S.O.S. Mathematics CyberBoard

[trekkiee]

Hi, 1st time poster here. I'm trying to figure out the average distance between nearest neighbors in a three dimensional randomly distributed sample of particles. My best guess is d_nearest neighbor_mean=(volume/n)^1/3 where n particles are randomly distributed in a 3 dimensional volume.

I'm not a student and this isn't homework, btw. The question originated when I wondered what was the average distance between stars in the solar neighborhood. atlasoftheuniverse.com gives 35 stars (including the Sun) within 12.5 light-years, and the above formula yields 6.16 ly as the avg distance from any given star to its closest neighbor. This seems a little high to me, since the distance from the Sun to its nearest neighbor (Proxima Centauri) is 4.4 ly. But perhaps the Sun has a closer-than-avg nearest neighbor, since, after all, the distribution should be random.

I originally thought it would be easy to figure this out, but after trying unsuccessfully for an hour to work out a satisfactory formula, then another hour trying to google one, I gave up. Thanks in advance

[Shadow]

[trekkiee]

Modest at

http://hypography.com/forums/physics-mathematics/21509-mean-nearest-neighbor-distance-3d.html

was helpful in pointing me to this link:

http://books.google.com/books?id=hpxpyOak1psC&pg=PA171&lpg=PA171&dq=%22probability+law+for+nearest-neighbor+distance+in+a+random+distribution+of+particles%22&source=bl&ots=T9lWLj9rLx&sig=q3ldVEWmqUBu3FWcWJ75S85-Wc8&hl=en&ei=n3MNS82wC4ikMJDlrNAC&sa=X&oi=book_result&ct=result&resnum=1&ved=0CAgQ6AEwAA#v=onepage&q=%22probability%20law%20for%20nearest-neighbor%20distance%20in%20a%20random%20distribution%20of%20particles%22&f=false

And now I need to integrate:

integral of x^3 exp(-a x^3) dx, with a = constant,

but I couldn't. Hopefully, it's an easy integral and and someone will figure it out.

In a 3-dimensional random distribution, the basic idea for finding the average distance from any given particle to its nearest neighbor begins with:

P(r)dr = [1 - integral from 0 to r of P(r)dr][4 pi r^2 pho dr] (1)

where
pho = average number of particles/unit volume
P(r) dr = the probability of a particle's nearest neighbor occurring in the interval [r,r+dr]
integral from 0 to r of P(r) dr = probablity that an arbitrary particle's nearest neighbor lies within a distance r of the particle
1 - integral from 0 to r of P(r) dr = the probability that the nearest neighbor is no closer than r.

differentiating & separating eq. 1:

dP/P = [2/r - 4 pi rho r^2] dr

integrating:

P = [constant] r^2 exp(- [4/3] pi rho r^3)

normalizing:

1 = integral from 0 to infinity of P(r) dr
1 = [constant] integral from 0 to infinity of r^2 exp(-[4/3] pi rho r^3) dr
1 = [constant] [-[1/(4 pi rho)] exp(-[4/3] pi rho r^3)] evlauated from 0 to infinity

gives the constant = 4 pi rho and P(r) = 4 pi rho r^2 exp(-[4/3] pi rho r^3)

The average distance from any given particle to its nearest neighbor in 3 dimensions is then the expectation value of r:

<r> = integral from 0 to infinity of r P(r) dr
<r> = [4 pi rho] [integral from 0 to infinity of r^3 exp(-[4/3] pi rho r^3) dr]

I was unable to do the last integral, but I'm sure someone can

[aswoods]

according to Gradshteyn and Ryzhik 3.381.10, and Wolfram Alpha seems to agree.

[trekkiee]

[trekkiee]

aswoods at

www.sosmath.com/CBB/viewtopic.php?p=197769#197769

helpfully pointed out that Gradshteyn and Ryzhik 3.381.10, and Wolfram Alpha agree that

the integral from 0 to infinity of x^3 e^(-ax^3)dx = 1/3 gamma(4/3) a^(-4/3)

so the mean nearest neighbor distance in 3d is:

<r> = [4 pi rho] integral from 0 to infinity r^3 e^(-[4/3] pi rho r^3) dr
<r> = 1/3 gamma(1/3) ([4/3] pi)^(-1/3) rho^(-1/3)
<r> = 0.55396 rho^(-1/3)
<r> = 3.93 light-years for the 23 star systems within 12.5 ly