Statistical Mechanics *

January 17, 2023January 20, 2023

202301171158 Problem 1.45

As an illustration of why it matters which variables you hold fixed when taking partial derivatives, consider the following mathematical example. Let $w=xy$ and $x=yz$ .

(a) Write $w$ purely in terms of $x$ and $z$ , and then purely in terms of $y$ and $z$ .
(b) Compute the partial derivatives

$\displaystyle{\bigg(\frac{\partial w}{\partial x}\bigg)_y}\quad\textrm{ and }\quad\displaystyle{\bigg(\frac{\partial w}{\partial x}\bigg)_z}$ ,

and show that they are not equal. (Hint: To compute $(\partial w/\partial x)_y$ , use a formula for $w$ in terms of $x$ and $y$ , not $z$ . Similarly, compute $(\partial w/\partial x)_z$ from a formula for $w$ in terms of only $x$ and $z$ .)
(c) Compute the other four partial derivatives of $w$ (two each with respect to $y$ and $z$ ), and show that it matters which variable is held fixed.

_{Extracted from D. V. Schroeder. (2000). An Introduction to Thermal Physics.}

Roughwork.

(a) Write on one hand

$\begin{aligned} w & = xy \\ & = x\bigg(\frac{x}{z}\bigg) \\ & = \frac{x^2}{z} \\ \end{aligned}$

and the other

$\begin{aligned} w & = xy \\ & = (yz)y \\ & = y^2z \\ \end{aligned}$

(b)

$\begin{aligned} \bigg(\frac{\partial w}{\partial x}\bigg)_y &=\frac{\partial}{\partial x}(w(x,y)) = \frac{\partial}{\partial x}(xy) = y \\ \bigg(\frac{\partial w}{\partial x}\bigg)_z & =\frac{\partial}{\partial x}(w(x,z)) = \frac{\partial}{\partial x}\bigg(\frac{x^2}{z}\bigg) = \frac{2x}{z} \\ \end{aligned}$

(c) Not to be attempted.

June 1, 2022October 24, 2022

202206011159 Problem 5E (Q25)

Test each of the following differentials to see whether they are exact, using two methods for each:

(a) $-y\sin x\,\mathrm{d}x+\cos x\,\mathrm{d}y$ ,

(b) $y\,\mathrm{d}x+x\,\mathrm{d}y$ ,
(c) $yx^3e^x\,\mathrm{d}x+x^3e^x\,\mathrm{d}y$ ,
(d) $(1+x)ye^x\,\mathrm{d}x+xe^x\,\mathrm{d}y$ ,
(e) $4x^3y^{-2}\,\mathrm{d}x-2x^4y^{-3}\,\mathrm{d}y$ .

_{K. S. Stowe. (2007). An Introduction to Thermodynamics and Statistical Mechanics}

Revision. (exact differentials)

The differential of a function is given by Eq. (5.8):

$\mathrm{d}F=\displaystyle{\frac{\partial F}{\partial x}\,\mathrm{d}x+\frac{\partial F}{\partial y}\,\mathrm{d}y}$ .

Therefore, one way to determine whether a differential given by Eq. (5.9):

$\mathrm{d}\Phi =g(x,y)\,\mathrm{d}x+h(x,y)\,\mathrm{d}y$

is exact is to see whether we can find some function $F(x,y)$ such that

$\displaystyle{\frac{\partial F}{\partial x}=g(x,y)}\qquad \textrm{and}\qquad\displaystyle{\frac{\partial F}{\partial y}=h(x,y)}$ .

If we can, the differential is exact, and if we can’t, it is inexact. Alternatively, we can use the identity

$\displaystyle{\frac{\partial^2F}{\partial y\partial x}=\frac{\partial^2F}{\partial x\partial y}}$ .

Combining this with equations (5.8) and (5.9), we can see that for exact differentials, Eq. (5.10):

$\displaystyle{\frac{\partial g}{\partial y}=\frac{\partial h}{\partial x}}$ .

_{Text on pg. 89, Sec. E, Ch. 5}

Roughwork.

