# Derivation of probability density of isolated polymers

(Crossposted on physics SE) I am reading Introduction to Polymer Physics by Doi, and in his proof for the probability distribution for ideal polymers of length $$N$$ and end-to-end vector $$\mathbf{R}$$, he does the following: \begin{align} P(\mathbf{R}-\mathbf{b}_i,N-1)=P(\mathbf{R},N)-\frac{\partial P}{\partial N}-\frac{\partial P}{\partial R_{\alpha}}b_{i\alpha}+\frac{1}{2}\cdot \frac{\partial ^2 P}{\partial R_{\alpha}\partial R_{\beta}}b_{i\alpha}b_{i\beta}\label{1}\tag{1} \end{align} where $$b_{i\alpha}$$ is the component of $$\mathbf{b}_i$$ in the $$\alpha$$ direction, $$R_{\alpha}$$ is a component of $$\mathbf{R}$$,for the lattice the polymer lies in. We are using the Einstein convention for summation in the above equation.

He makes the statement, $$\frac{1}{z}\sum _{i=1}^z b_{i\alpha}b_{i\beta} = \frac{\delta _{\alpha\beta}b^2}{3}\label{2}\tag{2}$$ where $$z$$ is the coordination number, $$\delta _{ij}$$ is the Kronecker delta, $$b$$ is the length of the lattice edge, so $$|\mathbf{b}_i| = b$$. I don't understand where this comes from.

What is the proof of this statement?

Since I have a problem with the Taylor expansion, I am going to write down what I am not understanding. \begin{align} P(\mathbf{R}-\mathbf{b}) = P([R_x-b_x, R_y-b_y]^T)\approx P([R_x,R_y]^T)-\frac{\partial P}{\partial R_x} b_x-\frac{\partial P}{\partial R_y}b_y+\frac{1}{2}\cdot \left(\frac{\partial ^2 P}{\partial R_x^2}b_x^2+\frac{\partial ^2 P}{\partial R_y^2}b_y^2 + 2\frac{\partial ^2 P}{\partial R_xR_y}b_xb_y \right) \end{align} This was derived using the standard higher-dimensional Taylor expansion formula for $$f(\mathbf{x})$$ where \begin{align} f(\mathbf{x}) = f(\mathbf{a}) + Df(\mathbf{x})(\mathbf{x}-\mathbf{a})^T+\frac{1}{2}(\mathbf{x}-\mathbf{a}) Hf(\mathbf{x})(\mathbf{x}-\mathbf{a})^T \end{align} where $$H$$ is the Hessian of $$f$$.

• +1. I suppose you know the Kronecker delta function? Also, it might help to define everything, including $b$, $\mathbf{b}_i$, etc. Jun 28 at 15:24
• Thank you @NikeDattani, i have made the changes Jun 28 at 15:37

For context to future readers, Doi starts with a model of an $$N$$ unit polymer formed by a random walk along a uniform grid of lattice length $$b$$.

It seems to be implicitly assumed in the described derivation that the lattice used is $$b\mathbb{Z}^3$$, that is the uniform 3D grid. Note there are other types of lattices for which Eq.$$\,(\ref{2})$$ is not true. Also note the factor of $$\frac{1}{3}$$ comes from the dimension, so this expression is also not true for the 2D grid they show as an example.

For the 3D uniform grid, the result arises fairly simply. We have $$z=6$$ corresponding to the vectors $$(\pm \mathbf{b}_1,\pm \mathbf{b}_2,\pm \mathbf{b}_3)$$. For simplicity, we can say these just align with the usual Cartesian axes $$(\pm x,\pm y,\pm z)$$, since Eq.$$\,(\ref{2})$$ should not be dependent on the orientation of the grid. Clearly if $$\alpha\neq\beta$$, the sum equals zero, as all of the $$\mathbf{b}_i$$ vectors only point along one Cartesian direction. For $$\alpha=\beta$$, only $$2$$ of the $$6$$ vectors (i.e. $$\frac{1}{3}$$) will have this component be nonzero and for these vectors the product is just $$b^2$$. So for any Cartesian, the sum will result in $$\frac{1}{3}zb^2$$, which when divided by $$z$$ gives the expression from Eq.$$\,(\ref{2})$$.

• +1, especially for going through the paper and figuring this one out so quickly! I was wondering why you added the equation labels and assumed you might have been in the process of writing an answer :) Jun 28 at 20:23
• @NikeDattani Does eqref show up unusually on your end? For me, it is displaying the label as enclosed in parenthesis and what look to be subscript periods. So something like $_.(2)_.$
– Tyberius
Jun 28 at 20:27
• It's showing the periods for me too, while ref is not giving the parentheses that we'd usually see in Journal of Chemical Physics, and the Physical Review series. I wonder if on the meta site for Math.SE there's any valuable information on how to fix this. Jun 28 at 20:29
• @Tyberius, so in a taylor expansion of this sort, we are taking the dot product of the vectors? Jun 29 at 14:43
• @Tyberius I have made some edits to the original post to expand on the problem I am having. Jun 29 at 15:44