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# Posts Tagged ‘entropy’

## A final pre-exam update of my notes compilation for ‘PHY452H1S Basic Statistical Mechanics’, Taught by Prof. Arun Paramekanti

Posted by peeterjoot on April 22, 2013

Here’s my third update of my notes compilation for this course, including all of the following:

April 21, 2013 Fermi function expansion for thermodynamic quantities

April 20, 2013 Relativistic Fermi Gas

April 10, 2013 Non integral binomial coefficient

April 10, 2013 energy distribution around mean energy

April 09, 2013 Velocity volume element to momentum volume element

April 04, 2013 Phonon modes

April 03, 2013 BEC and phonons

April 03, 2013 Max entropy, fugacity, and Fermi gas

April 02, 2013 Bosons

April 02, 2013 Relativisitic density of states

March 28, 2013 Bosons

plus everything detailed in the description of my previous update and before.

## PHY452H1S Basic Statistical Mechanics. Problem Set 6: Max entropy, fugacity, and Fermi gas

Posted by peeterjoot on April 3, 2013

[Click here for a PDF of this post with nicer formatting (especially if my latex to wordpress script has left FORMULA DOES NOT PARSE errors.)]

# Disclaimer

## Question: Maximum entropy principle

Consider the “Gibbs entropy”

\begin{aligned}S = - k_{\mathrm{B}} \sum_i p_i \ln p_i\end{aligned} \hspace{\stretch{1}}(1.1)

where $p_i$ is the equilibrium probability of occurrence of a microstate $i$ in the ensemble.

For a microcanonical ensemble with $\Omega$ configurations (each having the same energy), assigning an equal probability $p_i= 1/\Omega$ to each microstate leads to $S = k_{\mathrm{B}} \ln \Omega$. Show that this result follows from maximizing the Gibbs entropy with respect to the parameters $p_i$ subject to the constraint of

\begin{aligned}\sum_i p_i = 1\end{aligned} \hspace{\stretch{1}}(1.2)

(for $p_i$ to be meaningful as probabilities). In order to do the minimization with this constraint, use the method of Lagrange multipliers – first, do an unconstrained minimization of the function

\begin{aligned}S - \alpha \sum_i p_i,\end{aligned} \hspace{\stretch{1}}(1.3)

then fix $\alpha$ by demanding that the constraint be satisfied.

For a canonical ensemble (no constraint on total energy, but all microstates having the same number of particles $N$), maximize the Gibbs entropy with respect to the parameters $p_i$ subject to the constraint of

\begin{aligned}\sum_i p_i = 1,\end{aligned} \hspace{\stretch{1}}(1.4)

(for $p_i$ to be meaningful as probabilities) and with a given fixed average energy

\begin{aligned}\left\langle{{E}}\right\rangle = \sum_i E_i p_i,\end{aligned} \hspace{\stretch{1}}(1.5)

where $E_i$ is the energy of microstate $i$. Use the method of Lagrange multipliers, doing an unconstrained minimization of the function

\begin{aligned}S - \alpha \sum_i p_i - \beta \sum_i E_i p_i,\end{aligned} \hspace{\stretch{1}}(1.6)

then fix $\alpha, \beta$ by demanding that the constraint be satisfied. What is the resulting $p_i$?

For a grand canonical ensemble (no constraint on total energy, or the number of particles), maximize the Gibbs entropy with respect to the parameters $p_i$ subject to the constraint of

\begin{aligned}\sum_i p_i = 1,\end{aligned} \hspace{\stretch{1}}(1.7)

(for $p_i$ to be meaningful as probabilities) and with a given fixed average energy

\begin{aligned}\left\langle{{E}}\right\rangle = \sum_i E_i p_i,\end{aligned} \hspace{\stretch{1}}(1.8)

and a given fixed average particle number

\begin{aligned}\left\langle{{N}}\right\rangle = \sum_i N_i p_i.\end{aligned} \hspace{\stretch{1}}(1.9)

Here $E_i, N_i$ represent the energy and number of particles in microstate $i$. Use the method of Lagrange multipliers, doing an unconstrained minimization of the function

\begin{aligned}S - \alpha \sum_i p_i - \beta \sum_i E_i p_i - \gamma \sum_i N_i p_i,\end{aligned} \hspace{\stretch{1}}(1.10)

then fix $\alpha, \beta, \gamma$ by demanding that the constrains be satisfied. What is the resulting $p_i$?

Writing

\begin{aligned}f = S - \alpha \sum_{j = 1}^\Omega p_j,= -\sum_{j = 1}^\Omega p_j \left( k_{\mathrm{B}} \ln p_j + \alpha \right),\end{aligned} \hspace{\stretch{1}}(1.11)

our unconstrained minimization requires

\begin{aligned}0 = \frac{\partial {f}}{\partial {p_i}}= -\left( k_{\mathrm{B}} \left( \ln p_i + 1 \right) + \alpha \right).\end{aligned} \hspace{\stretch{1}}(1.12)

Solving for $p_i$ we have

\begin{aligned}p_i = e^{-\alpha/k_{\mathrm{B}} - 1}.\end{aligned} \hspace{\stretch{1}}(1.13)

The probabilities for each state are constant. To fix that constant we employ our constraint

\begin{aligned}1 = \sum_{j = 1}^\Omega p_j= \sum_{j = 1}^\Omega e^{-\alpha/k_{\mathrm{B}} - 1}= \Omega e^{-\alpha/k_{\mathrm{B}} - 1},\end{aligned} \hspace{\stretch{1}}(1.14)

or

\begin{aligned}\alpha/k_{\mathrm{B}} + 1 = \ln \Omega.\end{aligned} \hspace{\stretch{1}}(1.15)

Inserting eq. 1.15 fixes the probability, giving us the first of the expected results

\begin{aligned}\boxed{p_i = e^{-\ln \Omega} = \frac{1}{{\Omega}}.}\end{aligned} \hspace{\stretch{1}}(1.0.16)

Using this we our Gibbs entropy can be summed easily

\begin{aligned}S &= -k_{\mathrm{B}} \sum_{j = 1}^\Omega p_j \ln p_j \\ &= -k_{\mathrm{B}} \sum_{j = 1}^\Omega \frac{1}{{\Omega}} \ln \frac{1}{{\Omega}} \\ &= -k_{\mathrm{B}} \frac{\Omega}{\Omega} \left( -\ln \Omega \right),\end{aligned} \hspace{\stretch{1}}(1.0.16)

or

\begin{aligned}\boxed{S = k_{\mathrm{B}} \ln \Omega.}\end{aligned} \hspace{\stretch{1}}(1.0.16)

For the “action” like quantity that we want to minimize, let’s write

\begin{aligned}f = S - \alpha \sum_j p_j - \beta \sum_j E_j p_j,\end{aligned} \hspace{\stretch{1}}(1.0.16)

for which we seek $\alpha$, $\beta$ such that

\begin{aligned}0 &= \frac{\partial {f}}{\partial {p_i}} \\ &= -\frac{\partial {}}{\partial {p_i}}\sum_j p_j\left( k_{\mathrm{B}} \ln p_j + \alpha + \beta E_j \right) \\ &= -k_{\mathrm{B}} (\ln p_i + 1) - \alpha - \beta E_i,\end{aligned} \hspace{\stretch{1}}(1.0.16)

or

\begin{aligned}p_i = \exp\left( - \left( \alpha - \beta E_i \right) /k_{\mathrm{B}} - 1 \right).\end{aligned} \hspace{\stretch{1}}(1.0.16)

Our probability constraint is

\begin{aligned}1 &= \sum_j \exp\left( - \left( \alpha - \beta E_j \right) /k_{\mathrm{B}} - 1 \right) \\ &= \exp\left( - \alpha/k_{\mathrm{B}} - 1 \right)\sum_j \exp\left( - \beta E_j/k_{\mathrm{B}} \right),\end{aligned} \hspace{\stretch{1}}(1.0.16)

or

\begin{aligned}\exp\left( \alpha/k_{\mathrm{B}} + 1 \right)=\sum_j \exp\left( - \beta E_j/k_{\mathrm{B}} \right).\end{aligned} \hspace{\stretch{1}}(1.0.16)

Taking logs we have

\begin{aligned}\alpha/k_{\mathrm{B}} + 1 = \ln \sum_j \exp\left( - \beta E_j/k_{\mathrm{B}} \right).\end{aligned} \hspace{\stretch{1}}(1.0.16)

We could continue to solve for $\alpha$ explicitly but don’t care any more than this. Plugging back into the probability eq. 1.0.16 obtained from the unconstrained minimization we have

\begin{aligned}p_i = \exp\left( -\ln \sum_j \exp\left( - \beta E_j/k_{\mathrm{B}} \right) \right)\exp\left( - \beta E_i/k_{\mathrm{B}} \right),\end{aligned} \hspace{\stretch{1}}(1.0.16)

or

\begin{aligned}\boxed{p_i = \frac{ \exp\left( - \beta E_i/k_{\mathrm{B}} \right)}{ \sum_j \exp\left( - \beta E_j/k_{\mathrm{B}} \right)}.}\end{aligned} \hspace{\stretch{1}}(1.0.16)

To determine $\beta$ we must look implicitly to the energy constraint, which is

\begin{aligned}\left\langle{{E}}\right\rangle &= \sum_i E_i p_i \\ &= \sum_iE_i\left( \frac{ \exp\left( - \beta E_i/k_{\mathrm{B}} \right) } { \sum_j \exp\left( - \beta E_j/k_{\mathrm{B}} \right) } \right),\end{aligned} \hspace{\stretch{1}}(1.0.16)

or

\begin{aligned}\boxed{\left\langle{{E}}\right\rangle = \frac{ \sum_i E_i \exp\left( - \beta E_i/k_{\mathrm{B}} \right)}{ \sum_j \exp\left( - \beta E_j/k_{\mathrm{B}} \right)}.}\end{aligned} \hspace{\stretch{1}}(1.0.16)

The constraint $\beta$ ($=1/T$) is given implicitly by this energy constraint.

Again write

\begin{aligned}f = S - \alpha \sum_j p_j - \beta \sum_j E_j p_j - \gamma \sum_j N_j p_j.\end{aligned} \hspace{\stretch{1}}(1.0.16)

The unconstrained minimization requires

\begin{aligned}0 &= \frac{\partial {f}}{\partial {p_i}} \\ &= -\frac{\partial {}}{\partial {p_i}}\left( k_{\mathrm{B}} (\ln p_i + 1) + \alpha + \beta E_i + \gamma N_i \right),\end{aligned} \hspace{\stretch{1}}(1.0.16)

or

\begin{aligned}p_i = \exp\left( -\alpha/k_{\mathrm{B}} - 1 \right) \exp\left( -(\beta E_i + \gamma N_i)/k_{\mathrm{B}} \right).\end{aligned} \hspace{\stretch{1}}(1.0.16)

The unit probability constraint requires

\begin{aligned}1 &= \sum_j p_j \\ &= \exp\left( -\alpha/k_{\mathrm{B}} - 1 \right) \sum_j\exp\left( -(\beta E_j + \gamma N_j)/k_{\mathrm{B}} \right),\end{aligned} \hspace{\stretch{1}}(1.0.16)

or

\begin{aligned}\exp\left( -\alpha/k_{\mathrm{B}} - 1 \right) =\frac{1}{{\sum_j\exp\left( -(\beta E_j + \gamma N_j)/k_{\mathrm{B}} \right)}}.\end{aligned} \hspace{\stretch{1}}(1.0.16)

Our probability is then

\begin{aligned}\boxed{p_i = \frac{\exp\left( -(\beta E_i + \gamma N_i)/k_{\mathrm{B}} \right)}{\sum_j\exp\left( -(\beta E_j + \gamma N_j)/k_{\mathrm{B}} \right)}.}\end{aligned} \hspace{\stretch{1}}(1.0.16)

The average energy $\left\langle{{E}}\right\rangle = \sum_j p_j E_j$ and average number of particles $\left\langle{{N}}\right\rangle = \sum_j p_j N_j$ are given by

\begin{aligned}\left\langle{{E}}\right\rangle = \frac{E_i\exp\left( -(\beta E_i + \gamma N_i)/k_{\mathrm{B}} \right)}{\sum_j\exp\left( -(\beta E_j + \gamma N_j)/k_{\mathrm{B}} \right)}\end{aligned} \hspace{\stretch{1}}(1.0.35.35)

\begin{aligned}\left\langle{{N}}\right\rangle = \frac{N_i\exp\left( -(\beta E_i + \gamma N_i)/k_{\mathrm{B}} \right)}{\sum_j\exp\left( -(\beta E_j + \gamma N_j)/k_{\mathrm{B}} \right)}.\end{aligned} \hspace{\stretch{1}}(1.0.35.35)

The values $\beta$ and $\gamma$ are fixed implicitly by requiring simultaneous solutions of these equations.

