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PHY452H1S Basic Statistical Mechanics. Problem Set 3: State counting

Posted by peeterjoot on February 10, 2013

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Disclaimer

This is an ungraded set of answers to the problems posed.

Question: Surface area of a sphere in d-dimensions

Consider a d-dimensional sphere with radius R. Derive the volume of the sphere V_d(R) and its surface area S_d(R) using S_d(R) = dV_d(R)/dR.

Answer

Let’s start with some terminology and notation, borrowing from [2]

  1. 1-sphere : 2D circle, with “volume” V_2
  2. 2-sphere : 3D sphere, with volume V_3
  3. 3-sphere : 4D Euclidean hypersphere, with “volume” V_4
  4. 4-sphere : 5D Euclidean hypersphere, with “volume” V_5
  5. \cdots

To calculate the volume, we require a parameterization allowing for expression of the volume element in an easy to integrate fashion. For the 1-sphere, we can use the usual circular and spherical coordinate volume elements

\begin{aligned}V_2(R) = 4 \int_0^R r dr \int_0^{\pi/2} d\theta = 4 \frac{R^2}{2} \frac{\pi}{2} = \pi R^2\end{aligned} \hspace{\stretch{1}}(1.0.1a)

\begin{aligned}V_3(R) = 8 \int_0^R r^2 dr \int_0^{\pi/2} \sin\theta d\theta \int_0^{\pi/2} d\phi= 8 \frac{R^3}{3} (1) \frac{\pi}{2}= \frac{4}{3} \pi R^3\end{aligned} \hspace{\stretch{1}}(1.0.1b)

Here, to simplify the integration ranges, we’ve calculated the “volume” integral for just one quadrant or octet of the circle and sphere respectively, as in (Fig 1)

Fig1: Integrating over just one quadrant of circle or octet of sphere

How will we generalize this to higher dimensions? To calculate the volume elements in a systematic fashion, we can introduce a parameterization and use a Jacobian to change from Cartesian coordinates. For the 1-volume and 2-volume cases those parameterizations were the familiar

\begin{aligned}\begin{aligned}x_1 &= r \cos\theta \\ x_0 &= r \sin\theta\end{aligned}\end{aligned} \hspace{\stretch{1}}(1.0.2a)

\begin{aligned}\begin{aligned}x_2 &= r \cos\theta \\ x_1 &= r \sin\theta \cos\phi \\ x_0 &= r \sin\theta \sin\phi\end{aligned}\end{aligned} \hspace{\stretch{1}}(1.0.2b)

Some reflection shows that this generalizes nicely. Let’s use shorthand

\begin{aligned}C_k = \cos\theta_k\end{aligned} \hspace{\stretch{1}}(1.0.3a)

\begin{aligned}S_k = \sin\theta_k,\end{aligned} \hspace{\stretch{1}}(1.0.3b)

and pick V_5, say, a dimension bigger than the 2D or 3D cases that we can do by inspection, we can parameterize with

\begin{aligned}\begin{aligned}x_4 &= r C_4 \\ x_3 &= r S_4 C_3 \\ x_2 &= r S_4 S_3 C_2 \\ x_1 &= r S_4 S_3 S_2 C_1 \\ x_0 &= r S_4 S_3 S_2 S_1.\end{aligned}\end{aligned} \hspace{\stretch{1}}(1.0.4)

Our volume element

\begin{aligned}dV_5 = \underbrace{\frac{\partial(x_4, x_3, x_2, x_1, x_0)}{\partial(r, \theta_4, \theta_3, \theta_2, \theta_1)} }_{\equiv J_5}dr d\theta_4 d\theta_3 d\theta_2 d\theta_1\end{aligned} \hspace{\stretch{1}}(1.0.5)

