Closest mass to the line of sight

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In a cosmological model with condensed objects of mass m occupying a fraction Ω of the critical density ρc, the average distance to the closest mass along the line of sight to a QSRS is derived. The expectation value for this distance, denoted as l, is shown to be approximately (m H0/Ω ρc)^(1/2). This relationship arises from geometric considerations involving the expected number of objects within a defined volume, specifically using a cone or cylindrical approach to calculate density. Setting the expected number of objects N to one allows for solving the angle φ, which is then related to the distance l. The discussion emphasizes the geometric nature of this result in cosmological contexts.
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Suppose we have a cosmological model where condensed objects of characteristic mass ##m## occupies some fraction ##\Omega## of the critical density ##\rho_c##. Take a QSRS to be at cosmological distance ##L \sim 1/H_0## (##c=1##). Its easy to the that the average distance between the condensed masses will now be given by ##\Omega \rho_c /m##.

QUESTION: why will the expectation value ##\bar l## for the distance of the closest mass to the line of sight (between us at the QSRS) be
$$ l \sim (m H_0/\Omega \rho_c)^{1/2}?$$

I've read this claim in "Method for detecting a cosmological density of condensed objects" (1973) by Press and Gunn, and would very much like to see an argument on why this is true.
 
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I get this as geometric result. Let ##\phi## be the angle between the line of sight and directions in the sky. What is the expected number of objects N in a cone from ##\phi=0## to ##\phi=\phi_0##? It is given by the volume and the density of objects: ##N \propto L \frac{\Omega\rho_c}{m} \phi^2##. Set N to 1, solve for ##\phi## (and relate it to l somehow), ignore constant prefactors and you get the posted relationship.
 
mfb said:
I get this as geometric result. Let ##\phi## be the angle between the line of sight and directions in the sky. What is the expected number of objects N in a cone from ##\phi=0## to ##\phi=\phi_0##? It is given by the volume and the density of objects: ##N \propto L \frac{\Omega\rho_c}{m} \phi^2##. Set N to 1, solve for ##\phi## (and relate it to l somehow), ignore constant prefactors and you get the posted relationship.
Thanks for the tip! Instead of using a cone, I used a cylinder with radius ##l## and denoted ##\bar l## by the cylinder for which there on average would be just 1 object within its volume coaxial with the line of sight!
 
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