Blog #37: Illustris Simulation

The Halo Mass Function

When I zoomed in on a random region of high density, this is what I found:

Because I am technologically inept, here is a hand-drawn histogram of the masses of the subhaloes that I zoomed in. It appears that low-mass haloes are more common than high-mass haloes.

Because I'm only mostly technologically inept, I was able to use Excel to calculate that approximately 91% on average of the mass of the subhaloes (or at least the subhaloes that Illustris provided me with) was stellar mass.


Exploring Structure and Reionization in the Illustris Simulation 

Large scale:

Dark matter density

Gas density 

Small scale:

Dark matter density

Gas density 

The dark matter appears to be more confined to the filamentary structure than the gas is. This could be because it is denser than the gas and therefore it is more favorable for it to aggregate along the filaments, whereas gas is less affected by gravity and therefore more free to diffuse throughout space. On the smaller scale, however, the gas appears to aggregate more tightly than the dark matter does.

In medium to large galaxies, the gas appears to be densest near the center of the disk. This makes sense given what we have learned about the structure of the Milky Way, that a higher density of matter, including gas, towards its center.

Gas density (blue is denser) relative to galaxies (circles)

The most massive galaxies tend to cluster, which makes sense given their gravitational attraction to each other. 

The clustering of galaxies (circles) 

After watching the video, it seems clear that the dark matter is leading the structure formation. The formation of filaments is apparent much earlier in the dark energy box. 

This is only a bit after the beginning of the "Epoch of Reionization," when hydrogen atoms become ionized (indicated by the color that is starting to appear in the gas box). According to the video, it appeared to start around a redshift of 7.7, or 0.7 billion years. 

One of the fastest rates of star formation appears to be at redshift of between 3.50-2.90 or so, although it seems that star formation goes through cycles of speeding up and slowing down. It also appears that new objects are formed when large objects break up, as there are explosions occurring in the gas box, which will eject matter away to form smaller objects later. 

Structures probably form along these filaments because that is where the dark matter is. The dark matter gravitationally attracts normal matter, which makes it more favorable for structures to form there instead of in random regions of space. 

Blog #36: WS12.1, #1&2d

1. Linear perturbation theory. In the early universe, the matter/radiation distribution of the universe is very homogenous 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 \[\delta(r,t)\equiv\frac{\rho(r,t)-\bar{\rho}(t)}{\bar{\rho}(t)}.\]In this exercise we focus on the linear theory, namely, the density contrast in the problem remains small enough so we only need to 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 and 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. 

We can check that this is a solution by plugging the third equation into the second.
\(\frac{d\delta}{dt}=\dot{D}(t)\tilde{\delta}(x)\)
\(\frac{d^2\delta}{dt^2}=\ddot{D}(t)\tilde{\delta}(x)\)
\(\ddot{D}(t)\tilde{\delta}(x)+\frac{2\dot{a}}{a}\dot{D}(t)\tilde{\delta}(x)=4\pi G\bar{\rho}D(t)\tilde{\delta}(x)\)
\(\ddot{D}(t)+\frac{2\dot{a}}{a}\dot{D}(t)=4\pi G\bar{\rho}D(t)\)
This is a solution to the 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, \(3H^2_0/8\pi G\) as in WS11.1. Recall that the behavior 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; you will end up with a quadratic equation for q, There are two solutions, 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 \(D_-(t)\). 

By this ansatz, we can say:
\(D(t)=t^q\)
\(\dot{D}(t)=qt^{q-1}\)
\(\ddot{D}(t)=q(q-1)t^{q-2}\)
We can then plug into the differential equation.
\((q)(q-1)t^{q-2}+\frac{2\dot{a}}{a}qt^{q-1}=4\pi G\bar{\rho}t^q\)
We can multiply by \(t^{2-q}\) and substitute in today's critical density.
\((q)(q-1)+\frac{2\dot{a}}{a}qt-4\pi Ga^{-3}\rho_{c,0}t^2=0\)
We plug in the critical density and \(\frac{2}{3t}\) for \(\frac{\dot{a}}{a}\).
\(q^2+\frac{1}{3}q-\frac{2}{3}=0\)
\(q=\frac{2}{3}, -1\)

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

Since this is a linear combination, it means that \(D(t)=D_+(t)+D_-(t)=At^{2/3}+Bt^{-1}\).
\(D_+(t)\propto t^{2/3}\propto a(t)\). This term will grow as time increases, since t's exponent is positive.
\(D_-(t)\propto t^{-1}\). This term will decrease as time increases, since t's exponent is negative.
As a result, \(D_+(t)\) will come to dominate as time moves forward.

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. However, in some very special situations, analytical treatment is possible and provides 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 (open, closed, and flat universe), and show that in the closed case, the particle turns around and collapses; 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. 

Closed case: blue
\(r=A(1-cos\eta)\)
\(t=B(\eta-sin\eta)\)
The particle's radius increases, but then decreases until the particle collapses to zero again.

Open case: red
\(r=A(cosh\eta-1)\)
\(t=B(sinh\eta-\eta)\)
The particle's radius appears to continue to increase.

Flat case: green
\(r=A\eta^2/2\)
\(t=B\eta^3/6\)
The particle's radius appears to increase with a decreasing velocity.

Zoomed in 

Zoomed out

The x-axis is time and the y-axis is radius. A and B were arbitrarily assigned values of 1, but in practice these values depend on the mass of the patch of over-density.

Blog #35: WS11.1, #3

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

a) Despite the fact that the CMB has a very low temperature, the number of photons is enormous. Let us estimate what that number is. Each photon has energy \(h\nu\). From the Planck spectrum, 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. Keep all factors and use the fact that \(\int^{\infty}_0\frac{x^2}{e^x-1}dx\approx2.4\). 

The number density is the energy density (\(u_{\nu}d\nu\)) divided by the energy per photon (\(h\nu\).
\(n_{\nu}=\int^{\infty}_0\frac{8\pi h\nu^3}{c^3}\frac{1}{e^{\frac{h_P\nu}{k_BT}}-1}\frac{1}{h\nu}d\nu=\frac{8\pi}{c^3}\int^{\infty}_0\frac{\nu^2}{e^{\frac{h_P\nu}{k_BT}}-1}d\nu\)
By doing u-substitution and using the fact provided above, we can finish the integration.
\(u=\frac{h_P\nu}{k_BT}\)
\(n_{\nu}=\frac{8\pi}{c^3}(\frac{k_BT}{h_P})^3\int^{\infty}_0\frac{u^2}{e^u-1}du=\frac{8\pi}{c^3}(\frac{k_BT}{h_P})^3 2.4\)

b) Using the following values for the constants: \(k_B=1.38\times10^{-16}erg/K\), \(c=3.00\times10^{10}cm/s\), \(h_P=6.62\times10^{-27}erg\cdot s\), and use the temperature of the CMB today that you have computed to calculate the number density of photons in our universe today. 

