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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. enter image description here

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.

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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 $N$.

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  • $\begingroup$ 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. $\endgroup$ Commented Dec 17, 2020 at 11:11
  • $\begingroup$ @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. $\endgroup$ Commented Dec 17, 2020 at 12:33
  • $\begingroup$ 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. $\endgroup$ Commented Dec 17, 2020 at 22:50

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