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()

Basically, matter is the rarest thing in the universe (or at least it used to be)

We are bounded in a nutshell of Infinite space: Blog Post #35, Worksheet # 11.1, Problem #3: Basically, matter is the rarest thing in the universe (or at least it used to be)

3. Baryon-to-photon ratio of our universe.

(a) Despite the fact that the CMB has a very low temperature (that you have calculated above), the number of photons is enormous. Let us estimate what that number is. Each photon has energy $h\nu$. From the Plank Spectrum Equation, figure out the number density, $n_\nu$, of the photon per frequency interval $d\nu$. Integrate over $d\nu$to get an expression for total number density of photon given temperature T. Now you need to keep all factors, and use the fact that $$\int_0^\infty \frac{x^2}{e^x - 1} dx \approx 2.4.$$

(b) Use the following values for the constants: $k_B = 1.38 \times 10^{-16} erg~ K^{-1}, c = 3.00 \times 10^{10} cm ~s^{-1}, h = 6.62 \times 10^{-27} erg~s$ , and use the temperature of CMB today that you have computed from 2d), to calculate the number density of photon today in our universe today (i.e. how many photons per cubic centimeter?)

(c) Let us calculate the average baryon number density today. In general, baryons refer to protons or neutrons. The present-day density (matter + radiation + dark energy) of our Universe is $9.2 \times 10^{-30} g~cm^{-3}$ . The baryon density is about 4% of it. The masses of proton and neutron are very similar $(\approx 1.7 \times  10^{-24} G )$.

What is the number density of baryons?

(d) Divide the above two numbers, you get the baryon-to-photon ratio. As you can see, our universe contains much more photons than baryons (proton and neutron).

(a) Knowing the energy of a single photon, a simple scale value defines all photon’s energy: $$E = h\nu$$ $$E_n = n\cdot h\nu$$ $$n_\nu = \frac{E_{n_\nu}}{h\nu},$$ which can be defined with the equations for energy density as: $$n_\nu = \frac{u_\nu}{h\nu},$$ so $$u_\nu d\nu = \frac{8\pi h_P \nu^3}{c^3} \frac{1}{e^{\frac{h_P \nu}{k_B T}} - 1} d\nu,$$ becomes: $$\frac{u_\nu}{h\nu} d\nu = \frac{1}{h\nu} \cdot \frac{8\pi h_P \nu^3}{c^3} \frac{1}{e^{\frac{h_P \nu}{k_B T}} - 1} d\nu,$$ $$n_\nu d\nu =  \frac{8\pi \nu^2}{c^3} \frac{1}{e^{\frac{h_P \nu}{k_B T}} - 1} d\nu,$$ and integrating it for all frequencies: $$\int_0 ^\infty n_\nu d\nu = \int_0 ^\infty \frac{8\pi \nu^2}{c^3} \frac{1}{e^{\frac{h_P \nu}{k_B T}} - 1} d\nu,$$ $$n_{all \nu} =  \frac{8\pi}{c^3} \left(\frac{k_B T}{h_P}\right)^3 \int_0^\infty \frac{\nu^2}{e^{\frac{h_P \nu}{k_B T}} - 1} d\nu,$$ and knowing a particular equality established by the problem: $$\int_0^\infty \frac{x^2}{e^x - 1} dx \approx 2.4 ,$$ we have the final description of the number of photons as: $$n_{all ~\nu} =  \frac{8\pi}{c^3} \left(\frac{k_B T}{h_P}\right)^3 (2.4)$$

(b) And now plugging in the values given in the problem, we have: $$n_{all~ \nu} =  \frac{8\pi}{c^3} \left(\frac{k_B T}{h_P}\right)^3 (2.4)$$ $$k_B = 1.38 \times 10^{-16} erg~ K^{-1},$$ $$c = 3.00 \times 10^{10} cm ~s^{-1},$$ $$h = 6.62 \times 10^{-27} erg~s,$$ so:  

$$n_{all~ \nu} =  \frac{8\pi}{3.00 \times (10^{10} cm ~s^{-1})^3} \left(\frac{1.38 \times 10^{-16} erg~ K^{-1} \cdot 2.7K}{ 6.62 \times 10^{-27} erg~s }\right)^3 (2.4),$$ and thus the density of photons is:  $$n_{all ~\nu} = 4 \times 10^2 ~photons/cm^3$$

(c)  Being given the density of mass in the universe, we can now solve for baryon density in particular: $$\rho_{universe} = 9.2 \times 10^{-30} g~cm^{-3}$$ $$\rho_{baryons} = 0.04 \times \rho_{universe} ,$$ and knowing the mass of a baryon, we can solve for the number density of baryons: $$\rho_{baryons} = 3.68*10^{-31} g~cm^{-3}$$ $$n_{baryons} = \rho_{baryons} / m_{baryon}$$ $$n_{baryons} =3.68*10^{-31} g~cm^{-3} / 1.7 * 10^{-24} \frac{g}{baryon}$$ $$n_{baryon} = 2.16*10^{-7} \frac{baryons}{cm^3}$$

(d) Simply doing what the problem establishes, we have: $$n_{all \nu} = 4 \times 10^2 ~photons/cm^3$$ $$n_{baryon} = 2.16471*10^{-7} \frac{baryons}{cm^3}$$ $$n_{all ~\nu}/ n_{baryon} = 1.85*10^9 \frac{photons}{baryons},$$ and indeed, there are billions of photons for every baryon. 