\(\frac{8\pi}{(3\times10^{10}cm/s)^3}[\frac{(1.38\times10^{-16}erg/K)(2.72K)}{6.62\times10^{-27}erg\cdot s}]^32.4=407/cm^3\)

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 \(9.2\times10^{-30}g/cm^3\). The baryon density is about 4% of it. The masses of proton and neutron are very similar (\(\approx1.7\times10^{-24}g\)). What is the number density of baryons? 

\(\frac{0.04(9.2\times10^{-30}g)}{1cm^3}\frac{1 baryon}{1.7\times10^{-24}g}=2.16\times10^{-7} baryons/cm^3\)

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

\(\frac{2.16\times10^{-7}baryons/cm^3}{407photons/cm^3}=5.32\times10^{-10}baryons/photon\)

Blog #34: WS11.1, #2

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 satisfies the Planck spectrum \[u_{\nu}d\nu=\frac{8\pi h_P\nu^3}{c^3}\frac{1}{e^{\frac{h_P\nu}{k_BT}}-1}d\nu\]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 their 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 photon was emitted at redshift z with frequency \(\nu\), what is its frequency \(\nu'\) today? 

Based on the redshift equation, we can find that \(z=\frac{\nu-\nu'}{\nu'}=\frac{\nu}{\nu'}-1\)
Solving for \(\nu'\), we get \(\nu'=\frac{\nu}{z+1}\).

b) If a photon at redshift z had the energy density \(u_{\nu}d\nu\), what is its energy density \(u_{\nu'}d\nu'\) today? 

The energy of the photon decreases as its wavelength gets stretched with the expanding of space. This is quantified by the changing frequency calculated above: the energy decreases by a factor of \(\frac{1}{z+1}\). The number density of photons also decreases as space expands. If we imagine a region of space expanding:
\(v\propto r^3\propto a^3r^3\propto(\frac{1}{z+1})^3r^3\), since the volume depends on the scale factor a, which is itself also proportional to \(\frac{1}{z+1}\).
The combined effect means that \(u_{\nu'}d\nu'=u_{\nu}d\nu(\frac{1}{z+1})^4\).

c) Plug in the relation between \(\nu\) and \(\nu'\) into the Planck spectrum and multiply it with the overall energy density dilution factor that you have just figured out to get the energy density today. Write the final expression as the form \(u_{\nu'}d\nu'\. What is \(u_{\nu'}\)? This is the spectrum we observe today. Show that it is exactly the same as the Planck spectrum, except that the temperature is now \(T'=T(1+z)^{-1}\). 

We know the following:
\(T=T'(z+1)\)
\(u_{\nu}d\nu=u_{\nu'}d\nu'(z+1)^4\)
\(\nu=\nu'(z+1)\), which means that \(d\nu=d\nu'(z+1)\)
We can then plug these into the Planck equation.
\(u_{\nu'}d\nu'(z+1)^4=\frac{8\pi h_P[\nu'(z+1)]^3}{c^3}\frac{1}{e^{\frac{h_P\nu'(z+1)}{k_BT'(z+1)}}}d\nu'(z+1)\)
\(u_{\nu'}d\nu'=\frac{8\pi h_P\nu'^3}{c^3}\frac{1}{e^{\frac{h_P\nu'}{k_BT'}}}d\nu'\)
This is the same form as the original Planck spectrum.
\(u_{\nu'}=\frac{8\pi h_P\nu'^3}{c^3}\frac{1}{e^{\frac{h_P\nu'}{k_BT'}}}\)

d) According to the 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 the temperature T' today? 

\(T'=T(z+1)^{-1}\)
\(T'=\frac{3000K}{1100+1}=2.72K\)

Blog #33: Why Thanksgiving is the Most Underrated Holiday

We all know those people who, as soon as Halloween is over, break out the Christmas music (not guilty) and the winter-themed socks (definitely guilty). And it's not like they're being discouraged from doing so; for example, Spotify has been giving me aggressive Christmas-themed ads for the past couple weeks at least. Don't get me wrong, I love Christmas carols, Elf, and candy canes just as much as the next girl (plus I wrote a Christmas-themed blog post in October so that might make me a worse offender than the Christmas-songs-in-November crowd), but I don't like it when people forget about the major holiday that exists between Halloween and Christmas--and I'm not talking about Black Friday.

Thanksgiving! There are so many wonderful things about Thanksgiving that I don't even know where to start, so I'll just repeat: Thanksgiving!!! When most people think about Thanksgiving, probably the first thing they think about is food (note: I just proved this scientifically by asking my roommate). I love food. Probably my only weakness is my inability to resist delicious food (along with my crappy sense of direction).

People always used to compare me to Pam but I didn't understand until seeing this.

My dad is an awesome cook and he spends the entire day and a half before Thanksgiving cooking. Bacon-covered turkey is a prime example and one of my favorites, but our guests are from all over the world so they always bring really interesting food too.

Another great thing about Thanksgiving is getting to spend time with family and friends. This is something I've definitely come to appreciate more since coming to college. I haven't seen my family since August, and it'll be great to be able to annoy my parents again in person and bug my little brothers about the presumably-directionless lives they've been living without my guidance. I also get to be reunited with some of my best friends from home, my pickup truck (which I probably like more than I like most people), and, perhaps most importantly, my cat (who I definitely like more than I like most people).

But wait!--you say. Both of those things are true of Christmas too!

Yes, but there are two other things that make Thanksgiving so amazing. First, the timing. It's getting to the point in the semester where I really just need a break (actually we had probably gotten to that point about 4 or 5 weeks ago, but I've made it since then with only a few minor crises). I tend to get really sick at the end of each spring semester, and I think Thanksgiving break is what rejuvenates me enough to not die at the end of each fall semester.

Me at the end of every semester.