(a)

We can see that this is indeed an exact differential of the function

$F=y\cos x+\textrm{constant}$ ,

because

$\displaystyle{\frac{\partial F}{\partial x}=-y\sin x}\qquad\textrm{and}\qquad\displaystyle{\frac{\partial F}{\partial y}=\cos x}$ .

Or we can use Eq. (5.10). For this example $g=-y\sin x$ and $h=\cos x$ , so

$\displaystyle{\frac{\partial g}{\partial y}=-\sin x}\qquad\textrm{and}\qquad\displaystyle{\frac{\partial h}{\partial x}=-\sin x}$ .

The two are the same, so the differential is exact.

Parts (b) to (e) are left to the reader as an exercise. Cf. analytic functions et Cauchy-Riemann equations.

October 11, 2021January 19, 2023

202110110958 Sidenote (Section 3.8.2)

Given that the binomial coefficient is so defined as

$\begin{pmatrix} N \\ n \end{pmatrix} \stackrel{\textrm{def}}{=} \displaystyle{\frac{N!}{n!(N-n)!}}$ .

Prove the identities (3.49), (3.50), and (3.51) in Section 3.8.2 Useful Identities for the Binomial Coefficients. Namely,

Eq. (3.49):

$\begin{pmatrix} N \\ 0 \end{pmatrix} = \begin{pmatrix} N \\ N \end{pmatrix} = 1$

Eq. (3.50):

$\begin{pmatrix} N - 1 \\ n \end{pmatrix} + \begin{pmatrix} N - 1 \\ n-1 \end{pmatrix} = \begin{pmatrix} N \\ n \end{pmatrix}$

Eq. (3.51):

$\begin{pmatrix} N \\ n+1 \end{pmatrix} = \displaystyle{\frac{N-n}{n+1}}\begin{pmatrix} N \\ n \end{pmatrix}$

_{R. H. Swendsen. (2012). An Introduction to Statistical Mechanics and Thermodynamics}

Proof of Eq. (3.49).

$\begin{aligned} \begin{pmatrix} N \\ 0 \end{pmatrix} & = \frac{N!}{0!(N-0)!} \\ & = \frac{N!}{1\times N!} \\ & = 1 \end{aligned}$
$\begin{aligned} \begin{pmatrix} N \\ N \end{pmatrix} & = \frac{N!}{N!(N-N)!} \\ & = \frac{N!}{N!\times 1} \\ & = 1 \end{aligned}$

Proof of Eq. (3.50).

$\begin{aligned} & \quad \begin{pmatrix} N-1 \\ n \end{pmatrix} + \begin{pmatrix} N-1 \\ n-1 \end{pmatrix} \\ & = \frac{(N-1)!}{(n)!(N-1-n)!} + \frac{(N-1)!}{(n-1)!\big((N-1)-(n-1)\big)!} \\ & = \frac{(N-1)!}{(n)!(N-1-n)!} + \frac{(N-1)!}{(n-1)!(N-n)!} \\ & = \frac{(N-n)(N-1)!+(n)(N-1)!}{(n)(n-1)!(N-n)(N-n-1)!} \\ & = \frac{N(N-1)!}{n!(N-n)!} \\ & = \frac{N!}{n!(N-n)!} \\ & = \begin{pmatrix} N \\ n \end{pmatrix} \\ \end{aligned}$

Proof of Eq. (3.51).

$\begin{aligned} &\quad \begin{pmatrix} N \\ n+1 \end{pmatrix} \\ & = \frac{N!}{(n+1)!\big(N-(n+1)\big) !} \\ & = \frac{(N!)(N-n)}{\big( (n+1)(n!)\big) \big( (N-n)(N-n-1)! \big)} \\ & = \bigg(\frac{N-n}{n+1}\bigg)\bigg(\frac{N!}{n!(N-n)!}\bigg) \\ & = \frac{N-n}{n+1}\begin{pmatrix} N \\n \end{pmatrix} \\ \end{aligned}$

QED