Energy that was there 13.68 billion years ago

We are bounded in a nutshell of Infinite space: Blog Post #34, Worksheet # 11.1, Problem #2: Energy that was there 13.68 billion years ago

2. Cosmic microwave background. One of the successful predictions of the Big Bang model is the cosmic microwave background (CMB) existing today. In this exercise let us figure out the spectrum and temperature of the CMB today.

In the Big Bang model, the universe started with a hot radiation-dominated soup in thermal equilibrium. In particular, the spectrum of the electromagnetic radiation (the particle content of the electromagnetic radiation is called photon) satisfies The Plank Spectrum. At about the redshift $z \approx 1100$ when the universe had the temperature $T \approx 3000K$, almost all the electrons and protons in our universe are combined and the universe becomes electromagnetically neutral. So the electromagnetic waves (photons) no longer get absorbed or scattered by the rest of the contents of the universe. They started to propagate freely in the universe until reaching our detectors. Interestingly, even though the photons are no longer in equilibrium with its environment, as we will see, the spectrum still maintains an identical form to the Planck spectrum, albeit characterized by a different temperature.

(a) If the photons was emitted at redshift z with frequency $\nu$, what is its frequency  $\nu^{\prime}$ today?

(b) If a photon at redshift z had the energy density $u_\nu d\nu$, what is its energy density $u_{\nu^\prime}d\nu^\prime$ today? (Hint, consider two effects: 1) the number density of photons is diluted; 2) each photon is redshifted so its energy, $E = hv$, is also redshifted.)

(c) Plug in the relation between $\nu$ and $\nu^\prime$ into the Planck spectrum: $$u_\nu d\nu = \frac{8\pi h_P \nu^3}{c^3} \frac{1}{e^{\frac{h_P \nu}{k_B T}} - 1} d\nu,$$ and also multiply it with the overall energy density dilution factor that you have just figured out above to get the energy density today. Write the final expression as the form $u_{\nu^\prime}d\nu^\prime$. What is $u_{\nu^\prime}$? This is the spectrum we observe today. Show that it is exactly the same Planck spectrum, except that the temperature is now $T^\prime = T(1+z)^{-1}$  .

(d) As you have just derived, according to Big Bang model, we should observe a black body radiation with temperature $T^\prime$ filled in the entire universe. This is the CMB. Using the information given at the beginning of this problem, what is this temperature $T^\prime$ today? (This was indeed observed first in 1964 by American radio astronomers Arno Penzias and Robert Wilson, who were awarded the 1978 Nobel Prize.)

(a) Starting off with the normal redshift equation, it can be turned into a redshift equation that describes the same value with photon frequencies: $$z = \frac{\lambda_{observed} - \lambda_{emitted}}{\lambda_{emitted}}$$ $$\lambda = \frac{c}{\nu}$$ $$z = \frac{\frac{c}{\nu_{observed}} - \frac{c}{\nu_{emitted}}}{\frac{c}{\nu_{emitted}}} ,$$ and simplyfing the equation a bit to find a sole value for the $\nu^\prime$: $$z = \frac{\nu_{emitted}} {\nu_{observed}} - 1$$ $$z + 1= \frac{\nu_{emitted}} {\nu^\prime}$$ $$\nu^\prime= \frac{\nu_{emitted}}{ z + 1},$$ so we now see how the frequency seen today is always larger than the original frequency with which the photon was emitted.

(b)If we know a basic tenant of cosmology, the way the scale factor changes with redshift: $$a =\frac{1}{1+z},$$ we can use it to describe the rest of the proportions which describe how the universe changes over periods of time. Knowing that $$\lambda \propto a(t)$$ from $$z = \frac{\lambda_{observed} - \lambda_{emitted}}{\lambda_{emitted}}$$ $$z + 1= \frac{\lambda_{observed}} {\lambda_{emitted}}$$ and focusing in on $\lambda_{observed}$, we know how   $$\lambda \propto a(t)$$ and how$\lambda = \frac{c}{\nu}$ so the relationship can be described as: $$\frac{1}{\nu} \propto a(t)$$ $$\nu \propto \frac{1}{a(t)}$$ which can be rewritten as: $$\nu^\prime \propto \frac{1}{a^\prime (t)} .$$ Another aspect that the problem describes is how volume is related to the scale factor, which is simply: $$V \sim a(t)^3,$$ and so $$V^\prime \sim a(t)^{\prime~ 3}.$$ With these, we can start doing some squiggle math to correctly describe the relationships of the variables: $$E\sim \nu$$ $$E^\prime\sim \nu^prime,$$ and the energy density of the universe is described as $$\rho_E = \frac{E}{V},$$ so it can also be: $$\rho_E \sim \frac{\nu}{V}$$ $$\rho_E \sim \frac{\frac{1}{a}}{\frac{1}{a^3}}$$ $$\rho_E = \frac{1}{a^4},$$ which is just scaled as anything else has for this problem: $$\rho_E^\prime = \frac{1}{a^{\prime 4}},$$ and since the scale factor in the present is one: $$\rho_E^\prime = 1 ,$$ and thus:  $$\frac{\rho_E}{\rho_E^\prime} \sim \frac{\frac{1}{a^4}}{1}$$ $$\frac{\rho_E}{\rho_E^\prime} \sim \frac{1}{a^4},$$ and applying the definition of the scale factor from redshift: $$\frac{\rho_E}{\rho_E^\prime} \sim (1+z)^4,$$ so the energy density is what the Plank Spectrum calculates: $$\frac{u_\nu d\nu}{u_{\nu^\prime} d\nu^\prime} \sim (1+z)^4$$ $$u_\nu d\nu \sim u_{\nu^\prime} d\nu^\prime (1+z)^4$$ $$u_{\nu^\prime} d\nu^\prime \sim \frac{u_\nu d\nu}{(1+z)^4},$$ which is how energy density scales with redshift (i.e. time).

