<p>That the probability of galaxy occupation is <span class="math-container">$\bar{n}\ dV$</span> will work as an approximation so long as you are considering volumes that are limited in size such that the probability is <span class="math-container">$\ll 1$</span>. i.e. You are ignoring any probability that there could be more than 1 galaxy in the volume.</p>
<p>If you have <span class="math-container">$N$</span> particles in box of volume <span class="math-container">$V$</span>, then <span class="math-container">$\bar{n} = N/V$</span>.
The probability of finding a particular particle in a subset of the volume <span class="math-container">$dV$</span> will be <span class="math-container">$dV/V$</span>. But if there are <span class="math-container">$N$</span> particles then the probability that <em>any</em> of them are in <span class="math-container">$dV$</span> will be <span class="math-container">$N\ dV/V = \bar{n}\ dV$</span>.</p>
<p>But this probability will include cases where there are 1, 2, 3 or more particles in the box.</p>
<p>The probability of getting 2 particles in <span class="math-container">$dV$</span> will be <span class="math-container">$N\ (dV/V) \times (N-1)\ (dV/V)$</span>. But if <span class="math-container">$N\ (dV/V) \ll 1$</span>, then the probability of finding a second particle (or more) in the box is <span class="math-container">$\lll 1$</span> and could be ignored.</p>
<p>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).</p>

CC BY-SA 4.0
2022-10-13T02:53:31.443Z
Thanks a lot for clarifying. BTW, what do you think of my answer to this question?
Wang Yun
CC BY-SA 4.0
2022-10-12T15:28:33.900Z
2022-10-12T15:28:33.900Z
ProfRob
<p><strong>Here are my thoughts on this issue and comments are welcome.</strong></p>
<p>If there are <span class="math-container">$M$</span> ensembles of volume elements, i.e. <span class="math-container">$dV$</span>, and there are <span class="math-container">$N$</span> volume elements containing particles, then the probability of finding an object in an infinitesimal volume is
<span class="math-container">$$
dP = N/M.
$$</span></p>
<p>Also, assuming that <span class="math-container">$n_i$</span> is the particle (or galaxy) number density of the <span class="math-container">$i$</span>-th volume element, then we get
<span class="math-container">$$
\begin{align}
dP &amp;= (\Sigma_{i=1}^{M}(n_i\cdot dV))/M \\
&amp;= (\Sigma_{i=1}^{M}n_i/M)\cdot dV \\
&amp;= \overline{n}dV
\end{align},
$$</span>
where <span class="math-container">$\overline{n}=(\Sigma_{i=1}^{M}n_i/M)$</span> is the mean number density.</p>

CC BY-SA 4.0
2022-10-12T14:57:39.983Z
2022-10-12T14:57:39.983Z
Wang Yun
<p>As pointed out by many cosmology lectures, such as <strong>Eq. (63)</strong> of <a href="https://www.mv.helsinki.fi/home/hkurkisu/Cosm8.pdf" rel="nofollow noreferrer">Cosmology II-8 Structure Formation</a>, and <strong>Eq. (3.1)</strong> of <a href="https://www2.physik.uni-bielefeld.de/fileadmin/user_upload/theory_e6/Master_Theses/MA_MarianBiermann.pdf" rel="nofollow noreferrer">A Detailed Look at Estimators for the Two-Point Correlation Function</a>, the probability of finding an
object in an infinitesimal volume is
<span class="math-container">$$
dP=\overline{n}dV,
$$</span>
where <span class="math-container">$\overline{n}$</span> is the mean number density. In my opinion, <span class="math-container">$\overline{n}dV$</span> is just the number of objects in <span class="math-container">$dV$</span> and can be greater than one. However, the probability should be less or equal to one. <strong>I don't understand why <span class="math-container">$\overline{n}dV$</span> is the probability of finding an object in the volume</strong>.</p>

CC BY-SA 4.0
2022-07-08T05:48:58.257Z
Note cross posting is very strongly discouraged and you have posted this on [Physics SE](https://astronomy.stackexchange.com/questions/49794/why-is-the-probability-dp-of-finding-an-object-e-g-galaxy-in-an-infinitesmal) as well.
StephenG - Help Ukraine
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2022-07-08T06:20:26.517Z
@StephenG-HelpUkraine I have deleted the post on Physics SE.
Wang Yun
CC BY-SA 4.0
2022-07-08T10:30:07.030Z
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.
GrapefruitIsAwesome
CC BY-SA 4.0
2022-07-08T12:38:04.347Z
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
CC BY-SA 4.0
2022-07-08T12:53:30.240Z
@pela Does "so small that there is at most one galaxy in it" means that the maximum of $dP$ is one?
Wang Yun
CC BY-SA 4.0
2022-07-08T13:49:14.503Z
I think it's more correct to say that it's so small that $dP\ll1$.
pela
CC BY-SA 4.0
2022-07-08T14:00:56.147Z
@pela Ok, I got it.
Wang Yun
<galaxy><cosmology><dark-matter>
Why is the probability dP of finding an object (e.g. galaxy) in an infinitesmal volume dV equal to \overline{n}dV?
Wang Yun
# Why is the probability dP of finding an object (e.g. galaxy) in an infinitesmal volume dV equal to \overline{n}dV?

(Asked by: username_0 on 2022-07-08)

As pointed out by many cosmology lectures, such as **Eq. (63)** of [Cosmology II-8 Structure Formation](https://www.mv.helsinki.fi/home/hkurkisu/Cosm8.pdf), and **Eq. (3.1)** of [A Detailed Look at Estimators for the Two-Point Correlation Function](https://www2.physik.uni-bielefeld.de/fileadmin/user_upload/theory_e6/Master_Theses/MA_MarianBiermann.pdf), 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**.

**Tags:** galaxy, cosmology, dark-matter
> **Comments:**
> **username_1**: Note cross posting is very strongly discouraged and you have posted this on [Physics SE](https://astronomy.stackexchange.com/questions/49794/why-is-the-probability-dp-of-finding-an-object-e-g-galaxy-in-an-infinitesmal) as well.  
> **username_0**: @StephenG-HelpUkraine I have deleted the post on Physics SE.  
> **username_2**: 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.  
> **username_3**: 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_."  
> **username_0**: @pela Does "so small that there is at most one galaxy in it" means that the maximum of $dP$ is one?  
> **username_3**: I think it's more correct to say that it's so small that $dP\ll1$.  
> **username_0**: @pela Ok, I got it.  

---

## 2 answers

### Answer by username_4 on 2022-10-12. Score: 2 | Accepted answer

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).

> **Comments:**
> **username_0**: Thanks a lot for clarifying. BTW, what do you think of my answer to this question?  

---

### Answer by username_0 on 2022-10-12. Score: 1

**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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