The central limit theorem

We shall discuss here the central limit theorem  as applied to the calculation of the distribution function of the end-to-end vector .

Consider a chain consisting of N independent bond vectors $\vec{r}_{i}$. By this we mean that the probability density in configuration space $\Psi (\vec{r}^{N})$ may be written as

$$\Psi(\vec{r}^{N}) = \prod_{i=1}^N \psi (\vec{r}_i).$$ (3.1)

Assume further that the bond vector probability density depends only on the length of the bond vector, and has zero mean. For the second moment we write

$$\langle r^2 \rangle = \int d^3r r^2 \psi (r) =b^2.$$ (3.2)

The distribution of the end-to-end vector may be calculated according to

$$\Omega (\vec{R}) = \langle \delta (\vec{R}-\sum_{i=1}^N \vec{r}_i) \rangle.$$ (3.3)

The central limit theorem then states that

$$\Omega (\vec{R}) = \left\{ \frac{3}{2\pi Nb^2} \right\}^{3/2} \exp\left\{ - \frac{3R^2}{2Nb^2} \right\}$$ (3.4)

i.e. that the end-to-end vector has Gaussian distribution.

In order to prove Eq. (3.4) we write

 

$$\Omega (\vec{R}) \frac{1}{(2\pi)^3} \int d^3k \langle \exp \{ i \vec{k} \cdot (\vec{R}-\sum_i \vec{r}_i ) \} \rangle$$ =

$$\frac{1}{(2\pi)^3} \int d^3k e^{i\vec{k}\cdot\vec{R}} \langle \exp \{ -i \vec{k}\cdot \sum_i \vec{r}_i \} \rangle$$ =

$$\frac{1}{(2\pi)^3} \int d^3 k e^{i \vec{k}\cdot\vec{R}} \{ \int d^3r e^{-i\vec{k}\cdot\vec{r}} \psi(r)\}^N .$$ = (3.5)

For k=0, the Fourier transform of $\psi (r)$ will be equal to one. Because $\psi (r)$ has zero mean and finite second moment, the Fourier transform of $\psi (r)$ will have its maximum around k=0 and go to zero for large values of k. Raising such a function to the N'th power leaves us with a function that differs from zero only very close to the origin, and which may be approximated by

$$\left\{ \int d^3r e^{-i \vec{k}\cdot\vec{r}} \psi (r) \right\}^N \approx \{ 1 -\frac{1}{2} \langle (\vec{k}\cdot\vec{r})^2 \rangle \}^N$$

$$\approx 1 -\frac{1}{2} N \langle (\vec{k}\cdot\vec{r})^2 \rangle$$

$$1 - \frac{1}{6} N k^2 b^2$$ = (3.6)

for small values of k, and by zero for the values of k where $1 - \frac{1 }{6} N k^2 b^2$ is negative. This again may be approximated by $\exp \{ - \frac{1}{6} N k^2 b^2 \}$ for all values of k. Then

 

$$\Omega (\vec{R}) \frac{1}{(2\pi)^3} \int d^3k \exp \{ i\vec{k}\cdot \vec{R}-\frac{1}{6}Nk^2b^2\}$$ =

= I(R_x) I(R_y) I(R_z) (3.7)

 

$$\frac{1}{2\pi} \int dk_x \exp \{ iR_x k_x - \frac{1}{6}Nb^2 k_x^2\}$$ I(R_x) =

$$\left\{ \frac{3}{2\pi Nb^2} \right\}^{1/2} \exp \left\{ - \frac{3R_x^2 }{2Nb^2} \right\} .$$ = (3.8)

Combining Eqs. (3.7) and (3.8) we get Eq. ( 3.4).

Now apply the above result to a general RIS chain . To this end we write

$$\vec{R}=\sum_{i=1}^{N}\vec{r}_{i}=\sum_{n=1}^{N/\lambda }(\su... ...\vec{r}_{(n-1)\lambda +i})=\sum_{n=1}^{N/\lambda } \vec{x}_{n}$$ (3.9)

i.e. we combine $\lambda$ bond vectors into one new vector as in Fig. ( 3.1) where $\lambda =5$. For $\lambda$ large enough the distribution of these new vectors meets the conditions on $\psi (\vec{r})$used to derive the central limit theorem. Notice that this holds true because there are no long range excluded volume interactions  in the RIS chain. Writing

$$\langle x^{2}\rangle =\lambda b^{2}$$ (3.10)

we find that the distribution of the end-to-end vector is again given by Eq. (3.4). We conclude that every model which does not incorporate the long range excluded volume effect will have a Gaussianly distributed end-to-end vector. It is an easy exercise to check that in this case Eq. (1.45) holds true.

  

Figure 3.1: Five bonds combined into one new vector.