(c) Knowing the Plank Spectrum equation: $$u_\nu d\nu = \frac{8\pi h_P \nu^3}{c^3} \frac{1}{e^{\frac{h_P \nu}{k_B T}} - 1} d\nu,$$ and knowing how it scales with time (redshift): $$z + 1= \frac{\nu_{emitted}} {\nu^\prime}$$ $$\nu_{emitted} =\nu^\prime (z + 1)$$   $$d\nu_{emitted} =d\nu^\prime (z + 1),$$ we can solve for how it changes: $$u_\nu d\nu = \frac{8\pi h_P (\nu^\prime (z + 1))^3}{c^3} \frac{1}{e^{\frac{h_P (\nu^\prime (z + 1))}{k_B T}} - 1} (d\nu^\prime (z + 1))$$ $$u_{\nu^\prime} d\nu^\prime \sim \frac{u_\nu d\nu}{(1+z)^4}$$ $$\frac{u_\nu d\nu}{(1+z)^4} = \frac{\frac{8\pi h_P (\nu^\prime (z + 1))^3}{c^3} \frac{1}{e^{\frac{h_P (\nu^\prime (z + 1))}{k_B T}} - 1} (d\nu^\prime (z + 1))}{ (1+z)^4},$$ $$u_{\nu^\prime} d\nu^\prime = \frac{8\pi h_P \nu^\prime3}{(z+1) c^3} \frac{1}{e^{\frac{h_P (\nu^\prime (z + 1))}{k_B T}} - 1} (d\nu^\prime (z + 1))$$   $$u_{\nu^\prime} d\nu^\prime = \frac{8\pi h_P \nu^\prime3}{c^3} \frac{1}{e^{\frac{h_P (\nu^\prime (z + 1)}{k_B T}} - 1} (d\nu^\prime ),$$ and after all that algebra, we can solve for the final (z +1) by incorporating the change in temperature equation: $$T^\prime = T(1+z)^{-1},$$ so the final equation becomes:   $$u_{\nu^\prime} d\nu^\prime = \frac{8\pi h_P \nu^\prime3}{c^3} \frac{1}{e^{\frac{h_P (\nu^\prime}{k_B T^\prime}} - 1} (d\nu^\prime)$$

(d) For the last part, we simply have to incorporate the definitions given in the beginning of the problem and the temperature change equation: $$T^\prime = T(1+z)^{-1}$$ $$T = 3000 K$$ $$z= 1100$$ $$T^\prime = 3000K \cdot (1+1100)^{-1}$$ $$T^\prime = 2.7 K,$$ which is the temperature of the Cosmic Microwave background in the present. 

Relativity, the math, yeah, really, learn it, cry over it, love it, not necessarily in that order

We are bounded in a nutshell of Infinite space: Blog Post #33, Free Form #5: Relativity, the math, yeah, really, learn it, cry over it, love it, not necessarily in that order.

Following up the earlier blog post in September (http://ay17-rcordova.blogspot.com/2015/09/relativity-yes-we-are-doing-this-it-is.html), relativity is an essential part to how we’ve come to understand how the universe works, especially when considering the fact that the largest objects and forces in the universe move at speeds comparable, if not at, the speed of light. So, we shall begin with the fundamentals of how Special Relativity, as told to you by one who learned it not a month ago.

In Special Relativity, there are two basic assumptions which make up the basis of all the equations:

1.       There are no preferred inertial frames

2.       The speed of light is the same in every inertial frame

From these two basic tenets, there are three Fundamental effects that can be observed, and have been consistently tested.

The First of these fundamental effects is the Loss of Simultaneity. Here, two events that coincide in one frame (that is moving) do not coincide in another frame of reference. Take the example of a moving train, in which two flashlights placed at the center send photons to the opposite ends of the train. Within the train, they hit the ends at the same time, but because of the additional speed of the train, an outside observer would perceive the photons hitting the front of the train before the back end. This effect is paralleled by the next one, and we’ll go into the math with it.

This loss of simultaneity leads us to the Second Fundamental effect, Time Dilation. If you click on the link to see the post from September, the classic example of this effect is described by the observer on the train’s perception of light and the observer on the ground’s perception.