That Jacobian is

\begin{aligned}J_5 = \left\lvert\begin{array}{lllll}C_4 & - r S_4 & \phantom{-} 0 & \phantom{-} 0 & \phantom{-} 0 \\ S_4 C_3 & \phantom{-} r C_4 C_3 & - r S_4 S_3 & \phantom{-} 0 & \phantom{-} 0 \\ S_4 S_3 C_2 & \phantom{-} r C_4 S_3 C_2 & \phantom{-} r S_4 C_3 C_2 & - r S_4 S_3 S_2 & \phantom{-} 0 \\ S_4 S_3 S_2 C_1 & \phantom{-} r C_4 S_3 S_2 C_1 & \phantom{-} r S_4 C_3 S_2 C_1 & \phantom{-} r S_4 S_3 C_2 C_1 & - r S_4 S_3 S_2 S_1 \\ S_4 S_3 S_2 S_1 & \phantom{-} r C_4 S_3 S_2 S_1 & \phantom{-} r S_4 C_3 S_2 S_1 & \phantom{-} r S_4 S_3 C_2 S_1 & \phantom{-} r S_4 S_3 S_2 C_1 \end{array}\right\rvert=r^4 S_4^3 S_3^2 S_2^1 S_1^0\left\lvert\begin{array}{lllll} \begin{array}{l} C_4 \end{array} \begin{array}{l} \phantom{S_3 S_2 S_1} \end{array} & \begin{array}{l} -S_4 \end{array} \begin{array}{l} \phantom{S_3 S_2 S_1} \end{array} & \begin{array}{l} \phantom{-}0 \end{array} & \begin{array}{l} \phantom{-}0 \end{array} & \begin{array}{l} \phantom{-}0 \end{array}\\ \begin{array}{l}S_4 \\ S_4 \\ S_4 \\ S_4 \end{array}\boxed{\begin{array}{l} C_3 \\ S_3 C_2 \\ S_3 S_2 C_1 \\ S_3 S_2 S_1 \end{array} }&\begin{array}{l}\phantom{-}C_4 \\ \phantom{-}C_4 \\ \phantom{-}C_4 \\ \phantom{-}C_4 \end{array}\boxed{\begin{array}{l}C_3 \\ S_3 C_2 \\ S_3 S_2 C_1 \\ S_3 S_2 S_1 \end{array}}&\begin{array}{l} - S_3 \\ \phantom{-} C_3 C_2 \\ \phantom{-} C_3 S_2 C_1 \\ \phantom{-} C_3 S_2 S_1 \end{array}&\begin{array}{l}\phantom{-} 0 \\ - S_2 \\ \phantom{-} C_2 C_1 \\ \phantom{-} C_2 S_1 \end{array}&\begin{array}{l}\phantom{-}0 \\ \phantom{-}0 \\ - S_1 \\ \phantom{-}C_1 \end{array}\end{array}\right\rvert\end{aligned} \hspace{\stretch{1}}(1.0.6)

Observe above that the cofactors of both the 1,1 and the 1,2 elements, when expanded along the first row, have a common factor. This allows us to work recursively

\begin{aligned}J_5 = r^4 S_4^3 S_3^2 S_2^1 S_1^0(C_4^2 - - S_4^2)\left\lvert\begin{array}{llll}C_3 & - S_3 & \phantom{-} 0 & \phantom{-} 0 \\ S_3 C_2 & \phantom{-} C_3 C_2 & - S_2 & \phantom{-} 0 \\ S_3 S_2 C_1 & \phantom{-} C_3 S_2 C_1 & \phantom{-} C_2 C_1 & - S_1 \\ S_3 S_2 S_1 & \phantom{-} C_3 S_2 S_1 & \phantom{-} C_2 S_1 & \phantom{-} C_1 \end{array}\right\rvert=r S_4^3 J_4\end{aligned} \hspace{\stretch{1}}(1.0.7)

Similarly for the 4D volume

\begin{aligned}J_4 = r^3 S_3^2 S_2^1 S_1^0\left\lvert\begin{array}{llll}C_3 & - S_3 & \phantom{-} 0 & \phantom{-} 0 \\ S_3 C_2 & \phantom{-} C_3 C_2 & - S_2 & \phantom{-} 0 \\ S_3 S_2 C_1 & \phantom{-} C_3 S_2 C_1 & \phantom{-} C_2 C_1 & - S_1 \\ S_3 S_2 S_1 & \phantom{-} C_3 S_2 S_1 & \phantom{-} C_2 S_1 & \phantom{-} C_1 \end{array}\right\rvert= r^3 S_3^2 S_2^1 S_1^0 (C_2^2 + S_2^2)\left\lvert\begin{array}{lll}\phantom{-} C_2 & - S_2 & \phantom{-} 0 \\ \phantom{-} S_2 C_1 & \phantom{-} C_2 C_1 & - S_1 \\ \phantom{-} S_2 S_1 & \phantom{-} C_2 S_1 & \phantom{-} C_1 \end{array}\right\rvert= r S_3^2 J_3\end{aligned} \hspace{\stretch{1}}(1.0.8)

