{"id":223,"date":"2024-08-25T21:08:14","date_gmt":"2024-08-25T13:08:14","guid":{"rendered":"https:\/\/www.shirui.me\/blog\/?p=223"},"modified":"2024-08-25T21:10:53","modified_gmt":"2024-08-25T13:10:53","slug":"anisotropy-of-ideal-chain","status":"publish","type":"post","link":"https:\/\/www.shirui.me\/blog\/2024\/08\/25\/anisotropy-of-ideal-chain\/","title":{"rendered":"Anisotropy of ideal chain"},"content":{"rendered":"\n

A Gaussian chain, when averaged over all conformations and orientations, exhibits isotropic behavior. This means it can be treated as a sphere with a radius equal to its radius of gyration, $R_g$. However, an exercise in the second chapter of Rubinstein’s Polymer Physics<\/em> demonstrates that $R_g^2$ becomes asymmetric when the coordinate frame is aligned with the end-to-end vector, $\\mathbf{R}_{ee}$. Specifically, if $\\mathbf{R}_{ee}$ is set as the x-axis, i.e., $\\mathbf{R}_{ee}=(R,0,0)^T$, the three components of $R_g^2$ are $\\frac{Nb^2}{36}$, $\\frac{Nb^2}{36}$ in the y and z directions, and $\\frac{Nb^2}{36}+\\frac{R^2}{12}$ in the x direction. Here’s a simple proof:<\/p>\n\n

Consider a Gaussian chain fixed between $(0,0,0)^T$ and $\\mathbf{R}_{ee}$. This forms a Brownian bridge, with a multivariate Gaussian distribution centered at $\\frac{i}{N}\\mathbf{R}$ and a variance of $\\frac{i(N-i)}{N}b^2$. The proof is straightforward:\n
\n
\n$P_{0\\mathbf{R}}(\\mathbf{r},n)=\\frac{G(\\mathbf{r},0,n)G(\\mathbf{R},\\mathbf{r},N-n)}{G(0,\\mathbf{R},N)}$\n
\n
\nwhere $G(a,b,n)$ represents the distribution of a Gaussian chain with segment length $n$ and ends at points $a$ and $b$. This expression represents the probability of a length-$n$ Gaussian chain starting at 0 and ending at $\\mathbf{r}$, connected to another length-$(N-n)$ Gaussian chain starting at $\\mathbf{R}$ and ending at $\\mathbf{r}$. The entire chain is a Gaussian chain with length $N$ and end-to-end vector $\\mathbf{R}_{ee}=\\mathbf{R}$. It’s easy to show that the distribution of this chain:\n
\n
\n$P_{0\\mathbf{R}}(\\mathbf{r},n)=G\\left(\\mathbf{r}-\\frac{n}{N}\\mathbf{R}, 0,\\frac{n(N-n)}{N}\\right)$\n
\n
\nis equivalent to a Gaussian chain segment with ends at $\\mathbf{r}$ and $\\frac{n}{N}\\mathbf{R}$, and an equivalent length of $\\frac{n(N-n)}{N}$. The $R_g^2$ is then:\n
\n
\n$\\begin{align}R_g^2=&\\frac{1}{2N^2}\\int_0^N\\langle(\\mathbf{r}_i-\\mathbf{r_j})^2\\rangle\\mathrm{d}i\\mathrm{d}j\\\\
\n=&\\frac{1}{N^2}\\int_0^N\\int_j^N \\frac{(i-j)^2 R^2}{N^2}+\\frac{(i-j)(N-(i-j))}{N}b^2\\mathrm{d}i\\mathrm{d}j\\\\ =&\\frac{R^2+Nb^2}{12}\\end{align}$\n
\n
\nIn this equivalent method, $\\langle (\\mathbf{r}_i-\\mathbf{r}_j)^2\\rangle$ is interpreted as $(\\frac{i}{N}-\\frac{j}{N})^2R^2+\\frac{|i-j|(N-|i-j|)}{N}$ from the equivalent Gaussian chain. This is because, despite the chain being ‘fixed,’ it remains a Gaussian chain, implying translation invariance. Therefore, $\\langle (\\mathbf{r}_i-\\mathbf{r}_j)^2\\rangle$ depends only on $|i-j|$, and $P_{0\\mathbf{R}}(\\mathbf{r},n)$ provides the probability of segment $\\mathbf{r}_n-\\mathbf{r}_0$, which represents any $n$-segment on the Gaussian chain. Directly calculating $P(\\mathbf{r}_i-\\mathbf{r}_j)$ from the convolution of $P_{0\\mathbf{R}}$ is incorrect because $\\mathbf{r}_i$ and $\\mathbf{r}_j$ are correlated. Convolution of $P_{0\\mathbf{R}}$ introduces dependence on $i$ and $j$ in $\\langle (\\mathbf{r}_i-\\mathbf{r}_j)^2\\rangle$, violating the translation invariance of the chain.\n<\/p>\n\n

If $R^2=Nb^2$, we find that $R_g^2=\\frac{1}{6}Nb^2$. Furthermore, we have $\\frac{Nb^2}{36}+\\frac{R^2_{x,y,z}}{12}=\\frac{Nb^2+R^2}{36}$ for each dimension. Specifically, if $\\mathbf{R}_{ee}=(R,0,0)^T$, we obtain $\\frac{Nb^2}{36}$ in the y and z directions and $\\frac{Nb^2}{36}+\\frac{R^2}{12}$ in the x-direction.\n<\/p>\n\n

Here’s a simple simulation code:<\/p>\n

rg2_ree = rg2 = 0\nfor _ in  range(5000):  \n    ch = np.random.normal(size=(1000,3))  \n    ch = np.append(np.zeros((1,3)),np.cumsum(ch, axis=0),axis=0)  \n    ch = ch - ch.mean(axis=0)  \n    ree = ch[-1]-ch[0]  \n    ree = ree\/np.linalg.norm(ree)  \n    rg2_ree += np.sum(ch.dot(ree)**2)  \n    rg2 += np.sum(ch**2)\nrg2_ree\/rg2\n# Result is 0.666...\n<\/code><\/pre>\n","protected":false},"excerpt":{"rendered":"
A Gaussian chain, when averaged over all conformations and orientations, exhibits isotropic behavior. This means it can be treated as a sphere with a radius equal to its radius of gyration, $R_g$. However, an exercise in the second chapter of Rubinstein’s Polymer Physics demonstrates that $R_g^2$ becomes asymmetric when the coordinate frame is aligned with […]<\/p>\n","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-223","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\/223","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=223"}],"version-history":[{"count":2,"href":"https:\/\/www.shirui.me\/blog\/wp-json\/wp\/v2\/posts\/223\/revisions"}],"predecessor-version":[{"id":225,"href":"https:\/\/www.shirui.me\/blog\/wp-json\/wp\/v2\/posts\/223\/revisions\/225"}],"wp:attachment":[{"href":"https:\/\/www.shirui.me\/blog\/wp-json\/wp\/v2\/media?parent=223"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.shirui.me\/blog\/wp-json\/wp\/v2\/categories?post=223"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.shirui.me\/blog\/wp-json\/wp\/v2\/tags?post=223"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}