The second thing is that Thanksgiving focuses on being happy about what you already have, whereas Christmas, in all of its consumerist glory, kind of ends up feeling the opposite. I always spend the few weeks before Christmas panicking about getting gifts for other people and panicking whenever anyone asks me what I want for Christmas, because I never have any idea. Then I spend the week after Christmas feeling vaguely annoyed and dissatisfied with whatever I've gotten, regardless of the fact that I didn't know what I wanted in the first place so why would anybody else know what to get me? (I apologize to anyone who has ever had the misfortune of having to buy me a gift.)

Anyway, I'm not saying that Thanksgiving is necessarily better than Christmas--I really love Christmas too. Basically the bottom line (so that I don't end on a cliché) is that people need to stop under-appreciating Thanksgiving because it's a pretty awesome holiday, and I'm super pumped to be on break in less than 24 hours (shoutout to my Tuesday classes for canceling lecture!).

Confession: I was actually ready for Thanksgiving in August.

Happy Thanksgiving from me and my dorky brothers!

Blog #32: WS10.1, #3

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 is a matter dominated universe in co-moving coordinates. 
a) First of all, why do we use the light to figure out the horizon size? 

Since light has a finite speed and finite time has passed since the Big Bang, the part of the universe that is observable to us can be no larger than the distance that light can travel in the time interval since the Big Bang.

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. 

The FRW metric is:
\(ds^2=-c^2dt^2+a^2(t)[\frac{dr^2}{1-kr^2}+r^2(d\theta^2+sin^2\theta d\phi^2)]\)
Since we know that \(ds^2=d\theta=d\phi=0\), it becomes:
\(0=-c^2dt^2+a^2(t)[\frac{dr^2}{1-kr^2}]\)
\(c^2dt^2=a^2(t)[\frac{dr^2}{1-kr^2}]\)
\(\frac{dr}{dt}=\frac{c}{a(t)}(\frac{1}{1-kr^2})^{-1/2}\)

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. Find the radius of the horizon today (at \(t=t_0\)). 
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 \(t_0\). 

\(\int^{t_0}_0\frac{cdt}{a(t)}=\int^{r_{horizon}}_0(\frac{1}{1-kr^2})^{1/2}dr\)
We know that k = 0, since this is a flat universe. We also know that \(a\propto t^{2/3}\) in a matter-dominated universe.
\(c\int^{t_0}_0t^{-2/3}=\int^{r_{horizon}}_0(\frac{1}{1})^{1/2}dr\)
\(3ct_0^{1/2}\propto r_{horizon}\)

Blog #31: WS10.1, #2

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 \[ds^2_{2d}=dr^2+r^2d\theta^2.\]The circumference is found by fixing the radial coordinate (r = R and dr = 0) and both sides of the equation (note that \(\theta\) is integrated from 0 to 2π). 
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 2π. 

Since dr = 0 and r = R, our equation becomes \(ds^2_{2d}=R^2d\theta^2\).
\(ds_{2d}=Rd\theta\)
We can then integrate. Since we want the equation for circumference, we integrate ds from 0 to the full circumference.
\(\int^{circ}_0ds_{2d}=R\int^{2\pi}_0d\theta\)
\(circumference=2\pi R\), as we know.
The ratio of circumference to radius would therefore be \(\frac{2\pi R}{R}=2\pi\).

b) For a closed geometry, we calculated the analogous two-dimensional part of the metric in problem 1. This can be written as: \[ds^2_{2d}=d\xi^2+sin^2\xi d\theta^2.\] 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? 

Since \(\xi\) essentially represents the radius of the circle in this geometry, we can again fix it such that \(d\xi=0\). The process is then very similar to that of part a:
\(ds^2_{2d}=sin^2\xi d\theta^2\)
\(ds_{2d}=sin\xi d\theta\)
\(\int^{circ}_0ds_{2s}=\int^{2\pi}_0sin\xi d\theta\)
\(circumference=2\pi sin\xi\)
The ratio of circumference to radius would be \(\frac{2\pi sin\xi}{\xi}\). This would be smaller than the Euclidean ratio, since \(\xi\), the denominator, would increase faster than \(\sin\xi\), the numerator, would.

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

Since the simplified metric is not provided, we have to start from the FRW metric. Since this is an open, hyperbolic geometry, we know that k = -1, \(r=sinh\xi\), and therefore \(dr=cosh\xi\). Since this is the 2-dimensional case, we can again set \(d\phi=0\).
\(ds^2_{3d}=\frac{dr^2}{1-kr^2}+r^2(d\theta^2+sin^2\theta d\phi^2)\)
\(ds^2_{2d}=\frac{cosh^2\xi}{1+sinh^2\xi}d\xi^2+sinh^2\xi d\theta^2\)
\(ds^2_{2d}=d\xi^2+sinh^2\xi d\theta^2\)
Since we are working with a constant radius, \(d\xi=0\).
\(ds^2_{2d}=sinh^2\xi d\theta^2\)

\(\int^{circ}_0ds_{2d}=\int^{2\pi}_0sinh\xi d\theta\)

\(circumference=2\pi sinh\xi\) 

The ratio of circumference to radius would be \(\frac{2\pi sinh\xi}{\xi}\). This would be larger than the Euclidean 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? 

At small radii, both the closed and open geometries will have ratios close to that of the flat geometry because \(\frac{sin(x)}{x}\) and \(\frac{sinh(x)}{x}\) both tend towards 1 as x approaches 0. This makes sense, because a circle drawn on a sphere, if much smaller than the sphere, will appear to be flat. The Earth is a good example--if you draw a circle on the floor, it will look flat even though it's technically on a sphere. 

Blog #30: Lab Intro

Giant molecular clouds (GMCs) are some of the largest structures in the Galaxy, composed mostly of hydrogen and helium atoms, though other atoms and molecules are present as well. GMCs are where matter comes together to form new planets and stars.

An example of a GMC

In this lab, our goal is to determine the radial velocities of various GMCs in our Galaxy to see how the Galaxy's rotation speed varies by radius. We'll be doing this by measuring the Doppler shifts (specifically by using the emission spectrum of CO) of the GMCs in the radio frequency. This will allow us to see past the gas and dust that would obscure the GMCs were we to look for them using the visible spectrum. We will be observing in the Galactic plane, at Galactic longitudes from 10˚-70˚.

