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This analysis focuses on the probability distribution function, $P_i(\mathbf{r}_i-\mathbf{r}_\mathrm{cm})$, of the $i$th segment relative to the center of mass in an ideal chain. An ideal chain is modeled as a multidimensional random walk with independent steps. The distribution of each step, with a mean length of $b$, is assumed to be Gaussian: $P(\mathbf{r})\sim\mathcal{N}(0,b^2)$.

Let $\mathbf{b}_i=\mathbf{r}_{i+1}-\mathbf{r}_{i}$ represent the $i$th bond vector. Then, the position of the $i$th segment is given by:

$\mathbf{r}_i=\sum_{j=1}^{i-1} \mathbf{b}_j$

The center of mass, $\mathbf{r}_\mathrm{cm}$, is calculated as:

$\mathbf{r}_\mathrm{cm}=\frac{1}{N}\sum_{i=1}^{N} \mathbf{r}_i = \frac{1}{N}\sum_{j=1}^{N-1}(N-j)\mathbf{b}_j$

Therefore, the displacement of the $i$th segment relative to the center of mass is:

$\mathbf{r}_i-\mathbf{r}_\mathrm{cm}=\sum_{j=1}^{i-1}\frac{j}{N}\mathbf{b}_j+\sum_{j=i}^{N-1}\frac{N-j}{N}\mathbf{b}_j$

If $X$ is a Gaussian random variable with variance $\sigma^2$, then $aX$ is also a Gaussian random variable with variance $a^2\sigma^2$. Using this property, we can write the characteristic function for $P_i(\mathbf{r}_i-\mathbf{r}_\mathrm{cm})$ of a $d$-dimensional ideal chain:

$\phi_i(\mathbf{q})=\Pi_{j} \phi_{\mathbf{b}_j’}(\mathbf{q})=\exp\left(-\frac{1}{2}\mathbf{q}^T\left(\sum_{j=1}^{N-1}\Sigma_j\right)\mathbf{q}\right)$

where $\mathbf{b}’_j=\frac{j}{N}\mathbf{b}_j$ for $j\le i-1$ and $\frac{N-j}{N}\mathbf{b}_j$ for $i\le j \le N-1$. $\phi_{\mathbf{b}_j’}=-\exp(-0.5\mathbf{q}^T\Sigma\mathbf{q})$ is the characteristic function of the probability distribution of bond $j$. $\Sigma_j=\frac{j^2 b^2}{d N^2}\mathbb{I}_d$ for $j\le i-1$ and $\Sigma_j=\frac{(N-j)^2 b^2}{d N^2}\mathbb{I}_d$ for $i\le j \le N-1$, where $\mathbb{I}_d$ is the $d$-dimensional identity matrix. Setting $b=1$ for convenience, we obtain:

$\phi_i(\mathbf{q})=$ $\exp \left(-\frac{\left(6 i^2-6 i (N+1)+2 N^2+3 N+1\right) \left(q_x^2+q_y^2+q_z^2\right)}{36 N}\right)$

The corresponding distribution of this characteristic function is still a Gaussian distribution with $\Sigma=\frac{b^2}{3} \mathbb{I}_3$, where the equivalent bond length $b^2=\frac{\left(6 i^2-6 i (N+1)+2 N^2+3 N+1\right)}{6 N}$. The 6th moment is calculated as $\frac{1}{N}\sum_{i=1}^N \langle(\mathbf{r}_i-\mathbf{r}_\mathrm{cm})^6\rangle=\frac{58 N^6-273 N^4+462 N^2-247}{1944 N^3}$. For large $N$, only the leading term, $\frac{29}{972} N^3$, is significant. For $N=20$, the result is $235.886$, which agrees with simulations. Another example is for $N=5$, where the $R_g^2$ is $0.8$ compared to the expected value of $5/6=0.8\dot{3}$.

