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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.

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    $\begingroup$ @StephenG-HelpUkraine I have deleted the post on Physics SE. $\endgroup$
    – Wang Yun
    Jul 8, 2022 at 6:20
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    $\begingroup$ 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. $\endgroup$ Jul 8, 2022 at 10:30
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    $\begingroup$ 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." $\endgroup$
    – pela
    Jul 8, 2022 at 12:38
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    $\begingroup$ I think it's more correct to say that it's so small that $dP\ll1$. $\endgroup$
    – pela
    Jul 8, 2022 at 13:49
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    $\begingroup$ @pela Ok, I got it. $\endgroup$
    – Wang Yun
    Jul 8, 2022 at 14:00

2 Answers 2

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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).

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  • $\begingroup$ Thanks a lot for clarifying. BTW, what do you think of my answer to this question? $\endgroup$
    – Wang Yun
    Oct 13, 2022 at 2:53
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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.

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