Our Universe a Supercomputer Created

We are bounded in a nutshell of Infinite space: Blog Post #37, Illustris Simulation Worksheet: Our Universe a Supercomputer Created

An incredible feat of astronomical research and data use, the Illustris Simulation attempts to demonstrate how the observable universe has developed over billions of years in order to create the superclusters and incredible webs of gas and dark matter we can detect, in some ways. With the simulation, we can observe how galaxies and other immense structures have halos, extensions of the individual galaxies and clusters made of dark matter which wrap around the objects we ordinarily associate with matter.

By going to the a Illustris Simulation website (http://www.illustris-project.org/explorer/) and going to its “The Explorer” tab, one could see how the universe looks like today, according to the simulation, and see how dark matter, gas density, gas velocity, temperature, X-ray emissions, all of it, interact with one another and are present in the same areas.  Furthermore, by selecting on the tab’s “Spatial Query on Click” option, we can identify specific details about a set of halos. From this data, we will focus on the halo and star mass data. Seeing a table similar to the one below, we can export this data (essentially copying and pasting it out) and get something like this:

Exporting this data as a CSV file (save as CSV) and importing into Python software, we can start to analyze it.

First off, we prepare the environment for the program:

import numpy as np

import matplotlib

import matplotlib.pyplot as plt

Then we upload the specific data:

file = np.loadtxt(fname='data.txt')

And name the specific columns (we eliminated them before exporting the data, since Python doesn’t recognize text as data):

id, xpos, ypos, halo_mass, star_mass, umag, bvcolor= file.T

We then define our data, what we are going to use, exactly, and prepare a histogram to understand the scatter of the halo masses:

halo_mass_data = halo_mass

log_mass_bins = np.arange(9, 12,0.1)

plt.xlabel("Log Halo Mass")

plt.ylabel("Number")

plt.hist(halo_mass_data, bins = log_mass_bins)

plt.show()

Now with these, we produce a table which looks like:

Thus indicating how there is a greater number of lower mass halos than there are high mass halos.

Using some simple division to see what percentage of the halo mass was stellar mass, and then averaging all these values to understand the standard, about 85% of the halo mass is stellar mass.

If we were to qualitatively analyze how Dark Matter Density and Gas Density compare to one another, we can see how, at a large scale, the densest dark matter regions correspond directly with the densest gas regions, they follow the same filament structures and appear to have the all the clusters of gas and dark matter in the same regions.

Gas Density

Dark Matter Density 

However, once the images are zoomed in towards the smaller scales, with individual galactic clusters, the similarities give way to stark contrasts. The most evident one is how the gas density image illustrates how gas is most dense around and in galaxies, as one would expect. But for dark matter, the presence of a galaxy and the presence of the greatest density of dark matter does not correspond precisely, as you can see from the images of the same area below:

 

Dark Matter Density Close Up

Gas Density Close Up 

Furthermore, by observing the large scale structures, it is evident dark matter is more closely confined to the filamentary structure, not the gas, which is clearly more spread out across the universe. We can reason this out by understanding how the mass of dark matter is what is holding galaxies and bigger structures together. These filaments are the pathways which regular matter follows, so it is attracted to it and thus coalesces around it, but is in the process of doing so, as shown of how all matter has slowly organized itself into filaments set by the dark matter. If you analyze specific galaxies, you will also find how the gas density of individual galaxies is highest near the nucleus, the area “near” the black hole at the center of the galaxy. Also, the largest galaxies are rarely found on their own in the less dense areas, rather they are clearly in the presence of many other galaxies and have developed a cluster around itself.   

Single Galaxy as seen in Gas Density Filter

Range of Galaxies in a Cluster, in the Visible Light Filter

Thereafter, if we look at this video (http://www.illustris-project.org/movies/illustris_movie_cube_sub_frame.mp4 ) and analyze the proceedings of the dark matter and gas temperature; we see how their evolution seems to coincide in many ways, the dark matter giving the structure of how the diffuse gas would gain its “shape” from the filaments first set by the dark matter. So, we can understand that the structure formation is led by the dark matter, it itself become more defined as time goes on, but it was still the first to have a basic structure that the baryons (gas) followed and thus created the structure we see at the end.