and for the 3D volume

\begin{aligned}J_3 = r^2 S_2^1 S_1^0\left\lvert\begin{array}{lll}\phantom{-} C_2 & - S_2 & \phantom{-} 0 \\ \phantom{-} S_2 C_1 & \phantom{-} C_2 C_1 & - S_1 \\ \phantom{-} S_2 S_1 & \phantom{-} C_2 S_1 & \phantom{-} C_1 \end{array}\right\rvert= r^2 S_2^1 S_1^0 (C_2^2 + S_2^2)\left\lvert\begin{array}{ll}C_1 & - S_1 \\ S_1 & \phantom{-} C_1 \end{array}\right\rvert= r S_2^1 J_2,\end{aligned} \hspace{\stretch{1}}(1.0.9)

and finally for the 2D volume

\begin{aligned}J_2= r S_1^0\left\lvert\begin{array}{ll}C_1 & - S_1 \\ S_1 & \phantom{-} C_1 \end{array}\right\rvert= r S_1^0.\end{aligned} \hspace{\stretch{1}}(1.0.10)

Putting all the bits together, the “volume” element in the n-D space is

\begin{aligned}dV_n = dr d\theta_{n-1} \cdots d\theta_{1} r^{n-1} \prod_{k=0}^{n-2} (\sin\theta_{k+1})^{k}, \end{aligned} \hspace{\stretch{1}}(1.0.11)

and the total volume is

\begin{aligned}V_n(R) = 2^n \int_0^R r^{n-1} dr \prod_{k=0}^{n-2} \int_0^{\pi/2} d\theta \sin^k \theta= 2^n \frac{R^{n}}{n}\prod_{k=0}^{n-2} \int_0^{\pi/2} d\theta \sin^k \theta= \frac{4 R^2}{n} (n-2) 2^{n-2} \frac{R^{n-2}}{n-2}\prod_{k=n-3}^{n-2} \int_0^{\pi/2} d\theta \sin^k \theta\prod_{k=0}^{n-4} \int_0^{\pi/2} d\theta \sin^k \theta=4 R^2 \frac{n-2}{n} V_{n-2}(R)\int_0^{\pi/2} d\theta \sin^{n-2} \theta\int_0^{\pi/2} d\theta \sin^{n-3} \theta=4 R^2 \frac{n-2}{n} V_{n-2}(R)\frac{\pi}{2 (n-2)}.\end{aligned} \hspace{\stretch{1}}(1.0.12)

Note that a single of these sine power integrals has a messy result (see: statMechProblemSet3.nb)

\begin{aligned}\int_0^{\pi/2} d\theta \sin^{k} \theta=\frac{\sqrt{\pi } \Gamma \left(\frac{k+1}{2}\right)}{2 \Gamma \left(\frac{k}{2}+1\right)},\end{aligned} \hspace{\stretch{1}}(1.0.13)

but the product is simple, and that result, computed in statMechProblemSet3.nb, is inserted above, providing an expression for the n-D volume element as a recurrence relation

\begin{aligned}\boxed{V_{n} = \frac{2 \pi}{n} R^2 V_{n-2}(R)}\end{aligned} \hspace{\stretch{1}}(1.0.14)

With this recurrence relation we can find the volume V_{2n} or V_{2n+1} in terms of V_2 and V_3 respectively. For the even powers we have

\begin{aligned}\begin{array}{lll}V_{2(2)} &= \frac{2 \pi R^2}{4} V_{2} &= \frac{2^1 \pi R^2}{2(2)} V_{2} \\ V_{2(3)} &= \frac{(2 \pi R^2)^2}{(6)(4)} V_{2} &= \frac{2^2 (\pi R^2)^2}{2^2 (3)(2)} V_{2} \\ V_{2(4)} &= \frac{(2 \pi R^2)^3}{(8)(6)(4)} V_{2} &= \frac{2^3 (\pi R^2)^3}{2^3 (4)(3)(2)} V_{2} \end{array}\end{aligned} \hspace{\stretch{1}}(1.0.15)

\begin{aligned}V_{2(n)} = \frac{2^n (\pi R^2)^{n-1}}{(2n)!!} V_{2} = \frac{(\pi R^2)^{n-1}}{(n)!} V_{2} = \frac{(\pi R^2)^{n-1}}{(n)!} \pi R^2\end{aligned} \hspace{\stretch{1}}(1.0.16)

