Astronomy Education and the Shape of the Universe

In science, half the battle is being able to communicate and share ideas and discoveries. It is an important skill to have with many fun methods of communication in the digital era. For my educational project, I decided to explain cosmological geometries, one of the only completely new subjects to me in astronomy 16 and 17 and, in my opinion, one of the most interesting. The video can be found at https://www.youtube.com/watch?v=Y1aODOgNXCU or watched below.

Youtube description: A basic cosmological introduction to the various theories of the shape of the universe. More explanations for the geometry of circles in different universes can be found here (http://ay16-dfrostig.blogspot.com/2015/11/geometry-of-universe-blog-31-worksheet.html) and more information on Friedmann Equations can be found here (http://ay16-dfrostig.blogspot.com/2015/11/more-friedmann-equations-blog-29.html).

Sources include http://csep10.phys.utk.edu/astr162/lect/cosmology/geometry.html, http://www.astro.ucla.edu/~wright/cosmo_03.htm, and Harvard's Astronomy 17 class worksheets.

Modeling the Entire Universe: Blog 37 - The Last Blog

From the very first week of Astronomy 16 to now, from estimating the earth's rotation to modeling the geometry of the universe, we have made it to the last homework blog post of the introductory astronomy sequence. And what better way to wrap up our exploration of all things big and small in the universe than to model the entire universe? 

This week in astronomy, we explored how small density changes in the early universe can determine the overall structure of our universe today. As we have seen in the blog posts, these ideas take a lot of math, and we have barely scratched the surface of all the math that can be done to guess at the entirety of the universe, most of which we cannot see. So what better way to tackle such a large problem than with powerful computers modeling the universe?

The Illustris Simulation models the dark matter, baryonic matter, and dark energy to show the evolution of the universe, or at least a part of it that fits into a 75 Mpc/h box. The simulation goes from a redshift of 127 (the early universe) to a redshift of zero (present day).

The simulation starts off with a view of the dark matter density of the universe, which is not evenly distributed. Not coincidentally, the dark matter demonstrates a similar pattern to to the dense clumps and connecting filaments associated with galaxy clusters. Using the "Spatial Query on Click" feature, we can explore a 400 kpc/h radius of the simulation and return information on up to 20 subhalos. 

Dark Matter Density of a 75 Mpc/h box

Analyzing a random selection of subhalos in an over dense region, we can see that low mass halos are more frequent than high mass halos. Additionally, on average, 9% of the mass of these halos is stellar mass, meaning that this mass is significantly composed of something other than visible matter.

A histogram plotting the frequency of halo masses in log(M) with bins of 2.5.

A histogram plotting the frequency of halo masses in log(M) with bins of width 0.5.

Beyond these snapshots in time, we can also explore our universe over time. This video speeds up the formation of the universe in both dark matter and gas temperature. Just from this one short video, we can learn a lot of about our early universe and its evolution.

The simulation starts off rather "dark" with the universe being dominated by dark matter until about z = 9 (about half a billion years after the big bang), when the very first hints of gas appear. This early period is called the "Dark Ages," when gas is neutral. The gas goes from blue to having some green around z=6 (less than a billion years after the big bang), symbolic of the "Epoch of Reionization," the official end of the Dark Ages when gas becomes increasingly ionized. This seems to be around the same time the first stars begin to form.

The early universe (z \( \approx\) 6). Dark matter is prominent and spread out while there is little prominence of hot gas.

Stars then begin to form rapidly, with growth even accelerating at times. The most rapid star growth then occurs around redshifts of 2.3 to about 0.6. Around this time we also being to see large bursts of energy in the gas temperature simulation.

The universe closer to today (z \( \approx\) 1). Dark matter has concentrated into dramatic over dense regions and star formation is occurring. 

Clear structures form throughout the simulation with smaller structures combining to form larger ones, rather than large objects breaking up. This hints at a trend of increasing gravity and decreasing gas pressure over time. With this increase in gravity, we seem the formation of over dense regions of dark matter and hot gas connected by filaments. As we explored in previous blog posts, this distribution is not uniform due to inconsistencies in pressure in the early universe that have been magnified with the passage of time. Theses dense regions collapse in on themselves due to the high levels of gravity found in dark matter.

The universe today (z=0) with many dense regions of dark matter and hot gasses.

These structures are not only found on a large scale, but are also surprisingly similar to a simulations at smaller scales. On both scales, dark matter and gas density tends to be clustered into regions connected by filaments. The dark matter is much more closely limited to these filaments. This follows along with our earlier observation that ionized gas forms as a consequence of dark matter. However, the gas is at a higher energy and is less massive, less confined by gravity, and therefore less limited to the filaments. On the smaller scales, all types of matter seem to be less limited to these filaments, most likely due to the smaller scales of mass and gravity.

Dark Matter Density                                  Gas Density        

Most of the medium or large galaxies are found in clusters and not in the field and within those galaxies, gas is densest towards the middle of the galaxies. This lines up exactly with what we learned at the beginning of the semester, when we explored the structure of the Milky Way. And within one of these over dense regions of the universe lies a galaxy cluster containing a very familiar spiral galaxy, with an average-sized star being orbited by an averaged-sized planet filled with astronomers looking up and out at our universe today.

Sources:
http://www.illustris-project.org/explorer/
http://earthsky.org/space/dark-matter-hairs-filaments-streams-gary-prezeau

Large Scale Structure of the Universe: Blog 36, Worksheet 12.1, Problems 1 and 2d.

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 ρ(x, t) and the relative density contrast as \[δ(r, t) = \frac{ \rho(r, t) - \bar{\rho} (t)}{\bar{\rho} (t)} \] 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 δ. 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 δ satisfies the following second-order differential equation \[ \frac{d^2δ}{dt^2} + \frac{2 \dot{a}dδ}{adt} = 4 \pi G \bar{\rho}δ \] 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 \[δ(x, t) =  D(t) \tilde{δ}(x) \] where ˜δ(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.

Plugging in the second equation to the first equation gives: \[ \frac{d^2}{dt^2} D(t) \tilde{δ}(x) + \frac{2 \dot{a}d}{adt} D(t) \tilde{δ}(x) - 4 \pi G \bar{\rho} D(t) \tilde{δ}(x) = 0 \] Which is \[  \tilde{δ}(x)( \ddot{D}(t) + \frac{2 \dot{a}}{a} \dot{D}(t) - 4 \pi G \bar{\rho} D(t)) = 0 \] When \( \tilde{δ}(x) = 0\) this is \[   \ddot{D}(t) + \frac{2 \dot{a}}{a} \dot{D}(t) - 4 \pi G \bar{\rho} D(t) = 0 \]

(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, \( \frac{3H_0^2}{8 \pi G} \) as in Worksheet 11.1 (aside: such a universe sometimes is called the Einstein-de Sitter model). Recall that the behaviour of the scale factor of this universe can be written \( a(t) = (3H_0t/2)^{2/3} \) , 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).

