## Scattering from polymers Polymers in scattering are often considered as monomers connected by bonds, then form a polymer chain. Consider the monomers, like $\rm \{CH_{2}\}_{n}$, $\rm \{C_{8}H_{8}\}_{n}$, $\rm \{C_{2}H_{4} O\}_{n}$, as a unit with the same scattering length, then we convert the polymer chains with different atoms with $b_{i}$ to a course grain model, considering only the scattering length density of monomers. ![[Drawing 2024-05-20 16.20.31.excalidraw.svg]] ### Gaussian chain model (random walk) Now consider a polymer chain with $N+1$ monomers. $\mathbf{r}_n = \mathbf{R}_n - \mathbf{R}_{n-1}$, $|\mathbf{r}_n|=b$, distance between monomers. Then the end to end vector is $\mathbf{R} = \mathbf{R}_N - \mathbf{R}_0 = \sum_{n=1}^{N} \mathbf{r}_n$. If we assume the chain orientation is random, we have $\langle \mathbf{r}_n \rangle = 0 \rightarrow \langle \mathbf{R} \rangle = 0$, and the mean square is $\langle \mathbf{R}^2 \rangle = \sum_{n,m=1}^{N} \langle \mathbf{r}_n \cdot \mathbf{r}_m \rangle = \sum_{n=1}^{N} \langle \mathbf{r}_n^2 \rangle + 2 \sum_{n < m} \langle \mathbf{r}_n \cdot \mathbf{r}_m \rangle = N b^2$. This is under the condition that monomers are having random walk. Now consider the scattering, the amplitude at a fixed time is $A(\mathbf{q}) = \sum_{m=1}^{N} e^{-i \mathbf{q} \cdot \mathbf{R}_m(t)}$ cause monomers have the same SLD (here neglected) and phase factor. And assume $\mathbf{R}_m - \mathbf{R}_n$ satisfy the Gaussian distribution, this is valid for big $N$, under central limit theorem. Then we have $I(\mathbf{q}) = \left\langle \left| A(\mathbf{q}) \right|^2 \right\rangle = \sum_{n,m=1}^{N} \left\langle e^{-i \mathbf{q} \cdot (\mathbf{R}_m - \mathbf{R}_n)} \right\rangle = \sum_{n,m=1}^{N} e^{-\frac{q^2}{6} \left\langle (\mathbf{R}_m - \mathbf{R}_n)^2 \right\rangle}=\sum_{n,m=1}^N e^{-\frac{q^2 b^2 |m - n|}{6}}$ >[!Proof]- >The detailed steps are >$\left\langle e^{-i \mathbf{q} \cdot (\mathbf{R}_m - \mathbf{R}_n)} \right\rangle = e^{-\frac{1}{2} q^2 \left\langle (\mathbf{R}_m - \mathbf{R}_n)^2 \right\rangle \cos^2 \theta}$ >$\left\langle (\mathbf{q} \cdot (\mathbf{R}_m - \mathbf{R}_n))^2 \right\rangle = q^2 \left\langle (\mathbf{R}_m - \mathbf{R}_n)^2 \right\rangle \cos^2 \theta = q^2 \cdot \frac{1}{3} \cdot |m - n| b^2$ >$\left\langle e^{-i \mathbf{q} \cdot (\mathbf{R}_m - \mathbf{R}_n)} \right\rangle = e^{-\frac{1}{2} q^2 \cdot \frac{1}{3} \cdot |m - n| b^2} = e^{-\frac{q^2 b^2 |m - n|}{6}}$ >$I(q) = \sum_{n,m=1}^N e^{-\frac{q^2 b^2 |m - n|}{6}}$ >>[!Note] >>For 1D Gaussian, we have >>$\begin{aligned} p(x) &= \frac{1}{\sqrt{2 \pi \sigma}} e^{-\frac{(x - \langle x \rangle)^2}{2 \sigma^2}} \\ \int_{-\infty}^{\infty} p(x) \, dx &= 1 \\ \int_{-\infty}^{\infty} x \, p(x) \, dx &= \langle x \rangle \\ \int_{-\infty}^{\infty} x^2 \, p(x) \, dx &= \langle x \rangle^2 + \sigma^2 \\ \sigma^2 &= \langle (\Delta x)^2 \rangle = \langle (x - \langle x \rangle)^2 \rangle = \langle x^2 \rangle - \langle x \rangle^2 \end{aligned}$ >>and for exponential form, we have >>$\begin{aligned} g(\alpha) &= \left\langle e^{\alpha x} \right\rangle \\ &= \int_{-\infty}^{\infty} e^{\alpha x} p(x) \, dx \\ &= e^{\alpha \langle x \rangle + \frac{1}{2} \alpha^2 \langle (\Delta x)^2 \rangle} \end{aligned}$ >>$\begin{aligned} &\text{If } \langle x \rangle = 0: && \left\langle e^{\alpha x} \right\rangle = e^{\frac{\alpha^2 \sigma^2}{2}} \\ &\text{If } \langle x \rangle = 0 \text{ and } \alpha = iq: && \left\langle e^{iq x} \right\rangle = e^{-\frac{q^2 \sigma^2}{2}} \end{aligned}$ >>This is for 1D so we have a factor $\frac{1}{2}$ from the $\cos$ integration. For 2D it would be $\frac{1}{4}$ and for 3D $\frac{1}{6}$ (which is our case). Therefore, we may write the [[Form factor]] as $P(q) = \frac{I(q)}{I(0)} = \frac{1}{N^2} \sum_{n,m=1}^{N} e^{-\frac{q^2 b^2 |n - m|}{6}}$ If $N$ is sufficiently large and above Gaussian distribution condition is satisfied, we may have the form factor as an integral form, $\begin{aligned} P(q)&=\frac{I(q)}{I(0)}\\ &= \frac{1}{N^2} \sum_{n,m=1}^{N} e^{-\frac{q^2 b^2 |n - m|}{6}} \\ &= \frac{1}{N^2} \int_{0}^{N} \mathrm{d}m \int_{0}^{N} \mathrm{d}n \, e^{-\frac{q^2 b^2 |n - m|}{6}} \\ &= \frac{1}{N^2} \int_{0}^{N} \mathrm{d}m \int_{0}^{m} \mathrm{d}n \, e^{-\frac{q^2 b^2 (m - n)}{6}} + \frac{1}{N^2} \int_{0}^{N} \mathrm{d}m \int_{m}^{N} \mathrm{d}n \, e^{-\frac{q^2 b^2 (n - m)}{6}} \\ &= \frac{2}{N^2} \int_{0}^{N} \mathrm{d}m \int_{0}^{m} \mathrm{d}n \, e^{-\frac{q^2 b^2 (m - n)}{6}} \\ &= \frac{2}{(q^2 \langle R_G^2 \rangle)^2} \left[ e^{-q^2 \langle R_G^2 \rangle} + q^2 \langle R_G^2 \rangle - 1 \right] \end{aligned}$ And this $\langle R_G^2 \rangle=\frac{1}{6} Nb^{2}$. This result is given by Debye in 1947. >[!Proof]- >Recall previous part [[Radius of gyration]], for discrete monomers (polymer chain), the $R_{g}$ would be >$R_g^2 = \frac{1}{N} \sum_{i=1}^N \langle (\mathbf{R}_i - \mathbf{R}_{\text{cm}})^2 \rangle$ >where the center of mass $\mathbf{R}_{cm}$ is >$\mathbf{R}_{\text{cm}} = \frac{1}{N} \sum_{i=1}^N \mathbf{R}_i$ >For an ideal Gaussian chain model, the mean square end-to-end distance is (also shown above): >$\langle R^2 \rangle = \langle (\mathbf{R}_N - \mathbf{R}_0)^2 \rangle = Nb^2$ >where $N$ is the number of segments in the chain and $b$ is the length of each segment. > >Then calculate $R_{g}$, >$\begin{aligned} >R_g^2 &= \frac{1}{N} \sum_{i=1}^N \langle (\mathbf{R}_i - \mathbf{R}_{\text{cm}})^2 \rangle \\ >&= \frac{1}{N} \sum_{i=1}^N \left\langle \left( \mathbf{R}_i - \frac{1}{N} \sum_{j=1}^N \mathbf{R}_j \right)^2 \right\rangle \\ >&= \frac{1}{N} \sum_{i=1}^N \left\langle \left( \mathbf{R}_i^2 - \frac{2}{N} \mathbf{R}_i \cdot \sum_{j=1}^N \mathbf{R}_j + \frac{1}{N^2} \left( \sum_{j=1}^N \mathbf{R}_j \right)^2 \right) \right\rangle \\ >&= \frac{1}{N} \sum_{i=1}^N \left\langle \mathbf{R}_i^2 \right\rangle - \frac{2}{N^2} \sum_{i,j=1}^N \left\langle \mathbf{R}_i \cdot \mathbf{R}_j \right\rangle + \frac{1}{N^3} \sum_{i,j,k=1}^N \left\langle \mathbf{R}_i \cdot \mathbf{R}_j \right\rangle \\ >\end{aligned}$ >Since for any $i \neq j$, >$\left\langle \mathbf{R}_i \cdot \mathbf{R}_j \right\rangle = 0$ >Thus >$\left\langle \mathbf{R}_i^2 \right\rangle = i b^2$ >Simplify, >$R_g^2 = \frac{1}{N} \left( \sum_{i=1}^N i b^2 \right)$ >We get >$R_g^2 = \frac{b^2}{N} \cdot \frac{N (N+1) (2N+1)}{6N}$ >For large $N$, >$R_g^2 \approx \frac{b^2}{N} \cdot \frac{N^3}{6N} = \frac{1}{6} N b^2$ >This gives the final result, >$R_g^2 = \frac{1}{6} N b^2$ >This is valid for random walk, Gaussian chain model. This model has the following limiting behavior: - For $q^2 \langle R_G^2 \rangle \ll 1$, $P(q) \approx 1 - \frac{1}{3} q^2 \langle R_G^2 \rangle$ - For $q^2 \langle R_G^2 \rangle \gg 1$, $P(q) \approx \frac{2}{q^2 \langle R_G^2 \rangle}$ The polymer is a fractal with [[Fractal dimensions|fractal dimension]] 2.