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# PhD question #4: calculate the value of M*

Some post ago I’ve written about M*, the typical non-linear mass collapsing at the redshift we are considering. Now I have to find a value for it.

I said that $M^*$ is the typical mass of a perturbation that, at the time we are looking, has the associated liner density contrast $\delta(\mathbf{x})\sim1$, or, in the formalism of the excursion set, pass the barrier of &nbsp;$\delta_c=1.686$. This means that we are looking for a perturbation with$\sigma\simeq1.686$ and trying to quantify the mass it contains.

## $\sigma$ and R

First of all we need to find the radius of a perturbation whose $\sigma$ reached the value of 1.686. To do this we can use the code developed to manage the CAMB files in order to find the matter power spectrum and its normalization. Then we add few lines to “sample” the $\sigma(R)$ distribution and find the radius of the perturbation reaching the excursion set barrier for the collapse.

## $M^*$

Once we have the radius for which $\sigma = \delta_c$ we need to know the mean density in the universe to find $M^*$ with:

$M^* = \frac{4}{3}\pi R_*^3\rho_m$

I don’t know why we only need to use $\rho_m$ and not $\rho_m\delta$ or something similar is not clear to me, but it’s correct.We use this formula because $M^*$ is a quantity related to the linear perturbations. It’s correct because the difference between a linear and a non-linear perturbation is the value of the density contrast, but the mass is the same. In other words, the mass of a perturbation is the same both in the linear and in the non-linear evolution, but linear perturbations have smaller density contrasts and larger radii, non-linear perturbations instead have larger density contrasts and smaller radii. To be precise, the previous equation can be written:
$M^* = \frac{4}{3}\pi R*^3\rho{bg}(1+1.686) = \frac{4}{3}\pi R{vir}^3\rho{bg}(1+200)$.
To obtain $\rho_m$ we find the value of the critical density $\rho_c$ and multiply it for $\Omega = \rho_m / \rho_c$. These two values can be obtained from books (Lucchin, Mo&White for example) or in the WMAP data page. In the second case we prefer to use the single data fit because it’s simpler to refer to it.

Here’s the code:

#!/use/bin/env python
import time
import numpy as np
import matplotlib.pyplot as plt
import random as rnd
from scipy import integrate

""" Calculate M* . M* is propto the mass contained in the radius for which
s_8=delta_c refer to:
http://www.brunettoziosi.eu/blog/wordpress/the-initial-conditions-saga/
http://www.brunettoziosi.eu/blog/wordpress/phd-question-3-calculate-the-value-of-m/
"""

t = time.time()

#===============================================================================
#    Compute sigma
#===============================================================================

### Load data from the nasa-CAMB file

# matterpower is the file with the k and the total matter spectrum
# transfer is the file with the k and the transfer function for the various
#species, the 6th column is for the total matter (baryons+DM)
transfer = np.genfromtxt('camb_88704620_transfer_out_z0.dat', usecols = (0,6))
#matterpower = np.genfromtxt('2012-01-30_data/camb_88704620_matterpower_z0.dat')

# CAMB CDM transfer output
camb_k = transfer[:,0]
camb_tf = transfer[:,1]

R = 8 #Mpc/h
s_8 = 0.9#0.8118405 #from WMAP7 but we need the values for the Millennium-2, so
# we use its s_8
sp_ind = 1
delta_c = 1.686

### Calculate the amplitude to normalize the spectrum:
### P(k) = Ak^nT^2(k)

# FT of the window function (spherical top-hat)
def FTW(R, k):
""" Return the Fourier transform of the window function
(spherical top-hat)
"""
return 3.*(np.sin(k*R)-k*R*np.cos(k*R)) / (k*R)**3

def spectrum():
"""Calculate the power spectrum given the transfer function and the FT of the window
function.
"""
# camb_k**(2+sp_ind) that is k^(2+n) because d^3k=4pi k^2dk
amp_integrand = camb_k**(2+sp_ind)*camb_tf**2 * FTW(R, camb_k)**2
amp_integral = integrate.trapz(amp_integrand, camb_k)
# Amplitude for s_8 = 1
amp_0 = 2*np.pi**2/amp_integral
# Amplitude
amp = amp_0*s_8**2  # 9.9197881817e-09
#print amp
# Calculate the power spectrum
return camb_k**sp_ind*camb_tf**2 * amp

# Calculate the power spectrum
ps = spectrum()

# Calculate sigma on the radii
def sigma(R):
"""Return the sigma for the current radius.
"""
return pow(integrate.trapz(camb_k**2 * ps * FTW(R, camb_k)**2, camb_k)/(2*np.pi**2), 0.5)

#===============================================================================
#    Find the radius containing M*
#===============================================================================

# Initialize some variables
neigh = np.ones(2) # two nearest neighbours sigmas
r_min = 10**(-2) # min r to sample
r_max = 10**2 # max r ti sample
i = 0 # loop counter