This gives us a closed form expression for the even powers for n \ge 2

\begin{aligned}\boxed{V_{2(n)} = \frac{(\pi R^2)^{n}}{n!}.}\end{aligned} \hspace{\stretch{1}}(1.0.17)

Observe that this also conveniently gives V_2 = \pi R^2, so is actually valid for n \ge 1. Now for the odd powers

\begin{aligned}\begin{array}{lll}V_{2(2)+1} &= \frac{2 \pi R^2}{5} V_{3} &= 3 \frac{2^1 \pi R^2}{5!!} V_{3} \\ V_{2(3)+1} &= 3 \frac{(2 \pi R^2)^2}{7 5 3} V_{3} &= 3 \frac{2^2 (\pi R^2)^2}{7!!} V_{2} \\ V_{2(4)+1} &= 3 \frac{(2 \pi R^2)^3}{(9)(7)(5)(3)} V_{2} &= 3 \frac{2^3 (\pi R^2)^3}{9!!} V_{2} \end{array}\end{aligned} \hspace{\stretch{1}}(1.0.18)

\begin{aligned}V_{2(n)+1} = 3 \frac{2^{n-1} (\pi R^2)^{n-1}}{(2n +1)!!} V_{2} = 3 \frac{2^{n-1} (\pi R^2)^{n-1}}{(2n +1)!!} \frac{4}{3} \pi R^3\end{aligned} \hspace{\stretch{1}}(1.0.19)

So, for n \ge 2 we have

\begin{aligned}\boxed{V_{2(n)+1} = \frac{2^{n+1} \pi^n R^{2 n + 1}}{(2n +1)!!}.}\end{aligned} \hspace{\stretch{1}}(1.0.20)

As with the even powered expression 1.0.17 we see that this is also good for n = 1, yielding V_3 = 4 \pi R^2/3 as required for the 3D spherical volume.

The even and odd power expressions don’t look quite different on the surface, but can be put into a consistent form, when expressed in terms of the gamma function.

For the even powers, using a substitution 2 n = m we have for even values of m \ge 2

\begin{aligned}V_{m} = \frac{\pi^{m/2} R^m}{\Gamma(n + 1)}= \frac{\pi^{m/2} R^m}{\Gamma(m/2 + 1)}.\end{aligned} \hspace{\stretch{1}}(1.0.21)

For the even powers, with the help of [1] we find

\begin{aligned}(2 n + 1)!! = \frac{2^{n+1} \Gamma\left(n + \frac{3}{2}\right)}{\sqrt{\pi}}.\end{aligned} \hspace{\stretch{1}}(1.0.22)

This gives us

\begin{aligned}V_{2(n)+1} = \frac{\pi^{n + 1/2} R^{2 n + 1}}{\Gamma\left(n + \frac{3}{2}\right)}= \frac{\pi^{n + 1/2} R^{2 n + 1}}{\Gamma\left(\frac{(2 n + 1) + 2}{2}\right)}.\end{aligned} \hspace{\stretch{1}}(1.0.23)

Writing m = 2 n + 1, or n = (m - 1)/2 we have for odd values of m \ge 3 a match with the even powered expression of 1.0.21

\begin{aligned}\boxed{V_{m} = \frac{ \pi^{m/2} R^{m} }{ \Gamma\left( m/2 + 1 \right)}.}\end{aligned} \hspace{\stretch{1}}(1.0.24)

We’ve shown that this is valid for any dimension m \ge 2.