Combining the given equations allows us to find a new expression for density: \[\bar{\rho}(t) =  \left( \frac{3H_0^2}{2} \right)^{-2} \frac{3H_0^2}{8 \pi G} \] We know from previous weeks that \[H_0 = \frac{ \dot{a}}{a} \] and \[ a \propto t^{\frac{2}{3}} \] which means \[ \dot{a} \propto \frac{2}{3} t^{- \frac{1}{3}}\] so \[ H_0 = \frac{ \dot{a}}{a} \propto \frac{2}{3t} \] Combining that with our answer from (a) gives:  \[ \ddot{D}(t) + \frac{4 }{3t} \dot{D}(t) - \frac{2}{3t^2} D(t) = 0 \] Using the anasatz \( D(t) \propto t^q \) simplifies this to \[q(q-1)t^{q-2} +  \frac{4 }{3t}qt^{q-1}  - \frac{2}{3t^2}t^q \] \[t^{q-2}(q^2 - q + \frac{4q}{3} - \frac{2}{3}) = 0 \] Using the quadratic equation, q can be -1 or 2/3 so the solutions to this problem are: \[D_-(t) = C_1 t^{-1} \] \[D_+(t) = C_2 t^{\frac{2}{3}}\]
(c) Explain why the D+ component is generically the dominant one in structure formation, and show that in the Einstein-de Sitter model, \(D_+(t) \propto a(t) \).

The \(D_+ \) component is generally the dominant one because it is proportional to \(t^{\frac{2}{3} }\), which, when t is large, grows much faster than the shriking \(D_- \propto t^{-1} \). Because both  \(D_+(t) \propto t^{\frac{2}{3}} \) and  \(a(t) \propto t^{\frac{2}{3}}\),  \(D_+(t) \propto a(t) \).

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.

We consider a spherical patch of uniform over-density. Let us study the motion of a particle in terms of its distance r from the center of the sphere as a function of time t. Recall that in Worksheet 9, from Newtonian dynamics, we have derived that this function satisfies the following equation \[ \frac{1}{ 2} \left( \frac{dr}{dt} \right)^2 - \frac{GM }{r}  = C \] where C is a constant. We can study this in a closed case where r = A(1  - cos η) and t =B(η -  sin η), the open case where r = A(cosh η - 1) and t =B(sin η - η ), and the flat case where r = \(Aη^2 /2\) and t =  \(Bη^3 /6 \).

We found that in the closed case, \[ C = \frac{-A^2}{2B^2} \] in the open case \[ C = \frac{A^2}{B^2} \] and in the flat case \[ C = 0\] Plotting these three together with r as a function of t (with 0 < η < \( 2 \pi \) )  where 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. We can see this in a closed universe:

An open universe:

And a flat universe:

Which is where we are theorized to live.

Cosmic microwave background: Blog 34, Worksheet 11.1, Problem 2

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 (1). At about the redshift z « 1100 when the universe had the temperature T = 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.

Source

Part A: If the photons was emitted at redshift z with frequency ν, what is its frequency ν' today?

We know from two weeks ago that \[ z = \frac{v}{c} = \frac{ \lambda_0 - \lambda_e }{ \lambda_e } \]  Combining this with the fact that \[ \lambda = \frac{c}{\nu} \]  \[ z = \frac{ \frac{c}{\nu'}  -  \frac{c}{\nu}  }{  \frac{c}{\nu}  }= \frac{ \nu }{\nu'} \] \[ \nu' = \frac{\nu}{z + 1} \]

Part B:  If a photon at redshift z had the energy density uνdν, what is its energy density uν'dν' today? (Hint, consider two effects: 1) the number density of photons is diluted; 2) each photon is redshifted so its energy, E = hν, is also redshifted.)

We know energy density is a function of energy and volume \(u_ν'dν' = \frac{E}{V} \) so we just need to figure out how energy and volume have changed from the past to today.

Once again, we know \[ E = h \nu \] so \[ E' = \frac{h \nu}{z + 1} \] based on the conversion found above. We also know distances scale to volume as \( R^3 ~ V \) so we have the relationship: \[V_{today} \propto a_0^3 ( 1 + z )^3 V_{tomorrow} \] This means \[ n = n_0 a_0^{-3} ( 1 + z )^{-3} \] Because \[u_νdν = E n = h \nu n \] \[u_ν'dν' = \frac{h \nu}{z + 1}n_0 a_0^{-3} ( 1 + z )^{-3} \] So the energy density today would be \[u_ν'dν' = a_0^{-3} ( 1 + z )^{-4} u_νdν \]

Part C: Plug in the relation between ν and ν' into the Planck spectrum: \[u_ν dν  = \frac{ 8 \pi h \nu^3 }{c^3} \frac{1}{e^{\frac{h \nu}{kT}} - 1 } d \nu \] 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 uν'dν' . What is uν'? This is the spectrum we observe today. Show that it is exactly the same Planck spectrum, except that the temperature is now T' =  T(1 + z)^-1 .

Plugging in our scaling relation and multiplying by our dilution factor give the rather complicated expression: \[u_ν' dν' = \frac{ 8 \pi h ( \nu'(z+1))^3}{c^3} \frac{1}{e^{\frac{h( \nu'(z+1))}{kT}} - 1 } d \nu' ( 1 + z) \cdot a_0^{-3} ( 1 + z )^{-4} \] This simplifies to \[u_ν' dν' = \frac{ 8 \pi h  \nu'^3}{c^3} \frac{1}{e^{\frac{h( \nu'(z+1))}{kT}} - 1 } d \nu' \] This is the same result that we would have obtainted if we replaced T with \(\frac{T}{1 + z} \).

Part D: As you have just derived, according to Big Bang model, we should observe a black body radiation with temperature T' filled in the entire universe. This is the CMB. Using the information given at the beginning of this problem, what is this temperature T' today? (This was indeed observed first in 1964 by American radio astronomers Arno Penzias and Robert Wilson, who were awarded the 1978 Nobel Prize.)

This problem is now just a matter of plugging in things we know: \[ z = 1100 \] \[ T = 3000 \: K \] \[ T' = \frac{T}{1 + z} \] So our final temperature of the CMB today is \[ T' = \frac{3000}{1101} = 2.725 \: K \]  This is pretty close to what Penzias and Wilson found.

Baryon-to-photon ratio of our universe: Blog 35, Worksheet 11.1, Problem 3

Part 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 hν. From equation (1), figure out the number density, nν, of the photon per frequency interval dν. Integrate over dν 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 \[ \int_0^{\infty} \frac{x^2}{e^x - 1} \approx 4.2 \] We already know \[ E = h \nu \] and \[ E n d \nu = \frac{ 8 \pi h \nu }{c^3} \frac{1}{e^{\frac{h \nu}{kT}} - 1 } d \nu \] The integral of this is equivalent to \[ \int E n d \nu = \int \frac{ 8 \pi h \nu }{c^3} \frac{1}{e^{\frac{h \nu}{kT}} - 1 } d \nu = E n \] Adding this with our definition of energy and rearranging the integral gives us: \[ n = \frac{ 8 \pi K^3 T^3 }{c^3 h^3} \int_0^{\infty} \frac{x^2}{e^x - 1}\] Where \[ x = \frac{ h\nu}{KT}\] and \[ d \nu = \frac{ KT dx}{h} \] With the approximation given above we get a cleaner answer for n \[ n_{\nu} = \frac{ 8 \pi K^3 T^3 }{c^3 h^3}(2.4) \]

Part B: Use the following values for the constants: kB =1.38 x 10^-16 erg/K, c = 3.00 x 10^10 cm/s, h =  6.62 x 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?)