## Question: Fugacity expansion ([3] Pathria, Appendix D, E)

The theory of the ideal Fermi or Bose gases often involves integrals of the form

\begin{aligned}f_\nu^\pm(z) = \frac{1}{{\Gamma(\nu)}} \int_0^\infty dx \frac{x^{\nu - 1}}{z^{-1} e^x \pm 1}\end{aligned} \hspace{\stretch{1}}(1.36)

where

\begin{aligned}\Gamma(\nu) = \int_0^\infty dy y^{\nu-1} e^{-y}\end{aligned} \hspace{\stretch{1}}(1.37)

denotes the gamma function.

Obtain the behavior of $f_\nu^\pm(z)$ for $z \rightarrow 0$ keeping the two leading terms in the expansion.

For Fermions, obtain the behavior of $f_\nu^\pm(z)$ for $z \rightarrow \infty$ again keeping the two leading terms.

For Bosons, we must have $z \le 1$ (why?), obtain the leading term of $f_\nu^-(z)$ for $z \rightarrow 1$.

For $z \rightarrow 0$ we can rewrite the integrand in a form that allows for series expansion

\begin{aligned}\frac{x^{\nu - 1}}{z^{-1} e^x \pm 1} &= \frac{z e^{-x} x^{\nu - 1}}{1 \pm z e^{-x}} \\ &= z e^{-x} x^{\nu - 1}\left( 1 \mp z e^{-x} + (z e^{-x})^2 \mp (z e^{-x})^3 + \cdots \right)\end{aligned} \hspace{\stretch{1}}(1.38)

For the $k$th power of $z e^{-x}$ in this series our integral is

\begin{aligned}\int_0^\infty dx z e^{-x} x^{\nu - 1} (z e^{-x})^k &= z^{k+1}\int_0^\infty dx x^{\nu - 1} e^{-(k + 1) x} \\ &= \frac{z^{k+1}}{(k+1)^\nu}\int_0^\infty du u^{\nu - 1} e^{- u} \\ &= \frac{z^{k+1}}{(k+1)^\nu} \Gamma(\nu)\end{aligned} \hspace{\stretch{1}}(1.39)

Putting everything back together we have for small $z$

\begin{aligned}\boxed{f_\nu^\pm(z) =z\mp\frac{z^{2}}{2^\nu}+\frac{z^{3}}{3^\nu}\mp\frac{z^{4}}{4^\nu}+ \cdots}\end{aligned} \hspace{\stretch{1}}(1.40)

We’ll expand $\Gamma(\nu) f_\nu^+(e^y)$ about $z = e^y$, writing

\begin{aligned}F_\nu(e^y) &= \Gamma(\nu) f_\nu^+(e^y) \\ &= \int_0^\infty dx \frac{x^{\nu - 1}}{e^{x - y} + 1} \\ &= \int_0^y dx \frac{x^{\nu - 1}}{e^{x - y} + 1}+\int_y^\infty dx \frac{x^{\nu - 1}}{e^{x - y} + 1}.\end{aligned} \hspace{\stretch{1}}(1.41)

The integral has been split into two since the behavior of the exponential in the denominator is quite different in the $x y$ ranges. Observe that in the first integral we have

\begin{aligned}\frac{1}{{2}} \le \frac{1}{e^{x - y} + 1} \le \frac{1}{{1 + e^{-y}}}.\end{aligned} \hspace{\stretch{1}}(1.0.42)

Since this term is of order 1, let’s consider the difference of this from $1$, writing

\begin{aligned}\frac{1}{e^{x - y} + 1} = 1 + u,\end{aligned} \hspace{\stretch{1}}(1.0.42)

or

\begin{aligned}u = \frac{1}{e^{x - y} + 1} - 1 &= \frac{1 -(e^{x - y} + 1)}{e^{x - y} + 1} \\ &= \frac{-e^{x - y} }{e^{x - y} + 1} \\ &= -\frac{1}{1 + e^{y - x}}.\end{aligned} \hspace{\stretch{1}}(1.0.42)

This gives us

\begin{aligned}F_\nu(e^y) &= \int_0^y dx x^{\nu - 1} \left( 1 - \frac{ 1 } { 1 + e^{y - x} } \right)+\int_y^\infty dx \frac{x^{\nu - 1}}{e^{x - y} + 1} \\ &= \frac{y^\nu}{\nu}-\int_0^y dx \frac{ x^{\nu - 1} } { 1 + e^{y - x} }+\int_y^\infty dx \frac{x^{\nu - 1}}{e^{x - y} + 1}.\end{aligned} \hspace{\stretch{1}}(1.0.42)

Now let’s make a change of variables $a = y - x$ in the first integral and $b = x - y$ in the second. This gives

\begin{aligned}F_\nu(e^y) = \frac{y^\nu}{\nu}-\int_0^\infty da \frac{ (y - a)^{\nu - 1} } { 1 + e^{a} }+\int_0^\infty db \frac{(y + b)^{\nu - 1}}{e^{b} + 1}.\end{aligned} \hspace{\stretch{1}}(1.0.42)

As $a$ gets large in the first integral the integrand is approximately $e^{-a} (y-a)^{\nu -1}$. The exponential dominates this integrand. Since we are considering large $y$, we can approximate the upper bound of the integral by extending it to $\infty$. Also expanding in series we have

\begin{aligned}F_\nu(e^y) &\approx \frac{y^\nu}{\nu}+\int_0^\infty da \frac{ (y + a)^{\nu - 1} -(y - a)^{\nu - 1} } { 1 + e^{a} } \\ &= \frac{y^\nu}{\nu}+\int_0^\infty da \frac{1}{{e^a + 1}}\left( \left( \frac{1}{{0!}} y^{\nu-1} a^0 + \frac{1}{{1!}} (\nu-1) y^{\nu-2} a^1 + \frac{1}{{2!}} (\nu-1) (\nu-2) y^{\nu-3} a^2 + \cdots \right) - \left( \frac{1}{{0!}} y^{\nu-1} (-a)^0 + \frac{1}{{1!}} (\nu-1) y^{\nu-2} (-a)^1 + \frac{1}{{2!}} (\nu-1) (\nu-2) y^{\nu-3} (-a)^2 + \cdots \right) \right) \\ &= \frac{y^\nu}{\nu}+ 2\int_0^\infty da \frac{1}{{e^a + 1}} \left( \frac{1}{{1!}} (\nu-1) y^{\nu-2} a^1 + \frac{1}{{3!}} (\nu-1) (\nu-2) (\nu - 3)y^{\nu-4} a^3 + \cdots \right) \\ &= \frac{y^\nu}{\nu}+ 2\sum_{j = 1, 3, 5, \cdots} \frac{y^{\nu - 1 - j}}{j!} \left( \prod_{k = 1}^j (\nu-k) \right)\int_0^\infty da \frac{a^j}{e^a + 1} \\ &= \frac{y^\nu}{\nu}+ 2\sum_{j = 1, 3, 5, \cdots} \frac{y^{\nu - 1 - j}}{j!} \frac{ \Gamma(\nu) } {\Gamma(\nu - j)}\int_0^\infty da \frac{a^j}{e^a + 1}.\end{aligned} \hspace{\stretch{1}}(1.0.42)

For the remaining integral, Mathematica gives

\begin{aligned}\int_0^\infty da \frac{a^j}{e^a + 1}=\left( 1 - 2^{-j} \right) j! \zeta (j+1),\end{aligned} \hspace{\stretch{1}}(1.0.42)

where for $s > 1$

\begin{aligned}\zeta(s) = \sum_{k=1}^{\infty} k^{-s}.\end{aligned} \hspace{\stretch{1}}(1.0.42)

This gives

\begin{aligned}F_\nu(e^y) \approx \frac{y^\nu}{\nu}+ 2\sum_{j = 1, 3, 5, \cdots} y^{\nu - 1 - j}\frac{ \Gamma(\nu) } {\Gamma(\nu - j)}\left( 1 - 2^{-j} \right) \zeta(j+1),\end{aligned} \hspace{\stretch{1}}(1.0.42)

or

\begin{aligned}f_\nu^+(e^y) &\approx y^\nu\left( \frac{1}{\nu \Gamma(\nu)} + 2 \sum_{j = 1, 3, 5, \cdots} \frac{ 1 } {\Gamma(\nu - j) y^{j + 1} } \left( 1 - 2^{-j} \right) \zeta(j+1) \right) \\ &= \frac{y^\nu}{\Gamma(\nu + 1)}\left( 1 + 2 \sum_{j = 1, 3, 5, \cdots} \frac{ \Gamma(\nu + 1) } {\Gamma(\nu - j) } \left( 1 - 2^{-j} \right) \frac{\zeta(j+1)}{ y^{j + 1} } \right),\end{aligned} \hspace{\stretch{1}}(1.0.42)

or

\begin{aligned}\boxed{f_\nu^+(e^y) \approx \frac{y^\nu}{\Gamma(\nu + 1)}\left( 1 + 2 \nu \sum_{j = 1, 3, 5, \cdots} (\nu-1) \cdots(\nu - j) \left( 1 - 2^{-j} \right) \frac{\zeta(j+1)}{ y^{j + 1} } \right).}\end{aligned} \hspace{\stretch{1}}(1.0.42)

Evaluating the numerical portions explicitly, with

\begin{aligned}c(j) = 2 \left(1-2^{-j}\right) \zeta (j+1),\end{aligned} \hspace{\stretch{1}}(1.0.42)

\begin{aligned}\begin{aligned}c(1) &= \frac{\pi^2}{6} \\ c(3) &= \frac{7 \pi^4}{360} \\ c(5) &= \frac{31 \pi^6}{15120} \\ c(7) &= \frac{127 \pi^8}{604800},\end{aligned}\end{aligned} \hspace{\stretch{1}}(1.0.42)

so to two terms ($j = 1, 3$), we have

\begin{aligned}\boxed{f_\nu^+(e^y) \approx \frac{y^\nu}{\Gamma(\nu + 1)}\left( 1 + \nu(\nu-1) \frac{\pi^2}{6 y^{2}} + \nu(\nu-1)(\nu-2)(\nu -3) \frac{7 \pi^4}{360 y^4} \right).}\end{aligned} \hspace{\stretch{1}}(1.0.42)