Tabulating some values of these for n \in [2, 7] we have respectively

\begin{aligned}\pi r^2,\frac{4 \pi r^3}{3},\frac{\pi ^2 r^4}{2},\frac{8 \pi ^2 r^5}{15},\frac{\pi ^3 r^6}{6},\frac{16 \pi ^3 r^7}{105},\end{aligned} \hspace{\stretch{1}}(1.0.25)

The only task left is computation of the surface area. That comes by inspection and is

\begin{aligned}S_{m}(R) =\frac{d V_m(R)}{d R}= \frac{\pi^{m/2} m R^{m-1}}{\Gamma\left( m/2 + 1 \right)}.\end{aligned} \hspace{\stretch{1}}(1.0.26)

Again for m \in [2, 7] we have

\begin{aligned}2 \pi r,4 \pi r^2,2 \pi ^2 r^3,\frac{8 \pi ^2 r^4}{3},\pi ^3 r^5,\frac{16 \pi ^3 r^6}{15}.\end{aligned} \hspace{\stretch{1}}(1.0.27)

Question: State counting – polymer

A typical protein is a long chain molecule made of very many elementary units called amino acids – it is an example of a class of such macromolecules called polymers. Consider a protein made N amino acids, and assume each amino acid is like a sphere of radius a. As a toy model assume that the protein configuration is like a random walk with each amino acid being one “step”, i.e., the center-to-center vector from one amino acid to the next is a random vector of length 2a and ignore any issues with overlapping spheres (so-called “excluded volume” constraints). Estimate the spatial extent of this protein in space. Typical proteins assume a compact form in order to be functional. In this case, taking the constraint of nonoverlapping spheres, estimate the radius of such a compact protein assuming it has an overall spherical shape with fully packed amino acids (ignore holes in the packing, and use only volume ratios to make this estimate). With N = 300 and a \approx 5 \text{\AA}, estimate these sizes for the random walk case as well as the compact globular case.

Answer

We are considering a geometry like that of (Fig 2), depicted in two dimensions for ease of illustration.

Fig2: Touching “spheres”

From the geometry, if \mathbf{c}_k is the vector to the center of the kth sphere, we have for some random unit vector \hat{\mathbf{r}}_k

\begin{aligned}\mathbf{c}_{k+1} = \mathbf{c}_k + 2 a \hat{\mathbf{r}}_k\end{aligned} \hspace{\stretch{1}}(1.0.28)

Proceeding recursively, writing d_N, and n = N -1 > 0, we have for the difference of the positions of the first and last centers of the chain

\begin{aligned}d_N = \left\lvert {\mathbf{c}_N - \mathbf{c}_1} \right\rvert = 2 a \left\lvert { \sum_{k = 1}^{n} \hat{\mathbf{r}}_k} \right\rvert= 2 a \left( \sum_{k,m = 1}^{n} \hat{\mathbf{r}}_k \cdot \hat{\mathbf{r}}_m \right)^{1/2}= 2 a \left( n + 2 \sum_{1 \le k < m \le n} \hat{\mathbf{r}}_k \cdot \hat{\mathbf{r}}_m \right)^{1/2}= 2 a \sqrt{n} \left( 1 + \frac{2}{n} \sum_{1 \le k < m \le n} \hat{\mathbf{r}}_k \cdot \hat{\mathbf{r}}_m \right)^{1/2}\end{aligned} \hspace{\stretch{1}}(1.0.29)

The \hat{\mathbf{r}}_k‘s clearly cannot be completely random since we have a constraint that \hat{\mathbf{r}}_k \cdot \hat{\mathbf{r}}_{k+1} > -\cos\pi/3, or else two adjacent spheres will overlap. There will also be overlapping constraints for longer portions of the chain that are harder to express. We are ignoring both such constraints, and seek the ensemble average of all systems of the form 1.0.29.

