{"id":219,"date":"2024-08-25T21:01:42","date_gmt":"2024-08-25T13:01:42","guid":{"rendered":"https:\/\/www.shirui.me\/blog\/?p=219"},"modified":"2024-08-25T21:05:24","modified_gmt":"2024-08-25T13:05:24","slug":"219","status":"publish","type":"post","link":"https:\/\/www.shirui.me\/blog\/2024\/08\/25\/219\/","title":{"rendered":"Flory Characteristic Ratio"},"content":{"rendered":"\n
    \n
  1. This exercise focuses on the Flory characteristic ratio from Rubinstein’s textbook on Polymer Physics<\/i>.<\/li>\n\n\n\n
  2. Definition of Flory Characteristic Ratio:<\/b>


    $C_\\infty:=\\lim_{n\\to\\infty}\\frac{\\langle \\mathbf{R}^2 \\rangle}{nb^2}$


    where $\\langle \\mathbf{R}^2 \\rangle$ represents the mean square end-to-end vector, $b$ is the Kuhn length of the polymer, and $n$ is the corresponding chain length.
    <\/li>\n\n\n\n
  3. Calculation of $\\langle \\mathbf{R}^2\\rangle$:<\/b>


    Let $\\mathbf{r}_i$ denote the bond vector between the $i$th and $(i-1)$th particle, and let $\\mathbf{r}_1=\\mathbf{R}_1-\\mathbf{R}_0\\equiv0$. For an $n$-bead chain:


    $$\\begin{align}\\langle \\mathbf{R}^2 \\rangle&=\\left\\langle\\left(\\sum_{i=1}^n \\mathbf{r}_i^2\\right)\\cdot\\left(\\sum_{j=1}^n \\mathbf{r}_j^2\\right)\\right\\rangle\\\\

    &=\\sum_i\\sum_j\\langle \\mathbf{r}_i\\cdot\\mathbf{r}_j\\rangle\\end{align}$$
    <\/li>\n\n\n\n
  4. Calculation of $\\langle\\mathbf{r}_i\\cdot\\mathbf{r}_j\\rangle$:<\/b>


    Consider a local coordinate frame where $\\mathbf{r}_i=(b,0,0)^T$. Then, $\\mathbf{r}_{i+1}$ can be viewed as a rotation of $\\mathbf{r}_i$ by $\\theta$ about the $z$-axis (bond angle) and by $\\phi_{i}$ about the $x$-axis (torsion). The rotation matrix is:


    $$\\mathbf{T}_i:=\\left(

    \\begin{array}{ccc}

    \\cos (\\theta ) & -\\sin (\\theta ) & 0 \\\\

    \\sin (\\theta ) \\cos (\\phi_i) & \\cos (\\theta ) \\cos (\\phi_i) & -\\sin (\\phi_i) \\\\

    \\sin (\\theta ) \\sin (\\phi_i ) & \\cos (\\theta ) \\sin (\\phi_i ) & \\cos (\\phi_i ) \\\\

    \\end{array}\\right)$$


    or $\\pi-\\theta$, $\\pi – \\phi$ based on the definitions of bond angle $\\theta$ and torsion angle $\\phi$. In each reference frame $R_i$, the bond vector is $(b,0,0)^T$, and in $R_i$, $\\mathbf{r}_{i+1}=\\mathbf{T}_i\\cdot\\mathbf{r}_i=(b \\cos (\\theta ),b \\sin (\\theta ) \\cos (\\phi_i ),b \\sin (\\theta ) \\sin (\\phi_i ))^T$.

    Now, consider the simple case of $\\mathbf{r}_i\\cdot\\mathbf{r}_{i+2}$. $\\mathbf{r}_{i+2}$ is calculated in the $(i+1)$th reference frame $R_{i+1}$ where $\\mathbf{r}_{i+1}$ is $(b,0,0)^T$ and $\\mathbf{r}_{i+2}=\\mathbf{T}_{i+1}\\cdot\\mathbf{r}_{i+1}$. To perform the inner product, we need to rotate one of the vectors into the other frame. This involves rotating $\\mathbf{r}_i=(b,0,0)^T$ from $R_i$ to $R_{i+1}$ or vice versa.

    Note that $\\mathbf{r}_{i+1}$ in the $R_i$ frame is $(b \\cos (\\theta ),b \\sin (\\theta ) \\cos (\\phi_i ),b \\sin (\\theta ) \\sin (\\phi_i ))^T$ and $(b,0,0)^T$ in the $R_{i+1}$ frame. Therefore, to transform a vector from $R_i$ to $R_{i+1}$, we need a transformation matrix that converts $(b \\cos (\\theta ),b \\sin (\\theta ) \\cos (\\phi_i ),b \\sin (\\theta ) \\sin (\\phi_i ))^T$ to $(b,0,0)^T$. This matrix is $\\mathbf{T}_i^{-1}$, or $\\mathbf{T}_i^T$, because in frame $R_i$, we obtain $\\mathbf{r}_{i+1}$ using $\\mathbf{T}_i\\cdot(b,0,0)^T$, and $\\mathbf{T}_i$ is an orthonormal rotation matrix.