The GMC moving fastest from us will be those whose orbit is tangential to our line of sight. It will be the one with the tightest orbit, and if it is at the tangent point, its entire velocity (\(V_r(max)\)) will be directly away from us. However, we also need to take into account that the Sun, because of its own rotation, will be moving towards that GMC. Therefore, the total circular velocity \(V_{cir}=V_r(max)+V_{\odot}sinl\), where the sin(l) term corrects for the direction of the Sun's rotation.

Our ultimate goal with this lab is to create a Galactic rotation curve. We have talked a bit about the Galactic rotation curve earlier in the class--specifically in the context of how we know about the existence of dark matter. Along with that, we'll calculate/plot the orbital period, the total mass enclosed within each GMC's radius, the number of stars interior to the Sun, and the Sun's Galactic age.

In this diagram, GMC (b) is the one moving fastest from us, so it corresponds to the rightmost peak in the emissions spectrum. (From WS9.2)

Sources: 
http://coolcosmos.ipac.caltech.edu/cosmic_classroom/cosmic_reference/molecular_clouds.html
http://coolcosmos.ipac.caltech.edu/cosmic_classroom/cosmic_reference/images/horsehead.jpg

Blog #29: WS9.1, #2

2. In Question 1, you 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. The first Friedmann equation (1) is: 
\[(\frac{\dot{a}}{a})^2=\frac{8\pi}{3}G\rho-\frac{kc^2}{a^2}+\frac{\Lambda}{3}\]
The second Friedmann equation (2) is: 
\[\frac{\ddot{a}}{a}=\frac{-4\pi G}{3c^2}(\rho c^2+3P)+\frac{\Lambda}{3}\]
In these equations, \(\rho\) and P are the density and pressure of the content, respectively. k is the curvature parameter; k = -1, 0, 1 for open, flat, and closed universe, respectively. \(\Lambda\) is the cosmological constant. Note that in GR, not only density but also pressure are the sources of energy. 
Starting from these two equations, derive the third Friedmann equation (3), which governs the way average density in the universe changes with time. 
\[\dot{\rho}c^2=-3\frac{\dot{a}}{a}(\rho c^2+P)\]
To derive this equation, first multiply \(a^2\) on both sides of (1) and then take time derivative on both sides; plug (2) into your expression to eliminate \(\ddot{a}\).

\(\dot{a}^2=\frac{8\pi}{3}G\rho a^2-kc^2+\frac{\Lambda}{3}a^2\)
\(2\dot{a}\ddot{a}=\frac{8\pi}{3}G[\dot{\rho}a^2+2\rho a\dot{a}]+\frac{2\Lambda}{3}a\dot{a}\)
\(2\dot{a}\frac{\ddot{a}}{a}=\frac{8\pi}{3}G[\dot{\rho}a+2\rho\dot{a}]+\frac{2\Lambda}{3}\dot{a}\)
\(2\dot{a}(\frac{-4\pi G}{3c^2}[\rho c^2+3P]+\frac{\Lambda}{3})=\frac{8\pi}{3}G[\dot{\rho}a+2\rho\dot{a}]+\frac{2\Lambda}{3}\dot{a}\)
Simplify to \(\dot{\rho}c^2=-3\frac{\dot{a}}{a}(\rho c^2+P)\)

b) If the matter is cold, its pressure P = 0, and the cosmological constant \(\Lambda\) = 0. Use the third Friedmann equation to derive the evolution of the density of the matter \(\rho\) as a function of the scale factor of the universe a. You can leave this equation in terms of \(\rho\), \(\rho_0\), \(a\) and \(a_0\), where \(\rho_0\) and \(a_0\) 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 \(\rho\) 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 this differentiation equation to show that \(a(t)\propto t^{2/3}\). This is the characteristic expansion history of the universe if it is dominated by matter.

Determining the density of matter as a function of the scale factor: 

\(\dot{\rho}c^2=-3\frac{\dot{a}}{a}\rho c^2\)

\(\frac{d\rho}{dt}=\frac{-3}{a}\frac{da}{dt}\rho\) 

\(\int^{\rho}_{\rho_0}\frac{d\rho'}{\rho'}=\int^{a}_{a_0}\frac{-3da'}{a'}\) 

\(ln(\frac{\rho}{\rho_0})=-3ln(\frac{a}{a_0})\)

Simplify to \(\rho(a)=\rho_0(\frac{a_0}{a})^3\) 

Determining the scale factor for the matter-dominated universe as a function of time: 

\((\frac{\dot{a}}{a})^2=\frac{8\pi}{3}G\rho-\frac{kc^2}{a^2}+\frac{\Lambda}{3}\)

\((\frac{\dot{a}}{a})^2=\frac{8\pi}{3}G\rho\) since \(\Lambda\) and k = 0 in the matter-dominated universe. 

\(\frac{da}{dt}=(\frac{8\pi}{3}G\rho_0\frac{a_0^3}{a})^{1/2}\)

The right side of the above equation is a constant.

\(\int^a_0a'^{1/2}da'=\int^t_0Cdt'\)

\(\frac{2}{3}a^{2/3}=Ct\) so \(a(t)\propto t^{2/3}\) 

c) Let us repeat the above exercise for a universe filled with radiation only. For radiation, \(P=\frac{1}{3}\rho c^2\) and \(\Lambda=0\). 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 \(\rho\) and a and the first Friedmann equation to show that \(a(t)\propto t^{1/2}\) for the radiation only universe.

Determining the density of radiation as a function of the scale factor: 

\(\dot{\rho}c^2=-3\frac{\dot{a}}{a}(\rho c^2+\frac{1}{3}\rho c^2\)

\(\dot{\rho}=-4\frac{\dot{a}}{a}\rho\)

\(\frac{d\rho}{dt}=\frac{-4}{a}\frac{da}{dt}\rho\)

\(\int^{\rho}_{\rho_0}\frac{d\rho'}{\rho'}=\int^{a}_{a_0}\frac{-4da'}{a'}\) 

Simplify to \(\rho(a)=\rho_0(\frac{a_0}{a})^4\) 

Determining the scale factor for the radiation-dominated universe as a function of time: 

\((\frac{\dot{a}}{a})^2=\frac{8\pi}{3}G\rho_0(\frac{a_0}{a})^4\)

\(\dot{a}^2=\frac{8\pi}{3}G\rho_0\frac{a_0^4}{a^2}\)

\(\frac{da}{dt}=(\frac{8\pi}{3}G\rho_0)^{1/2}\frac{a_0^2}{a}\)

\(\int^a_0a'da'=\int^t_0Cdt'\)

\(\frac{1}{2}a^2=Ct\) so \(a(t)\propto t^{1/2}\) 

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 \(\rho=0\) and P = 0 and only keep \(\Lambda\) 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 \(P=-\rho c^2\). 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?
Hint: While calculating the scale factor as a function of time, you will find that setting \(a(0)=0\) leads to a negative infinity. Feel free to ignore this term to show the dependence.