Employing random azimuthal and polar angular variables \theta_k, and \phi_k, we have

\begin{aligned}\hat{\mathbf{r}}_k = \begin{bmatrix}\sin\theta_k \cos\phi_k \\ \sin\theta_k \sin\phi_k \\ \cos\theta_k,\end{bmatrix}\end{aligned} \hspace{\stretch{1}}(1.0.30)

so that the average polymer length is

\begin{aligned}\begin{aligned}\left\langle{{d_N }}\right\rangle &=2 a \sqrt{n} \left( \frac{1}{{(2 \pi)(\pi)}} \right)^{n} \int_{\theta_j \in [0, \pi]}d\theta_1 d\theta_2 \cdots d\theta_n \int_{\phi_j \in [0, 2 \pi]}d\phi_1 d\phi_2 \cdots d\phi_n \times \\ &\quad \left( 1 + \frac{2}{n} \sum_{1 \le k < m \le n} \sin\theta_k \cos\phi_k \sin\theta_m \cos\phi_m + \sin\theta_k \sin\phi_k \sin\theta_m \sin\phi_m + \cos\theta_k \cos\theta_m \right)^{1/2}\end{aligned}\end{aligned} \hspace{\stretch{1}}(1.0.31)

Observing that even \int \sqrt{1 + a \cos\theta} d\theta is an elliptic integral, we don’t have any hope of evaluating this in closed form. However, to first order, we have

\begin{aligned}\begin{aligned}d_N&\approx2 a \sqrt{n} \left( \frac{1}{{(2 \pi)(\pi)}} \right)^{n} \int_{\theta_j \in [0, \pi]}d\theta_1 d\theta_2 \cdots d\theta_n \int_{\phi_j \in [0, 2 \pi]}d\phi_1 d\phi_2 \cdots d\phi_n \times \\ &\quad \left( 1 + \frac{1}{n} \sum_{1 \le k < m \le n} \sin\theta_k \cos\phi_k \sin\theta_m \cos\phi_m + \sin\theta_k \sin\phi_k \sin\theta_m \sin\phi_m + \cos\theta_k \cos\theta_m \right) \\ &=2 a \sqrt{n} \left( \frac{1}{{(2 \pi)(\pi)}} \right)^{n} \left( 2 \pi^2 \right)^n \\ &\quad +2 a \sqrt{n} \left( \frac{1}{{(2 \pi)(\pi)}} \right)^{n} \left( 2 \pi^2 \right)^{n - 2} \left( \frac{1}{{2}} n(n+1) \right) \frac{1}{{n}} \times \\ &\quad \int_0^{\pi}d\theta \int_0^{\pi}d\theta'\int_0^{2 \pi}d\phi\int_0^{2 \pi}d\phi'\left( \sin\theta \cos\phi \sin\theta' \cos\phi' + \sin\theta \sin\phi \sin\theta' \sin\phi' + \cos\theta \cos\theta' \right)\end{aligned}\end{aligned} \hspace{\stretch{1}}(1.0.32)

The \int_0^{2 \pi} \cos\phi integrals kill off the first term, the \int_0^{2 \pi} \sin\phi integrals kill of the second, and the \int_0^\pi \cos\theta integral kill of the last term, and we are left with just

\begin{aligned}d_N \approx 2 a \sqrt{N-1}.\end{aligned} \hspace{\stretch{1}}(1.0.33)

Ignoring the extra 2 \times a of the end points, and assuming that for large N we have \sqrt{N-1} \approx \sqrt{N}, the spatial extent of the polymer chain is

\begin{aligned}\boxed{d_N \approx 2 a \sqrt{N}.}\end{aligned} \hspace{\stretch{1}}(1.0.34)

Spherical packing

Assuming the densest possible spherical packing, a face centered cubic [3], as in (Fig 3), we see that the density of such a spherical packing is

Fig3: Element of a face centered cubic

\begin{aligned}\eta_{\mathrm{FCC}} =\frac{\left( 8 \times \frac{1}{{8}} + 6 \times \frac{1}{{2}} \right) \frac{4}{3} \pi r^3}{ \left( \sqrt{8 r^2} \right)^3 }= \frac{\pi}{\sqrt{18}}.\end{aligned} \hspace{\stretch{1}}(1.0.35)

With a globular radius of R and an atomic radius of a, and density \eta we have

\begin{aligned}\eta \frac{4}{3} \pi R^3 = N \frac{4}{3} \pi a^3,\end{aligned} \hspace{\stretch{1}}(1.0.36)

so that the globular radius R is

\begin{aligned}\boxed{R_N(\eta) = a \sqrt[3]{\frac{N}{\eta}}.}\end{aligned} \hspace{\stretch{1}}(1.0.37)