Here is the simulation code:

ch = np.random.normal(size=(100000,20,3),scale=1/3**0.5)
ch[:,0,:]=0
ch = ch.cumsum(axis=1)
ch -= ch.mean(axis=1,keepdims=1)
m6 = np.mean(np.linalg.norm(ch,axis=-1)**6)

A circulant matrix is defined as:

$C=\begin{bmatrix}c_{0}&c_{n-1}&\dots &c_{2}&c_{1}\\c_{1}&c_{0}&c_{n-1}&&c_{2}\\\vdots &c_{1}&c_{0}&\ddots &\vdots \\c_{n-2}&&\ddots &\ddots &c_{n-1}\\c_{n-1}&c_{n-2}&\dots &c_{1}&c_{0}\\\end{bmatrix}$

where $C_{j, k}=c_{j-k \mod n}$. The $k$-th eigenvalue $\lambda_k$ and eigenvector $x_k$ satisfy $C\cdot x_k=\lambda_k x_k$, which can be expressed as $n$ equations:

$\sum_{j=0}^{m-1}c_{m-j}x_j+\sum_{j=m}^{n-1}c_{n-j+m}x_j=\lambda_k x_m\quad m=0,1,\dots,n-1$

with $c_n=c_0$, where $x_m$ is the $m$-th component of the eigenvector $x_k$. By changing the dummy summing variables ($j\to m-j$ and $j\to n-j+m$), we obtain:

$\sum_{j=1}^{m}c_j x_{m-j} +\sum_{j=m+1}^{n}c_j x_{n+m-j}=\lambda_k x_m$

with $m=0,1,2,\dots,n-1$. We can “guess” a solution where $x_j=\omega^j$, which transforms the equation into:

$\begin{align}&\sum_{j=1}^{m}c_j \omega^{m-j} +\sum_{j=m+1}^{n}c_j \omega^{n+m-j}=\lambda_k \omega^m\\ \leftrightarrow &\sum_{j=1}^{m}c_j \omega^{-j} +\omega^{n}\sum_{j=m+1}^{n}c_j \omega^{-j}=\lambda_k\end{align}$

Let $\omega$ be one of the $n$-th roots of unity, i.e., $\omega = \exp(-2\pi\mathbb{i}/n)$, then $\omega^{n}=1$. This leads to the eigenvalue:

$\lambda = \sum_{j=0}^{n-1}\omega^{-j} c_j$

with the corresponding eigenvector:

$x= (1, \exp(2\pi\mathbb{i}/n), \exp(2\pi\mathbb{i}/n)^2,\dots,\exp(2\pi\mathbb{i}/n)^{n-1})^T$

The $k$-th eigenvalue is generated from $\omega^{-k} = \exp(2\pi k\mathbb{i}/n)$ with $k\in [0,n-1]$, resulting in the $k$-th eigenvalue and eigenvector:

$\lambda_k = \sum_{j=0}^{n-1}c_j \exp(2\pi k\mathbb{i}/n)$

and

$x_k = (1, \exp(2\pi k\mathbb{i}/n), \exp(2\pi k\mathbb{i}/n)^2,\dots,\exp(2\pi k\mathbb{i}/n)^{n-1})^T$

The eigenspace is simply the DFT matrix, and all circulant matrices share the same eigenspace. It is easy to verify that circulant matrices possess the following properties:

If $A$ and $B$ are circulant matrices, then:

1. $AB=BA=W^\ast\Gamma W$ where $W$ is the DFT matrix and $\Gamma=\Gamma_A\Gamma_B$, with $\Gamma_i$ representing the diagonal matrix consisting of the eigenvalues of $i$; $\Gamma_A = WAW^\ast$.
2. $B+A=A+B=W^\ast\Omega W$, where $\Omega=\Gamma_A+\Gamma_B$.
3. If $\mathrm{det}(A)\ne 0$, then $A^{-1}=W^\ast \Gamma_A^{-1}W$.

The proof is straightforward:

1. $AB=W^\ast \Gamma_AW W^\ast\Gamma_BW=$
   $W^\ast\Gamma_A\Gamma_BW=W^\ast \Gamma_B\Gamma_AW=BA$
2. $W(A+B)W^\ast=\Gamma_A+\Gamma_B$.
3. $AW^\ast \Gamma_A^{-1}W=W^\ast \Gamma_AW W^\ast \Gamma_A^{-1}W=\mathbb{I}$