Distribution of segments on Gaussian chain

The probability distribution function of $P_i(\mathbf{r}_i-\mathbf{r}_\mathrm{cm})$ of ideal chain is evaluated, for $P_i$ represents the probability distribution fucntion of $i$th segment with respect to its centre of mass on the ideal chain. An ideal chain is modeled as multidimensional random walk, where the steps are independent, and the distribution of a step at mean length $b$ is given by $P(\mathbf{r})\sim\mathcal{N}(0,b^2)$. Let $\mathbf{b}_i=\mathbf{r}_{i+1}-\mathbf{r}_{i}$ be the $i$th bond vector, then we have

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

and the centre of mass $\mathbf{r}_\mathrm{cm}=\frac{1}{N}\sum_i \mathbf{r}_i$ is




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

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

with $\mathbf{b}'_j=\frac{j}{N}\mathbf{b}_j$ 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 fucntion 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 $i\le j \le N-1$, and $\mathbb{I}_d$ is the $d$-dimensional identical matrix, and therefore: (For convenience, I will set $b=1$ in the later calculations.)

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

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 $\frac{1}{N}\sum_{i=1}^N \langle(\mathbf{r}_i-\mathbf{r}_\mathrm{cm})^6\rangle=\frac{58 N^6-273 N^4+462 N^2-247}{1944 N^3}$. At large $N$, only leading term matters, which is $\frac{29}{972} N^3$. For $N=20$, the result is $235.886$, which is in accord with the simulation. Another example is for $N=5$, the $R_g^2$ is $0.8$ rather than $5/6=0.8\dot{3}$ in this case.

Here is the simulation code:
ch = np.random.normal(size=(100000,20,3),scale=1/3**0.5)
ch = ch.cumsum(axis=1)
ch -= ch.mean(axis=1,keepdims=1)
m6 = np.mean(np.linalg.norm(ch,axis=-1)**6)

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