Some numbers

With N = 300 and a \approx 5 \text{\AA}, and ignoring spaces (i.e. \eta = 1, for a non-physical infinite packing), our globular diameter is approximately

\begin{aligned}2 \times 5 \text{\AA} \sqrt[3]{300} \approx 67 \text{\AA}.\end{aligned} \hspace{\stretch{1}}(1.0.38)

This is actually not much different than the maximum spherical packing of an FCC lattice, which results a slightly larger globular cluster diameter

\begin{aligned}2 \times 5 \text{\AA} \sqrt[3]{300 \sqrt{18}/\pi} \approx 74 \text{\AA}.\end{aligned} \hspace{\stretch{1}}(1.0.39)

Both however, are much less than the end to end length of the random walk polymer chain

\begin{aligned}2 (5 \text{\AA}) \sqrt{300} \approx 173 \text{\AA}.\end{aligned} \hspace{\stretch{1}}(1.0.40)

Question: State counting – spins

Consider a toy model of a magnet where the net magnetization arises from electronic spins on each atom which can point in one of only two possible directions – Up/North or Down/South. If we have a system with N spins, and if the magnetization can only take on values \pm 1 (Up = +1, Down = -1), how many configurations are there which have a total magnetization m, where m = (N_\uparrow - N_\downarrow)/N (note that N_\uparrow + N_\downarrow = N)? Simplify this result assuming N \gg 1 and a generic m (assume we are not interested in the extreme case of a fully magnetized system where m = \pm 1). Find the value of the magnetization m for which the number of such microscopic states is a maximum. For N = 20, make a numerical plot of the number of states as a function of the magnetization (note: -1 \le m \le 1) without making the N \gg 1 assumption.

Answer

For the first couple values of N, lets enumerate the spin sample spaces, their magnetization.

N = 1

  1. \uparrow : m = 1
  2. \downarrow, m = -1

N = 2

  1. \uparrow \uparrow : m = 1
  2. \uparrow \downarrow : m = 0
  3. \downarrow \uparrow, m = 0
  4. \downarrow \downarrow, m = -1

N = 3

  1. \uparrow \uparrow \uparrow : m = 1
  2. \uparrow \uparrow \downarrow : m = 1/3
  3. \uparrow \downarrow \uparrow : m = 1/3
  4. \uparrow \downarrow \downarrow : m = -1/3
  5. \downarrow \uparrow \uparrow : m = 1/3
  6. \downarrow \uparrow \downarrow : m = -1/3
  7. \downarrow \downarrow \uparrow : m = -1/3
  8. \downarrow \downarrow \downarrow : m = 1

The respective multiplicities for N = 1,2,3 are \{1\}, \{1, 2, 1\}, \{1, 3, 3, 1\}. It’s clear that these are just the binomial coefficients. Let’s write for the multiplicities

\begin{aligned}g(N, m) = \binom{N}{i(m)}\end{aligned} \hspace{\stretch{1}}(1.0.41)

where i(m) is a function that maps from the magnetization values m to the integers [0, N]. Assuming

\begin{aligned}i(m) = a m + b,\end{aligned} \hspace{\stretch{1}}(1.0.42)

where i(-1) = 0 and i(1) = N, we solve

\begin{aligned}\begin{aligned}a (-1) + b &= 0 \\ a (1) + b &= N,\end{aligned}\end{aligned} \hspace{\stretch{1}}(1.0.43)

so

\begin{aligned}i(m) = \frac{N}{2}(m + 1)\end{aligned} \hspace{\stretch{1}}(1.0.44)

and

\begin{aligned}g(N, m) = \binom{N}{\frac{N}{2}(1 + m)} = \frac{N!}{\left(\frac{N}{2}(1 + m)\right)!\left(\frac{N}{2}(1 - m)\right)!}\end{aligned} \hspace{\stretch{1}}(1.0.45)

From

\begin{aligned}\begin{aligned}2 m &= N_{\uparrow} - N_{\downarrow} \\ N &= N_{\uparrow} + N_{\downarrow},\end{aligned}\end{aligned} \hspace{\stretch{1}}(1.0.46)

we see that this can also be written

\begin{aligned}\boxed{g(N, m) = \frac{N!}{(N_{\uparrow})!(N_{\downarrow})!}}\end{aligned} \hspace{\stretch{1}}(1.0.47)

