I am more confused how you define volume occupied in the first case, how do you intend to do that? — Tristan Maxson Oct 10 at 18:04

@TristanMaxson Volume is a measure of the space that a material occupies. So in the 1st case, the solid has a defined fixed shape/geometrical region in which all the atoms are contained. — Magic_Number Oct 10 at 20:22

I agree with @TristanMaxson, it's the first case I don't understand. The density of your actual system is N/V, it doesn't matter how the atoms are distributed within V. Calculating the volume "of occupied 3D space" is tricky, but occasionally needed; I usually use the inverse of what you're doing, and use V_occ = N/bulk density. — Phil Hasnip Oct 11 at 12:25

As a first pass approximation, you could use the bulk value per atom scaled by some average bond length, but you should strongly consider what you really want to do here. On a side note, please do answer your own question if you come up with a good solution yourself because I am interested myself. — Tristan Maxson Oct 12 at 5:01

@PhilHasnip If we put an ice cube in a closed container, do we say the density of the ice cube is N/V_of_container? — Magic_Number Oct 13 at 7:27

@TristanMaxson Bulk density is well defined, I agree, but to me, it doesn't seem right to apply the same logic for bulk density to a finite system. For the volume of a dimer of atoms, maybe we can introduce Van der Waals surface? For the 2nd system, I was thinking to differentiate the solid and gas atoms based on some criteria and then calculate Total density summing up their number fraction times indivual density. — Magic_Number Oct 13 at 7:35

@Magic_Number no, but we do say the density of the container is N/V_of_container, which is what you ask in your question. Defining the density of an ice cube is much better defined, since even an uncertainty of 1 A in length makes a negligible difference to the volume. In your example, the uncertainty is a substantial fraction of the actual volume. — Phil Hasnip Oct 14 at 9:11

@PhilHasnip Ok maybe I could have phrased my question in a much better and clear way, but when in the 1st case I say density of my system, I mean to say is the density of the solid nanoparticle (imagine the nanoparticle is in a vacuum). by constructing a local density profile, it is rather easy to find the size of the system, as when the nanoparticle ends, the profile goes to zero. So calculating the occupied volume is not really difficult in the 1st case. — Magic_Number Oct 14 at 15:41

@PhilHasnip But that is rather difficult for the 2nd case, as the nanoparticle atoms occupy the whole box. But if I consider the volume of box V, in both cases, then since the box size is not changing, density remains the same. That to me seems rather counter-intuitive. — Magic_Number Oct 14 at 15:41

How are you defining the density profile? I think actually the main problem here is that I don't understand why you want to know the density, which makes it difficult to think what the appropriate volume definition is. — Phil Hasnip Oct 15 at 21:53

@PhilHasnip I am constructing the local density profile by using this formula, $\rho_N(r)=\frac{\left\langle N(r) \right\rangle}{\Delta V(r)}=\frac{\left\langle N(r) \right\rangle}{\frac{4}{3}\pi\left[ (r+\delta r)^3 - r^3 \right]}$. here I am basically constructing spherical shell volumes between $r$ and $r+\delta r$ and finding the average number of atoms in that shell and plotting that against the distance from the center of the NP. In this way, I will reach a distance when inside this spherical shell it won't find any atoms. from this, I can conclude the size of my nanoparticle and so on. — Magic_Number Oct 17 at 6:59

@PhilHasnip I somehow want to see in my (finite) system, how the density changes with temperature, but I am struggling to find a good definition of density for the case when there is a mix of 2 phases in the simulation box — Magic_Number Oct 17 at 7:21