The dark energy is a property of spacetime itself, so it remains constant through time. 

Determining the scale factor for the dark energy-dominated universe as a function of time: 

\((\frac{\dot{a}}{a})^2=\frac{8\pi}{3}G\rho-\frac{kc^2}{a^2}+\frac{\Lambda}{3}\)

\((\frac{\dot{a}}{a})^2=\frac{\Lambda}{3}\) since \(\rho\) and k = 0 in the dark energy-dominated universe. 

\(\dot{a}=a(\frac{\Lambda}{3})^{1/2}\)

\(\int^a_0\frac{da'}{a'}=\int^t_0(\frac{\Lambda}{3})^{1/2}dt'\)

\(a(t)=exp(t(\frac{\Lambda}{3})^{1/2})\)

Determining the Hubble parameter: 

\(H(t)=\frac{\dot{a}}{a}\) so \(H(t)=(\frac{\Lambda}{3})^{1/2}\) 

(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?

Matter will become the dominant component in this universe as time goes on. The radiation component loses energy not only because the number density decreases as the volume increases, but also because the energy of each individual photon decreases as its wavelength is stretched. This can also be seen mathematically, since at large values for t, the value of the scale factor in the matter-dominated universe (\(a(t)\propto t^{2/3}\)) will be larger than that of the radiation-dominated universe (\(a(t)\propto t^{1/2}\)).

(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?

The energy of dark energy remains constant as the universe expands, whereas the energy of matter decreases as volume increases (the matter gets diluted). Therefore, as the universe expands and matter gets increasingly diluted, dark energy will come to dominate. Our universe will eventually come to be dominated by dark energy as the universe expands enough to make the contributions from matter and radiation negligible. 

Blog #28: WS9.1, #1

1. In this exercise, we will derive the first and second Friedmann equations of a homogenous, isotropic, and matter-only universe. We use the Newtonian approach. 
Consider a universe filled with matter which has a mass density \(\rho(t)\). Note that as the universe expands or contracts, the density of the matter changes with time, which is why it's a function of time. 
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 (homogenous 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 \(\dot{v}\) to avoid confusion with the scale factor a. 

Since F=ma, we can also say that \(F=m\dot{v}\). The only force in this system is gravity, so we can equate this with the gravitational force.
\(\frac{-GMm}{R^2}=m\dot{v}\)
\(\dot{v}=\frac{-GM}{R^2}\)

b) To derive an energy equation, it is a common trick to multiply both sides of your acceleration equation by v. Turn your velocity into \(\frac{dR}{dt}\), cancel dt, and integrate both sides of your equation. This is an indefinite integral, so you will have a constant of integration; combine these integration constants (C). You should arrive at the following equation: 
\[\frac{1}{2}\dot{R}^2-\frac{GM}{R}=C\]
This has units of energy per unit mass. 

\(\dot{v}\frac{dR}{dt}=\frac{-GM}{R^2}\frac{dR}{dt}\)
We can replace \(\dot{v}\) with \(\ddot{R}\) since velocity is the time derivative of the radius:
\(\int\ddot{R}d\dot{R}=-GM\int\frac{1}{R^2}dR\)
\(\frac{1}{2}\dot{R}^2+C=\frac{GM}{R}+C\)
\(\frac{1}{2}\dot{R}^2-\frac{GM}{R}=C\)

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 \((\frac{\dot{R}}{R})^2\), where \(\dot{R}\) is equal to \(\frac{dR}{dt}\).

The mass of the sphere can be expressed as \(M=\frac{4}{3}\pi R^3\rho(t)\)
\(\frac{1}{2}\dot{R}^2-\frac{4G\pi R^2\rho(t)}{3}=C\)
\((\frac{\dot{R}}{R})^2=\frac{8G\pi R^2\rho(t)}{3}+\frac{2C}{R^2}\)

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 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). 

If \(R=a(t)r\), then \(\dot{R}=\dot{a}(t)r\).
\((\frac{\dot{a}(t)}{a(t)})^2=\frac{8G\pi\rho(t)}{3}+\frac{2C}{a^2(t)r^2}\)

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

\((\frac{\dot{a}}{a})^2=\frac{8G\pi\rho(t)}{3}+\frac{2C}{a^2(t)r^2}\)

f) Derive the first Friedmann Equation: We know that \(H(t)=\frac{\dot{a}}{a}\). Plugging this relation into your above result and identifying the constant \(2C/r^2=-kc^2\) where k is the "curvature" parameter, you will get the first Friedmann equation. The Friedmann equation tells us about how the shell expands or contracts; in other words, it tells us about the Hubble expansion (or contraction) rate of the universe. 

\(H^2(t)=\frac{8G\pi\rho(t)}{3}-\frac{kc^2}{a^2(t)}\)

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 \(\frac{\ddot{a}}{a}=-\frac{4\pi}{3}G\rho\), which is known as the second Friedmann equation. 

\(\dot{v}=\frac{-GM}{R}=\frac{-4}{3}G\pi R\rho(t)\)
Since \(\dot{v}=\ddot{R}\) and \(\dot{R}=\dot{a}(t)r\), then \(\dot{v}=\ddot{a}(t)\).
\(\ddot{a}(t)=\frac{-4}{3}G\pi\rho(t)a(t)\)
\(\frac{\ddot{a}}{a}=\frac{-4}{3}G\pi\rho(t)\)

The more complete second Friedmann equation actually has another term involving the pressure following from Einstein's general relative, 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. 

