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I am trying to find the expression for the rate of change of comoving volume with respect to redshift, that is $\frac{\mathrm{d}V_c}{\mathrm{d}z}$. In this paper (Hogg, David W.), the comoving volume differential is given as follows:

$$\mathrm{d}V_c=D_\mathrm{H}\frac{(1+z)^2D_\mathrm{A}^2}{E(z)}\mathrm{d}\Omega\mathrm{d}z$$

$$\implies\frac{\mathrm{d}V_c}{\mathrm{d}z}=D_\mathrm{H}\frac{(1+z)^2D_\mathrm{A}^2}{E(z)}\mathrm{d}\Omega$$

Where $D_\mathrm{H}$ is the Hubble distance, $D_\mathrm{A}$ is the angular diameter distance. These expressions can be expressed as follows:

$$D_\mathrm{H}=\frac{c}{H_0}$$

Where $c$ is the speed of light in vacuum, and $H_0$ is the Hubble's constant.

$$D_\mathrm{A}=\frac{D_\mathrm{M}}{1+z}$$

Where $D_\mathrm{M}$ is the transverse comoving distance given by:

$$D_\mathrm{M}=D_\mathrm{C},\textrm{ }\Omega_k=0$$

Where $\Omega_k=0$ implies spatial curvature 0, and $D_\mathrm{C}$ is the comoving distance given by:

$$D_\mathrm{C}=D_\mathrm{H}\int_0^z\frac{\,\mathrm{d}z'}{E(z')}$$

Now, this derivation can be substituted in $\frac{\mathrm{d}V_c}{\mathrm{d}z}=4\pi D_\mathrm{H}\frac{(1+z)^2D_\mathrm{A}^2}{E(z)}$, where $4\pi$ substitutes the solid angle element $\mathrm{d}\Omega$. Now, I want to plot this result as a function of redshift, for which I have written the following code in Python:

import numpy as np
import matplotlib.pyplot as plt
from astropy.cosmology import FlatLambdaCDM

cosmopar = FlatLambdaCDM(H0 = 67.8, Om0 = 0.3) #Defining the Cosmological model

speed_light = 3*10**5 #km/s
D_H = (speed_light/cosmopar.H(0).value)/10**3 #Gpc

Omega_m = 0.3
Omega_k = 0
Omega_Lambda = 1-Omega_m

def E(z_m): #Redshift z_m
    return math.sqrt((1+z_m)^2*Omega_m+(1+z_m)^2*Omega_k+Omega_Lambda)

def Comov_vol(z_m): #in Gpc^3
    return 4*math.pi*D_H*(((1+z_m)^2*(cosmopar.angular_diameter_distance(z_m).value/10**3)^2)/(E(z_m)))

#Plotting dVc_dz as a function of z:

ztrial = np.linspace(0,10,100) #Creating an array of redshift
comovol_arr = np.array([Comov_vol(v) for v in ztrial]) #Creating an array for corresponding comoving volume per redshift

plt.xlabel(r'Redshift')
plt.ylabel(r'$\frac{\mathrm{d}V_c}{\mathrm{d}z}$')

plt.plot(ztrial, comovol_arr)

This gives me the following plot:

My plot for comoving volume per unit redshift vs redshift

Now, as one might notice, the magnitude of $\frac{\mathrm{d}V_c}{\mathrm{d}z}$ (in Gpc$^3$) quite large as compared to literature (Karathanasis, Christos, et. al.). I am referring to this paper for a project I am working on and the comoving volume per unit redshift vs redshift graph provided in this paper is as follows:

enter image description here

This graph is again given in Gpc$^3$, and the shape of both my plot and this literature plot is the same; however, the magnitude of $\frac{\mathrm{d}V_c}{\mathrm{d}z}$ seems to differ vastly. Can someone please help me understand where I am going wrong?

For context: I am new to both Astronomy and Coding. As an undergraduate Physics student, I am working on a project as a part of my summer internship and the paper which I refer to for the literature plot is the one required by my mentor, thank you.

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In the paper, the plot is showing volume distribution to understand the distribution of their selected sample. The plot is probably normalized to have unit area (integral under the line is 1). Notice that the mean is at about .2 and it goes from z = 0 to 5, so .2*5 = 1.

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    $\begingroup$ The plot in question comes in the middle of 5 pages of appendices all devoted to calculating Bayesian probabilities based on a set of expected distributions. Only a normalized distribution makes any sense here. Plus, the question itself goes into complete detail about how to calculate it in units of Gpc$^3$ per z. If you want me to move it to a comment, I can. But, it answers the question, and it would be a waste of time for others to work on it. $\endgroup$
    – eshaya
    Commented Jul 29, 2023 at 23:01
  • $\begingroup$ ...and... sold! $\endgroup$
    – uhoh
    Commented Jul 29, 2023 at 23:46

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