In order for the Boson occupation numbers to be non-singular we require $\mu$ less than all $\epsilon$. If that lowest energy level is set to zero, this is equivalent to $z < 1$. Given this restriction, a $z = e^{-\alpha}$ substitution is convenient for investigation of the $z \rightarrow 1$ case. Following the text, we'll write

\begin{aligned}G_\nu(e^{-\alpha})=\Gamma(\nu)f_\nu^-(e^{-\alpha}) = \int_0^\infty dx \frac{x^{\nu - 1}}{e^{x + \alpha} - 1}.\end{aligned} \hspace{\stretch{1}}(1.0.42)

For $\nu = 1$, this is integrable

\begin{aligned}\frac{d}{dx} \ln\left( 1 - e^{-x - \alpha} \right) &= \frac{e^{-x - \alpha}}{ 1 - e^{-x - \alpha} } \\ &= \frac{1}{{ e^{x + \alpha} - 1}},\end{aligned} \hspace{\stretch{1}}(1.0.42)

so that

\begin{aligned}G_1(e^{-\alpha}) &= \int_0^\infty dx \frac{1}{e^{x + \alpha} - 1} \\ &= {\ln \left( 1 - e^{-x - \alpha} \right)}_{0}^{\infty} \\ &= \ln 1 - \ln \left( 1 - e^{- \alpha} \right) \\ &= -\ln \left( 1 - e^{- \alpha} \right).\end{aligned} \hspace{\stretch{1}}(1.0.42)

Taylor expanding $1 - e^{-\alpha}$ we have

\begin{aligned}1 - e^{-\alpha} = 1 - \left( 1 - \alpha + \alpha^2/2 - \cdots \right).\end{aligned} \hspace{\stretch{1}}(1.0.42)

Noting that $\Gamma(1) = 1$, we have for the limit

\begin{aligned}\lim_{\alpha \rightarrow 0} G_1(e^{-\alpha}) \rightarrow - \ln \alpha,\end{aligned} \hspace{\stretch{1}}(1.0.42)

or

\begin{aligned}\lim_{z\rightarrow 1} f_\nu^-(z)= -\ln (-\ln z).\end{aligned} \hspace{\stretch{1}}(1.0.42)

For values of $\nu \ne 1$, the denominator is

\begin{aligned}e^{\alpha + x} - 1 = (\alpha + x) + (\alpha + x)^2/2 + \cdots\end{aligned} \hspace{\stretch{1}}(1.0.42)

To first order this gives us

\begin{aligned}f_\nu^-( e^{-\alpha} ) \approx \frac{1}{{\Gamma(\nu)}} \int_0^\infty dx \frac{1}{x + \alpha}.\end{aligned} \hspace{\stretch{1}}(1.0.42)

Of this integral Mathematica says it can be evaluated for $0 < \nu < 1$, and has the value

\begin{aligned}\frac{1}{{\Gamma(\nu)}} \int_0^\infty dx \frac{1}{x + \alpha}=\frac{\pi}{\sin(\pi\nu)} \frac{1}{\alpha^{1 - \nu} \Gamma (\nu )}.\end{aligned} \hspace{\stretch{1}}(1.0.42)

From [1] 6.1.17 we find

\begin{aligned}\Gamma(z) \Gamma(1-z) = \frac{\pi}{\sin(\pi z)},\end{aligned} \hspace{\stretch{1}}(1.0.42)

with which we can write

\begin{aligned}\boxed{f_\nu^-( e^{-\alpha} ) \approx \frac{ \Gamma(1 - \nu)}{ \alpha^{1 - \nu} }.}\end{aligned} \hspace{\stretch{1}}(1.0.42)

## Question: Nuclear matter ([2], prob 9.2)

Consider a heavy nucleus of mass number $A$. i.e., having $A$ total nucleons including neutrons and protons. Assume that the number of neutrons and protons is equal, and recall that each of them has spin-$1/2$ (so possessing two spin states). Treating these nucleons as a free ideal Fermi gas of uniform density contained in a radius $R = r_0 A^{1/3}$, where $r_0 = 1.4 \times 10^{-13} \text{cm}$, calculate the Fermi energy and the average energy per nucleon in MeV.

Our nucleon particle density is

\begin{aligned}\rho &= \frac{N}{V} \\ &= \frac{A}{\frac{4 \pi}{3} R^3} \\ &= \frac{3 A}{4 \pi r_0^3 A} \\ &= \frac{3}{4 \pi r_0^3} \\ &= \frac{3}{4 \pi (1.4 \times 10^{-13} \text{cm})^3} \\ &= 8.7 \times 10^{37} (\text{cm})^{-3} \\ &= 8.7 \times 10^{43} (\text{m})^{-3}\end{aligned} \hspace{\stretch{1}}(1.67)

With $m$ for the mass of either the proton or the neutron, and $\rho_m = \rho_p = \rho/2$, the Fermi energy for these particles is

\begin{aligned}\epsilon_{\mathrm{F}} = \frac{\hbar^2}{2m} \left( \frac{6 \pi (\rho/2)}{2 S + 1} \right)^{2/3},\end{aligned} \hspace{\stretch{1}}(1.68)

With $S = 1/2$, and $2 S + 1 = 2(1/2) + 1 = 2$ for either the proton or the neutron, this is

\begin{aligned}\epsilon_{\mathrm{F}} = \frac{\hbar^2}{2 m} \left( \frac{3 \pi^2 \rho}{2} \right)^{2/3}.\end{aligned} \hspace{\stretch{1}}(1.69)

\begin{aligned}\begin{aligned}\hbar &= 1.05 \times 10^{-34} \,\text{m^2 kg s^{-1}} \\ m &= 1.67 \times 10^{-27} \,\text{kg}\end{aligned}.\end{aligned} \hspace{\stretch{1}}(1.70)

This gives us

\begin{aligned}\epsilon_{\mathrm{F}} &= \frac{(1.05 \times 10^{-34})^2}{2 \times 1.67 \times 10^{-27}} \left( \frac{3 \pi^2 }{2} \frac{8.7 \times 10^{43} }{2} \right)^{2/3}\text{m}^4 \frac{\text{kg}^2}{s^2} \frac{1}{{\text{kg}}} \frac{1}{{\text{m}^2}} \\ &= 3.9 \times 10^{-12} \,\text{J} \times \left( 6.241509 \times 10^{12} \frac{\text{MeV}}{J} \right) \approx 24 \text{MeV}\end{aligned} \hspace{\stretch{1}}(1.71)

In lecture 16

we found that the total average energy for a Fermion gas of $N$ particles was

\begin{aligned}E = \frac{3}{5} N \epsilon_{\mathrm{F}},\end{aligned} \hspace{\stretch{1}}(1.72)

so the average energy per nucleon is approximately

\begin{aligned}\frac{3}{5} \epsilon_{\mathrm{F}} \approx 15 \,\text{MeV}.\end{aligned} \hspace{\stretch{1}}(1.73)

## Question: Neutron star ([2], prob 9.5)

Model a neutron star as an ideal Fermi gas of neutrons at $T = 0$ moving in the gravitational field of a heavy point mass $M$ at the center. Show that the pressure $P$ obeys the equation

\begin{aligned}\frac{dP}{dr} = - \gamma M \frac{\rho(r)}{r^2},\end{aligned} \hspace{\stretch{1}}(1.74)

where $\gamma$ is the gravitational constant, $r$ is the distance from the center, and $\rho(r)$ is the density which only depends on distance from the center.

In the grand canonical scheme the pressure for a Fermion system is given by

\begin{aligned}\beta P V &= \ln Z_{\mathrm{G}} \\ &= \ln \prod_\epsilon \sum_{n = 0}^1 \left( z e^{-\beta \epsilon} \right)^n \\ &= \sum_\epsilon \ln \left( 1 + z e^{-\beta \epsilon} \right).\end{aligned} \hspace{\stretch{1}}(1.75)

The kinetic energy of the particle is adjusted by the gravitational potential

\begin{aligned}\epsilon &= \epsilon_\mathbf{k}- \frac{\gamma m M}{r} \\ &= \frac{\hbar^2 \mathbf{k}^2}{2m}- \frac{\gamma m M}{r}.\end{aligned} \hspace{\stretch{1}}(1.76)

Differentiating eq. 1.75 with respect to the radius, we have

\begin{aligned}\beta V \frac{\partial {P}}{\partial {r}} &= -\beta \frac{\partial {\epsilon}}{\partial {r}}\sum_\epsilon \frac{z e^{-\beta \epsilon}}{ 1 + z e^{-\beta \epsilon} } \\ &= -\beta\left( \frac{\gamma m M}{r^2} \right)\sum_\epsilon \frac{1}{ z^{-1} e^{\beta \epsilon} + 1} \\ &= -\beta\left( \frac{\gamma m M}{r^2} \right)\left\langle{{N}}\right\rangle.\end{aligned} \hspace{\stretch{1}}(1.77)

Noting that $\left\langle{{N}}\right\rangle m/V$ is the average density of the particles, presumed radial, we have

\begin{aligned}\boxed{\frac{\partial P}{\partial r} &= -\frac{\gamma M}{r^2} \frac{m \left\langle N \right\rangle}{V} \\ &= -\frac{\gamma M}{r^2} \rho(r).}\end{aligned} \hspace{\stretch{1}}(1.0.78)

# References

[1] M. Abramowitz and I.A. Stegun. \emph{Handbook of mathematical functions with formulas, graphs, and mathematical tables}, volume 55. Dover publications, 1964.

[2] Kerson Huang. Introduction to statistical physics. CRC Press, 2001.

[3] RK Pathria. Statistical mechanics. Butterworth Heinemann, Oxford, UK, 1996.

## An updated compilation of notes, for ‘PHY452H1S Basic Statistical Mechanics’, Taught by Prof. Arun Paramekanti

Posted by peeterjoot on March 27, 2013

Here’s my second update of my notes compilation for this course, including all of the following:

March 27, 2013 Fermi gas

March 26, 2013 Fermi gas thermodynamics

March 26, 2013 Fermi gas thermodynamics

March 23, 2013 Relativisitic generalization of statistical mechanics

March 21, 2013 Kittel Zipper problem

March 18, 2013 Pathria chapter 4 diatomic molecule problem

March 17, 2013 Gibbs sum for a two level system

March 16, 2013 open system variance of N

March 16, 2013 probability forms of entropy

March 14, 2013 Grand Canonical/Fermion-Bosons

March 13, 2013 Quantum anharmonic oscillator

March 12, 2013 Grand canonical ensemble

March 11, 2013 Heat capacity of perturbed harmonic oscillator

March 10, 2013 Langevin small approximation

March 10, 2013 Addition of two one half spins

March 10, 2013 Midterm II reflection

March 07, 2013 Thermodynamic identities

March 06, 2013 Temperature

March 05, 2013 Interacting spin

plus everything detailed in the description of my first update and before.

## probability forms of entropy

Posted by peeterjoot on March 16, 2013

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## Question: Entropy as probability

[1] points out that entropy can be written as

\begin{aligned}S = - k_{\mathrm{B}} \sum_i P_i \ln P_i\end{aligned} \hspace{\stretch{1}}(1.0.1)

where

\begin{aligned}P_i = \frac{e^{-\beta E_i}}{Z}\end{aligned} \hspace{\stretch{1}}(1.0.2a)

\begin{aligned}Z = \sum_i e^{-\beta E_i}.\end{aligned} \hspace{\stretch{1}}(1.0.2b)

Show that this follows from the free energy $F = U - T S = -k_{\mathrm{B}} \ln Z$.