In problem 2b, we found the temperature of CMB today to be about 2.725 Kelvin. Plugging this in with the constants specified above should give us: \[ n_{\nu} = \frac{ 8 \pi K^3 T^3 }{c^3 h^3}(2.4) = \frac{ 8 \pi (1.38 \times 10^{-16})^3 2.725^3 }{( 3 \times 10^{10})^3 (6.62 \times 10^{-27})^3}(2.4) \approx 170.62 \frac{photons}{cm^3}\]

Part 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 x 10^-30g/cm^3 . The baryon density is about 4% of it. The masses of proton and neutron are very similar (= 1.7 x 10^-24g). What is the number density of baryons?

This means that the density of our universe that is baryons is \[ 9.2 \times 10^{-30} \times 0.04 = 3.68 \times 10^{-31} \] Dividing this number by the mass of a typical baryon gives us the density: \[ n_b = \frac{\rho}{m} = \frac{ 3.68 \times 10^{-31}}{1.7 \times 10^{-24} } \approx 2.16 \times 10^{-7} \]

Part 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).
\[ \frac{photons}{baryons} = \frac{2.16 \times 10^{-7} }{170.62} = 1.26 \times 10^{-9} \]

Blog 33: Free-form blog post

Last spring, during one very long night in the eighth floor of the science center, Barra, Sean, April and I remotely operated the Minerva telescope, loudly sang disney songs, and ate way too much junk food in order to obtain three light curves. Two of those light curves were of exoplanets transiting other stars (we wanted to be very sure of our grade in the class). The third exoplanet was not transiting a star at all, but a white dwarf. One of our TFs last semester, Andrew Vanderburg, had asked us to take a light curve of this curious system for a project he was working on. Fast forward about half a year, and that project is now been published in Nature, Sky and Telescope, and a few other astronomical publications. In light if its recent publicity, I figured I would actually read the Nature paper so I can gain a deeper understanding of what exactly we were observing that night atop the science center.

The object in question is a white dwarf, one of the most common fates for a star, in the Virgo region, about 570 light years away from Earth. White dwarf atmospheres are generally composed of lighter elements, with a carbon and oxygen core and a helium and hydrogen outer shell. However, a quarter to a half of all white dwarfs observed contain heavier elements in their atmospheric spectrum, a puzzling fact knowing heavy elements should sink to the core. This discrepancy caused astronomers to suspect that outside materials had been introduced to the white dwarf's atmosphere, either through surrounding dust or disrupted asteroids, but no evidence had been found.

This white dwarf in question, WD 1145 + 017, was being observed for a different reason. The object exhibited a telltale, periodic, dip in its light curve. This occurred about once every 4.5 hours and could obscure up to 40% of the white dwarf's light. This behavior is a telling sign of a transiting exoplanet, though no exoplanets had ever been discovered around a white dwarf before.

An artist's rendering of a disintegrating rock planet around a white dwarf.

However, the periodic dip in the light curve was not a typical one either. Unlike a consistently transiting exoplanet around a star, this light curve was more erratic. The results were puzzling, until researchers saw two asymmetric light curves that are indicative of a disintegrating planet, which had been observed on main sequence stars before. The asymmetry was evidence for a comet-like tail trailing the planet and the variable transit depths indicated a disintegrating system. Combining this with data showing evidence of heavy elements, like nickel and iron, indicated to researchers that the disintegrating planet was rocky.

Some of the main periodic light curves of WD 1145 + 017.

The planet is about twice the distance from the Earth to the Moon away from the white dwarf, and about the mass of Ceres. The origin of the planet is unclear; it was most likely disrupted from a previous orbit by the gravity of the white dwarf. There is still a lot to learn about this unique system. It is currently the best evidence we have for external white dwarf "pollution" and exoplanets transiting white dwarfs. This is definitely something I will keep an eye on in the future.

Sources:
https://www.cfa.harvard.edu/~avanderb/page1.html
http://www.nature.com/articles/nature15527.epdf?referrer_access_token=JQgS2CFJ63QHiqwEPtPBJNRgN0jAjWel9jnR3ZoTv0OabmFzXPDJ8_WF8JNyTjQSuCoJ78UVkpUoy_k_5w1_QoSdPQRTejjUKTX0cRCWjHcKxRGYqDGUtTqkJgFEKN4xBCrjmjgKIdaoaOP99aALkSQAZ5OmnFbTUeiSGH9Dk_EhDjR0a7Z65xPZ34YmCVHX7DtNfxjg2G-X9-H-TMLmdg%3D%3D&tracking_referrer=www.nature.com

Size of the Observable Universe: Blog 32, Worksheet 10.1, Problem 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?

Light is both the fastest known phenomena in the universe and our way of observing the universe. All we can see is determined by how quickly light can reach our eyes, so we are limited to a particle horizon determined by the speed of light.

(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 dθ = dφ = 0, find the differential equation in terms of the coordinates t and r only.
\[ds^2 = -c^2dt^2 + a^2(t) [\frac{dr^2}{1-kr^2} + r^2 (d \theta + sin^2( \theta) d \theta ]  \] With the conditions for light and comoving radii this becomes \[0= -c^2dt^2 + a^2(t) [\frac{dr^2}{1-kr^2} + 0 ]  \] \[ \frac{c^2dt^2}{ a^2(t)} = \frac{dr^2}{1-kr^2} \]

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 \(t_0 \).) \[ \frac{c^2dt^2}{ a^2(t)} = \frac{dr^2}{1-kr^2} \] In a flat universe, this simplifies to \[ \frac{c}{ a(t)} dt = dr \] The integral is \[ \int_0^{t_0} \frac{c}{ a(t)} dt = \int_0^{r_H} dr \] We know from a previous blog post that \( a(t) \propto  t^{\frac{2}{3}} \) so the integral becomes \[ \int_0^{t_0} \beta  \frac{c}{  \left( \frac{t}{t_0}\right)^{\frac{2}{3}}} dt = \int_0^{r_H} dr \] where beta is a constant. So we have \[ r_H = 3 \beta c t_0 \]

Correction: we can solve this problem more accurately by replacing the relation \( a(t) \propto  t^{\frac{2}{3}} \) with the more accurate \( a(t) \propto \left( \frac{ t}{t_0} \right)^{\frac{2}{3}} \) in this way we can redefine beta in terms of \(a_0 \text{ and } t_0 \) where \(a_0 \) is 1 so our final answer cleans up to just be \[ r_H = 3  c t_0 \]

Geometry of the Universe: Blog 31: Worksheet 10.1, Question 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 dφ = 0 because a circle encloses a two-dimensional surface. For the flat case, this part is just \[ ds_{2d}^2 = dr^2 + r^2 dθ^2\] The circumference is found by fixing the radial coordinate (r = R and dr = 0) and both sides of the equation (note that θ is integrated from 0 to 2π). The radius is found by fixing the angular coordinate (θ, dθ = 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 2π.