Also, if we were to read the time and redshift data, we can see there is a range of time when the gas structure begins to “brighten”, which in this diagram means become energized. We know the most common element in the universe is hydrogen, so we also know that it requires a minimum amount of energy to ionize and move its electrons into higher orbits. From this understanding of a bit of atomic physics and quantum mechanics, we can recognize ionized hydrogen by its wavelength (and shift caused by redshift). Furthermore, the fact we can see the hydrogen is also an important fact, for it signaled the end of the “Dark Ages”, the time between the big bang and the first light emitted from ionized hydrogen, and the beginning “Epoch of Reionization”. This, according to the video, occurs at approximately 0.5 billion years after the Big Bang and at redshift 9.5-9.8. Also, if we focus on how quickly stellar mas is forming (how many stars are beginning to ignite) we see how there is a definite range in which stellar mass begins to develop very rapidly. Here, in the period between 4.5 and 7 billion years, the stellar mass increases by approximately 27 billion solar masses, the most sustained rapid growth seen in the simulation, although there were some “incredibly quick” periods every few billion years afterwards that also had this rapid increment.

Another aspect of the simulation to consider is how, near the beginning, small structures were coming together to form the largest structures, following the filament structure laid out by dark matter .But these large structures eventually star breaking up and creating a more diffuse structure, although it is more highly energized than it was before the explosions. This high energization allows smaller structures to form and combine again and lead to more explosion, a cycle of destruction and creation. This pattern is likely caused by the force of gravity attracting the large masses together, tugging them along towards the filaments of dark matter, which we know has to emit a rather large gravitational force. This is the reason the structures form along filaments, the gravity that binds the dark matter together in the way that it has maintained itself for over 13 billion years continues to control how the baryons reorganize themselves. The filaments are the basic gravitational structure of the universe, and normal matter adheres to it by the forces that act on it over the span of billions of years.

Images, videos, and data taken from:

http://www.illustris-project.org/explorer/

http://www.illustris-project.org/movies/illustris_movie_cube_sub_frame.mp4

https://www.youtube.com/watch?v=NjSFR40SY58  

Just how stretched out can the universe get?

We are bounded in a nutshell of Infinite space: Blog Post #36, Worksheet # 12.1, Problem #1 & #2d: Just how stretched out can the universe get?  

1. Linear perturbation theory. In this and the next exercise we study how small fluctuations in the initial condition of the universe evolve with time, using some basic fluid dynamics. In the early universe, the matter/radiation distribution of the universe is very homogeneous and isotropic. At any given time, let us denote the average density of the universe as \(\bar{\rho}(t)\). Nonetheless, there are some tiny fluctuations and not everywhere exactly the same. So let us define the density at comoving position r and time t as \(\rho (x,t)\) and the relative density contrast as \[\delta(r,t) = \frac{\rho(r,t) - \bar{\rho}(t)}{ \bar{\rho}(t)}\]. In this exercise we focus on the linear theory, namely, the density contrast in the problem remains small enough so we only need consider terms linear in \(\delta\). We assume that cold dark matter, which behaves like dust (that is, it is pressureless) dominates the content of the universe at the early epoch. The absence of pressure simplifies the fluid dynamics equations used to characterize the problem.

(a) In the linear theory, it turns out that the fluid equations simplify such that the density contrast \(\delta\) satisfies the following second-order differential equation \[\frac{d^2\delta}{dt^2} + \frac{2\dot{a}}{a} \frac{d\delta}{dt} = 4\pi G \bar{\rho}\delta,\] where \(a(t)\) is the scale factor of the universe. Notice that remarkably in the linear theory this equation does not contain spatial derivatives. Show that this means that the spatial shape of the density fluctuations is frozen in comoving coordinates, only their amplitude changes. Namely this means that we can factorize \[\delta(x,t) = D(t)\tilde{\delta}(x)\] , where \(\tilde{\delta}(x)\) is arbitrary and independent of time, and \(D(t)\) is a function of time and valid for all x. \(D(t)\)  is not arbitrary and must satisfy a differential equation. Derive this differential equation.

(b) Now let us consider a matter dominated flat universe, so that \(\bar{\rho}(t) = a^{-3} \rho_{c,0}\) where \(\rho_{c,0}\) is the critical density today, \(3H_0 ^2/8\pi G\) as in Worksheet 11.1 (aside: such a universe sometimes is called the Einstein-de Sitter model). Recall that the behavior of the scale factor of this universe can be written \(a(t) = (3H_0 t /2)^{2/3}\) , which you learned in previous worksheets, and solve the differential equation for \(D(t)\). Hint: you can use the ansatz \(D(t) \propto t^q\)  and plug it into the equation that you derived above; and you will end up with a quadratic equation for q. There are two solutions for q, and the general solution for D is a linear combination of two components: One gives you a growing function in t, denoting it as \(D_+ (t)\); another decreasing function in t, denoting it as \(D_- (t)\).

(c) Explain why the \(D)_+\) component is generically the dominant one in structure formation, and show that in the Einstein-de Sitter model, \(D_+ (t) \propto a(t)\).