In terms of the free and average energies, we have

\begin{aligned}\frac{S}{k_{\mathrm{B}}} &= \frac{U - F}{k_{\mathrm{B}} T} \\ &= \beta \left( -\frac{\partial {\ln Z}}{\partial {\beta}} \right) - \beta \left( -k_{\mathrm{B}} T \ln Z \right) \\ &= \frac{\sum_i \beta E_i e^{-\beta E_i}}{Z} +\ln Z \\ &= -\sum_i P_i \ln e^{-\beta E_i} + \sum_i P_i \ln Z \\ &= -\sum_i P_i \ln \frac{e^{-\beta E_i}}{Z} P_i \\ &= -\sum_i P_i \ln P_i.\end{aligned} \hspace{\stretch{1}}(1.0.3)

## Question: Entropy in terms of grand partition probabilites ( [2] pr 4.1)

Generalize \cref{pr:entropyProbabilityForm:1} to the grand canonical scheme, where we have

\begin{aligned}P_{r, s} = \frac{e^{-\alpha N_r - \beta E_s}}{Z_{\mathrm{G}}}\end{aligned} \hspace{\stretch{1}}(1.0.4a)

\begin{aligned}Z_{\mathrm{G}} = \sum_{r,s} e^{-\alpha N_r - \beta E_s}\end{aligned} \hspace{\stretch{1}}(1.0.4b)

\begin{aligned}z = e^{-\alpha} = e^{\mu \beta}\end{aligned} \hspace{\stretch{1}}(1.0.4c)

\begin{aligned}q = \ln Z_{\mathrm{G}},\end{aligned} \hspace{\stretch{1}}(1.0.4d)

and show

\begin{aligned}S = - k_{\mathrm{B}} \sum_{r,s} P_{r,s} \ln P_{r,s}.\end{aligned} \hspace{\stretch{1}}(1.0.5)

With

\begin{aligned}\beta P V = q,\end{aligned} \hspace{\stretch{1}}(1.0.6)

the free energy takes the form

\begin{aligned}F = N \mu - P V = N \mu - q/\beta,\end{aligned} \hspace{\stretch{1}}(1.0.7)

so that the entropy (scaled by $k_{\mathrm{B}}$) leads us to the desired result

\begin{aligned}\frac{S}{k_{\mathrm{B}}} &= \beta U - N \mu \beta + q/(\beta k_{\mathrm{B}} T) \\ &= -\beta \frac{\partial {q}}{\partial {\beta}} - z \mu \beta \frac{\partial {q}}{\partial {z}} + q \\ &= \frac{1}{{Z_{\mathrm{G}}}}\sum_{r, s}\left( -\beta (-E_s) - \mu \beta N_r \right) e^{-\alpha N_r - \beta E_s}+ \ln Z_{\mathrm{G}} \\ &= \sum_{r, s} \ln e^{ \alpha N_r + \beta E_s } P_{r,s} + \left( \sum_{r, s} P_{r, s} \right)\ln Z_{\mathrm{G}} \\ &= -\sum_{r, s} \ln \frac{e^{ -\alpha N_r - \beta E_s }}{Z_{\mathrm{G}}} P_{r,s} \\ &= -\sum_{r, s} P_{r, s} \ln P_{r, s}\end{aligned} \hspace{\stretch{1}}(1.0.8)

# References

[1] E.A. Jackson. Equilibrium statistical mechanics. Dover Pubns, 2000.

[2] RK Pathria. Statistical mechanics. Butterworth Heinemann, Oxford, UK, 1996.

## PHY452H1S Basic Statistical Mechanics. Lecture 14: Grand canonical ensemble. Taught by Prof. Arun Paramekanti

Posted by peeterjoot on March 13, 2013

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# Disclaimer

Peeter’s lecture notes from class. May not be entirely coherent.

This lecture had a large amount of spoken content not captured in these notes. Reference to section 4 [1] was made for additional details.

# Grand canonical ensemble

Fig 1.1: Ensemble pictures

We are now going to allow particles to move to and from the system and the reservoir. The total number of states in the system is

\begin{aligned}\Omega_tot (N, V, E) =\sum_{N_S, E_S} \Omega_S(N_S, V_S, E_S)\Omega_R(N - N_S, V_R, E - E_S),\end{aligned} \hspace{\stretch{1}}(1.2.1)

so for $N_S \ll N$, and $E_S \ll E$, we have

\begin{aligned}\Omega_R &= \exp\left( \frac{1}{{k_{\mathrm{B}}}} S_R(N- N_S, V_R, E - E_S) \right) \\ &\approx \exp\left( \frac{1}{{k_{\mathrm{B}}}} S_R(N, V_R, E) - \frac{N_S}{k_{\mathrm{B}}} \left({\partial {S_R}}/{\partial {N}}\right)_{{V, E}} - \frac{E_S}{k_{\mathrm{B}}} \left({\partial {S_R}}/{\partial {E}}\right)_{{N, V}} \right) \\ &\propto \Omega_S(N_S, V_S, E_S)e^{-\frac{\mu}{k_{\mathrm{B}} T} N_S}e^{-\frac{E_S}{k_{\mathrm{B}} T} },\end{aligned} \hspace{\stretch{1}}(1.2.2)

where the chemical potential \index{chemical potential} and temperature \index{temperature} are defined respectively as

\begin{aligned}\frac{\mu}{T} = -\left({\partial {S_R}}/{\partial {N}}\right)_{{V,E}}\end{aligned} \hspace{\stretch{1}}(1.0.3a)

\begin{aligned}\frac{1}{T} = \left({\partial {S_R}}/{\partial {E}}\right)_{{N,V}}.\end{aligned} \hspace{\stretch{1}}(1.0.3b)

\begin{aligned}\mathcal{P} \propto e^{\frac{\mu}{k_{\mathrm{B}} T} N_S}e^{-\frac{E_S}{k_{\mathrm{B}} T} }.\end{aligned} \hspace{\stretch{1}}(1.0.4)

With $\{c\}$ as the set of all possible configuration pairs $\{N_S, E_S\}$, we define the grand partition function

\begin{aligned}Z_{\mathrm{G}} = \sum_{\{c\}}e^{\frac{\mu}{k_{\mathrm{B}} T} N_S}e^{-\frac{E_S}{k_{\mathrm{B}} T} }.\end{aligned} \hspace{\stretch{1}}(1.0.5)

So that the probability of finding a given state with energy and particle numbers $\{E_S, N_S\}$ is

\begin{aligned}\mathcal{P}(E_S, N_S) = \frac{e^{\frac{\mu}{k_{\mathrm{B}} T} N_S}e^{-\frac{E_S}{k_{\mathrm{B}} T} }}{Z_{\mathrm{G}}}.\end{aligned} \hspace{\stretch{1}}(1.0.6)

For a classical system we have

\begin{aligned}\{ c \} \rightarrow \{ x \} \{ p \},\end{aligned} \hspace{\stretch{1}}(1.0.7)

whereas in a quantum content we have

\begin{aligned}\{ c \} \rightarrow \text{eigenstate}.\end{aligned} \hspace{\stretch{1}}(1.0.8)

\begin{aligned}Z_{\mathrm{G}}^{\mathrm{\mathrm{\mathrm{QM}}}} = {\text{Tr}}_{\{\text{energy}, N\}} \left( e^{ -\beta (\hat{H} - \mu \hat{N} } \right).\end{aligned} \hspace{\stretch{1}}(1.0.9)

We want to do this because the calculation of the number of states

\begin{aligned}\int_{\{ x \} \{ p \}} \delta\left( \frac{p_1^2}{2m} + \frac{p_2^2}{2m} + \cdots + m g x_1 + m g x_2 + \cdots \right),\end{aligned} \hspace{\stretch{1}}(1.0.10)

can quickly become intractable. We want to go to the canonical ensemble was because the partition function

\begin{aligned}Z_c = \int_{\{ x \} \{ p \}}e^{-\beta \left( \frac{p_1^2}{2m} + \frac{p_2^2}{2m} + \cdots + m g x_1 + m g x_2 + \cdots \right)},\end{aligned} \hspace{\stretch{1}}(1.0.11)

yields the same results, but can be much easier to compute. We have a similar reason to go to the grand canonical ensemble, because this computation, once we allow the number of particles to vary also becomes very hard.

We are now going to define a notion of equilibrium so that it includes

1. All forces are equal (mechanical equilibrium)
2. Temperatures are equal (no net heat flow)
3. Chemical potentials are equal (no net particle flow)

We’ll isolate a subsystem, containing a large number of particles fig. 1.2.

Fig 1.2: A subsystem to and from which particle motion is allowed

When we think about Fermions we have to respect the “Pauli exclusion” principle \index{Pauli exclusion principle}.

Suppose we have just a one dimensional Fermion system for some potential as in fig. 1.3.

Fig 1.3: Energy level filling in a quantum system

For every momentum $k$ there are two possible occupation numbers $n_k \in \{0, 1\}$

our partition function is

\begin{aligned}Z_c = \sum_{n_k,\sum_k n_k = N} e^{-\beta \sum_k \epsilon_k n_k}.\end{aligned} \hspace{\stretch{1}}(1.0.12)

We’d find that this calculation with this $\sum_k n_k = N$ constraint becomes essentially impossible.

We’ll see that relaxing this constraint will allow this calculation to become tractable.

# References

[1] RK Pathria. Statistical mechanics. Butterworth Heinemann, Oxford, UK, 1996.

## PHY452H1S Basic Statistical Mechanics. Problem Set 5: Temperature

Posted by peeterjoot on March 10, 2013

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# Disclaimer

## Question: Polymer stretching – “entropic forces” (2013 problem set 5, p1)

Consider a toy model of a polymer in one dimension which is made of $N$ steps (amino acids) of unit length, going left or right like a random walk. Let one end of this polymer be at the origin and the other end be at a point $X = \sqrt{N}$ (viz. the rms size of the polymer) , so $1 \ll X \ll N$. We have previously calculated the number of configurations corresponding to this condition (approximate the binomial distribution by a Gaussian).

### Part a

Using this, find the entropy of this polymer as $S = k_{\mathrm{B}} \ln \Omega$. The free energy of this polymer, even in the absence of any other interactions, thus has an entropic contribution, $F = -T S$. If we stretch this polymer, we expect to have fewer available configurations, and thus a smaller entropy and a higher free energy.

### Part b

Find the change in free energy of this polymer if we stretch this polymer from its end being at $X$ to a larger distance $X + \Delta X$.

### Part c

Show that the change in free energy is linear in the displacement for small $\Delta X$, and hence find the temperature dependent “entropic spring constant” of this polymer. (This entropic force is important to overcome for packing DNA into the nucleus, and in many biological processes.)

Typo correction (via email):
You need to show that the change in free energy is quadratic in the displacement $\Delta X$, not linear in $\Delta X$. The force is linear in $\Delta X$. (Exactly as for a “spring”.)