“The classic example Einstein gives in his papers on Special and General Relativity is the case of two persons, one on a train heading in a direction towards a point where a lightning bolt just hit the ground, and another person stands some distance away and can observe both the train and the lightning bolt.  In this scenario, the person on the train is moving at a speed v, who will hit the rays of light with this speed and thus (one would expect to) perceive the speed of photons as the intrinsic speed of light minus the speed of the person. This would differ from the person outside the train, seeing the lightning bolt come and thus (expect) to perceive light at its normal speed.

However, the speed of light must be CONSTANT at all times, so the person on the train must have something change in order for him to experience the speed of light at the correct value. What happens is, as Einstein describes in Special Relativity, that the person moving with speed v experiences time (and length and differently, it slows down in his (moving) reference frame and so the speed of light he perceives is maintained at the constant rate. For the person outside the train, he would see the light coming from the lightning bolt at its normal speed, without any special considerations to be taken into account. But as he sees the light going towards the train in the distance, he would clearly see the light in direction of the train has the same speed as the light that reached his reference frame, maintaining the constant speed of light and set consolidated with the perception of the person on the train, instead of two people experiencing two different speeds of light.”

Therefore, you can assume each person had a clock on them at the time of the lighting strike, and each perfectly measured the time at which they saw the light, and each would have different measurement. You can also picture this by seeing how two different paths light takes take different times to be completed, indicating that the longer path, the one where the frame itself was also moving. This is illustrated by:

                                 

As you can see, the initial case projects a different (and shorter) path once in the reference frame of A, than in the frame of B. So if the original distance was simply 2h, and the speed travelled c, so the time for the photons to traverse is $$\frac{2h}{c}.$$ In the case of B, the total vertical distance the photons travel is 2h, but the speed component is different, being (by use of a bit of reasoning with the Pythagorean Theorem): $$\sqrt{c^2 – v^2},$$ so the time would be: $$\frac{2h}{ \sqrt{c^2 – v^2}}.$$ These two times can be compared and set as: $$t_A \sim t_B,$$ $$\frac{2h}{c} = \frac{2h}{ \sqrt{c^2 – v^2}},$$ $$1 = \frac{c}{ \sqrt{c^2 – v^2}},$$ $$t_A = t_B \frac{1}{ \sqrt{1 – \frac{v^2}{c^2}}},$$ where the additional factor for $t_B$ is what is called $\gamma_v$: $$\gamma_v = \frac{1}{\sqrt{1- \frac{v^2}{c^2}}},$$ a common “conversion” factor which allows the switching of frames in Special Relativity. So the final time dilation equation is $$t_A = \gamma_v~ t_B,$$ or $$t_B = \gamma_v~ t_A,$$ depending on the reference from which the action is being observed.

This principle of time dilation, the factor $\gamma_v$, establishes the base for most Relativistic problems (as long as they don’t involve gravity). 

The Third Fundamental Effect is Length Contraction. This effect establishes how an object moving at a speed is contracted proportionally to the increase in speed. This is described by $$L^\prime = \frac{L}{\gamma_v},$$ describing how length is inversely proportional to speed.

These three fundamental effects are at the core of relativistic situations, which are then expanded upon by Einstein in 1915 with his treatise on General Relativity (100 Years of General Relativity has been floating around Science Media the last few months, if you hadn’t heard), where the impact of gravity is coalesced with relativity to describe myriad situations like Gravitational Lensing and Microlensing (see the posts about that by going back to the September group). These equations have become the basis for modern physics and astrophysics, forming part of the ongoing search to unite quantum mechanics and General Relativity, the greatest descriptions of the universe humans have ever had the ability to (sort of) prove. 

Just beyond the horizon, but no way we can see it

We are bounded in a nutshell of Infinite space: Blog Post #32, Worksheet # 10.1, Problem #3: Just beyond the horizon, but no way we can see it

3. Observable Universe. It is important to realize that in our Big Bang universe, at any given time, the size of the observable universe is finite. The limit of this observable part of the universe is called the horizon (or particle horizon to be precise, since there are other definitions of horizon.) In this problem, we’ll compute the horizon size in a matter dominated universe in co-moving coordinates. To compute the size of the horizon, let us compute how far the light can travel since the Big Bang.

(a) First of all - Why do we use the light to figure out the horizon size?

(b) Light satisfies the statement that $ds^2 = 0$. Using the FRW metric, write down the differential equation that describes the path light takes. We call this path the geodesic for a photon. Choosing convenient coordinates in which the light travels in the radial direction so that we can set $d\theta = d\phi = 0$, find the differential equation in terms of the coordinates t and r only.

(c) Suppose we consider a flat universe. Let’s consider a matter dominated universe so that $a(t)$ as a function of time is known (in the last worksheet). Find the radius of the horizon today (at $t= t_0$).

(Hint: move all terms with variable r to the LHS and t to the RHS. Integrate both sides, namely r from 0 to $r_{horizon}$ and t from 0 to $t_0$.)