# While stops when the computed sigma is less then 0.001 from delta_c
while np.abs(np.amin(delta_c+neigh)) &gt; 0.001:
print "Loop ", i
i+=1
print "Condition start ", np.abs(np.amin(delta_c+neigh))
r = np.linspace(r_min, r_max, num=100)
# Compute sigma for those radii, the minus sign is to avoid resorting of the
# array to be used by np.searchsorted
s_r = -np.asarray(map(sigma, r))
# Find the two nearest neighbours
neigh[0] = np.amax(s_r[s_r  -delta_c])
r_min = r[np.searchsorted(s_r, neigh).min()]
r_max = r[np.searchsorted(s_r, neigh).max()]
print "Sigmas", -neigh[0], -neigh[1]
print "Radii [Mpc/h] ", r_min, r_max
print "Condition end ", np.abs(np.amin(delta_c+neigh))

# Selected values
s_star = neigh[np.argmin(delta_c+neigh)]
r_star = r[np.searchsorted(s_r, s_star)]
deviation = np.abs(np.amin(delta_c+neigh))

print "############################################"

print "Selected sigma ", -s_star
print "Selected radius [Mpc/h] ", r_star

print "Calculate M* using:"
print "Gt4.299 x 10^(-9) Mpc /M_sun (km/s)^2tfrom Mo&amp;White"
print "Ht100*h^2"
print "Omega_mt0.25tfrom the Millennium-2 simulation"
print "Omega_mt0.241tfrom WMAP7"

#===============================================================================
#    Cosmological parameters and find the mean density in the Universe
#===============================================================================

H = 100
h = 0.732 #WMAP http://lambda.gsfc.nasa.gov/product/map/dr2/params/lcdm_wmap.cfm
G = 4.299*10**(-9)
omega_m_mill = 0.25
omega_m_WMAP = 0.241

# Until here it's correct
rho_c = 3*H**2/(8*np.pi*G) # 2.7766040316101764 * h**2 x 10^11 M_sun/Mpc^3
# 2.778 from Lucchin book
# 2.775 from Mo&amp;White book

rho_mean_mill = rho_c * omega_m_mill # 6.94151007903 * h**2 x 10^10 M_sun/Mpc^3 = 3.71942769658 x 10^10 M_sun/Mpc^3
rho_mean_WMAP = rho_c * omega_m_WMAP # 6.69161571618 * h**2 x 10^10 M_sun/Mpc^3 = 3.58552829951 x 10^10 M_sun/Mpc^3

print "rho_c = 3H^2/8 pi G = ", rho_c," h^2 M_sun/Mpc^3"
print "Millennium-2 rho_mean = omega_m_mill * rho_c = ", rho_mean_mill, " h^2 M_sun/Mpc^3 = ", rho_mean_mill*h**2
print "WMAP7 rho_mean = omega_m_WMAP * rho_c = ", rho_mean_WMAP, " h^2 M_sun/Mpc^3 = ", rho_mean_WMAP*h**2

#===============================================================================
#    Compute M*
#===============================================================================

M_star_mill = np.pi * r_star**3 * rho_mean_mill * 4./3#* (delta_c + 1)
M_star_WMAP = np.pi * r_star**3 * rho_mean_WMAP * 4./3#* (delta_c + 1)

print "M* Millennium-2 ", M_star_mill, " M_sun/h" # 4.81467115575e+12 M_sun/h
print "M* WMAP ", M_star_WMAP, " M_sun/h" # 4.64134299414e+12 M_sun/h

#===============================================================================
#    Compare with Hayashi&amp;White 2008 article
#===============================================================================

print "Hayashi&amp;White's value:"

M_star_white = 6.15*10**(12)
omega_m_white = 3.*M_star_white/(4*np.pi*r_star**3*rho_c)

print "M*: ", M_star_white
print "Omega_m", omega_m_white

r_s = pow(3.*M_star_white/(4*np.pi*rho_mean_mill), 1./3)
s_white = sigma(r_s)

print "As alternative:"
print "R* ", r_s
print "Sigma ", s_white

#===============================================================================
#    Summary
#===============================================================================

print ""
print "##############################################################################################################"
print "SUMMARY"
print "##############################################################################################################"
print ""
print "WhottttM*ttR*ttOmegattSigmattSigma-delta_c"
print "--------------------------------------------------------------------------------------------------------------"
print "Hayashi&amp;White given Rtt{:e}t{:e}t{:e}t{:e}t{:e}".format(M_star_white,r_star,omega_m_white,-s_star,deviation)
print "Hayashi&amp;White given Omegat{:e}t{:e}t{:e}t{:e}t{:e}".format(M_star_white,r_s,omega_m_mill,s_white,np.abs(s_white-delta_c))
print 'Me WMAP datattt{:e}t{:e}t{:e}t{:e}t{:e}'.format(M_star_WMAP,r_star,omega_m_WMAP,-s_star,deviation)
print "Me Millennium-2 datatt{:e}t{:e}t{:e}t{:e}t{:e}".format(M_star_mill,r_star,omega_m_mill,-s_star,deviation)
print ""
print "##############################################################################################################"
print "##############################################################################################################"
print ""
print "Done in ", time.time()-t


The differences between the different values of $M^*$ are acceptable and probably depends on different integration boundaries for $\sigma$.