### Entropy.

In lecture 2 probabilities for the sums of fair coin tosses were considered. Assigning $\pm 1$ to the events $Y_k$ for heads and tails coin tosses respectively, a random variable $Y = \sum_k Y_k$ for the total of $N$ such events was found to have the form

\begin{aligned}P_N(Y) = \left\{\begin{array}{l l}\left(\frac{1}{{2}}\right)^N \frac{N!}{\left(\frac{N-Y}{2}\right)!\left(\frac{N+Y}{2}\right)!}& \quad \mbox{if Y and N have same parity} \\ 0& \quad \mbox{otherwise} \end{array}\right.\end{aligned} \hspace{\stretch{1}}(1.1.1)

For an individual coin tosses we have averages $\left\langle{{Y_1}}\right\rangle = 0$, and $\left\langle{{Y_1^2}}\right\rangle = 1$, so the central limit theorem provides us with a large $N$ Gaussian approximation for this distribution

\begin{aligned}P_N(Y) \approx\frac{2}{\sqrt{2 \pi N}} \exp\left( -\frac{Y^2}{2N} \right).\end{aligned} \hspace{\stretch{1}}(1.1.2)

This fair coin toss problem can also be thought of as describing the coordinate of the end point of a one dimensional polymer with the beginning point of the polymer is fixed at the origin. Writing $\Omega(N, Y)$ for the total number of configurations that have an end point at coordinate $Y$ we have

\begin{aligned}P_N(Y) = \frac{\Omega(N, Y)}{2^N},\end{aligned} \hspace{\stretch{1}}(1.1.3)

From this, the total number of configurations that have, say, length $X = \left\lvert {Y} \right\rvert$, in the large $N$ Gaussian approximation, is

\begin{aligned}\Omega(N, X) &= 2^N \left( P_N(+X) +P_N(-X) \right) \\ &= \frac{2^{N + 2}}{\sqrt{2 \pi N}} \exp\left( -\frac{X^2}{2N} \right).\end{aligned} \hspace{\stretch{1}}(1.1.4)

The entropy associated with a one dimensional polymer of length $X$ is therefore

\begin{aligned}S_N(X) &= - k_{\mathrm{B}} \frac{X^2}{2N} + k_{\mathrm{B}} \ln \frac{2^{N + 2}}{\sqrt{2 \pi N}} \\ &= - k_{\mathrm{B}} \frac{X^2}{2N} + \text{constant}.\end{aligned} \hspace{\stretch{1}}(1.1.5)

Writing $S_0$ for this constant the free energy is

\begin{aligned}\boxed{F = U - T S = U + k_{\mathrm{B}} T \frac{X^2}{2N} + S_0 T.}\end{aligned} \hspace{\stretch{1}}(1.1.6)

### Change in free energy.

At constant temperature, stretching the polymer from its end being at $X$ to a larger distance $X + \Delta X$, results in a free energy change of

\begin{aligned}\Delta F &= F( X + \Delta X ) - F(X) \\ &= \frac{k_{\mathrm{B}} T}{2N} \left( (X + \Delta X)^2 - X^2 \right) \\ &= \frac{k_{\mathrm{B}} T}{2N} \left( 2 X \Delta X + (\Delta X)^2 \right)\end{aligned} \hspace{\stretch{1}}(1.1.7)

If $\Delta X$ is assumed small, our constant temperature change in free energy $\Delta F \approx (\partial F/\partial X)_T \Delta X$ is

\begin{aligned}\boxed{\Delta F = \frac{k_{\mathrm{B}} T}{N} X \Delta X.}\end{aligned} \hspace{\stretch{1}}(1.1.8)

### Temperature dependent spring constant.

I found the statement and subsequent correction of the problem statement somewhat confusing. To figure this all out, I thought it was reasonable to step back and relate free energy to the entropic force explicitly.

Consider temporarily a general thermodynamic system, for which we have by definition free energy and thermodynamic identity respectively

\begin{aligned}F = U - T S,\end{aligned} \hspace{\stretch{1}}(1.0.9a)

\begin{aligned}dU = T dS - P dV.\end{aligned} \hspace{\stretch{1}}(1.0.9b)

The differential of the free energy is

\begin{aligned}dF &= dU - T dS - S dT \\ &= -P dV - S dT \\ &= \left( \frac{\partial {F}}{\partial {T}} \right)_V dT+\left( \frac{\partial {F}}{\partial {V}} \right)_T dV.\end{aligned} \hspace{\stretch{1}}(1.0.10)

Forming the wedge product with $dT$, we arrive at the two form

\begin{aligned}0 &= \left( \left( P + \left( \frac{\partial {F}}{\partial {V}} \right)_T \right) dV + \left( S + \left( \frac{\partial {F}}{\partial {T}} \right)_V \right) dT \right)\wedge dT \\ &= \left( P + \left( \frac{\partial {F}}{\partial {V}} \right)_T \right) dV \wedge dT,\end{aligned} \hspace{\stretch{1}}(1.0.11)

This provides the relation between free energy and the “pressure” for the system

\begin{aligned}P = - \left( \frac{\partial {F}}{\partial {V}} \right)_T.\end{aligned} \hspace{\stretch{1}}(1.0.12)

For a system with a constant cross section $\Delta A$, $dV = \Delta A dX$, so the force associated with the system is

\begin{aligned}f &= P \Delta A \\ &= - \frac{1}{{\Delta A}} \left( \frac{\partial {F}}{\partial {X}} \right)_T \Delta A,\end{aligned} \hspace{\stretch{1}}(1.0.13)

or

\begin{aligned}f = - \left( \frac{\partial {F}}{\partial {X}} \right)_T.\end{aligned} \hspace{\stretch{1}}(1.0.14)

Okay, now we have a relation between the force and the rate of change of the free energy

\begin{aligned}f(X) = -\frac{k_{\mathrm{B}} T}{N} X.\end{aligned} \hspace{\stretch{1}}(1.0.15)

Our temperature dependent “entropic spring constant” in analogy with $f = -k X$, is therefore

\begin{aligned}\boxed{k = \frac{k_{\mathrm{B}} T}{N}.}\end{aligned} \hspace{\stretch{1}}(1.0.16)

## Question: Independent one-dimensional harmonic oscillators (2013 problem set 5, p2)

Consider a set of $N$ independent classical harmonic oscillators, each having a frequency $\omega$.

### Part a

Find the canonical partition at a temperature $T$ for this system of oscillators keeping track of correction factors of Planck constant. (Note that the oscillators are distinguishable, and we do not need $1/N!$ correction factor.)

### Part b

Using this, derive the mean energy and the specific heat at temperature $T$.

### Part c

For quantum oscillators, the partition function of each oscillator is simply $\sum_n e^{-\beta E_n}$ where $E_n$ are the (discrete) energy levels given by $(n + 1/2)\hbar \omega$, with $n = 0,1,2,\cdots$. Hence, find the canonical partition function for $N$ independent distinguishable quantum oscillators, and find the mean energy and specific heat at temperature $T$.

### Part d

Show that the quantum results go over into the classical results at high temperature $k_{\mathrm{B}} T \gg \hbar \omega$, and comment on why this makes sense.

### Part e

Also find the low temperature behavior of the specific heat in both classical and quantum cases when $k_{\mathrm{B}} T \ll \hbar \omega$.

### Classical partition function

For a single particle in one dimension our partition function is

\begin{aligned}Z_1 = \frac{1}{{h}} \int dp dq e^{-\beta \left( \frac{1}{{2 m}} p^2 + \frac{1}{{2}} m \omega^2 q^2 \right)},\end{aligned} \hspace{\stretch{1}}(1.0.17)

with

\begin{aligned}a = \sqrt{\frac{\beta}{2 m}} p\end{aligned} \hspace{\stretch{1}}(1.0.18a)

\begin{aligned}b = \sqrt{\frac{\beta m}{2}} \omega q,\end{aligned} \hspace{\stretch{1}}(1.0.18b)

we have

\begin{aligned}Z_1 &= \frac{1}{{h \omega}} \sqrt{\frac{2 m}{\beta}} \sqrt{\frac{2}{\beta m}} \int da db e^{-a^2 - b^2} \\ &= \frac{2}{\beta h \omega}2 \pi \int_0^\infty r e^{-r^2} \\ &= \frac{2 \pi}{\beta h \omega} \\ &= \frac{1}{\beta \hbar \omega}.\end{aligned} \hspace{\stretch{1}}(1.0.19)

So for $N$ distinguishable classical one dimensional harmonic oscillators we have

\begin{aligned}\boxed{Z_N(T) = Z_1^N = \left( \frac{k_{\mathrm{B}} T}{\hbar \omega} \right)^N.}\end{aligned} \hspace{\stretch{1}}(1.0.20)

### Classical mean energy and heat capacity

From the free energy

\begin{aligned}F = -k_{\mathrm{B}} T \ln Z_N = N k_{\mathrm{B}} T \ln (\beta \hbar \omega),\end{aligned} \hspace{\stretch{1}}(1.0.21)

we can compute the mean energy

\begin{aligned}U &= \frac{1}{{k_{\mathrm{B}}}} \frac{\partial {}}{\partial {\beta}} \left( \frac{F}{T} \right) \\ &= N \frac{\partial {}}{\partial {\beta}} \ln (\beta \hbar \omega) \\ &= \frac{N }{\beta},\end{aligned} \hspace{\stretch{1}}(1.0.22)

or

\begin{aligned}\boxed{U = N k_{\mathrm{B}} T.}\end{aligned} \hspace{\stretch{1}}(1.0.23)

The specific heat follows immediately

\begin{aligned}\boxed{C_{\mathrm{V}} = \frac{\partial {U}}{\partial {T}} = N k_{\mathrm{B}}.}\end{aligned} \hspace{\stretch{1}}(1.0.24)

### Quantum partition function, mean energy and heat capacity

For a single one dimensional quantum oscillator, our partition function is

\begin{aligned}Z_1 &= \sum_{n = 0}^\infty e^{-\beta \hbar \omega \left( n + \frac{1}{{2}} \right)} \\ &= e^{-\beta \hbar \omega/2}\sum_{n = 0}^\infty e^{-\beta \hbar \omega n} \\ &= \frac{e^{-\beta \hbar \omega/2}}{1 - e^{-\beta \hbar \omega}} \\ &= \frac{1}{e^{\beta \hbar \omega/2} - e^{-\beta \hbar \omega/2}} \\ &= \frac{1}{{\sinh(\beta \hbar \omega/2)}}.\end{aligned} \hspace{\stretch{1}}(1.0.25)

Assuming distinguishable quantum oscillators, our $N$ particle partition function is

\begin{aligned}\boxed{Z_N(\beta) = \frac{1}{{\sinh^N(\beta \hbar \omega/2)}}.}\end{aligned} \hspace{\stretch{1}}(1.0.26)

This time we don’t add the $1/\hbar$ correction factor, nor the $N!$ indistinguishability correction factor.

Our free energy is

\begin{aligned}F = N k_{\mathrm{B}} T \ln \sinh(\beta \hbar \omega/2),\end{aligned} \hspace{\stretch{1}}(1.0.27)

our mean energy is

\begin{aligned}U &= \frac{1}{{k_{\mathrm{B}}}} \frac{\partial {}}{\partial {\beta}} \frac{F}{T} \\ &= N \frac{\partial {}}{\partial {\beta}}\ln \sinh(\beta \hbar \omega/2) \\ &= N \frac{\cosh( \beta \hbar \omega/2 )}{\sinh(\beta \hbar \omega/2)} \frac{\hbar \omega}{2},\end{aligned} \hspace{\stretch{1}}(1.0.28)

or

\begin{aligned}\boxed{U(T)= \frac{N \hbar \omega}{2} \coth \left( \frac{\hbar \omega}{2 k_{\mathrm{B}} T} \right).}\end{aligned} \hspace{\stretch{1}}(1.0.29)

This is plotted in fig. 1.1.