Blog #27: WS8.1, #3

3. It is not strictly correct to associate this ubiquitous distance-dependent redshift we observe with the velocity of galaxies (at very large separations, Hubble's Law gives 'velocities' that exceed 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 center of this expansion. But, 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 \(a(t_0)\equiv 1\). 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. 

a) Assume the rate of expansion, \(\dot{a}\equiv da/dt\), 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 the rate of expansion is constant, we can figure out a as a function of t.
\(\frac{da}{dt}=c\)
\(\int_0^{a(t)}da=\int_0^tdt'\)
\(a(t)=ct\)
We can then plug this into the equation for the Hubble Constant, with \(\frac{da}{dt}\) simplified to a constant.
\(H(t)=\frac{1}{ct}c\)
\(H(t)=\frac{1}{t}\)
Now we just need to plug in the Hubble Constant (68.6 (km/s)/Mpc, calculated in #2) in for H(t) to calculate the Hubble Time.
\(68.6\frac{km/s}{Mpc}=\frac{1}{t}\)
\(t=0.0146\frac{Mpc}{km/s}\rightarrow14.28Gyr\)
The universe is about 2Gyr older than the oldest globular clusters.

b) What is the size of the observable universe? What you will calculate is known as the Hubble Length.

The radius of the observable universe should be the distance that light can travel over the Hubble Time, so we can just multiply the Hubble Time by the speed of light.
\(r=(0.0146\frac{Mpc}{km/s})(3.0*10^5km/s)=4380Mpc\)

Blog #26: WS8.1, #1

1. Pretend that there is an infinitely long series of balls sitting in a row. Imagine that during a time interval \(\Delta t\) the space between each ball increases by \(\Delta x\). 

a) Look at Ball C in the figure above. Imagine that Ball C is sitting still (so we are in the reference frame of Ball C). What is the distance to Ball D after time \(\Delta t\)? What about Ball B? 

If the initial distance between Ball C and Ball B (or D) is \(x_0\), the distance after the time interval should be \(d=x_0+\Delta x\). 

b) What are the distances from Ball C to Ball A and Ball E? 

If we assume that the balls are evenly spaced to begin with, the distance to Ball A should be \(d=2(x_0+\Delta x)\), to account for both the distance that Ball B moves from Ball C and the distance that Ball A moves from Ball B. The same is true for Ball E. 

c) Write a general expression for the distance to a ball N balls away from Ball C after time \(\Delta t\). Interpret your finding. 

\(d=N(x_0+\Delta x)\) 

\(d(t)=N\frac{\Delta x}{\Delta t}\Delta t+Nx_0\) 

This means that the distance between a ball N balls away from Ball C increases linearly with time. 

d) Write the velocity of a ball N balls away from Ball C during \(\Delta t\). Interpret your finding. 

The velocity would be the slope of the above equation: \(N\frac{\Delta x}{\Delta t}\). This means that the velocity is not only dependent on the rate of change of the distance, but also on how many balls are between that ball and Ball C. Farther balls will move away from Ball C at higher velocities than balls close to Ball C. 

Blog #25: WS7.2, #5

5. You may have also noticed some weak "dips" (or absorption features) in the spectrum: 

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. 

These dips are probably due to the absorption of photons by the gaseous material between the quasar and the Earth. The wavelengths probably correspond to the photons at appropriate energies to excite electrons from the ground state to the second, third, or higher energy levels. There is decreased flux at these wavelengths because some of those photons will be absorbed by the gas and not reach the Earth. 

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 the object? 

The observed wavelength appears to be about 5600Å. 

\(z=\frac{5600Å-1216Å}{1216Å}=3.61\) 

c) What is the most noticeable difference between this spectrum and the spectrum of 3C 273 (below)? 

What conclusion might we draw regarding the incidence of gas in the early Universe as compared to the nearby Universe? 

One noticeable feature is that the other spectrum is located at much greater wavelengths than the 3C 273 spectrum. This means that the redshift is much greater (as we calculated in part b), and therefore the object is farther away than 3C 273. We also notice all of the dips to the left of the Lyα peak. These are probably caused by the absorption of the quasar's radiation by dust and gas. This implies that gas, specifically neutral Hydrogen gas, was very prevalent in the early Universe. 

Blog #24: WS7.2, #4

4. One prominent feature of quasar spectra are the strong, broad emission lines. Here's a close-up of one: 

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 (called the ground state), labeled n = 1.

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. 

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 10.19eV, corresponding to a photon wavelength of \(\lambda\) = 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 the strongest emission feature you observed in the quasar spectrum above arises from Lyα emission from material orbiting around the central black hole. 

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. 

Looking at the spectrum, the peak appears to be at about 1405 Angstroms. Since we know the actual wavelength, we can just plug the numbers in to calculate the redshift. 

\(z=\frac{1405Å-1216Å}{1216Å}=0.155\) 

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 \(10^8M_\odot\) black hole extends to a radius of \(r=10^{15}m\). 

The Virial Theorem can be used to show that \(M\approx\frac{\sigma^2R}{G}\). We then need to find \(\sigma\) (the stellar velocity dispersion) so that we can calculate the mass. We can do this by looking at the wavelength at half the peak flux (which appears to be about 1400Å and 1415Å). Taking the average difference between these value and the peak wavelength (1405Å) gives us 7.5Å. We can divide this wavelength by the peak wavelength to calculate the dispersion. 

\(\frac{7.5Å}{1405Å}=0.0053\approx\frac{\sigma}{c}\)

\(\sigma=0.0053*c=1.59*10^6m/s\) 

Finally, we can plug this into the equation, using the typical radius. 

\(M=\frac{(1.59*10^6m/s)^2(10^{15}m)}{6.674*10^{-14}m^3g^{-1}s^{-2}}=3.79*10^{40}g\rightarrow1.90*10^7M_\odot\)

Blog #23: Astronomy & the Bible

In the beginning of the New Testament in the Bible, the events leading up to and following the birth of Jesus are recorded. One of the events that is described is the visit of the wise men, or magi. Matthew 2:9 says that a star in the night sky "went on before them until it came and stood over the place where the Child was" -- essentially guiding the wise men to the exact location where Jesus was born. While this is definitely a cool way for the magi to figure out where they're going, based on a modern understanding of astronomy, it seems a little weird. Sure, the North Star and the Southern Cross were often used in navigation, but in more of an orienting fashion. What kind of star (or other celestial body) appears out of nowhere, guides you where you need to go, and then disappears when it's no longer useful -- and where can I get one?