With these new definitions of R and dr, our relationship is:  \[ ds_{2d}^2 = 0 + R^2 dθ^2\] or \[ ds_{2d} = R dθ\] Taking the integral of each side gives us an expression for circumference:  \[ \int_0^C ds = \int_0^{2 \pi} R dθ\] \[ C = 2 \pi R \] This looks pretty familiar, now we can move on to less familiar geometries. \[ \frac{C}{R} = \frac{2 \pi R}{R} = 2 \pi \]

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

Radius:
With these new definitions of R and dr, the metric goes from: \[ds_{2d}^2 = dξ^2 + sin(ξ)^2 (dθ^2 + sin(θ)^2 d \phi^2) \] to \[ds_{2d}^2 = dξ^2 \]  or \[ ds_{2d} = dξ\] Taking the indefinite integral of each side gives us an expression for radius: \[ \int ds_{2d} = \int dξ\] \[ s =  ξ = R \]
Circumference:
Once again, our relationship simplifies nicely. \[ds_{2d}^2 = dξ^2 + sin(ξ)^2 dθ^2\]  \[ds_{2d}^2 = sin(ξ)^2 dθ^2 \] \[ds_{2d} = sin(ξ) dθ \] Taking the integral with the outlined bounds to find circumference:   \[ \int_0^C ds = \int_0^{2 \pi} sin(ξ)  dθ\] \[ C = 2 \pi sin(ξ)  \] This expression is less familiar to us, but does resemble \( C = 2 \pi R\) with our definition of R in this case.
Ratio: \[ \frac{C}{R} = \frac{2 \pi sin(ξ)}{ξ}  \] Because this circumference is dependent on a sine term, the circumference will never be greater than \( 2 \pi \) or less than \( -2 \pi \) so this ratio will always be equal to or less than the ratio found in Euclidian Space in part a.

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

Radius:
In an open geometry, the procedure for radius is the same. \[ds_{2d}^2 = dξ^2 + sin(ξ)^2 (dθ^2 + sin(θ)^2 d \phi^2) \] \[ ds_{2d} = dξ\]  \[ \int ds_{2d} = \int dξ\] \[ s =  ξ = R \] Circumference: In the metric for open geometry is R = sinh(x) so the metric simplifies to  \[ds_{2d}^2 = sinh(ξ)^2 dθ^2 \]  \[ \int_0^C ds = \int_0^{2 \pi} sinh(ξ)  dθ\] \[ C = 2 \pi sinh(ξ)  \] Ratio:  \[ \frac{C}{R} = \frac{2 \pi sinh(ξ)}{ξ}  \] This ratio will always be larger or equal to the ratio in the flat, and closed universe.

(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?

The two ratios we found are :  \[ \frac{C}{R} = \frac{2 \pi sin(ξ)}{ξ}  \] \[ \frac{C}{R} = \frac{2 \pi sinh(ξ)}{ξ}  \] As we approach 0, we can use the small angle approximation for both of these functions, (e.g. sin(ξ) is ξ) so as the limit goes to zero, both ratios go to zero, approaching the flat case (this is easy to visualize in the graph above). In class, Ashley compared this geometry to us being small human beings on a large Earth. On the whole, we know the Earth is a three-dimensional sphere, but at small scales, as the limit approaches zero, the Earth looks flat.

Representations of the three types on universes - Source

Blog 30: Galactic Rotation

Next week we will begin our second lab of the semester. We will use the millimeter-wave telescope, a radio telescope, in order to observe giant molecular clouds in the Milky Way. This is a lot of new information at once, so we are going to answer some questions before we actually head to lab.

Why? - We are observing GMCs in order to find their redshifts. From redshift, we can find the velocity of a GMC along our line of sight. With that velocity, we can find the circular velocity of a GMC rotating in the Milky Way using \[ V_{cir} = V_r(max) + V_{\odot} sin(l) \] where Vr is the radial velocity of the GMC moving away from us and the second term is the velocity of the sun in the direction of the cloud.

In previous blog posts  we explored why the rotation curve of the galaxy is important in astronomy. The basic idea is that we observe objects to be rotating too quickly, too far out in the galaxy just based on visible matter. This is our main indicator that there is another type of dark matter in the universe. Additionally, we can learn more about the Milky Way's visible matter from this experiment. For example, we can estimate the number of stars in the galaxy. 

How? - The telescope will detect radiation at the wavelength near 2.6 mm, or a frequency of 115.271 GHz. This wavelength corresponds with a strong emission line of Carbon Monoxide. We will obtain this spectrum from the telescope's spectrometer, one of many of the telescope's parts among an antenna for collecting and focusing the signal, a liquid helium cooler, and a computer for recording and interpreting data.

What is going to be the typical integration time per point? - What integration time would be needed to detect the peak of 12CO with SNR = 10 if I use a filter bank with 256 channels that are 0.5 MHz wide. TA for CO is about 2-3 K and Tsys = 500K?

In worksheet 9.2 we found \[ \tau = \frac{ SNR^2 T_{sys}^2 }{T_A^2 \Delta \nu } \] \[ \tau = \frac{ (10)^2 (500)^2 }{(2.5)^2 (0.5 \times 10^6) } \approx 8 \: seconds \]

Over what range of longitude do you plan to observe? - We are planning on observing from \(10^{\circ} \: \text{ to } 70^{\circ} \). 

How many positions do you plan to observe? - We plan on observing 4 GMCs, so we will most likely observe 4 positions.

At what LST are you going to start observing? At what EST? - I have been assigned to Monday's lab from 11:00 - 3:00 EST. Let's say that means we start observing at 11:30 EST. The LST on that date and time at our longitude (71.11 degrees west) would be 15:55.

Additionally, we can solve question number two on the worksheet to learn more about the experiment.  

Resolution of a single dish radio telescope: The spatial resolution of a telescope is θ = 1.2 λ/D , where D is the diameter of the dish, and λ is the observing wavelength. In radio astronomy, θ 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 ν in GHz, and diameter D in meters. Use a calculator to determine this to 1 decimal place. Write an equation of the form \[ \theta = degrees \left( \frac{ \nu}{GHz} \right)^{-1} \left( \frac{D}{m} \right)^{-1} \] We can solve this by redefining theta: \[ \theta = \frac{c}{\nu D} = \frac{1.2 (3 \times 10^8) }{(1 \times 10^9)(1)} = 0.36 \: rad \left( \frac{180^{\circ}}{ \pi rad } \right) = 20.6^{\circ} \] \[ \theta = 20.6^{\circ} \left( \frac{ \nu}{GHz} \right)^{-1} \left( \frac{D}{m} \right)^{-1} \] 

(b) What is the beamwidth, in arcminutes, for the CfA 1.2 m telescope at the 12CO frequency?

Using the formula we just found, we know \[ \theta = 20.6^{\circ} \left( \frac{ \nu}{GHz} \right)^{-1} \left( \frac{D}{m} \right)^{-1} \] \[ \theta = 20.6^{\circ} \left( \frac{ 115.271 \: GHz}{GHz} \right)^{-1} \left( \frac{1.2 \: m}{m} \right)^{-1} = 0.149^{\circ} = 8.95 \: arcminutes \] 

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

We can call on our trusty parallax formula for this problem: \[ d = \frac{1}{ \theta} \] Where 8.95 arcminutes is 537 arcseconds, so \[ d = \frac{1}{ 537} = 1.9 \times 10^{-3} \: pc \] Now we should be all set to start lab.