Fig 1.1: Mean energy for N one dimensional quantum harmonic oscillators

With $\coth'(x) = -1/\sinh^2(x)$, our specific heat is

\begin{aligned}C_{\mathrm{V}} &= \frac{\partial {U}}{\partial {T}} \\ &= \frac{N \hbar \omega}{2} \frac{-1}{\sinh^2 \left( \frac{\hbar \omega}{2 k_{\mathrm{B}} T} \right)} \frac{\hbar \omega}{2 k_{\mathrm{B}}} \left( \frac{-1}{T^2} \right),\end{aligned} \hspace{\stretch{1}}(1.0.30)

or

\begin{aligned}\boxed{C_{\mathrm{V}} = N k_{\mathrm{B}}\left( \frac{\hbar \omega}{2 k_{\mathrm{B}} T \sinh \left( \frac{\hbar \omega}{2 k_{\mathrm{B}} T} \right) } \right)^2.}\end{aligned} \hspace{\stretch{1}}(1.0.31)

### Classical limits

In the high temperature limit $1 \gg \hbar \omega/k_{\mathrm{B}} T$, we have

\begin{aligned}\cosh \left( \frac{\hbar \omega}{2 k_{\mathrm{B}} T} \right)\approx 1\end{aligned} \hspace{\stretch{1}}(1.0.32)

\begin{aligned}\sinh \left( \frac{\hbar \omega}{2 k_{\mathrm{B}} T} \right)\approx \frac{\hbar \omega}{2 k_{\mathrm{B}} T},\end{aligned} \hspace{\stretch{1}}(1.0.33)

so

\begin{aligned}U \approx N \frac{\not{{\hbar \omega}}}{\not{{2}}} \frac{\not{{2}} k_{\mathrm{B}} T}{\not{{\hbar \omega}}},\end{aligned} \hspace{\stretch{1}}(1.0.34)

or

\begin{aligned}U(T) \approx N k_{\mathrm{B}} T,\end{aligned} \hspace{\stretch{1}}(1.0.35)

matching the classical result of eq. 1.0.23. Similarly from the quantum specific heat result of eq. 1.0.31, we have

\begin{aligned}C_{\mathrm{V}}(T) \approx N k_{\mathrm{B}}\left( \frac{\hbar \omega}{2 k_{\mathrm{B}} T \left( \frac{\hbar \omega}{2 k_{\mathrm{B}} T} \right) } \right)^2= N k_{\mathrm{B}}.\end{aligned} \hspace{\stretch{1}}(1.0.36)

This matches our classical result from eq. 1.0.24. We expect this equivalence at high temperatures since our quantum harmonic partition function eq. 1.0.26 is approximately

\begin{aligned}Z_N \approx \frac{2}{\beta \hbar \omega},\end{aligned} \hspace{\stretch{1}}(1.0.37)

This differs from the classical partition function only by this factor of $2$. While this alters the free energy by $k_{\mathrm{B}} T \ln 2$, it doesn’t change the mean energy since ${\partial {(k_{\mathrm{B}} \ln 2)}}/{\partial {\beta}} = 0$. At high temperatures the mean energy are large enough that the quantum nature of the system has no significant effect.

### Low temperature limits

For the classical case the heat capacity was constant ($C_{\mathrm{V}} = N k_{\mathrm{B}}$), all the way down to zero. For the quantum case the heat capacity drops to zero for low temperatures. We can see that via L’hopitals rule. With $x = \hbar \omega \beta/2$ the low temperature limit is

\begin{aligned}\lim_{T \rightarrow 0} C_{\mathrm{V}} &= N k_{\mathrm{B}} \lim_{x \rightarrow \infty} \frac{x^2}{\sinh^2 x} \\ &= N k_{\mathrm{B}} \lim_{x \rightarrow \infty} \frac{2x }{2 \sinh x \cosh x} \\ &= N k_{\mathrm{B}} \lim_{x \rightarrow \infty} \frac{1 }{\cosh^2 x + \sinh^2 x} \\ &= N k_{\mathrm{B}} \lim_{x \rightarrow \infty} \frac{1 }{\cosh (2 x) } \\ &= 0.\end{aligned} \hspace{\stretch{1}}(1.0.38)

We also see this in the plot of fig. 1.2.

Fig 1.2: Specific heat for N quantum oscillators

## Question: Quantum electric dipole (2013 problem set 5, p3)

A quantum electric dipole at a fixed space point has its energy determined by two parts – a part which comes from its angular motion and a part coming from its interaction with an applied electric field $\mathcal{E}$. This leads to a quantum Hamiltonian

\begin{aligned}H = \frac{\mathbf{L} \cdot \mathbf{L}}{2 I} - \mu \mathcal{E} L_z,\end{aligned} \hspace{\stretch{1}}(1.0.39)

where $I$ is the moment of inertia, and we have assumed an electric field $\mathcal{E} = \mathcal{E} \hat{\mathbf{z}}$. This Hamiltonian has eigenstates described by spherical harmonics $Y_{l, m}(\theta, \phi)$, with $m$ taking on $2l+1$ possible integral values, $m = -l, -l + 1, \cdots, l -1, l$. The corresponding eigenvalues are

\begin{aligned}\lambda_{l, m} = \frac{l(l+1) \hbar^2}{2I} - \mu \mathcal{E} m \hbar.\end{aligned} \hspace{\stretch{1}}(1.0.40)

(Recall that $l$ is the total angular momentum eigenvalue, while $m$ is the eigenvalue corresponding to $L_z$.)

### Part a

Schematically sketch these eigenvalues as a function of $\mathcal{E}$ for $l = 0,1,2$.

### Part b

Find the quantum partition function, assuming only $l = 0$ and $l = 1$ contribute to the sum.

### Part c

Using this partition function, find the average dipole moment $\mu \left\langle{{L_z}}\right\rangle$ as a function of the electric field and temperature for small electric fields, commenting on its behavior at very high temperature and very low temperature.

### Part d

Estimate the temperature above which discarding higher angular momentum states, with $l \ge 2$, is not a good approximation.

### Sketch the energy eigenvalues

Let’s summarize the values of the energy eigenvalues $\lambda_{l,m}$ for $l = 0, 1, 2$ before attempting to plot them.

$l = 0$

For $l = 0$, the azimuthal quantum number can only take the value $m = 0$, so we have

\begin{aligned}\lambda_{0,0} = 0.\end{aligned} \hspace{\stretch{1}}(1.0.41)

$l = 1$

For $l = 1$ we have

\begin{aligned}\frac{l(l+1)}{2} = 1(2)/2 = 1,\end{aligned} \hspace{\stretch{1}}(1.0.42)

so we have

\begin{aligned}\lambda_{1,0} = \frac{\hbar^2}{I} \end{aligned} \hspace{\stretch{1}}(1.0.43a)

\begin{aligned}\lambda_{1,\pm 1} = \frac{\hbar^2}{I} \mp \mu \mathcal{E} \hbar.\end{aligned} \hspace{\stretch{1}}(1.0.43b)

$l = 2$

For $l = 2$ we have

\begin{aligned}\frac{l(l+1)}{2} = 2(3)/2 = 3,\end{aligned} \hspace{\stretch{1}}(1.0.44)

so we have

\begin{aligned}\lambda_{2,0} = \frac{3 \hbar^2}{I} \end{aligned} \hspace{\stretch{1}}(1.0.45a)

\begin{aligned}\lambda_{2,\pm 1} = \frac{3 \hbar^2}{I} \mp \mu \mathcal{E} \hbar\end{aligned} \hspace{\stretch{1}}(1.0.45b)

\begin{aligned}\lambda_{2,\pm 2} = \frac{3 \hbar^2}{I} \mp 2 \mu \mathcal{E} \hbar.\end{aligned} \hspace{\stretch{1}}(1.0.45c)

These are sketched as a function of $\mathcal{E}$ in fig. 1.3.

Fig 1.3: Energy eigenvalues for l = 0,1, 2

### Partition function

Our partition function, in general, is

\begin{aligned}Z &= \sum_{l = 0}^\infty \sum_{m = -l}^l e^{-\lambda_{l,m} \beta} \\ &= \sum_{l = 0}^\infty \exp\left( -\frac{l (l+1) \hbar^2 \beta}{2 I} \right)\sum_{m = -l}^l e^{ m \mu \hbar \mathcal{E} \beta}.\end{aligned} \hspace{\stretch{1}}(1.0.46)

Dropping all but $l = 0, 1$ terms this is

\begin{aligned}Z \approx 1 + e^{-\hbar^2 \beta/I} \left( 1 + e^{- \mu \hbar \mathcal{E} \beta } + e^{ \mu \hbar \mathcal{E} \beta} \right),\end{aligned} \hspace{\stretch{1}}(1.0.47)

or

\begin{aligned}\boxed{Z \approx 1 + e^{-\hbar^2 \beta/I} (1 + 2 \cosh\left( \mu \hbar \mathcal{E} \beta \right)).}\end{aligned} \hspace{\stretch{1}}(1.0.48)

### Average dipole moment

For the average dipole moment, averaging over both the states and the partitions, we have

\begin{aligned}Z \left\langle{{ \mu L_z }}\right\rangle &= \sum_{l = 0}^\infty \sum_{m = -l}^l {\left\langle {l m} \right\rvert} \mu L_z {\left\lvert {l m} \right\rangle} e^{-\beta \lambda_{l, m}} \\ &= \sum_{l = 0}^\infty \sum_{m = -l}^l \mu {\left\langle {l m} \right\rvert} m \hbar {\left\lvert {l m} \right\rangle} e^{-\beta \lambda_{l, m}} \\ &= \mu \hbar \sum_{l = 0}^\infty \exp\left( -\frac{l (l+1) \hbar^2 \beta}{2 I} \right)\sum_{m = -l}^l m e^{ \mu m \hbar \mathcal{E} \beta} \\ &= \mu \hbar \sum_{l = 0}^\infty \exp\left( -\frac{l (l+1) \hbar^2 \beta}{2 I} \right)\sum_{m = 1}^l m \left( e^{ \mu m \hbar \mathcal{E} \beta} -e^{-\mu m \hbar \mathcal{E} \beta} \right) \\ &= 2 \mu \hbar \sum_{l = 0}^\infty \exp\left( -\frac{l (l+1) \hbar^2 \beta}{2 I} \right)\sum_{m = 1}^l m \sinh (\mu m \hbar \mathcal{E} \beta).\end{aligned} \hspace{\stretch{1}}(1.0.49)

For the cap of $l = 1$ we have

\begin{aligned}\left\langle{{ \mu L_z }}\right\rangle \approx\frac{2 \mu \hbar }{Z}\left( 1 (0) + e^{-\hbar^2 \beta/ I} \sinh (\mu \hbar \mathcal{E} \beta) \right)\approx2 \mu \hbar \frac{e^{-\hbar^2 \beta/ I} \sinh (\mu \hbar \mathcal{E} \beta) }{1 + e^{-\hbar^2 \beta/I} \left( 1 + 2 \cosh( \mu \hbar \mathcal{E} \beta) \right)},\end{aligned} \hspace{\stretch{1}}(1.0.50)

or

\begin{aligned}\boxed{\left\langle{{ \mu L_z }}\right\rangle \approx\frac{2 \mu \hbar \sinh (\mu \hbar \mathcal{E} \beta) }{e^{\hbar^2 \beta/I} + 1 + 2 \cosh( \mu \hbar \mathcal{E} \beta)}.}\end{aligned} \hspace{\stretch{1}}(1.0.51)

This is plotted in fig. 1.4.

