.6) -- (\x+0.8,0.6);
  \draw (\x,1.2) -- (\x+0.8,1.2);
  \node at (\x+0.4,-0.4) {\small state of \lab\ particle};
}
\filldraw (0.4,0.6) circle (2pt);
\filldraw (2.0,0) circle (2pt);
\filldraw (3.6,1.2) circle (2pt);
\node at (2,1.8) {\small same configuration $\rightarrow$ level};
\draw[->] (4.4,0.8) -- (5.2,0.8);
\foreach \x/\lab in {5.6/1,7.2/2,8.8/3} {
  \draw (\x,0) -- (\x+0.8,0);
  \draw (\x,0.6) -- (\x+0.8,0.6);
  \draw (\x,1.2) -- (\x+0.8,1.2);
}
\node at (6.0,-0.4) {\small state};
\node at (7.6,-0.4) {\small 2nd particle};
\node at (9.2,-0.4) {\small 3rd particle};
\node at (6.0,1.4) {$\times$};
\node at (7.6,0.2) {$\times$};
\node at (9.2,0.8) {$\times$};
\end{tikzpicture}

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

In quantum statistics we will correct this mistake. Instead of asking where in which level particle 1 is.. and 2 is.., we ask how many particles are in ground state, 1\(^{\text{st}}\) excited state, and so on....

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

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

and,

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

So, \(S = -\left(\dfrac{\partial F}{\partial T}\right)_{V,N}\)

\[ = Nk_B\ln\left(\frac{V}{h^3}(2\pi m k_BT)^{3/2}\right) + \frac{Nk_BT}{\frac{V}{h^3}(2\pi mk_BT)^{3/2}}\cdot\frac{V}{h^3}\cdot\frac{3}{2}(2\pi mk_BT)^{1/2}(2\pi m k_B) \] \[ = Nk_B\ln\left(\frac{V}{h^3}(2\pi m k_BT)^{3/2}\right) + \frac{3}{2}Nk_B \]

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

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

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

Let us find

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

This formula is also not sensible....

Even the formula for Free energy \(F = -Nk_BT\ln\left(\dfrac{V}{h^3}(\;)^{3/2}\right)\)

as, we expect free energy to be extensive quantity.

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

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

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

Correction to partition function

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

ex. 3 particles.

\usetikzlibrary{arrows.meta,decorations.pathreplacing,calc,angles,quotes}
\begin{tikzpicture}
\draw (0,0) rectangle (1,1) node[pos=.5]{3};
\draw (1,0) rectangle (2,1) node[pos=.5]{2};
\draw (2,0) rectangle (3,1) node[pos=.5]{1};
\node at (4,0.5) {$= 6$ ways.};
\node at (0.5,-0.4) {\small 1st};
\node at (1.5,-0.4) {\small 2nd};
\node at (2.5,-0.4) {\small 3rd};
\node at (1.5,-0.8) {\small particle};
\end{tikzpicture}

So a rough way to correct \(Z\) is to divide by \((N!)\)

but that is only true if no two particles attain same `state'

\usetikzlibrary{arrows.meta,decorations.pathreplacing,calc,angles,quotes}
\begin{tikzpicture}[scale=0.7]
\draw (0,0) -- (2,0);
\draw (0,0.8) -- (2,0.8);
\draw (0,1.6) -- (2,1.6);
\filldraw (0.6,1.6) circle (2pt);
\filldraw (1.4,0.8) circle (2pt);
\filldraw (0.6,0) circle (2pt);
\filldraw (1.4,0) circle (2pt);
\end{tikzpicture}

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

per particle is much larger than number of particles. and indeed that happens in one case of classical ideal gas.

In general, \(i,j,k,\dots\) are energy levels of particle \(i\).

\[ Z = \frac{1}{N!}\sum_{i\neq j\neq k} \;+\; \frac{1}{2!(N-2)!}\sum_{i=j\neq k \neq l} \;+\; \frac{1}{3!(N-3)!}\sum_{(i=j=k\neq\cdots)} + \cdots \]

case if any two particles attains same level. prob. that any 3 particles have are in same level.

If number of states \((i,j,k\dots)\) are much larger than no. of particles it is very less likely that any two or any three particles attains same state.