(a) The key for this problem is solving to eliminate the x, which would make the differential equation true for all space and would change for a specific time. Thus, starting off with the original equation \[\frac{d^2\delta}{dt^2} + \frac{2\dot{a}}{a} \frac{d\delta}{dt} = 4\pi G \bar{\rho}\delta,\] and the equivalency of the space-time factor with its specific parts separated into time and space: \[\delta(x,t) = D(t)\tilde{\delta}(x)\] we just have to plug into the first equation and take the time derivative to solve: \[ \ddot{D}(t)\tilde{\delta}(x) + \frac{2\dot{a}}{a} \dot{D}(t)\tilde{\delta}(x)= 4\pi G \bar{\rho} D(t)\tilde{\delta}(x),\] and since we can show with this that the space dimension is not affected, the common factor on both sides can be eliminated: \[ \ddot{D}(t) + \frac{2\dot{a}}{a} \dot{D}(t) = 4\pi G \bar{\rho} D(t),\] and we thus have a differential equation valid for all x.

(b) First off, we need to establish the correct expressions for the scale factor and the critical density: \[ a(t) = \left(\frac{3H_0 t}{2}\right)^{2/3},\] and if we were to take the first derivative: \[ \dot{a}=\frac{2}{3} \left(\frac{3H_0 t}{2}\right)^{-1/3} \frac{3h_0}{2} \] \[\dot{a} =H_0 \left(\frac{3H_0 t}{2}\right)^{-1/3} ,\] and then divided by the original definition of the scale factor: \[\frac{\dot{a}}{a}= \frac{ H_0 \left(\frac{3H_0 t}{2}\right)^{-1/3}}{\left(\frac{3H_0 t}{2}\right)^{2/3}}\] \[\frac{\dot{a}}{a} = \frac{H_0}{\frac{3H_0 t}{2}}, \] we now have: \[\frac{\dot{a}}{a} = \frac{2}{3t}.\]

As for the critical density: \[\bar{\rho}(t) = a^{-3} \rho_{c,0},\] so the value of a can simply be put into the density expression: \[\bar{\rho}(t) = \left[\left(\frac{3H_0 t}{2}\right)^{2/3}\right]^{-3} \rho_{c,0}\] \[ \rho_{c,0} = 3H_0 ^2/8\pi G\] \[\bar{\rho}(t) = \frac{4}{9 H_0 ^2 t^2} \frac{3H_0 ^2 }{8\pi G} \]  \[\bar{\rho}(t) = \frac{1}{6\pi G t^2}.\]

Taking the differential equation we solved for all space, we can place the values we have just found like: \[ \ddot{D}(t) + \frac{2\dot{a}}{a} \dot{D}(t) = 4\pi G \bar{\rho} D(t),\] \[ \ddot{D}(t) + \frac{2\dot{a}}{a} \dot{D}(t) = 4\pi G D(t) \frac{1}{6\pi G t^2},\] \[ \ddot{D}(t) + 2\left(\frac{2}{3t}\right) \dot{D}(t) = \frac{2 D(t)}{3 t^2},\] \[ \ddot{D}(t) + \frac{4}{3t} \dot{D}(t) - \frac{2}{3t^2} D(t) = 0 .\] And now that the equation has been simplified as much as it can be, we add in some squiggle math, knowing an important relationship between density with respect to time and time to a variable q:\[D(t) \propto t^q\] and taking the necessary derivatives, we have: \[ q(q-1) t^{q-2} + \frac{4}{3t} q t^{q-1}- \frac{2}{3t^2} t^q  \sim 0 .\] \[ q(q-1) t^{q-2} + \frac{4}{3t} q t^{q-1}- \frac{2}{3t^2} t^q  \sim 0 .\] \[ (q^2 - q) t^{q-2} + \frac{4}{3} q t^{q-2}- \frac{2}{3} t^{q-2}  \sim 0 .\] So after simplifying the equation a bit, we can take out a couple of common factor and be left with a solvable polynomial: \[ t^{q-2} [(q^2 - q) + \frac{4}{3} q - \frac{2}{3t}] \sim 0 .\] \[ \frac{t^{q-2}}{3} [ 3q^2  + q - 2 \sim 0 ,\] and now we can use the quadratic formula to solve for q \[ q = \frac{-1 + \sqrt{1-4(3)(-2)}}{2(3)}\] \[ q = \frac{2}{3} ~,~ -1\] with these two values for q, they correspond to the increasing and decreasing functions the problem talks about, as: \[D_+ (t) \propto t^{2/3} ~,~ D_- (t) \propto \frac{1}{t} ,\] which means that a combination of these two expressions describes the entirety of the density with respect to time function: \[D(t) \approx D_+ (t) + D_- (t) .\]

(c) Because of the nature of \( D_+ (t) \) and its value we now know, it is clearly the dominant factor in establishing the development of density I the universe over time, for its growth is much more sustained and clear than that of \( D_- (t) \), which actually tends towards 0. However, the \(D_- (t)\) part of the function once was the dominant figure, at very low/small t, which makes it an important factor to consider. Furthermore, the \(D_+ (t)\) model closely resembles (it is actually identical to) the description of the expansion of the universe in a matter dominated universe, as we saw in a previous post (http://ay17-rcordova.blogspot.com/2015/11/cosmology-101-part-2.html), which indicates how a flat universe expands is identical to the scale factor for a matter dominated universe. 