Fig 1.4: Dipole moment

For high temperatures $\mu \hbar \mathcal{E} \beta \ll 1$ or $k_{\mathrm{B}} T \gg \mu \hbar \mathcal{E}$, expanding the hyperbolic sine and cosines to first and second order respectively and the exponential to first order we have

\begin{aligned}\left\langle{{ \mu L_z }}\right\rangle &\approx 2 \mu \hbar \frac{ \frac{\mu \hbar \mathcal{E}}{k_{\mathrm{B}} T}}{ 4 + \frac{h^2}{I k_{\mathrm{B}} T} + \left( \frac{\mu \hbar \mathcal{E}}{k_{\mathrm{B}} T} \right)^2}=\frac{2 (\mu \hbar)^2 \mathcal{E} k_{\mathrm{B}} T}{4 (k_{\mathrm{B}} T)^2 + \hbar^2 k_{\mathrm{B}} T/I + (\mu \hbar \mathcal{E})^2 } \\ &\approx\frac{(\mu \hbar)^2 \mathcal{E}}{4 k_{\mathrm{B}} T}.\end{aligned} \hspace{\stretch{1}}(1.0.52)

Our dipole moment tends to zero approximately inversely proportional to temperature. These last two respective approximations are plotted along with the all temperature range result in fig. 1.5.

Fig 1.5: High temperature approximations to dipole moments

For low temperatures $k_{\mathrm{B}} T \ll \mu \hbar \mathcal{E}$, where $\mu \hbar \mathcal{E} \beta \gg 1$ we have

\begin{aligned}\left\langle{{ \mu L_z }}\right\rangle \approx\frac{ 2 \mu \hbar e^{\mu \hbar \mathcal{E} \beta} }{ e^{\hbar^2 \beta/I} + e^{\mu \hbar \mathcal{E} \beta} }=\frac{ 2 \mu \hbar }{ 1 + e^{ (\hbar^2 \beta/I - \mu \hbar \mathcal{E})/{k_{\mathrm{B}} T} } }.\end{aligned} \hspace{\stretch{1}}(1.0.53)

Provided the electric field is small enough (which means here that $\mathcal{E} < \hbar/\mu I$) this will look something like fig. 1.6.

Fig 1.6: Low temperature dipole moment behavior

### Approximation validation

In order to validate the approximation, let’s first put the partition function and the numerator of the dipole moment into a tidier closed form, evaluating the sums over the radial indices $l$. First let’s sum the exponentials for the partition function, making an $n = m + l$

\begin{aligned}\sum_{m = -l}^l a^m &= a^{-l} \sum_{n=0}^{2l} a^n \\ &= a^{-l} \frac{a^{2l + 1} - 1}{a - 1} \\ &= \frac{a^{l + 1} - a^{-l}}{a - 1} \\ &= \frac{a^{l + 1/2} - a^{-(l+1/2)}}{a^{1/2} - a^{-1/2}}.\end{aligned} \hspace{\stretch{1}}(1.0.54)

With a substitution of $a = e^b$, we have

\begin{aligned}\boxed{\sum_{m = -l}^l e^{b m}=\frac{\sinh(b(l + 1/2))}{\sinh(b/2)}.}\end{aligned} \hspace{\stretch{1}}(1.0.55)

Now we can sum the azimuthal exponentials for the dipole moment. This sum is of the form

\begin{aligned}\sum_{m = -l}^l m a^m &= a \left( \sum_{m = 1}^l + \sum_{m = -l}^{-1} \right)m a^{m-1} \\ &= a \frac{d}{da}\sum_{m = 1}^l\left( a^{m} + a^{-m} \right) \\ &= a \frac{d}{da}\left( \sum_{m = -l}^l a^m - \not{{1}} \right) \\ &= a \frac{d}{da}\left( \frac{a^{l + 1/2} - a^{-(l+1/2)}}{a^{1/2} - a^{-1/2}} \right).\end{aligned} \hspace{\stretch{1}}(1.0.56)

With $a = e^{b}$, and $1 = a db/da$, we have

\begin{aligned}a \frac{d}{da} = a \frac{db}{da} \frac{d}{db} = \frac{d}{db},\end{aligned} \hspace{\stretch{1}}(1.0.57)

we have

\begin{aligned}\sum_{m = -l}^l m e^{b m}= \frac{d}{db}\left( \frac{ \sinh(b(l + 1/2)) }{ \sinh(b/2) } \right).\end{aligned} \hspace{\stretch{1}}(1.0.58)

With a little help from Mathematica to simplify that result we have

\begin{aligned}\boxed{\sum_{m = -l}^l m e^{b m}=\frac{l \sinh(b (l+1)) - (l+1) \sinh(b l) }{2 \sinh^2(b/2)}.}\end{aligned} \hspace{\stretch{1}}(1.0.59)

We can now express the average dipole moment with only sums over radial indices $l$

\begin{aligned}\left\langle{{ \mu L_z }}\right\rangle &= \mu \hbar \frac{ \sum_{l = 0}^\infty \exp\left( -\frac{l (l+1) \hbar^2 \beta}{2 I} \right) \sum_{m = -l}^l m e^{ \mu m \hbar \mathcal{E} \beta}}{ \sum_{l = 0}^\infty \exp\left( -\frac{l (l+1) \hbar^2 \beta}{2 I} \right) \sum_{m = -l}^l e^{ m \mu \hbar \mathcal{E} \beta}} \\ &= \mu \hbar\frac{ \sum_{l = 0}^\infty \exp\left( -\frac{l (l+1) \hbar^2 \beta}{2 I} \right) \frac { l \sinh(\mu \hbar \mathcal{E} \beta (l+1)) - (l+1) \sinh(\mu \hbar \mathcal{E} \beta l) } { 2 \sinh^2(\mu \hbar \mathcal{E} \beta/2) }}{\sum_{l = 0}^\infty \exp\left( -\frac{l (l+1) \hbar^2 \beta}{2 I} \right) \frac { \sinh(\mu \hbar \mathcal{E} \beta(l + 1/2)) } { \sinh(\mu \hbar \mathcal{E} \beta/2) }}.\end{aligned} \hspace{\stretch{1}}(1.0.60)

So our average dipole moment is

\begin{aligned}\boxed{\left\langle{{ \mu L_z }}\right\rangle = \frac{\mu \hbar }{2 \sinh(\mu \hbar \mathcal{E} \beta/2)}\frac{ \sum_{l = 0}^\infty \exp\left( -\frac{l (l+1) \hbar^2 \beta}{2 I} \right)\left( l \sinh(\mu \hbar \mathcal{E} \beta (l+1)) - (l+1) \sinh(\mu \hbar \mathcal{E} \beta l) \right)}{ \sum_{l = 0}^\infty \exp\left( -\frac{l (l+1) \hbar^2 \beta}{2 I} \right) \sinh(\mu \hbar \mathcal{E} \beta(l + 1/2))}.}\end{aligned} \hspace{\stretch{1}}(1.0.61)

The hyperbolic sine in the denominator from the partition function and the difference of hyperbolic sines in the numerator both grow fast. This is illustrated in fig. 1.7.

Fig 1.7: Hyperbolic sine plots for dipole moment

Let’s look at the order of these hyperbolic sines for large arguments. For the numerator we have a difference of the form

\begin{aligned}x \sinh( x + 1 ) - (x + 1) \sinh ( x ) &= \frac{1}{{2}} \left( x \left( e^{x + 1} - e^{-x - 1} \right) -(x +1 ) \left( e^{x } - e^{-x } \right) \right)\approx\frac{1}{{2}} \left( x e^{x + 1} -(x +1 ) e^{x } \right) \\ &= \frac{1}{{2}} \left( x e^{x} ( e - 1 ) - e^x \right) \\ &= O(x e^x).\end{aligned} \hspace{\stretch{1}}(1.0.62)

For the hyperbolic sine from the partition function we have for large $x$

\begin{aligned}\sinh( x + 1/2) = \frac{1}{{2}} \left( e^{x + 1/2} - e^{-x - 1/2} \right)\approx \frac{\sqrt{e}}{2} e^{x}= O(e^x).\end{aligned} \hspace{\stretch{1}}(1.0.63)

While these hyperbolic sines increase without bound as $l$ increases, we have a negative quadratic dependence on $l$ in the $\mathbf{L}^2$ contribution to these sums, provided that is small enough we can neglect the linear growth of the hyperbolic sines. We wish for that factor to be large enough that it dominates for all $l$. That is

\begin{aligned}\frac{l(l+1) \hbar^2}{2 I k_{\mathrm{B}} T} \gg 1,\end{aligned} \hspace{\stretch{1}}(1.0.64)

or

\begin{aligned}T \ll \frac{l(l+1) \hbar^2}{2 I k_{\mathrm{B}} T}.\end{aligned} \hspace{\stretch{1}}(1.0.65)

Observe that the RHS of this inequality, for $l = 1, 2, 3, 4, \cdots$ satisfies

\begin{aligned}\frac{\hbar^2 }{I k_{\mathrm{B}}}<\frac{3 \hbar^2 }{I k_{\mathrm{B}}}<\frac{6 \hbar^2 }{I k_{\mathrm{B}}}<\frac{10 \hbar^2 }{I k_{\mathrm{B}}}< \cdots\end{aligned} \hspace{\stretch{1}}(1.0.66)

So, for small electric fields, our approximation should be valid provided our temperature is constrained by

\begin{aligned}\boxed{T \ll \frac{\hbar^2 }{I k_{\mathrm{B}}}.}\end{aligned} \hspace{\stretch{1}}(1.0.67)

## Thermodynamic identities

Posted by peeterjoot on March 7, 2013

[Click here for a PDF of this post with nicer formatting (especially if my latex to wordpress script has left FORMULA DOES NOT PARSE errors.)]

Impressed with the clarity of Baez’s entropic force discussion on differential forms [1], let’s use that methodology to find all the possible identities that we can get from the thermodynamic identity (for now assuming $N$ is fixed, ignoring the chemical potential.)

This isn’t actually that much work to do, since a bit of editor regular expression magic can do most of the work.

Our starting point is the thermodynamic identity

\begin{aligned}dU = d Q + d W = T dS - P dV,\end{aligned} \hspace{\stretch{1}}(1.0.1)

or

\begin{aligned}0 = dU - T dS + P dV.\end{aligned} \hspace{\stretch{1}}(1.0.2)

It’s quite likely that many of the identities that can be obtained will be useful, but this should at least provide a handy reference of possible conversions.