Several explanations have been presented in light of our present-day knowledge of astronomy. One idea is that the "star" could have been an alignment of Jupiter with Saturn and/or Mars, which would create a transient star-looking object. Since planets at this time were thought of as "wandering stars," it's possible that as they moved, they would appear to guide the wise men on their journey. A nova or a supernova is another candidate, since it would appear like a bright star and then eventually fade away.

A comet may seem to "point" at a specific location

Since we know the approximate date range of Jesus's birth (somewhere between 4-7BC), the location of his birth (Bethlehem), and where the magi were coming from (from east of Bethlehem), some of these suggestions can actually be cross-checked with observations that were recorded at the time. For example, one suggestion is that the celestial object was actually a comet rather than a star. A comet with the appropriate orientation was indeed observed by Chinese astronomers in 5BC -- although comets are not particularly rare, and they were actually considered bad omens at the time, making it unlikely that the magi would have followed this one.

One of the cool things about astronomy is that it's such an old field, and people have been making observations for millennia. Although we may never know the exact truth behind the magi's star, I still think it's awesome that we can think about it in astronomical terms, both theoretically, and in light of 2000-year-old observations.

Sincere apologies for writing this post way before the Christmas season

Sources: 

https://en.wikipedia.org/wiki/Biblical_Magi

http://earthsky.org/astronomy-essentials/was-the-christmas-star-real

http://www.bbc.com/news/magazine-20730828

http://il4.picdn.net/shutterstock/videos/3148000/thumb/1.jpg

https://upload.wikimedia.org/wikipedia/commons/d/d4/Comet_P1_McNaught02_-_23-01-07-edited.jpg

Blog #22: WS7.1, #5

5. Gamma rays from the radioactive decay of nickel into iron drive most of the optical luminosity of a Type-Ia supernova. The process is given by \[^{56}N\longrightarrow ^{56}Co+\gamma\longrightarrow ^{56}Fe+\gamma\] where \(\gamma\) represents photons. The total atomic weights of \(^{56}N\) and \(^{56}Fe\) are 55.942135 and 55.934941, respectively. Let's calculate the total energy radiated in the optical wavelengths during the event, given that the characteristic times for the two decay processes are 8.8 days and 111 days, respectively. 
a) Let's balance the decay process from \(^{56}N\) to \(^{56}Fe\) for a single atom, ignoring the intermediate step. According to the first law of thermodynamics, energy cannot be created or destroyed. Use the fact that \(E=mc^2\) to balance the equation. 

The overall equation is \(^{56}N\longrightarrow^{56}Fe+\gamma\), but since the masses of \(^{56}N\) and \(^{56}Fe\) aren't the same, some mass must be converted to energy. We can figure out the amount of energy that is produced by balancing the equation and first finding the amount of "missing" mass.

\(55.942135g/mol=55.934941g/mol+m_{missing}\)
\(m_{missing}=0.007194g/mol\rightarrow7.194*10^{-6}kg\)

We can then use the missing mass to calculate the amount of energy released.

\(E=(7.194*10^{-6}kg)(3.0*10^8m/s)^2=6.47*10^{11}J/mol\)

We then divide by Avogadro's Number to calculate the amount of energy per atom.

\(E=\frac{6.47*10^{11}J/mol}{6.02*10^{23}atoms/mol}=1.08*10^{-12}J/atom\)

b) How many nickel atoms are there in the white dwarf? Use this number to estimate the total energy emitted in photons. 

Since we know the mass of the white dwarf (\(1.4M_\odot\)) and the atomic mass of nickel-56, this is just a matter of dimensional analysis.

\(\frac{1.4M_\odot}{1}\frac{2*10^{33}g}{1M_\odot}\frac{1mol^{56}N}{55.942135g}\frac{6.02*10^{23}atoms}{1mol^{56}N}=3.0*10^{55}atoms\)

Since we already solved for the energy per atom, we can easily calculate the total energy.

\(\frac{3.0*10^{55}atoms}{1}\frac{1.08*10^{-12}J}{1atom}=3.25*10^{43}J\)

c) Now combine the characteristic times for the two processes to find a total characteristic time. Divide the energy found in part (b) by this time scale to find a characteristic luminosity.

Total time = 119.8 days = \(1.04*10^7s\).
\(L=\frac{3.25*10^{43}J}{1.04*10^7s}=3.14*10^{36}J/s\)

Blog #21: WS7.1, #4

4. Calculate the total energy output, in ergs, of the explosion, assuming that the white dwarf's mass is converted to output energy via fusion of carbon into nickel. Note that the process of carbon fusion is not entirely efficient, and only about 0.1% of this mass will be radiated away as electromagnetic radiation (light). 

Since we're converting mass to energy, we can use \(E=mc^2\), but we have to multiply by 0.001 since it's only 0.1% efficient.
\(E=(1.4*2*10^{30}kg)(3.0*10^8m/s)^2*0.001=2.52*10^{44}J\rightarrow 2.52*10^{51}ergs\)

How does this compare to the total binding energy, in ergs, of the original white dwarf? Does the white dwarf completely explode, or is some mass left over in the form of a highly concentrated remnant? 

To calculate the binding energy of the star, we can use the potential energy as given by the Virial Theorem: \(E_{bind}=U=\frac{-GM^2}{R}\). Since \(M=2M_\odot\) and \(R=12*10^6\), we can simply plug those values in.
\(E_{bind}=U=\frac{-(6.674*10^{-11}m^2kg^{-1}s^{-2})(1.4*2*10^{30}kg)^2}{12*10^6m}=4.36*10^{43}J\rightarrow 4.36*10^{50}ergs\)
Since the energy released is greater than the binding energy, the bonds holding the star's atoms together are broken. These atoms will be blown away, and there won't be anything left over.

Blog #20: Hubble Tuning Fork

The Hubble Tuning Fork is a method of categorizing galaxies based on their shape. It breaks galaxies up into three major categories: elliptical, spiral, and barred spiral galaxies.

Elliptical galaxies are classified based on their ellipticities, with more elliptical galaxies closer to the fork. These galaxies are obviously characterized by their elliptical or round shape, but also by their more incoherent rotations. They tend to be older (and therefore redder) galaxies, which is part of the reason why their rotations are more random. Often, they don't have active areas of star formation.