Correction: The parallax is not a terribly useful number in this case. Rather the question is asking for resolvable size, which we could find with simple trigonometry. \[sin( \theta) = \frac{l}{8.5 \: kpc} \] With the small angle approximation, this is \[ l \approx 0.149 \cdot \frac{\pi}{180} \cdot 8,500 \approx 22 \: pc\]

More Friedmann Equations: Blog 29, Worksheet 9.1, Problem 2

GR modification to Newtonian Friedmann Equation:

In Question 1, you have derived the Friedmann Equation in a matter-only universe in the Newtonian approach. That is, you now have an equation that describes the rate of change of the size of the universe, should the universe be made of matter (this includes stars, gas, and dark matter) and nothing else. Of course, the universe is not quite so simple. In this question we’ll introduce the full Friedmann equation which describes a universe that contains matter, radiation and/or dark energy. We will also see some correction terms to the Newtonian derivation.

A) The full Friedmann equations follow from Einstein’s GR, which we will not go through in this course. Analogous to the equations that we derived in Question 1, the full Friedmann equations express the expansion/contraction rate of the scale factor of the universe in terms of the properties of the content in the universe, such as the density, pressure and cosmological constant. We will directly quote the equations below and study some important consequences. Use the first Friedmann equation: \[ \left( \frac{\dot{a}}{a} \right)^2 = \frac{8 \pi}{3} G \rho + \frac{ \kappa c^2 }{a^2} + \frac{ \Lambda }{3} \] and the second Friedmann equation: \[ \frac{\ddot{a}}{a}  = - \frac{4 \pi G}{3c^2} ( \rho c^2 + 3P) + \frac{ \Lambda }{3} \] To find the third equation in a flat universe \[ \dot{ \rho} c^2 = -3 \frac{\dot{a}}{a}( \rho c^2 + P) \]

With the universe being flat and multiplying the first equation by acceleration squared gives us: \[ \ddot{a}^2 = \frac{8 \pi}{3} G \rho a^2+ \frac{ \Lambda }{3} a^2 \] Taking the time derivative gives us \[ 2 \dot{a} \ddot{a} = \frac{8 \pi}{3} G \dot{\rho} a^2 + \frac{16 \pi}{3} G \rho \dot{a} a   + \frac{2  \Lambda }{3} \dot{a} a \] Rearranging this allows us to set it equal to the second equation: \[  \frac{ \ddot{a} }{a} = \frac{4 \pi a}{3  \dot{a} } G \dot{\rho}  + \frac{ 8 \pi}{3} G \rho + \frac{ \Lambda }{3} =  - \frac{4 \pi G}{3c^2} ( \rho c^2 + 3P) + \frac{ \Lambda }{3} \] Which simplifies with a few cancellations to: \[ \dot{ \rho} c^2 = -3 \frac{\dot{a}}{a}( \rho c^2 + P) \]

B) Cold matter dominated universe. If the matter is cold, its pressure P = 0, and the cosmological constant Λ = 0. Use the third Friedmann equation to derive the evolution of the density of the matter ρ as a function of the scale factor of the universe a. You can leave this equation in terms of ρ, ρ0, a and a0, where ρ0 and a0 are current values of the mass density and scale factor. The result you got has the following simple interpretation. The cold matter behaves like “cosmological dust” and it is pressureless (not to be confused with warm/hot dust in the interstellar medium!). As the universe expands, the mass of each dust particle is fixed, but the number density of the dust is diluted - inversely proportional to the volume. Using the relation between ρ and a that you just derived and the first Friedmann equation, derive the differential equation for the scale factor a for the matter dominated universe. Solve 2 Astronomy 17 - Galactic and Extragalactic Astronomy Fall 2015 this differentiation equation to show that \(a(t) \propto t^{\frac{3}{2}}\) . This is the characteristic expansion history of the universe if it is dominated by matter.

Without pressure, the third equation can be written as \[ \frac{\dot{\rho}}{\rho} = -3 \frac{\dot{a}}{a}\] Taking the derivative of both sides with respect to time starting at \( \rho_0 \: \text{and} \: a_0 \) produces:  \[ \frac{\rho}{\rho_0} = \left(\frac{a}{a_0} \right)^{-3}\] Plugging this relationship into the first equation gives: \[ \left( \frac{\dot{a}}{a} \right)^2 = \frac{8 \pi}{3} G \left( \frac{a_0^3 \rho_0}{a^3} \right) \] Taking the derivative of each side gives \[ a^{\frac{1}{2}} = c \: dt \] Where c is a constant (because we just want a relationship between a and t). Taking the integral of each side gives \[ \frac{2}{3} a^{\frac{3}{2} } = c t \] So \[ a(t) \propto t^{\frac{3}{2}} \]

C) Radiation dominated universe. Let us repeat the above exercise for a universe filled with radiation only. For radiation, \(P = \frac{1}{ 3} ρc^2 \)and Λ = 0. Again, use the third Friedmann equation to see how the density of the radiation changes as a function of scale factor. The result also has a simple interpretation. Imagine the radiation being a collection of photons. Similar to the matter case, the number density of the photon is diluted, inversely proportional to the volume. Now the difference is that, in contrast to the dust particle, each photon can be thought of as wave. As you learned last week, the wavelength of the photon is also stretched as the universe expands, proportional to the scale factor of the universe. According to quantum mechanics, the energy of each photon is inversely proportional to its wavelength: E = hν. Unlike the dust case where each particle has a fixed energy. So in an expanding universe, the energy of each photon is decreasing inversely proportional to the scale factor. Check that this understanding is consistent with the result you got. Again using the relation between ρ and a and the first Friedmann equation to show that \(a(t) \propto t^{\frac{1}{2}}\) for the radiation only universe.

With the new pressure, the third equation can be written as \[ \frac{\dot{\rho}}{\rho} = -4 \frac{\dot{a}}{a}\] Taking the derivative of both sides with respect to time starting at \( \rho_0 \: \text{and} \: a_0 \) produces:  \[ \frac{\rho}{\rho_0} = \left(\frac{a}{a_0} \right)^{-4}\] Plugging this relationship into the first equation gives: \[ \left( \frac{\dot{a}}{a} \right)^2 = \frac{8 \pi}{3} G \left( \frac{a_0^4 \rho_0}{a^4} \right) \] Taking the derivative of each side gives \[ a = c \: dt \] Where c is a constant (because we just want a relationship between a and t). Taking the integral of each side gives \[ \frac{1}{2} a^{2 } = c t \] So \[ a(t) \propto t^{\frac{1}{2}} \]

D) Cosmological constant/dark energy dominated universe. Imagine a universe dominated by the cosmological-constant-like term. Namely in the Friedmann equation, we can set ρ = 0 and P = 0 and only keep Λ nonzero. As a digression, notice that we said “cosmological-constant-like” term. This is because the effect of the cosmological constant may be mimicked by a special content of the universe which has a negative pressure \(P = - ρc^2 \). Check that the effect of this content on the right-hand-side of third Friedmann equation is exactly like that of the cosmological constant. To be general we call this content the Dark Energy. How does the energy density of the dark energy change in time? Show that the scale factor of the cosmological-constant-dominated universe expands exponentially in time. What is the Hubble parameter of this universe?