Differentials in $P, V$

This first case illustrates the method.

\begin{aligned}0 &= dU - T dS + P dV \\ &= \left( \frac{\partial {U}}{\partial {P}} \right)_{V} dP +\left( \frac{\partial {U}}{\partial {V}} \right)_{P} dV- T\left( \left( \frac{\partial {S}}{\partial {P}} \right)_{V} dP + \left( \frac{\partial {S}}{\partial {V}} \right)_{P} dV \right)+ P dV \\ &= dP \left( \left( \frac{\partial {U}}{\partial {P}} \right)_{V} - T \left( \frac{\partial {S}}{\partial {P}} \right)_{V} \right)+dV \left( \left( \frac{\partial {U}}{\partial {V}} \right)_{P} - T \left( \frac{\partial {S}}{\partial {V}} \right)_{P} + P \right).\end{aligned} \hspace{\stretch{1}}(1.0.3)

Taking wedge products with $dV$ and $dP$ respectively, we form two two forms

\begin{aligned}0 = dP \wedge dV \left( \left( \frac{\partial {U}}{\partial {P}} \right)_{V} - T \left( \frac{\partial {S}}{\partial {P}} \right)_{V} \right)\end{aligned} \hspace{\stretch{1}}(1.0.4a)

\begin{aligned}0 = dV \wedge dP \left( \left( \frac{\partial {U}}{\partial {V}} \right)_{P} - T \left( \frac{\partial {S}}{\partial {V}} \right)_{P} + P \right).\end{aligned} \hspace{\stretch{1}}(1.0.4b)

Since these must both be zero we find

\begin{aligned}\left( \frac{\partial {U}}{\partial {P}} \right)_{V} = T \left( \frac{\partial {S}}{\partial {P}} \right)_{V}\end{aligned} \hspace{\stretch{1}}(1.0.5a)

\begin{aligned}P =-\left( \frac{\partial {U}}{\partial {V}} \right)_{P}- T \left( \frac{\partial {S}}{\partial {V}} \right)_{P}.\end{aligned} \hspace{\stretch{1}}(1.0.5b)

Differentials in $P, T$

\begin{aligned}0 &= dU - T dS + P dV \\ &= \left( \frac{\partial {U}}{\partial {P}} \right)_{T} dP + \left( \frac{\partial {U}}{\partial {T}} \right)_{P} dT-T \left( \left( \frac{\partial {S}}{\partial {P}} \right)_{T} dP + \left( \frac{\partial {S}}{\partial {T}} \right)_{P} dT \right)+\left( \frac{\partial {V}}{\partial {P}} \right)_{T} dP + \left( \frac{\partial {V}}{\partial {T}} \right)_{P} dT,\end{aligned} \hspace{\stretch{1}}(1.0.6)

or

\begin{aligned}0 = \left( \frac{\partial {U}}{\partial {P}} \right)_{T} -T \left( \frac{\partial {S}}{\partial {P}} \right)_{T} + \left( \frac{\partial {V}}{\partial {P}} \right)_{T}\end{aligned} \hspace{\stretch{1}}(1.0.7a)

\begin{aligned}0 = \left( \frac{\partial {U}}{\partial {T}} \right)_{P} -T \left( \frac{\partial {S}}{\partial {T}} \right)_{P} + \left( \frac{\partial {V}}{\partial {T}} \right)_{P}.\end{aligned} \hspace{\stretch{1}}(1.0.7b)

Differentials in $P, S$

\begin{aligned}0 &= dU - T dS + P dV \\ &= \left( \frac{\partial {U}}{\partial {P}} \right)_{S} dP + \left( \frac{\partial {U}}{\partial {S}} \right)_{P} dS- T dS+ P \left( \left( \frac{\partial {V}}{\partial {P}} \right)_{S} dP + \left( \frac{\partial {V}}{\partial {S}} \right)_{P} dS \right),\end{aligned} \hspace{\stretch{1}}(1.0.8)

or

\begin{aligned}\left( \frac{\partial {U}}{\partial {P}} \right)_{S} = -P \left( \frac{\partial {V}}{\partial {P}} \right)_{S}\end{aligned} \hspace{\stretch{1}}(1.0.9a)

\begin{aligned}T = \left( \frac{\partial {U}}{\partial {S}} \right)_{P} + P \left( \frac{\partial {V}}{\partial {S}} \right)_{P}.\end{aligned} \hspace{\stretch{1}}(1.0.9b)

Differentials in $P, U$

\begin{aligned}0 &= dU - T dS + P dV \\ &= dU - T \left( \left( \frac{\partial {S}}{\partial {P}} \right)_{U} dP + \left( \frac{\partial {S}}{\partial {U}} \right)_{P} dU \right)+ P\left( \left( \frac{\partial {V}}{\partial {P}} \right)_{U} dP + \left( \frac{\partial {V}}{\partial {U}} \right)_{P} dU \right),\end{aligned} \hspace{\stretch{1}}(1.0.10)

or

\begin{aligned}0 = 1 - T \left( \frac{\partial {S}}{\partial {U}} \right)_{P} + P \left( \frac{\partial {V}}{\partial {U}} \right)_{P} \end{aligned} \hspace{\stretch{1}}(1.0.11a)

\begin{aligned}T \left( \frac{\partial {S}}{\partial {P}} \right)_{U} = P \left( \frac{\partial {V}}{\partial {P}} \right)_{U}.\end{aligned} \hspace{\stretch{1}}(1.0.11b)

Differentials in $V, T$

\begin{aligned}0 &= dU - T dS + P dV \\ &= \left( \frac{\partial {U}}{\partial {V}} \right)_{T} dV + \left( \frac{\partial {U}}{\partial {T}} \right)_{V} dT - T \left( \left( \frac{\partial {S}}{\partial {V}} \right)_{T} dV + \left( \frac{\partial {S}}{\partial {T}} \right)_{V} dT \right)+ P dV,\end{aligned} \hspace{\stretch{1}}(1.0.12)

or

\begin{aligned}0 = \left( \frac{\partial {U}}{\partial {V}} \right)_{T} - T \left( \frac{\partial {S}}{\partial {V}} \right)_{T} + P \end{aligned} \hspace{\stretch{1}}(1.0.13a)

\begin{aligned}\left( \frac{\partial {U}}{\partial {T}} \right)_{V} = T \left( \frac{\partial {S}}{\partial {T}} \right)_{V}.\end{aligned} \hspace{\stretch{1}}(1.0.13b)

Differentials in $V, S$

\begin{aligned}0 &= dU - T dS + P dV \\ &= \left( \frac{\partial {U}}{\partial {V}} \right)_{S} dV + \left( \frac{\partial {U}}{\partial {S}} \right)_{V} dS - T dS+ P dV,\end{aligned} \hspace{\stretch{1}}(1.0.14)

or

\begin{aligned}P = -\left( \frac{\partial {U}}{\partial {V}} \right)_{S}\end{aligned} \hspace{\stretch{1}}(1.0.15a)

\begin{aligned}T = \left( \frac{\partial {U}}{\partial {S}} \right)_{V} .\end{aligned} \hspace{\stretch{1}}(1.0.15b)

Differentials in $V, U$

\begin{aligned}0 &= dU - T dS + P dV \\ &= dU- T \left( \left( \frac{\partial {S}}{\partial {V}} \right)_{U} dV + \left( \frac{\partial {S}}{\partial {U}} \right)_{V} dU \right)+ P \left( \left( \frac{\partial {V}}{\partial {V}} \right)_{U} dV + \left( \frac{\partial {V}}{\partial {U}} \right)_{V} dU \right)\end{aligned} \hspace{\stretch{1}}(1.0.16)

or

\begin{aligned}0 = 1 - T \left( \frac{\partial {S}}{\partial {U}} \right)_{V} + P \left( \frac{\partial {V}}{\partial {U}} \right)_{V} \end{aligned} \hspace{\stretch{1}}(1.0.17a)

\begin{aligned}T \left( \frac{\partial {S}}{\partial {V}} \right)_{U} = P \left( \frac{\partial {V}}{\partial {V}} \right)_{U}.\end{aligned} \hspace{\stretch{1}}(1.0.17b)

Differentials in $S, T$

\begin{aligned}0 &= dU - T dS + P dV \\ &= \left( \left( \frac{\partial {U}}{\partial {S}} \right)_{T} dS + \left( \frac{\partial {U}}{\partial {T}} \right)_{S} dT \right)- T dS+ P \left( \left( \frac{\partial {V}}{\partial {S}} \right)_{T} dS + \left( \frac{\partial {V}}{\partial {T}} \right)_{S} dT \right),\end{aligned} \hspace{\stretch{1}}(1.0.18)

or

\begin{aligned}0 = \left( \frac{\partial {U}}{\partial {S}} \right)_{T} - T + P \left( \frac{\partial {V}}{\partial {S}} \right)_{T} \end{aligned} \hspace{\stretch{1}}(1.0.19a)

\begin{aligned}0 = \left( \frac{\partial {U}}{\partial {T}} \right)_{S} + P \left( \frac{\partial {V}}{\partial {T}} \right)_{S}.\end{aligned} \hspace{\stretch{1}}(1.0.19b)

Differentials in $S, U$

\begin{aligned}0 &= dU - T dS + P dV \\ &= dU - T dS+ P \left( \left( \frac{\partial {V}}{\partial {S}} \right)_{U} dS + \left( \frac{\partial {V}}{\partial {U}} \right)_{S} dU \right)\end{aligned} \hspace{\stretch{1}}(1.0.20)

or

\begin{aligned}\frac{1}{{P}} = - \left( \frac{\partial {V}}{\partial {U}} \right)_{S} \end{aligned} \hspace{\stretch{1}}(1.0.21a)

\begin{aligned}T = P \left( \frac{\partial {V}}{\partial {S}} \right)_{U}.\end{aligned} \hspace{\stretch{1}}(1.0.21b)

Differentials in $T, U$

\begin{aligned}0 &= dU - T dS + P dV \\ &= dU - T \left( \left( \frac{\partial {S}}{\partial {T}} \right)_{U} dT + \left( \frac{\partial {S}}{\partial {U}} \right)_{T} dU \right)+ P\left( \left( \frac{\partial {V}}{\partial {T}} \right)_{U} dT + \left( \frac{\partial {V}}{\partial {U}} \right)_{T} dU \right),\end{aligned} \hspace{\stretch{1}}(1.0.22)

or

\begin{aligned}0 = 1 - T \left( \frac{\partial {S}}{\partial {U}} \right)_{T} + P \left( \frac{\partial {V}}{\partial {U}} \right)_{T} \end{aligned} \hspace{\stretch{1}}(1.0.23a)

\begin{aligned}T \left( \frac{\partial {S}}{\partial {T}} \right)_{U} = P \left( \frac{\partial {V}}{\partial {T}} \right)_{U}.\end{aligned} \hspace{\stretch{1}}(1.0.23b)

# References

[1] John Baez. Entropic forces, 2012. URL http://johncarlosbaez.wordpress.com/2012/02/01/entropic-forces/. [Online; accessed 07-March-2013].

## An updated compilation of notes, for ‘PHY452H1S Basic Statistical Mechanics’, Taught by Prof. Arun Paramekanti

Posted by peeterjoot on March 3, 2013

That compilation now all of the following too (no further updates will be made to any of these) :

February 28, 2013 Rotation of diatomic molecules

February 28, 2013 Helmholtz free energy

February 26, 2013 Statistical and thermodynamic connection

February 24, 2013 Ideal gas

February 16, 2013 One dimensional well problem from Pathria chapter II

February 15, 2013 1D pendulum problem in phase space

February 14, 2013 Continuing review of thermodynamics

February 13, 2013 Lightning review of thermodynamics

February 11, 2013 Cartesian to spherical change of variables in 3d phase space

February 10, 2013 n SHO particle phase space volume

February 10, 2013 Change of variables in 2d phase space

February 10, 2013 Some problems from Kittel chapter 3

February 07, 2013 Midterm review, thermodynamics

February 06, 2013 Limit of unfair coin distribution, the hard way

February 05, 2013 Ideal gas and SHO phase space volume calculations

February 03, 2013 One dimensional random walk

February 02, 2013 1D SHO phase space

February 02, 2013 Application of the central limit theorem to a product of random vars

January 31, 2013 Liouville’s theorem questions on density and current

January 30, 2013 State counting