(a) As we all know, every modern theory of the large scale universe is based on several constants which are assumed and tested repeatedly to check for the consistency of the value. This is the case of the speed of light, c, the constant of the universe Einstein used to create special and general relativity, and the consequences of this knowledge have allowed us to understand the deeper throws of the universe. So, in order to find the radius of the horizon of the universe, we can use the only object which has reached us from the earliest point in time we can see, photons. Light has simply gotten to us before anything else, so it is the best representation of the horizon we are going to have.

(b) Simply put, we are going to find how the base FWR Metric: $$ds^2 = -cdt^2 + a^2(t) \left[\frac{dr^2}{1-kr^2} + r^2 (d\theta^2 + \sin^2\theta d\phi^2)\right],$$  can become a differential equation once we set $ds^2$ to 0: $$cdt^2 = a^2(t) \left[\frac{dr^2}{1-kr^2} + r^2 (d\theta^2 + \sin^2\theta d\phi^2)\right],$$ $$cdt^2 = \frac{a^2(t) (dr^2)}{1-kr^2},$$ $$\frac{cdt^2 }{a^2(t)} = \frac{ dr^2}{1-kr^2}.$$ After all this rearrangement of the terms initially found, with both derivatives of angles set to 0, the equation becomes a relatively simple expression to describe the path light travels.

 (c) Furthermore, by establishing a flat universe, we have a definitive value for k, 0 in this case, so the equation becomes: $$\frac{cdt^2 }{a^2(t)} = \frac{ dr^2}{1-kr^2},$$   $$\frac{cdt^2 }{a^2(t)} = { dr^2},$$   which can now be set up to be solved in a way similar to the Friedmann Equations, using a parameter they established to complete the problem. So, from the Friedmann equation for a Matter Dominated Universe(http://ay17-rcordova.blogspot.com/2015/11/cosmology-101-part-2.html), the relation of the universal scale factor and time is: $$a(t) = a_0 \left(\frac{t}{t_0}\right)^{2/3} ,$$ which can be placed into our derived form of the FRW metric and get: $$dr^2= \frac{cdt^2 }{ a_0^2 \left(\frac{t}{t_0}\right)^{4/3}} ,$$ which can be rewritten and analyzed in a manner similar to problem #2 of this Worksheet:   $$\int_0^{r_{horizon}} dr= \int_0^{t_0} \frac{cdt }{ a_0\left(\frac{t}{t_0}\right)^{2/3}} ,$$ and after some integration and separation of variables, as well as recognizing $a_0$ and $t_0$ as constants, we solve and get: $$r_{horizon}= \frac{c t_0^{2/3}}{ a_0} \cdot 3t_0^{1/3},$$ $$r_{horizon}=3 \frac{c }{ a_0} t_0,$$ which is the description for the radius of horizon today. 

Turns out Euclid was right about math after all

We are bounded in a nutshell of Infinite space: Blog Post #31, Worksheet # 10.1, Problem #2: Turns out Euclid was right about math after all

2. Ratio of circumference to radius. Let’s continue to study the difference between closed, flat and open geometries by computing the ratio between the circumference and radius of a circle.

(a) To compute the radius and circumference of a circle, we look at the spatial part of the metric and concentrate on the two-dimensional part by setting $d\phi =0$  because a circle encloses a two-dimensional surface. For the flat case, this part is just $$ds_{2d}^2= dr^2 + r^2d\theta^2.$$

The circumference is found by fixing the radial coordinate ($r=\theta$ and $dr=0$) and both sides of the equation (note that $\theta$ is integrated from 0 to $2\pi$ ).

The radius is found by fixing the angular coordinate $(\theta , d\theta = 0 )$ and integrating both sides (note that dr is integrated from 0 to R).

Compute the circumference and radius to reproduce the famous Euclidean ratio $2\pi$.

(b) For a closed geometry, we calculated the analogous two-dimensional part of the metric in Problem (1). This can be written as: $$ds_{2d}^2 = d\xi^2 + \sin\xi^2 d\theta^2.$$

Repeat the same calculation above and derive the ratio for the closed geometry. Compare your results to the flat (Euclidean) case; which ratio is larger? (You can try some arbitrary values of ξ to get some examples.)

(c) Repeat the same analyses for the open geometry, and comparing to the flat case.

(d) You may have noticed that, except for the flat case, this ratio is not a constant value. However, in both the open and closed case, there is a limit where the ratio approaches the flat case. Which limit is that?

(a) From the Friedman-Robertson-Walker Metric, an equation we will be looking on later in in this problem, we can derive a few basic equations that describes the physical phenomenon in universes with different configurations. Thought of either flat, open, or closed, the universe’s configuration describes how we perceive reality, where circles are the same in any location and their circumference is always $2\pi R$, but this could be different for open or closed universes (even though astronomers observe the universe is most likely flat).

For a flat universe, we can prove the traditional Euclidean description of a circle with a few simple calculations: assuming $$d\phi = 0 ,$$ we get: $$ds_{2d}^2= dr^2 + r^2d\theta^2.$$    This is a version of the FRW Metric with the $d\phi$ set to 0.

Once we have this equation, we simply have to establish some parameters such as $r = R, dr= 0$, and so the equation becomes: $$ds_{2d} = Rd\theta,$$ , and by setting the integrals like the problem asks, we solve: $$\int_0^{Circumference}ds_{2d}=  \int_0^{2\pi}Rd\theta,$$ $$Circumference = 2\pi R,$$ which is exactly the traditional Euclidean case.