From the first equation we have: \[ \left( \frac{\dot{a}}{a} \right)^2 =  \frac{ \Lambda }{3} \] Taking the derivative gives: \[ \frac{da}{a} = \sqrt{ \frac{ \Lambda }{3} } dt \] Ignoring constants, the integral of this relationship turns into \[ ln(a) \propto t \] \[ a \propto e^t \] And because \[ H=  \frac{\dot{a}}{a} \] The Hubble constant in this universe is  \[ H = \sqrt{ \frac{ \Lambda }{3} } \]

E) Suppose the energy density of a universe at its very early time is dominated by half matter and half radiation. (This is in fact the case for our universe 13.7 billion years ago and only 60 thousand years after the Big Bang.) As the universe keeps expanding, which content, radiation or matter, will become the dominant component? Why?

Radiation will dominate because it falls off at a slower rate ( \(a(t) \propto t^{\frac{1}{2}}\) ) than matter ( \(a(t) \propto t^{\frac{2}{3}}\) ).

F) Suppose the energy density of a universe is dominated by similar amount of matter and dark energy. (This is the case for our universe today. Today our universe is roughly 68% in dark energy and 32% in matter, including 28% dark matter and 5% usual matter, which is why it is acceleratedly expanding today.) As the universe keeps expanding, which content, matter or the dark energy, will become the dominant component? Why? What is the fate of our universe?

Dark energy will be dominant as the universe expands and matter is scattered. Energy, and therefore normal matter, cannot be created or destroyed, by the law of conservation of energy. We have also shown that dark energy will increase at a faster rate ( \(a \propto e^t\) ) than radiation ( \(a(t) \propto t^{\frac{2}{3}}\) ). Eventually, our universe will be a very sparse, vast expanse of dark energy with few pockets of matter spread throughout.

Cosmology: Blog 28, Worksheet 9.1, Problem 1

A Matter-only Model of the Universe in Newtonian Approach 

In this exercise, we will derive the first and second Friedmann equations of a homogeneous, isotropic and matter-only universe. We use the Newtonian approach. Consider a universe filled with matter which has a mass density ρ(t). Note that as the universe expands or contracts, the density of the matter changes with time, which is why it is a function of time t. Now consider a mass shell of radius R within this universe. The total mass of the matter enclosed by this shell is M. In the case we consider (homogeneous and isotropic universe), there is no shell crossing, so M is a constant.

(a) What is the acceleration of this shell? Express the acceleration as the time derivative of velocity,  \( \dot{v} \)(pronounced v-dot) to avoid confusion with the scale factor a (which you learned about last week).

Because this acceleration is due to gravity, we know \[ ma  = F_g \] \[ m\dot{v} = - \frac{GMm}{R^2} \] \[ \dot{v} = - \frac{GM}{R^2} \]

(b) To derive an energy equation, it is a common trick to multiply both sides of your acceleration equation by v. You should arrive at the following equation: \[ \frac{1}{2} \dot{R}^2 - \frac{GM}{R} = C \]

Using what we found in a: \[ \frac{dR}{dt} \dot{v} = - \frac{GM}{R^2} \frac{dR}{dt} \] Which simplifies to \[v \: dv = - \frac{GM}{R^2} dR \] The integral of this is \[ \frac{1}{2} v^2 = \frac{GM}{R} + C \] Where C is a constant. Substituting \( v = \dot{R} \) we have: \[ \frac{1}{2} \dot{R}^2 - \frac{GM}{R} = C \]

(c) Express the total mass M using the mass density, and plug it into the above equation. Rearrange your equation to give an expression for \( \left( \frac{ \dot{R}}{R} \right)^2 \), where \(  \dot{R} = \frac{dR}{dt}  \). \[ M = \rho V = \frac{4}{3} \pi R^3 \rho \] Plugging this into our expression from part B gives us:  \[ \frac{1}{2} \dot{R}^2 - \frac{G \left( \frac{4}{3} \pi R^3 \rho \right)}{R} = C \] Rearranging this makes it: \[ \left( \frac{\dot{R}}{R} \right)^2 = \frac{2C}{R^2} + \frac{8 \pi G}{3} \rho (t) \]

(d) R is the physical radius of the sphere. It is often convenient to express R as R = a(t)r, where r is the comoving radius of the sphere. The comoving coordinate for a fixed shell remains constant in time. The time dependence of R is captured by the scale factor a(t). The comoving radius equals to the physical radius at the epoch when a(t) = 1. Rewrite your equation in terms of the comoving radius, R, and the scale factor, a(t). \[  \frac{\dot{R}}{R} =  \frac{\dot{a}(t) r}{a(t) r} =  \frac{\dot{a}}{a} \] So our expression is  \[ \left( \frac{\dot{a}}{a} \right)^2 = \frac{2C}{R^2} + \frac{8 \pi G}{3} \rho (t) \]

(e) Rewrite the above expression so that \( \left( \frac{\dot{a}}{a} \right)^2 \) appears alone on the left side of the equation. \[ \checkmark \]

(f) Derive the first Friedmann Equation: From the previous worksheet, we know that \( H(t) =  \frac{\dot{a}}{a} \). Plugging this relation into your above result and identifying the constant \( \frac{2C}{r} = \kappa c^2 \) where k is the “curvature” parameter, you will get the first Friedmann equation. The Friedmann equation tells us about how the shell of expands or contracts; in other words, it tells us about the Hubble expansion (or contracration) rate of the universe. \[ \left( \frac{\dot{a}}{a} \right)^2 = \frac{8 \pi}{3} G \rho - \frac{ \kappa c^2 }{a^2} \]

(g) Derive the second Friedmann Equation: Now express the acceleration of the shell in terms of the density of the universe, and replace R with R = a(t)r. You should see that a: \( \frac{\ddot{a}}{a}  = - \frac{4 \pi G}{3c^2} \rho \), which is known as the second Friedmann equation. The more complete second Friedmann equation actually has another term involving the pressure following from Einstein’s general relativity (GR), which is not captured in the Newtonian derivation. If the matter is cold, its pressure is zero. Otherwise, if it is warm or hot, we will need to consider the effect of the pressure.
\[ \dot{v} = - \frac{GM}{R^2} = -\frac{4}{3} \pi G \rho (t) a(t) r \] We know \( \dot{v} = \ddot{R} \) so \(  \dot{v} = \ddot{a}r \) and \[\ddot{a}r = -\frac{4}{3} \pi G \rho (t) a(t) r \]  \[ \frac{\ddot{a}}{a}  = - \frac{4 \pi G}{3c^2} ( \rho c^2 + 3P) + \frac{ \Lambda }{3} \]

