Appendix B

In the case of the freely jointed chain  both the angles $\vartheta$ and $\varphi$ are completely random. Then  

$$\langle R^{2}\rangle =Nl^{2}$$ (3.45)

$$\frac{1}{(N+1)^{2}}\sum_{i=0}^{N-1}\sum_{j=i+1}^{N}(j-i)l^{2}=\frac{1}{ (N+1)^{2}}\sum_{i=0}^{N}\sum_{k=1}^{N-i}kl^{2}$$ R_g² =

$$\frac{1}{(N+1)^{2}}\sum_{i=0}^{N-1}\frac{1}{2}(N-i)(N-i+1)l^{2}=\frac{1}{ (N+1)^{2}}\sum_{k=1}^{N}\frac{1}{2}k(k+1)l^{2}$$ =

$$\frac{N+2}{N+1}\frac{1}{6}Nl^{2}.$$ = (3.46)

A simple trick exists to calculate all the moments of the chain. To this end, look at the Fourier transform

 

$$\hat{\Omega}(\vec{k}) \int d^3R e ^{-i\vec{k}\cdot\vec{R}} \Omega( \vec{R})$$ =

$$\sum_n \frac{1}{n!} \int d^3R (-i\vec{k}\cdot\vec{R})^n \Omega(\vec{ R})$$ =

$$\sum_n \frac{(-ik)^n}{n!} \int d^3R (\cos\vartheta)^n R^n \Omega(R)$$ =

$$\sum_n \frac{(-ik)^n}{n!} 2\pi \int_{-1}^1 d\cos\vartheta \: (\cos\vartheta)^n \int dR R^2 R^n \Omega (R)$$ =

$$\sum_n ^{\prime}\frac{(-ik)^n}{(n+1)!} 4\pi \int dR R^2 R^n \Omega (R)$$ =

$$\sum_n (-1)^n \frac{k^{2n}}{(2n+1)!} \langle R^{2n}\rangle$$ = (3.47)

The prime at the summation sign in the last but one line indicates that only terms with even n should occur in the sum. Eq. (3.47) tells us that

$$\langle R^{2n} \rangle = (2n+1) (-1)^n \frac{d^{2n}\hat{\Omega}}{dk^{2n}} \vert _{k=0}$$ (3.48)

In order to use this formula we need to know $\hat{\Omega}(k)$. Inverting Eq. (3.5) we get

$$\hat{\Omega}(k) \langle e^{-i\vec{k}\cdot\vec{r}} \rangle^N$$ =

$$\left\{ \frac{\sin kl}{kl} \right\}^N .$$ = (3.49)

Although the trick is nice, the task to evaluate the derivations is rather tedious.