For the radius, we have to establish how $\theta, d\theta= 0$ and the equation now becomes: $$ds_{2d}^2= dr^2 + r^2d\theta^2,$$ and following the same process as before for the circumference: $$ds_{2d}^2= dr^2,$$   $$\int_0^rds_{2d}= \int_0^R dr ,$$ and so a true statement appears for the description of our universe: $$r= R.$$

Knowing the circumference and radius, we can use both of these to establish a ratio between them that becomes a standard ratio for comparison with other physical universe interpretations: $$\frac{Circumference}{Radius} = \frac{2\pi R}{R} = 2\pi,$$ which is the Euclidean model the problem asked for.

(b) For a closed universe, the equation is altered slightly because of the base FRW metric used, so we now have: $$ds_{2d}^2 = d\xi^2 + \sin\xi^2 d\theta^2,$$ and with the alterations and limits the problem is establishing in order to better describe a closed universe, with  $$d\xi = 0 ,$$ the equation now becomes $$ds_{2d} = \sin\xi d\theta.$$ Next, using the hints the problem gives, we set up the integration: $$\int_0^{Circumference}ds_{2d}=  \int_0^{2\pi}\sin\xi d\theta,$$ and now have $$Circumference = 2\pi \sin\xi.$$

Next we find the radius, with a process similar to the circumference, and the problem already set up how $\theta, d\theta= 0$, so the original equation now becomes: $$ds_{2d}^2 = d\xi^2 + \sin\xi^2 d\theta^2,$$    $$ds_{2d}^2= d\xi^2$$ and integrating with the correct parameters: $$\int_0^rds_{2d}= \int_0^{\xi} d\xi ,$$ we now have the radius: $$r= \xi.$$

Finally, establishing the ratio of circumference to ratio, we have: $$\frac{Circumference}{Radius} = \frac{2\pi \sin\xi}{\xi},$$ which is identical to the flat universe ratio multiplied by the factor of $$\frac{\sin\xi}{\xi}.$$

(c) Taking the FRW metric as the basis, we can derive how: $$ds^2 = \frac{dr^2}{1-kr^2} + r^2 (d\theta^2 + \sin^2\theta d\phi^2),$$ becomes $$ds^2 = \frac{dr^2}{1-kr^2} + r^2 d\theta^2,$$ when $d\phi = 0$. Then in the case of an open universe where k = -1, the equation becomes: $$ds^2 = \frac{dr^2}{1+r^2} + r^2 d\theta^2,$$ so, taking a recommendation from an earlier problem of establishing that $r = \sinh\xi$, the equation can now be solved to find a simpler solution which is similar to the other ratios. Now, the equation becomes: $$ds^2 = \frac{\cosh^2\xi \cdot d\xi^2}{1+\sinh^2\xi} + (\sinh\xi)^2 d\theta^2,$$ and we can do this since an identity of hyperbolic functions like $\sinh\xi$ establishes that its derivative is $\cosh\xi d\xi$. Next, we simply use another identity says $\cosh^2x – \sinh^2x = 1$, so the equation becomes: $$ds^2 = \frac{\cosh^2\xi \cdot d\xi^2}{\cosh^2\xi } + \sinh^2\xi d\theta^2,$$ $$ds^2 = d\xi^2 + \sinh^2\xi d\theta^2.$$ Using the exact same process as earlier in part b, we find that the ratio of circumference to radius is: $$\frac{Circumference}{Radius}=\frac{2\pi \sinh\xi}{\xi},$$ which is the same as the flat universe ratio multiplied by the factor of $$\frac{\sinh\xi}{\xi}.$$

(d) From these ratios, a clear patter can be discerned once compared to the original, flat universe, case. With all the ratios illustrated clearly: (flat, closed, and open, respectively) $$2\pi , ~ \frac{2\pi \sin\xi}{\xi} , ~ \frac{2\pi \sinh\xi}{\xi}$$ there is a clear change. Both open and closed universe models exhibit the trigonometric and hyperbolic function, which both happen to have a similar Taylor Expansion (a way to approximate for small numbers), which turns the ratios into: $$2\pi , ~ \frac{2\pi \xi}{\xi} , ~ \frac{2\pi \xi}{\xi}$$ $$2\pi , ~2\pi , ~ 2\pi.$$ Therefore, as $\xi$ becomes small, all universes look more and more like each other, as a result of the limiting case as $$\xi \to 0 .$$

Telescopes in the night, but actually used in the day

We are bounded in a nutshell of Infinite space: Blog Post #30, Worksheet # 9.2: Telescopes in the night, but actually used in the day

1. Noise properties of a single dish radio telescope

Radio telescopes (or radiometers) work like a standard radio antenna or satellite dish, converting an incoming radio signal into a pattern of oscillating voltages through the interaction of the wave with the electrons in the antenna. A radiometer receives a large number of signals, but filters allow only a certain chunk of the spectrum through. Our telescope is what has called a “heterodyne” receiver, which means that it takes the incoming signal at $\nu_0$, and mixes it down with a local oscillator of frequency $\nu_{LO}$ to produce a signal centered at a lower new frequency $\nu_{IF} = \nu_0 - \nu_{LO}$. The system is designed so that νif is a constant. So, when we “tune” the telescope, we are really setting $\nu_{LO}$. This signal is then fed into the filter bank and split into $N_{chan}$  channels of width $\Delta\nu_{RF}$. The power in each channel is plotted, and this is called a spectrum. Note that $T_{sys}$ is called the “system temperature” and has units of Kelvin; however, you should think of $T_{sys}$ as a measurement of noise and not a physical temperature.

