# Normalization of a pair-wise radial radial distribution function with solid angle dependence

I have 3D position data of particles. I need to calculate g(r, $$\theta$$). Where g(r, $$\theta$$) is defined as:

$$g(r, \theta) = \frac{H(r, \theta)}{N \cdot n(r,\theta)}$$

r is a Euclidean distance between particles and theta is the polar angle from particle moving direction to r (Explained in attached picture). H(r,theta) is the histogram for all N pairs.

Usually normalization will be performed by dividing H by N. But I also need to normalize by Volume around the particle. As further will go away from the particle more chances there are to find a particle there. I have found normalization function of the form of n(r, theta): $$n(r,\theta) = \frac{2}{3} \pi [(r + \delta r)^{3} - r^{3}][cos(\theta) - cos(\theta + \delta\theta)]$$

First of all I do not understand how this function have come into be. Secondly, is this even correct?. Ideally it should be equal to 1 for particles moving in random direction.

I would not call a function $$g(r,\theta)$$ a radial distribution function. It does not depend only on the radial distance $$r$$. I also think that your normalization is not fully consistent with the definition of this function. It should be such that $$\rho g(r,\theta) dV$$ counts the number of particles present in a system of average number density $$\rho$$ in a volume $$dV$$ characterized by a distance $$r$$ and solid angle $$2 \pi \sin(\theta)\mathrm{d}\theta$$ with respect to a particle fixed at the origin. Therefore, if the value of the histogram $$H(r,\theta)$$ counts the number of particles in such a volume, it should be normalized by the number of particles that would be in the same volume if the density would be uniform. I.e. you should divide by $$\rho \Delta V(r,\theta)$$.
Based on its meaning, the explicit finite difference expression for $$\Delta V(r,\theta)$$ can be obtained as the difference of differences of the volumes of four spherical sectors (a look at the picture of a spherical sector would help). If $$V_s(r,\theta)= \frac{2 \pi}{3}r^3(1-cos(\theta))$$ is the volume of the spherical sector of radius $$r$$ and angle $$\theta$$,
\begin{align} \Delta V(r,\theta)&= \left[ V_s(r+\delta r, \theta +\delta \theta) - V_s(r+\delta r, \theta ) \right] - \left[ V_s(r, \theta +\delta \theta) - V_s(r, \theta ) \right]\\ &=\frac{2 \pi}{3}[(r + \delta r)^{3} - r^{3}][cos(\theta) - cos(\theta + \delta\theta)] \end{align} which is precisely the formula you quote for your $$n(r,\theta)$$. Notice however that (also for dimensional reasons) it should be multiplied by $$\rho$$ and not by $$N$$.
• Thank you for the explanation. How would one acquire the $\rho$ value when I only have position values. Would it be $\frac{n}{V}$ while n is number of atom and V will be Total volume under consideration? It would be Total number of Particles in system divided by $\frac{4}{3}\pi r_{max}^3$. I am taking $r_{max}$ here because my system is constraint and total volume would depend upon maximum radius where I am looking for the particle. Commented Dec 17, 2020 at 11:11
• @JunaidMehmood $\rho$ is the number density. If the number of particles is fixed, the number of atoms in a box of volume $V$ divided by $V$. If it is a gran-canonical ensemble, the number of particles should be intended as the average number. Commented Dec 17, 2020 at 12:33
• I have discrete time steps so I have taken an average of histogram and average number density to calculate g(r,$\theta$). It is working out fine now. My mistake was to not take average of histogram for each particle. Commented Dec 17, 2020 at 22:50