    By transforming $\\mathbf{r}_i$ into $R_{i+1}$, the inner product $\\mathbf{r}_i\\cdot\\mathbf{r}_{i+2}$ becomes:


    $$\\langle\\mathbf{T}_i^T\\mathbf{r}_i,\\mathbf{T}_{i+1}\\mathbf{r}_{i+1}\\rangle=$$
    $$\\langle\\mathbf{r}_i,\\mathbf{T}_i\\mathbf{T}_{i+1}\\mathbf{r}_{i+1}\\rangle=(b,0,0)\\mathbf{T}_i\\mathbf{T}_{i+1}(b,0,0)^T$$

    By induction, $\\mathbf{r}_i\\cdot\\mathbf{r}_j=(b,0,0)\\mathbf{T}_i\\mathbf{T}_{i+1}\\mathbf{T}_{i+2}\\cdots\\mathbf{T}_{j-1}(b,0,0)^T$. Since $\\theta$ is the same for all bonds and $\\phi_{1,2,\\dots,n}$ are independently distributed, let $\\mathbf{T}$ represent the average transformation matrix of ${\\mathbf{T}_i}$:


    $$\\begin{align}\\mathbf{T}&:=\\frac{\\int_{-\\pi}^{\\pi}\\mathbf{T}_i(\\phi)\\exp{(-U(\\phi)\/k_BT)}\\mathrm{d}\\phi}{\\int_{-\\pi}^{\\pi}\\exp{(-U(\\phi)\/k_BT)}\\mathrm{d}\\phi}\\\\

    &=\\left(

    \\begin{array}{ccc}

    \\cos (\\theta ) & -\\sin (\\theta ) & 0 \\\\

    \\sin (\\theta ) \\langle\\cos (\\phi) \\rangle & \\cos (\\theta )\\langle \\cos (\\phi)\\rangle &0 \\\\

    0& 0& \\langle\\cos (\\phi)\\rangle \\\\

    \\end{array}\\right)\\end{align}$$

    Therefore, we have $\\langle\\mathbf{r}_i\\cdot\\mathbf{r}_j\\rangle=(b,0,0)\\mathbf{T}^{|j-i|}(b,0,0^T)=b^2(\\mathbf{T}^{|j-i|})_{11}$. Every $\\sin(\\phi)$ term vanishes in the integral because $U$ is even.
    <\/li>\n\n\n\n
  5. Flory Characteristic Ratio:<\/b>


    $$\\begin{align}C_\\infty&=\\lim_{n\\to\\infty}\\frac{1}{nb^2}\\sum_i\\sum_j \\mathbf{r}_i\\cdot\\mathbf{r}_j\\\\

    &=\\lim_{n\\to\\infty}\\frac{1}{n}(\\sum_i\\sum_j T^{|j-i|})_{11}\\\\

    &=\\lim_{n\\to\\infty}\\left(-\\frac{1}{n}\\frac{-2 \\mathbf{T}^{n+1}+\\mathbf{T}^2 n+2 \\mathbf{T}-n\\mathbf{I}}{(\\mathbf{I}-\\mathbf{T})^2}\\right)_{11}\\\\

    &=\\left(\\frac{\\mathbf{I}+\\mathbf{T}}{\\mathbf{I}-\\mathbf{T}}\\right)_{11}\\\\

    &=\\frac{1+\\cos(\\theta)}{1-\\cos(\\theta)}\\frac{1-\\langle\\cos(\\phi)\\rangle}{1+\\langle\\cos(\\phi)\\rangle}\\end{align}$$

    Here, $\\frac{\\bullet}{n}=0$ and $\\mathbf{T}^n=0$ as $n\\to\\infty$ because $\\mathbf{T}$ is constant, and the correlation between two beads approaches zero as the distance between them increases.
    <\/li>\n<\/ol>\n","protected":false},"excerpt":{"rendered":"","protected":false},"author":1,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[9],"tags":[29],"class_list":["post-219","post","type-post","status-publish","format-standard","hentry","category-notes","tag-polymer-physics"],"_links":{"self":[{"href":"https:\/\/www.shirui.me\/blog\/wp-json\/wp\/v2\/posts\/219","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/www.shirui.me\/blog\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/www.shirui.me\/blog\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/www.shirui.me\/blog\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/www.shirui.me\/blog\/wp-json\/wp\/v2\/comments?post=219"}],"version-history":[{"count":3,"href":"https:\/\/www.shirui.me\/blog\/wp-json\/wp\/v2\/posts\/219\/revisions"}],"predecessor-version":[{"id":222,"href":"https:\/\/www.shirui.me\/blog\/wp-json\/wp\/v2\/posts\/219\/revisions\/222"}],"wp:attachment":[{"href":"https:\/\/www.shirui.me\/blog\/wp-json\/wp\/v2\/media?parent=219"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.shirui.me\/blog\/wp-json\/wp\/v2\/categories?post=219"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.shirui.me\/blog\/wp-json\/wp\/v2\/tags?post=219"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}