When doing this, the signal has very particular noise properties. Recall that the root-mean-square (rms) error decreases as a function of the number of samples $\sigma \sim 1/\sqrt{N}$. For spectral line observations like ours, in a radiometer, each sample has noise $T_{sys}$, and the number of samples in a channel goes up with increasing channel width and integration time. The noise is characterized by the ideal radiometer equation.

Precisely, if I have a receiver with a system temperature $T_{sys}$, channel width $\Delta\nu$, and integration time $\tau$ , the radiometer equation is $$\sigma = \frac{T_{sys}}{\sqrt{\tau ~\Delta \nu}}.$$

(a) What is the expression for the signal to noise ratio (SNR)? The signal has a brightness temperature $T_{A}$.

(b) Find an equation for the velocity resolution $\Delta\nu$ km/s that corresponds to a channel width of $\Delta\nu$ MHz at the frequency of $^{12}CO, \nu =$115.271 GHz (2.6 mm). Write this in the form $$\Delta\nu = \_\_\_\_ km/s \left(\frac{\Delta\nu}{1~MHz}\right)$$  

(c) What integration time would be needed to detect the peak of $^{12}CO$ with SNR = 10 if I use a filter bank with 256 channels that are 0.5 MHz wide. $T_{A}$ for CO is about 2-3 K and $T_{sys} = 500 K$. (Hint: The peak appears in 1 channel. Do the algebra first to get the final expression and then put in the numbers.)

2. Resolution of a single dish radio telescope

The spatial resolution of a telescope is $\theta = 1.2\frac{\lambda}{D}$ , where D is the diameter of the dish, and $\lambda$ is the observing wavelength. In radio astronomy, $\theta$ is also known as the half-power beam width (or full-width half-max of the beam).

(a) Find an equation for the beam width, in arcminutes, of a single-dish radio in terms of frequency $\nu$ in GHz, and diameter D in meters. Use a calculator to determine this to 1 decimal place. Write an equation of the form $$\theta_{HPBW} = \_\_\_\_ degree \left(\frac{\nu}{GHz}\right)^{-1} \left(\frac{D}{m}\right)^{-1}$$

(b) What is the beamwidth, in arcminutes, for the CfA 1.2 m telescope at the $^{12}CO$ frequency. (to 1 decimal place)

(c) What linear dimension in pc does this correspond to at the Galactic center (8.5 kpc)?

3. Geometry of the experiment

Below is a schematic of the experiment. A molecular cloud a distance R from the galactic center has a velocity $V_{cir}$. When we observe this cloud, we detect a different velocity, $V_{obs}$, which is the combination of our velocity towards it and its velocity away. Note: We are using R to denote the distances because the sources are some radius from the Galactic center.

(a) For our experiment we will take a spectra which will have velocity components from multiple clouds along the line of sight. Looking at the figure above, which cloud will have the highest velocity and why?

(b) Plot, very roughly, the velocity of the clouds versus their distance along the line sight, what would the shape be?

(c) The velocity which we measure using the $^{12}CO$ line is the combination of the cloud’s velocity around the galactic center and the projection of the Sun’s velocity along our line of sight. What is the equation for the velocity we measure $V_{max}$?

(d) Write the equations for the rotational velocity, $V_{cir}$and $R_{tan}$in terms of the quantities we know - $V_{\odot}$, $V_{\odot}$, l, and $V_{max}$ .

(e) What is the equation for the mass profile of the galaxy? The ultimate goals of this lab are to estimate the total gravitational mass of the Galaxy within the Suns orbit and to infer something about how that mass is distributed in Galactic radius. To do this we will determine the so-called Galactic rotation curve by measuring the velocities of molecular clouds orbiting in the Galactic potential.

1. (a) For this post, we will be exploring the nature of a radio telescope and how these can be used and their slight limits corrected by knowing exactly what to look for and what effects to mitigate. First of all, we’ll find the ratio of the signal to the noise. So if the noise has already been defined as $$\sigma = \frac{T_{sys}}{\sqrt{\tau ~\Delta \nu}},$$ and the signal is defined as $T_A$, then it is simply a matter of dividing one by the other. So the Signal to Noise Ratio is simply: $$\frac{T_A}{\sigma}$$ $$\frac{T_A}{T_{sys}} \sqrt{\tau~\Delta\nu}.$$