The Age and Size of the Universe: Blog 27: Worksheet 8.1, Problem 3

It is not strictly correct to associate this ubiquitous distance-dependent redshift we observe with the velocity of the galaxies (at very large separations, Hubble’s Law gives ‘velocities’ that exceeds the speed of light and becomes poorly defined). What we have measured is the cosmological redshift, which is actually due to the overall expansion of the universe itself. This phenomenon is dubbed the Hubble Flow, and it is due to space itself being stretched in an expanding universe. Since everything seems to be getting away from us, you might be tempted to imagine we are located at the centre of this expansion. But, as you explored in the opening thought experiment, in actuality, everything is rushing away from everything else, everywhere in the universe, in the same way. So, an alien astronomer observing the motion of galaxies in its locality would arrive at the same conclusions we do. In cosmology, the scale factor, a(t), is a dimensionless parameter that characterizes the size of the universe and the amount of space in between grid points in the universe at time t. In the current epoch, t = \(t_0\) and \( a(t)_0 = 1\) a(t) is a function of time. It changes over time, and it was smaller in the past (since the universe is expanding). This means that two galaxies in the Hubble Flow separated by distance \(d_0 = d(t_0) \) in the present were \(d(t) = a(t)d_0 \)apart at time t. The Hubble Constant is also a function of time, and is defined so as to characterize the fractional rate of change of the scale factor: \[H(t) = \frac{1}{a(t)} \frac{ da}{ dt}|_t \] and the Hubble Law is locally valid for any t: \[v = H(t)d\] where v is the relative recessional velocity between two points and d the distance that separates them.

Part A: Assume the rate of expansion, a = da/dt, has been constant for all time. How long ago was the Big Bang (i.e. when a(t=0) = 0)? How does this compare with the age of the oldest globular clusters (= 12 Gyr)? What you will calculate is known as the Hubble Time.

Since we know \[v = H(t)d\] \[H(t) = \frac{v}{d} = \frac{1}{t} \] So \[ t_0 = \frac{1}{H(t)} = \frac{1}{68.816 \frac{km/s}{Mpc}} = 0.01453 \frac{Mpc \cdot s}{km} \] This is not a very useful number for us, so we can convert it to years: \[ 0.01453 \frac{Mpc \cdot s}{km} \cdot \frac{ 3.086 \times 10^{19} \: km}{Mpc} \cdot \frac{ 1 year }{3.15 \times 10^7}  = 1.385 \times 10^{10} \: years \] This is pretty close! The actual age of the universe is 13.82 billion years. This means the oldest globular clusters formed less than 2 billion years after the big bang.

Part B:  What is the size of the observable universe? What you will calculate is known as the Hubble Length.

Once again, we know \[ v = H(t)d\] So, if the universe is expanding at the speed of light and the speed of light in km/s is \( 3 \times 10^{5} \). \[ d = \frac{c}{H(t)} \] \[ d = \frac{3 \times 10^{5}}{68.816} = 4,356 Mpc \approx 4 \times 10^{3} Mpc \] The actual size of the universe is about \( 3 \times 10^{3} \) Mpc, so we aren't too far off.

Our old friends, Lyman-alphas: Blog 24 & 25, Worksheet 7.2, Problem 4 & 5

This week, we learned about active galactic nuclei and all of their fun applications. One of those applications in particular is especially dear to me, because the summer before tenth grade, I did a research project (and wrote a blog post about it) in a lab where I calculated the distance to ancient galaxies using redshift from Lyman-alpha emitting active galactic nuclei. And wouldn't you know, that's what we're doing in class this week:

Problem 4

One feature you surely noticed in a spectrum was the strong, broad emission lines. Here is a closer look at the strongest emission line in the spectrum:

This feature arises from hydrogen gas in the accretion disk. The photons radiated during the accretion process are constantly ionizing nearby hydrogen atoms. So there are many free protons and electrons in the disk. When one of these protons comes close enough to an electron, they recombine into a new hydrogen atom, and the electron will lose energy until it reaches the lowest allowed energy state, labeled n = 1 in the model of the hydrogen atom shown below (and called the ground state):

On its way to the ground state, the electron passes through other allowed energy states (called excited states). Technically speaking, atoms have an infinite number of allowed energy states, but electrons spend most of their time occupying those of lowest energies, and so only the n = 2 and n = 3 excited states are shown above for simplicity. Because the difference in energy between, e.g., the n = 2 and n = 1 states are always the same, the electron always loses the same amount of energy when it passes between them. Thus, the photon it emits during this process will always have the same wavelength. For the hydrogen atom, the energy difference between the n = 2 and n = 1 energy levels is 10.19 eV, corresponding to a photon wavelength of λ = 1215.67 Angstroms. This is the most commonly-observed atomic transition in all of astronomy, as hydrogen is by far the most abundant element in the Universe. It is referred to as the Lyman α transition (or Lyα for short). It turns out that that strongest emission feature you observed in the quasar spectrum above arises from Lyα emission from material orbiting around the central black hole.

Part A: Recall the Doppler equation: \[ \frac{ \lambda_{observed} - \lambda_{emitted}}{\lambda_{emitted}} = z \approx \frac{v}{c} \]Using the data provided, calculate the redshift of this quasar. \[ z =  \frac{ 1410 - 1215.67}{1215.67} \approx \frac{1}{6}  \] So that \[ v \approx \frac{c}{6}\]

Part B: Again using the data provided, along with the Virial Theorem, estimate the mass of the black hole in this quasar. It will help to know that the typical accretion disk around a \(10^8 M_{\odot} \) black hole extends to a radius of r = \(10^{15} \) m. 

Ideally, our emission lines would be infinitely thin and tall, but they have width. This happens because there is some dispersion due to the rotation of the galaxy. This means some of the galaxy will be redshifted and some will be blueshifted. 

So we can use this to find rotational velocity.

\[ z = \frac{ \lambda_{observed} - \lambda_{emitted}}{\lambda_{emitted}} = \frac{1415 - 1406}{1406} \approx 0.00640 \]

\[ z = \frac{1415 - 1406}{1406} = 0.0064 \] \[ v \approx 0.0064 c \]

Now we can use the virial theorem: \[ K = - \frac{1}{2} U \] \[ Mv^2 = \frac{G M^2}{R}\] \[M_{BH} = \frac{ v^2 R}{G} = \frac{( 0.0064 \cdot 3 \times 10^{10})^2 (10^{17})}{6.67 \times 10^{-8}} = 5.5 \times 10^{40} \: g \approx 3 \times 10^7 M_{\odot} \] 

Problem 5

You may also have noticed some weak “dips” (or absorption features) in the spectrum:

Part A: Suggest some plausible origins for these features. By way of inspiration, you may want to consider what might occur if the bright light from this quasar’s accretion disk encounters some gaseous material on its way to Earth. That gaseous material will definitely contain hydrogen, and those hydrogen atoms will probably have electrons occupying the lowest allowed energy state.

As the question implies, we are probably looking at instances where light is being absorbed or emitted by hydrogen gas on its way to us. These dips themselves, occurring before the peak, imply that this gas is between the galaxy and us, therefore they are not as redshifted as the distant galaxy. 

Part B: A spectrum of a different quasar is shown below. Assuming the strongest emission line you see here is due to Lyα, what is the approximate redshift of this object?

\[ \frac{ \lambda_{observed} - \lambda_{emitted}}{\lambda_{emitted}} = z \approx \frac{v}{c} \]

\[ z = \frac{5650 - 1215.67}{1215.67} = 3.347 \approx 3\]

Part C: What is the most noticeable difference between this spectrum and the spectrum of 3C 273? What conclusion might we draw regarding the incidence of gas in the early Universe as compared to the nearby Universe?