Simplification for large N

Using Stirlings approximation

\begin{aligned}\begin{aligned}\ln g(N, m) &= \ln N! - \ln N_{\downarrow}! - \ln N_{\uparrow}! \\ &\approx \left( N + \frac{1}{{2}} \right) \ln N - \not{{N}} + \not{{ \frac{1}{{2}} \ln 2 \pi }} \\ &\quad-\left( N_{\uparrow} + \frac{1}{{2}} \right) \ln N_{\uparrow} + \not{{N_{\uparrow}}} - \not{{\frac{1}{{2}} \ln 2 \pi}} \\ &\quad-\left( N_{\downarrow} + \frac{1}{{2}} \right) \ln N_{\downarrow} + \not{{N_{\downarrow}}} - \frac{1}{{2}} \ln 2 \pi \\ &=\left( N_{\uparrow} + N_{\downarrow} + \frac{1}{{2}} + \underbrace{\frac{1}{{2}}}_{\text{add}} \right) \ln N -\frac{1}{{2}} \ln 2 \pi - \underbrace{\frac{1}{{2}} \ln N}_{\text{then subtract}} \\ &\quad -\left( N_{\uparrow} + \frac{1}{{2}} \right) \ln N_{\uparrow} \quad -\left( N_{\downarrow} + \frac{1}{{2}} \right) \ln N_{\downarrow} \\ &=-\left( N_{\uparrow} + \frac{1}{{2}} \right) \ln N_{\uparrow}/N-\left( N_{\downarrow} + \frac{1}{{2}} \right) \ln N_{\downarrow}/N- \frac{1}{{2}} \ln 2 \pi N \\ &=-\left( N_{\uparrow} + \frac{1}{{2}} \right) \ln \frac{1}{{2}}( 1 + m )-\left( N_{\downarrow} + \frac{1}{{2}} \right) \ln \frac{1}{{2}} ( 1 - m )- \frac{1}{{2}} \ln 2 \pi N \\ &\approx\left( N_{\uparrow} + \frac{1}{{2}} \right) \ln 2+\left( N_{\downarrow} + \frac{1}{{2}} \right) \ln 2 - \frac{1}{{2}} \ln 2 \pi N \\ &\quad -\left( N_{\uparrow} + \frac{1}{{2}} \right) \left(m - \frac{1}{{2}} m^2 \right) \\ &\quad -\left( N_{\downarrow} + \frac{1}{{2}} \right) \left( -m - \frac{1}{{2}} m^2 \right) \\ &= (N + 1) \ln 2 - \frac{1}{{2}} \ln 2 \pi N+ m ( N_{\downarrow} - N_{\uparrow} )+ \frac{1}{{2}} m^2 ( N + 1 ) \\ &= (N + 1) \ln 2 - \frac{1}{{2}} \ln 2 \pi N- N m^2+ \frac{1}{{2}} m^2 ( N + 1 ) \\ &\approx(N + 1) \ln 2 - \frac{1}{{2}} \ln 2 \pi N- \frac{1}{{2}} m^2 N\end{aligned}\end{aligned} \hspace{\stretch{1}}(1.0.48)

This gives us for large N

\begin{aligned}g(N, m) \approx 2^N \sqrt{\frac{2}{ \pi N}} e^{ -m^2 N/2 }\end{aligned} \hspace{\stretch{1}}(1.0.49)

For N = 20 this approximation and the exact expression are plotted in (Fig 4).

Fig4: Distribution of number of configurations for N = 20 magnets as a function of magnetization

With the large scales of this extremely peaked function, no visible difference can be seen. That difference does exist, and is plotted in (Fig 5)

Fig5: N = 20 differences between the exact binomial expression and the Gaussian approximation

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] Wikipedia. N-sphere — Wikipedia, The Free Encyclopedia, 2013\natexlab{a}. URL http://en.wikipedia.org/w/index.php?title=N-sphere&oldid=534164100. [Online; accessed 26-January-2013].

[3] Wikipedia. Sphere packing — wikipedia, the free encyclopedia, 2013\natexlab{b}. URL http://en.wikipedia.org/w/index.php?title=Sphere_packing&oldid=535578971. [Online; accessed 31-January-2013].

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