# Why is the probability dP of finding an object (e.g. galaxy) in an infinitesmal volume dV equal to \overline{n}dV?

As pointed out by many cosmology lectures, such as Eq. (63) of Cosmology II-8 Structure Formation, and Eq. (3.1) of A Detailed Look at Estimators for the Two-Point Correlation Function, the probability of finding an object in an infinitesimal volume is $$dP=\overline{n}dV,$$ where $$\overline{n}$$ is the mean number density. In my opinion, $$\overline{n}dV$$ is just the number of objects in $$dV$$ and can be greater than one. However, the probability should be less or equal to one. I don't understand why $$\overline{n}dV$$ is the probability of finding an object in the volume.

• @StephenG-HelpUkraine I have deleted the post on Physics SE. Commented Jul 8, 2022 at 6:20
• n-bar can be grater than 1, but dV is an infinitesimal, so n-bar x dV doesn't really even have a "value", you really need to integrate it over a volume to have it make sense. Commented Jul 8, 2022 at 10:30
• In addition to what @GrapefruitIsAwesome says, note that (between eqs. 61 and 62) it is assumed that $dV$ is "so small that there is at most one galaxy in it."
– pela
Commented Jul 8, 2022 at 12:38
• I think it's more correct to say that it's so small that $dP\ll1$.
– pela
Commented Jul 8, 2022 at 13:49
• @pela Ok, I got it. Commented Jul 8, 2022 at 14:00

That the probability of galaxy occupation is $$\bar{n}\ dV$$ will work as an approximation so long as you are considering volumes that are limited in size such that the probability is $$\ll 1$$. i.e. You are ignoring any probability that there could be more than 1 galaxy in the volume.

If you have $$N$$ particles in box of volume $$V$$, then $$\bar{n} = N/V$$. The probability of finding a particular particle in a subset of the volume $$dV$$ will be $$dV/V$$. But if there are $$N$$ particles then the probability that any of them are in $$dV$$ will be $$N\ dV/V = \bar{n}\ dV$$.

But this probability will include cases where there are 1, 2, 3 or more particles in the box.

The probability of getting 2 particles in $$dV$$ will be $$N\ (dV/V) \times (N-1)\ (dV/V)$$. But if $$N\ (dV/V) \ll 1$$, then the probability of finding a second particle (or more) in the box is $$\lll 1$$ and could be ignored.

The approximation will break down as the probability rises, even to a small fraction, because there is a tendency to find galaxies in groups and clusters (thanks to gravity).

• Thanks a lot for clarifying. BTW, what do you think of my answer to this question? Commented Oct 13, 2022 at 2:53

Here are my thoughts on this issue and comments are welcome.

If there are $$M$$ ensembles of volume elements, i.e. $$dV$$, and there are $$N$$ volume elements containing particles, then the probability of finding an object in an infinitesimal volume is $$dP = N/M.$$

Also, assuming that $$n_i$$ is the particle (or galaxy) number density of the $$i$$-th volume element, then we get \begin{align} dP &= (\Sigma_{i=1}^{M}(n_i\cdot dV))/M \\ &= (\Sigma_{i=1}^{M}n_i/M)\cdot dV \\ &= \overline{n}dV \end{align}, where $$\overline{n}=(\Sigma_{i=1}^{M}n_i/M)$$ is the mean number density.