(b) Next, we are asked to find the relationship between velocity and frequency, something that sounds very familiar to red shift. In fact, from the red shift equation, with a bit changed, we have: $$\frac{\nu_{obs} - \nu_{em} }{\nu_{em}} = \frac{v}{c} ,$$ which can be interpreted as: $$\Delta v = c\frac{\Delta\nu}{\nu}.$$ So taking this equation and plugging in the necessary values as seen above, we have: $$\Delta v = 3 \times 10^5 km/s \cdot \frac{\Delta\nu}{115.271 \times 10^9 Hz} ,$$ which yields: $$\Delta v = 2.6 \times 10^{-6} \Delta\nu \frac{km/s}{Hz} \times \frac{10^6 Hz}{MHz},$$ adding in the conversion factor to achieve the correct units as asked, we finally know that: $$\Delta v = 2.6 \frac{km/s}{MHz} \Delta\nu .$$

(c) Now, knowing how the velocity resolution equation works and how the SNR operates, we can use them to find the next factor, the time of integration for $^{12} CO$, $\tau$. Using the SNR equation, we have:    $$SNR = \frac{T_A}{T_{sys}} \sqrt{\tau~\Delta\nu},$$ and solving for the integration time, $$\tau =\frac{SNR^2\frac{T_{sys}^2}{T_A^2}}{\Delta \nu} .$$ By plugging in the values given to us by the question, we have: $$\tau = \frac{10^2\frac{500K^2}{2.5K^2}}{0.5 \times 10^6 Hz},$$ which simplifies down to: $$\tau = 8 s.$$

2.(a) Here, we are given a basic piece of information that essentially tells us what to do in order to find the final angle. Taking into consideration the $\theta = 1.2\frac{\lambda}{D}$, it can be turned into $\theta = 1.2\frac{\frac{c}{\nu}}{D}$, so by plugging this into the equation given, we can make it in the correct dimensions by multiplying by the correct orders of magnitude. So: $$\theta_{HPBW} = \_\_\_\_ degree \left(\frac{\nu}{GHz}\right)^{-1} \left(\frac{D}{m}\right)^{-1} ,$$ $$\theta_{HPBW} = 1.2\frac{\frac{c}{\nu}}{D} \left(\frac{\nu}{GHz}\right)^{-1} \left(\frac{D}{m}\right)^{-1} ,$$   $$\theta_{HPBW} = \frac{1.2 \cdot 3\times 10^8 m/s }{ \nu D} \times \frac{1~GHz}{10^{-9} Hz}  \times \frac{180~degrees}{\pi ~radians} ,$$ which it all simplifies to: $$\theta_{HPBW} = \frac{20.36~degrees~ m\cdot GHz}{\nu~D} .$$

(b) Now we take the equation we just found and plug in the values of D and $\nu$ they give. So the equation turns into: $$\theta_{HPBW} = \frac{20.36~degrees}{\nu~D} ,$$ $$\theta_{HPBW} = \frac{20.36~degrees~ m\cdot GHz}{1.2 m \cdot 115.271 GHz},$$ $$\theta_{HPBW} = 0.149 ~degrees = 8.9 ~arcminutes.$$

(c) Now that we know the angle with which we are observing, we can use it in the basic angle to distance relation that is parallax. From the information given by the question, we know that the distance from the object is 8.5 kpc, and if we have the distance and the angle, then we can know the width of our field of view. From earlier problems, we found how $$angle_{(in~ arcseconds)} =\frac{1 AU}{D_{in ~pc}},$$ but the relationship can be rewritten as: $$angle_{(in~ arcseconds)} \cdot D_{(in~pc)} = width ~of ~view ~in~ AU ,$$ and we know the values that need to be plugged in and the conversion factors to be used: $$(8.9 ~arcminutes \times 60) \cdot (8.5 \times 10^3 pc) = width ,$$ $$4.539 \times 10^6 AU  \ times \frac{1 ~pc }{2.05 \times 10^5 AU}= width,$$ $$Linear~Dimension_{(Width)} = 22.14 pc.$$

3. (a) After looking at the diagram for a bit, it becomes obvious that the object with the highest speed away from us, the object with the greatest Red Shift, is undoubtedly object B, since all the velocity we observe is pointed tangentially to its orbit path, thus making it the fastest among all the objects in our line of sight.

(b)

(c) Taking into consideration all the factors that are changing how we are perceiving the objects on the other side of the galaxy,  the best equation that relates the max speed of any object is: $$V_{Max} = V_{cir} - V_\odot \sin(l) ,$$ which recognizes how our velocity impacts the way we perceive the velocity of the observed object.

(d) By simply rewriting the above equation, we can solve for the circular velocity, which is simply: $$V_{cir} = V_{Max} + V_\odot \sin(l) ,$$ and the tangential radius is simply the component of the radius from the galactic core to the sun, such that: $$R_{tan} = R_\odot \sin(l) .$$

(e) And finally, to understand the mass profile for the galaxy, we must go back to a post from a couple of months ago ( http://ay17-rcordova.blogspot.com/2015/09/the-milky-way-is-now-sphere-trust-me.html ) where the velocity profile is described as: $$v(r) = \left(\frac{G M_{enc}}{r}\right)^{1/2} ,$$ and to find the mass profile, a bit of rearranging is all that is necessary: $$M_{enc} = \frac{V_{cir}^2 R_{tan}}{G}.$$