2. Spherical collapse. Gravitational instability makes initial small density contrasts grow in time. When the density perturbation grows large enough, the linear theory, such as the one presented in the above exercise, breaks down. Generically speaking, non-linear and non-perturbative evolution of the density contrast have to be dealt with in numerical calculations. We will look at some amazingly numerical results later in this worksheet. However, in some very special situations, analytical treatment is possible and provide some insights to some important natures of gravitational collapse. In this exercise we study such an example.

(d) Plot r as a function of t for all three cases (i.e. use y-axis for r and x-axis for t), and show that in the closed case, the particle turns around and collapse; in the open case, the particle keeps expanding with some asymptotically positive velocity; and in the flat case, the particle reaches an infinite radius but with a velocity that approaches zero.

(d)Taking key facts from the other parts of this problem, we know that for a closed universe, the equations for the distance from the origin r and the time that defines this are: \[r =A(1-\cos\eta),\] \[t = B(\eta - \sin\eta), ~~(0 \leq \eta \leq 2\pi),\] whereas for an open universe, the equations are: \[r = A(\cosh\eta - 1),\] \[t = B(\sinh\eta - \eta), ~~(0\leq \eta \leq \infty). \] Finally, for a flat universe, the equations become: \[ r = A\eta^2 / 2,\] \[t = B\eta^3 / 6, ~~ (0\leq \eta \leq \infty),\] where one can be expressed in terms of the other as: \[t = B\eta^3 / 6\] \[\eta = \frac{\sqrt[3]{6t}}{B}\] \[ r = A\eta^2 / 2\] \[ r = \frac{A \left(\frac{\sqrt[3]{6t}}{B}\right)^2}{ 2},\] \[ r = \frac{A\left(\frac{6t}{B}\right)^{2/3}}{2}.\] Knowing these values, limits and expressions, we can plot these into Python and create a good model to describe how particles act in these descriptions of the universe.

First, we just have to create the environment for the program:

import numpy as np

import matplotlib

import matplotlib.pyplot as plt

Next, we start describing each of the r and t and the \(\eta\) limits, for each system, while setting the 

constants to 1 since these are only scale factor and do not directly influence the overall shape the graph takes:

#closed

n1=np.arange(0,2.*np.pi, 0.1)

r1=1.*(1.-np.cos(n1))

t1=1.*(n1-np.sin(n1))

plt.plot(t1,r1,label="Closed")

We do this process again with the open universe, first describing the limits (scaled here so they all fit in one graph at the end) and then the equations with the constants once again set to 1.

#open

n2=np.arange(0,2.7,0.1)

r2=1.*(np.cosh(n2)-1.)

t2=1.*(np.sinh(n2)-n2)

plt.plot(t2,r2,label="Open")

And here, for the flat case, we place the solved equation for r in terms of t we derived earlier and give t the limits as we did in the open universe description.

#flat

t3=np.arange(0,2.*np.pi,0.1)

r3= (((6.*t3)**(2./3.))/2.)

plt.plot(t3,r3,label="Flat")

Now with the final programs to define the plot with labels and legends:

plt.legend(loc=2)

plt.xlabel('Scaled Time')

plt.ylabel('Scaled Growth of the Radius of the Universe')

plt.show()

We have:

Full code used:

import numpy as np

import matplotlib

import matplotlib.pyplot as plt

#closed

n1=np.arange(0,2.*np.pi, 0.1)

r1=1.*(1.-np.cos(n1))

t1=1.*(n1-np.sin(n1))

plt.plot(t1,r1,label="Closed")

#open

n2=np.arange(0,2.7,0.1)

r2=1.*(np.cosh(n2)-1.)

t2=1.*(np.sinh(n2)-n2)

plt.plot(t2,r2,label="Open")

#flat

t3=np.arange(0,2.*np.pi,0.1)

r3= (((6.*t3)**(2./3.))/2.)

plt.plot(t3,r3,label="Flat")

plt.legend(loc=2)

plt.xlabel('Scaled Time')

plt.ylabel('Scaled Growth of the Radius of the Universe')

plt.show()