Particle densities vs mean distances

[shadron]

Do any of the wise men on our forum have a derivation that relates particle density (in terms of particles per volume) to mean distance between particles? I could find only one reference where such a derivation was drawn, in terms of droplet densities (as in a cloud) at http://www.atmos.millersville.edu/~adecaria/DERIVATIONS/Mean_Distance_Between_Cloud_Drops.pdf.

I can compute a star density of .09 stars/cubic parsec for the volume around the sun (375 stars within a 10 parsec radius) which yields a mean distance between stars of what seems to be a ridiculously low figure of .77 light years.

Anyone up for a straight-forward calculation, as opposed to quantum arcania? :)

[Ziggurat]

The problem is that this question is somewhat ambiguous. Are you looking for the average distance between any two objects, or between nearest neighbor pairs? Furthermore, the answers can vary even with a given density, depending on the distribution. For example, consider a simple cubic lattice versus a face-centered cubic lattice (where nearest neighbor distances are well defined and constant). For the same nearest-neighbor separation distance, the latter will be denser than the former, which means that at the same density, the simple cubic lattice will have closer nearest-neighbor distances. Go to a diamond lattice and you get even closer distances for the same density. I'm not sure where a random arrangement would lie, but I'd guess somewhere between diamond lattice and FCC.

But one can get an approximate value rather easily. If one doesn't know the arrangement, or if it's random (like a gas), one can just treat it as being on average like a simple cubic arrangement (which lies somewhere between FCC and diamond), and that may be good enough. So you take your density (objects/volume), invert it (volume/object, where the denominator is a pure number), and take the cube root of it to get the size of your "unit cell", as it were. That will be, roughly, the average spacing between your objects. There are probably more sophisticated approaches, and maybe even proofs of how good an approximation this is, but it's good enough for many purposes.

Oh, and here's a toy to play around with:
http://demonstrations.wolfram.com/DistanceDistributionsInFiniteUniformlyRandomPointP rocesses/

[Dilb]

You might just want to check your math. At 0.09 stars/cubic parsec, that's 10.66 cubic parsecs per star, or a cube with length 2.20 parsecs per star. Since it's one cube to the next star, that would be 7.3 light years for the average separation.

From the paper, they suggest 0.55n-1/3, which works out to 4 light years, though it seems to me that's rather close together. It suggests that uniform random distributions are very inefficient at packing.

Using the most efficient packing FCC/HCP, the separation goes up 8.2 ly, while a less efficient packing like diamond is goes down to 5.9 ly. If they were randomly packed like oranges (p.f. around 0.637), it would actually be more like 7.8 ly, more efficient than a simple cubic distribution.

The calculation with the packing factor is just

Spherical volume per star = total volume*p.f./number of stars

since

V = 4/3 * Pi * r3,

find r, then double it to get to where the next star is. Or since this is a spherical volume we're finding start in anyway, cancel the factors for comparing the volume o

r3 = distance3 * p.f. / number of stars

where again you need to double r at the end to get to the next star.

[shadron]

Thanks, Zig. That's exactly the same result as the cited paper gives, though he derives a constant of .554 based on random placement. Essentially, he says

r = .554 / n ^ 1/3

where r is the mean distance (among closest pairs) and n is the average density. Working it out without the math error I made before results in r = 3.8 light years in the vicinity of the sun, which seems a very good value.

[shadron]

You might just want to check your math. At 0.09 stars/cubic parsec, that's 10.66 cubic parsecs per star, or a cube with length 2.20 parsecs per star. Since it's one cube to the next star, that would be 7.3 light years for the average separation.

Yeah, you're right - I just did, and got 3.8 ly average distance, which seems a good value to me.

From the paper, they suggest 0.55n-1/3, which works out to 4 light years, though it seems to me that's rather close together. It suggests that uniform random distributions are very inefficient at packing.

I don't think that's too close; our distance to Alpha Centauri is 3.2 ly. Indeed; I suppose it is fortunate for us that there is no forcing function driving efficient packing of the galaxy :).

Using the most efficient packing FCC/HCP, the separation goes up 8.2 ly, while a less efficient packing like diamond is goes down to 5.9 ly. If they were randomly packed like oranges (p.f. around 0.637), it would actually be more like 7.8 ly, more efficient than a simple cubic distribution.

The calculation with the packing factor is just

Spherical volume per star = total volume*p.f./number of stars

since

V = 4/3 * Pi * r3,

find r, then double it to get to where the next star is. Or since this is a spherical volume we're finding start in anyway, cancel the factors for comparing the volume o

r3 = distance3 * p.f. / number of stars

where again you need to double r at the end to get to the next star.Thanks for your help.

[jasonpatterson]

I don't think that's too close; our distance to Alpha Centauri is 3.2 ly.


Minor correction: 4.2 ly to proxima centauri, the closest star (beyond the sun, of course.)

[shadron]

Thanks. Update the brain here. :)