In this galaxy, there are many more variation, especially a larger number of deeper "dips." Based on the large redshift of the galaxy, we know it is quite old and far away. This means the light had to have traveled through a lot more gas to get here, and some of that gas had hydrogen atoms occupying the lowest energy state. Additionally, this could tell us that there was a lot more gas in the early universe. This fits in well with our current ideas of the early universe. 

Deriving Hubble's Law: Blog 26: Worksheet 8.1: Problem 2

In 1929, astronomer Edwin Hubble discovered that almost all distant galaxies exhibit a positive redshift. Furthermore, it appeared that the farther the galaxy, the larger its redshift. Here we will rediscover Hubble’s Law using modern spectroscopic data and supernovae Ia as our distance indicator to these galaxies. The data we will use come from the Sloan Digital Sky Survey (SDSS), a project that aims to comprehensively map the universe. You can access the relevant data products for this exercise at http://goo.gl/fmIvqc

Part A: Below is a list of supernovae observed between 2004 and 2007 and their positions in RA and Dec. You can find the images and spectra of their host galaxies by entering their coordinates in the respective fields. Explore the functions available, including magnifying the image, reading off the photometric measurements (magnitudes in wavebands u, g, r, i, z) of your selected object, and using the ‘Explore’ button to access more quantitative measurements for these objects. In particular, familiarize yourself with the ‘interactive spectrum’ feature.

For each galaxy, we have a pretty extensive spectrum.

Part B: One of the features for determining distances to Type Ia supernovae is its peak absolute magnitude. You explored the peak bolometric luminosities of SN Ia’s in Worksheet 7.1. The peak V-band magnitude for SN Ia’s is about -19.3. Use the apparent peak magnitudes given in the table above to calculate the distance of these supernovae in unit of Mpc.

To find distance, we can use our trusty distance formula :\[d = 10^{ \frac{m - M + 5}{5}} \] for each magnitude. The fourth galaxy did not have a spectrum associated with it, so we did not analyze it.

Part C: We can use the absorption or emission lines of the host galaxy to find their redshifts which, as you found in Question 1), roughly equals the recessional velocity as a fraction of the speed of light. To measure the redshift to each host galaxies, click on ‘Explore’ and then on the link ‘Interactive Spectrum’. Uncheck the boxes Best Fit and Mark Emission Lines. Zoom in on the absorption line labeled Hα, and move your mouse over to the center of the line to read the observed wavelength in Angstroms. The Hα has a rest (i.e. emitted) wavelength of 6563.0 Angstroms. Calculate the redshift, and then derive the radial velocity in kilometers per second, using the relation z = v/c. How close does your redshift measurements compare to the one SDSS reports in the table under the Interactive Spectrum link? Repeat for all the galaxies.

We found redshift and velocity using \[ \frac{ \lambda_{observed} - \lambda_{emitted}}{\lambda_{emitted}} = z \approx \frac{v}{c} \] to find: 

Part D: Make a plot of your findings, with distance on the x-axis and velocity on the y-axis. Report the slope of your line in appropriate units. This is the Hubble Constant, H0.

Result: \(H_0 = 68.816 \frac{km/s}{Mpc} \)

Part E: Write an equation for this line in the form of v = ___ D, where v is an object’s recessional velocity and D is the distance to that object. Express your Hubble Constant in terms of units km/s/Mpc. Congratulations, you have arrived at Hubble’s Law!
\[ v = 68.816 \frac{km/s}{Mpc} D \] Now we have a basic law of the universe (kinda) down!

Blog 23: Astronomy and the Harvard Art Museums

This week, we are going to explore the intersection of Astronomy and another field because while Astronomy is the study of the universe, it is not the only subject to study in the universe. Last semester, I talked about Astronomy and Mystery-Solving. This semester I started a job as a tour guide at the Harvard Art Museums (HAM). At HAM, tour guides have to design their own tours based on a central theme connecting 3-5 objects on display in the museums. Themes vary from political to religious to technical. For this blog post I am going to try to propose an astronomy-themed tour based on our collection. All images and information about the objects can be found at the museum's website.

Admittedly, astronomy is not one of the most centralizing themes in art history. However, it has been a constant part of our lives before, throughout, and long after history. The night sky has always been round us, both inciting our imaginations and giving us the tools to understand the world we live in. The oldest astronomy-themed object on this tour is the Long-Case Musical Clock, created by the Dutch craftsman Otto Van Meurs in 1750. The grand, imposing clock sits in a seconds floor gallery of the museum, still keeping accurate time - with a little adjusting from conservationists every set number of years. That is not all the clock does; it also tracks tides, the current position of the sun and moon, gives the date, and plays one of dozens of charming tunes every half hour. Mechanical clocks like these are one of the first iterations of modern computers. This "computer" in particular, heavily showcases society's uses for the skies above. Our entire conception of time, of calendars and schedules, of past, present, and future is based on the motion of stars and planets. We can see that in this piece, a representation of years of evolution of the clock and calendar.

Yet this piece isn't purely functional. It is highly symbolic, covered in classical allusions such as Atlas carrying Earth at the very top of the clock. This piece, used to show the high societal status of its owner, is inherently intertwined with modern cultural norms. This culminates in a symbolic, functional, and aesthetic piece that dominates any room it is in.

More than a century later, Paul Manship (considered the most prominent American Art Deco artist) created about a dozen celestial spheres as a preparation for a grand memorial to Woodrow Wilson. This historically influential tool was also highly functional, allowing for navigation, charting, and seasonal predictions. Even today, it is our foremost model of the night sky.

Celestial Sphere - Paul Manship

However, this piece itself is also not purely functional, it is not even very useful. Its chief function is to memorialize a political figure and does so with many cultural symbols. The constellations and base figures are a mix of Chinese, Babylonian, Assyrian, Latin, and Greek zodiacs to emphasize the international political harmony the artist wanted a viewer to associate with Woodrow Wilson. Once again, Astronomy is used here as a seamless part of history and modern society, and not just for one culture, but for many. 

There are definitely cultural levels of astronomy that move beyond the scientific into the even more cultural and aesthetic. For example, the second floor has a beautiful painting with a puzzling composition that signifies the coming of the dawn, the daily transition from dark to light and a pagan goddess. The astronomical phenomena has become a personified deity, with her own gossip, lovers, and enemies.

The Dawn - John LaFarge

Even Christian religions have used Astronomy as cultural markers. In this stained glass window, angels are represented by the morning stars in a reference to a biblical passage. The stars symbolized heavenly light, purity, and truth.

When the Morning Stars Sang Together and All the Sons of God Shouted for Joy

Even in contemporary art today, these themes persist. De Kooning's piece, Untitled (The Cow Jumps Over the Moon), plays upon a common cultural anchor, a childhood nursery rhyme. From there, De Kooning's questions a viewer's perspective of traditional art and accepted ways of portraying cows and emotions alike. 

Willem de Kooning - Untitled (The Cow Jumps Over the Moon)

Even in art's evolution, Astronomy is used as a grounding point. The night sky has become irrevocably intertwined with our perception of the world, and what better way is there to portray our world than through art?