You may assume any easy model without collisions/attractions, molecule diameters of 0 and perfect mixing.

If you have 500 copies per micro-m^3 of molecule A , then each molecule A has a box of 125 nm x 125 nm x 125 nm. However, according to my intuition, the expected distance to the closest neighbor should be significantly less than 125 nm, as the shapes of the theoretical volumes will not be perfectly cubic/spherical.

So my question is:

  • What is the expected distance between molecules A (to the closest molecule A)?

Background According to this source, we have 1^6 proteins per micro m^3 in huma cells. If we assume 20.000 genes and that my protein is ~10x more expressed than average genes (which might not be expressed at all), then this would be 500 copies per micro-m^3.

EDIT: I asked the wrong question

Actually I need to know the expected distance between 1 RNA to the closest of 500 proteins within the cubic micron (center to center, no collusion). I did a small 3d random-walk simulation, but the result was quite different from @whuber result. I wonder if the reason is the slightly different question or that my simulation is inaccurate.

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  • $\begingroup$ This sounds like a homework, test, or similar self-study question. If so, please read the information on how we handle these questions, and add the self-study tag to your question. $\endgroup$
    – EdM
    Jan 22, 2021 at 16:16
  • $\begingroup$ This is for a paper. I want to compare the expected distance between copies of a certain protein of interest to the distance they assume in presence of an adapter protein. $\endgroup$
    – KaPy3141
    Jan 22, 2021 at 16:21
  • $\begingroup$ If you are looking for references en.m.wikipedia.org/wiki/… $\endgroup$ Mar 12, 2021 at 18:06

1 Answer 1


Consider $d$ dimensions. The distribution to the nearest neighbor of any point can be approximated by supposing $N$ neighbors are independently, uniformly, and randomly situated within a radius of one unit from that point (where the distance unit and $N$ are chosen to reproduce the molecular density; preferably $N$ is large).

The chance that one given neighbor is further than a distance $r$ (for $0\le r \le 1$) is the relative volume of the spherical shell between the balls of radii $r$ and $1,$ equal to

$$S_{d}(r) = 1 - r^d.$$

Since the neighbor positions are independent, the chance that all are further than distance $r$ is

$$S_{N;d}(r) = (1-r^d)^N.$$

This (the survival function) determines the distribution of the distance $R$ to the nearest neighbor. Its expectation is the integral of the survival function,

$$E[R] = \int_0^1 S_{N;d}(r)\,\mathrm{d}r = \int_0^1 (1-r^d)^N\,\mathrm{d}r = B(N+1,1/d)/d$$

where $B$ is the Beta function

$$B(a,b) = \frac{\Gamma(a)\Gamma(b)}{\Gamma(a+b)}.$$

For instance, here is a histogram of 50,000 simulated values of $R$ with $N=500-1$ neighbors.


On it I have superimposed the graph of the density function $-d/dr\, S_{499;3}(r)$ in red to show the agreement and I have plotted a vertical line to show $E[R].$

  • $\begingroup$ Nice approach! Thanks for this solution! Would the result change, if you did the calculation with different absolute molecule counts at constant concentration? $\endgroup$
    – KaPy3141
    Jan 22, 2021 at 18:11
  • 2
    $\begingroup$ That's a good question. You can see that the distribution is concentrated at small values of $R.$ In this range, $(1-r^d)^N$ is closely approximated by $\exp(-Nr^d).$ The argument $Nr^d$ is the expected number of points a distance $r$ away. Thus, once $N$ exceeds a fairly small value, the distribution essentially stabilizes as a generalized exponential. I originally thought of making this approximation--which I'm sure countless physicists have done--the availability of a simple exact solution was more appealing. $\endgroup$
    – whuber
    Jan 22, 2021 at 18:35
  • $\begingroup$ Dear @whuber, would the distance of a molecule B to the closest of 500 molecules A, be the same as your simulation of 500 molecules A, but with N= 501? $\endgroup$
    – KaPy3141
    Mar 12, 2021 at 17:52

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