So,

\[ Z = \frac{1}{N!}\left(\sum e^{-\beta\varepsilon}\right)^N \qquad \text{(summed over states of one particle)} \]

"This was done before advent of Q.M. by Sackur-Tetrode"

\[ \left(\ln Z = \ln \frac{1}{N!}\left[\frac{V(2\pi m k_BT)^{3/2}}{h^3}\right]^N\right) \]

We know Sterling formula

\[ N! \simeq N^N e^{-N}\sqrt{2\pi N} \qquad \text{(for very large $N$)} \] \[ N! \simeq N^N e^{-N} \]

So then, volume:

\[ Z \simeq \left[\left(\frac{V}{N}\right)\frac{(2\pi m k_BT)^{3/2}}{h^3}\cdot e\right]^N \]

So all non-sensible formulas of \((\mu, S)\) are corrected by this. Of course \(S(T=0)\) won't be explained by ideal gas.

Lecture 25 (continued): Criterion for Classical Statistics

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

Criterion for classical statistics to be valid

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

\usetikzlibrary{arrows.meta,decorations.pathreplacing,calc,angles,quotes}
\begin{tikzpicture}[scale=0.9]
\draw[domain=-1.5:1.5,smooth,variable=\x] plot ({\x},{1.4*exp(-2*\x*\x)});
\node at (0,-0.4) {Particle 1};
\node at (0,1.7) {wave packet};
\begin{scope}[xshift=5cm]
\draw[domain=-1.5:1.5,smooth,variable=\x] plot ({\x},{1.4*exp(-2*\x*\x)});
\node at (0,-0.4) {Particle 2};
\node at (0,1.7) {wave packet};
\end{scope}
\end{tikzpicture}

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

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

\usetikzlibrary{arrows.meta,decorations.pathreplacing,calc,angles,quotes}
\begin{tikzpicture}[scale=0.8]
\draw[->] (0,0) -- (3,1.5);
\draw[->] (3,1.5) -- (5,0.3) node[right] {2};
\node at (-0.3,0) {1};
\node at (3,1.7) {$\times$};
\node[right] at (5,1.5) {or};
\begin{scope}[xshift=6cm]
\draw[->] (0,0.3) -- (2,1.5);
\draw[->] (2,1.5) -- (5,0);
\node at (-0.3,0.3) {1};
\node at (5.3,0) {2};
\end{scope}
\end{tikzpicture}

(particles are exchanged).

So, for classical statistics to work,

mean interparticle separation \(\gg\gg \lambda_{\text{deBroglie}}\)

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

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

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

\[ \left(\frac{V}{N}\right)^{1/3} \gg\gg \frac{h}{p_{rms}} \qquad \left(\lambda_{deB} = \frac{h}{p}\right) \] \[ \left\langle\frac{p^2}{2m}\right\rangle = \frac{3}{2}k_BT \quad \text{for one particle} \qquad p_{rms} = \sqrt{3mk_BT} \] \[ \sqrt{\langle p^2\rangle} = p_{rms} \;\Rightarrow\; = (3mk_BT)^{1/2} \text{ or } (mk_BT)^{1/2} \] \[ \left(\frac{V}{N}\right)^{1/3} \gg\gg \frac{h}{(mk_BT)^{1/2}} \] \[ \left(\frac{V}{N}\right) \gg\gg \frac{h^3}{(mk_BT)^{3/2}} \qquad \text{or} \qquad \left(\frac{N}{V}\right) \gg\gg \frac{h^3}{(mk_BT)^{3/2}} \]

or

\[ 1 \gg\gg \frac{nh^3}{(mk_BT)^{3/2}} \qquad \frac{nh^3}{(mk_BT)^{3/2}} = \text{degeneracy factor} \]

So,

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

is criterion for statistics to be valid. (\(n=\) number density of particles)

Let's check for Nitrogen at N.T.P.: mass \(m \sim 10^{-26}\)kg, \(h = 10^{-34}\), \(n = 10^{24}\) per cubic meter, \(T = 300\)K, \(k_B = 1.38\times10^{-23}\).

\[ \frac{10^{24}\cdot(10^{-34})^3}{(10^{-26}\cdot 10^{-23}\cdot 300)^{3/2}} = \frac{10^{-78}}{(10^{-51})^{3/2}} \approx \frac{10^{-78}}{10^{-75}} \approx 10^{-3} \ll 1. \]

So classical statistics works.

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

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

Lecture 26

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

(After) (10:00)

Translate this page