Messier 87, an E0 galaxy

Maffei 1, an E3 galaxy

Messier 59, an E5 galaxy

An E7 galaxy

Spiral galaxies have identifiable spiral arms and a central bulge. They are categorized first by the presence or absence of a bar of stars through the center, and then by the tightness of the spiral, with looser spirals towards the ends of the fork. Spiral galaxies tend to be younger galaxies with more uniform rotation than elliptical galaxies. They also have active star formation in their gas and dust lanes, and therefore tend to be bluer than elliptical galaxies.

The Sombrero Galaxy, an Sa galaxy

The Sunflower Galaxy, an Sb galaxy

The Triangulum Galaxy, an Sc galaxy

NGC 4314, an SBa galaxy

The Andromeda Galaxy, which is argued to be an SBb galaxy (the Milky Way is another example)

NGC 7479, an SBc galaxy

The final type of galaxy on the Hubble Tuning Fork is the S0, or lenticular galaxy. These galaxies don't have as well-defined of a structure as spiral galaxies do, but like spiral galaxies, they have a central bulge and a disk. 

NGC 6861, an S0 galaxy

Although not included on the Tuning Fork, irregular galaxies also exist. These tend to be galaxies that have formed through the collision or gravitational interactions of multiple other galaxies, so they don't really have standard features in the same way elliptical or spiral galaxies do.

NGC 1427A, an irregular galaxy (plus a bonus spiral galaxy photobomber!)

Blog #19: Simulating Our Universe

In this article, a new way of simulating the universe, called Illustris, is described. In the past, many simulations have had problems with resolution--it's been difficult to be detailed on the scale of galactic clusters while still maintaining detail on the scale of single stars. However, by adjusting the resolution of the simulation based on the density of matter in each particular location, Illustris brings high resolution where it's needed without having high resolution where it would be useless. By doing this, the simulation has been able to model over 40,000 galaxies--and apparently pretty accurately!

From the original paper.

Aside from the issues with resolution, a lot of other simulations have had disagreements with theory-based predictions (in part because of the resolution problem). Illustris, while not perfect, has been more able to produce results that are consistent with theory in several problem areas. For example, by increasing the resolution and including more detailed and accurate models, Illustris produced more accurate simulations of satellite galaxies in galactic clusters, the distribution of neutral hydrogen, and the distribution of metallicity throughout the simulated universe.

The simulations and data from Illustris so far have matched observations really well. What's more, the Illustris team has striven to make the simulation and its data accessible to everyone, so it should be really exciting to see the directions the project goes in the future.

Blog #18: WS6.1, #4

4. Over time, from measurements of the photometric and kinematic properties of normal galaxies, it became apparent that there exist correlations between the amount of motion of objects in the galaxy and the galaxy's luminosity. In this problem, we'll explore one of these relationships. 
a) Spiral galaxies obey the Tully-Fisher Relation: \(L\sim v^4_{max}\), where L is total luminosity, and \(v_{max}\) is the maximum observed rotational velocity. This relation was initially discovered observationally, but it's not hard to derive: 
i. Assume that \(v_{max}~\sigma\). Given what you know about the Virial Theorem, how should \(v_{max}\) relate to the mass and radius of the Galaxy? 

The Virial Theorem states that \(M\approx\frac{\sigma^2R}{G}\). Since we're assuming that \(v_{max}\sim\sigma\) (where \(\sigma^2\) is the velocity scatter of the Galaxy), we can just plug it into the equation.
\(M\approx\frac{v^2_{max}R}{G}\)
\(v^2_{max}\propto \frac{M}{R}\)

ii. To proceed from here, you need some handy observational facts. First, all spiral galaxies have similar disk surface brightnesses (\(\langle I\rangle=L/R^2\)) (Freeman's Law). Second, they also have similar total mass-to-light ratios (M/L). 
iii. Use some squiggle math to find the Tully-Fisher relationship.

\(v^2_{max}\sim\frac{M}{R}\)
\(v^4_{max}\sim\frac{M^2}{R^2}\)
Through Freeman's Law, \(v^4_{max}\sim\frac{M^2}{(L/I)}\)
\(v^4_{max}\sim\frac{M^2I}{L}\)
We can cancel M/L because the total mass-to-light ratio is constant between galaxies, so \(v^4_{max}\sim MI\)
Since M~L, \(v^4_{max}\sim L\), which brings us to the Tully-Fisher relationship.

iv. It turns out the Tully-Fisher relationship is so well-obeyed that it can be used as a standard candle, just like Cepheids and Supernova Ia. In the B-band (blue light), this relation is approximately: 
\(M_B=-10log(\frac{v_{max}}{km/s})+3\) 
Suppose you observe a spiral galaxy with apparent, extinction-corrected magnitude B=13. You perform longslit optical spectroscopy, obtaining a maximum rotational velocity of 400km/s for this galaxy. How distant do you infer this spiral galaxy to be? 

First we plug the observed velocity into the Tully-Fisher relationship to determine its absolute magnitude.
\(M_B=-10log(400)+3=-23\)
We can then use the distance modulus to determine its distance.
\(13-(-23)=5log(\frac{d}{10pc})\)
\(d=1585kpc\)

b) It turns out our Galaxy is not unique in hosting a supermassive black hole (SMBH) in its center. Peering deeply into the galaxies around us, we find that having a central compact region occupied by a SMBH is a common phenomenon. The SMBH appears to be deeply connected with the overall nature of its galaxy. The \(M-\sigma\) relation: \(M_{SMBH}\propto\sigma^4_e\), correlates the mass of the black hole (\(M_{SMBH}\)) with the velocity dispersion of the galactic bulge (\(\sigma_e\)). For elliptical galxies, \(\sigma_e\) refers to that of the velocity dispersion of the entire galaxy. The \(M-\sigma\) relation is: 
\(M_{SMBH}=1.35*10^8M_\odot[\frac{\sigma_e}{200km/s}]^4\) 

i. Using the rotation velocity profile in the inner 1", estimate the mass enclosed. Recall Andromeda is 770kpc from us. 

Looking at the graph above for R = 1.0, v looks to be about 200km/s. We also need to convert the radius to parsecs, which we can do using the small-angle approximation.
\(1as=4.8*10^{-6}\)
\(tan(\frac{x}{770kpc})=\frac{x}{770}=4.8*10^{-6}\)
\(x=3.7pc\)
We can now use Virial Theorem to determine the mass: \(M_{tot}=\frac{v^2_{max}R}{G}\).
\(M_{tot}=\frac{(200km/s)^2(3.7pc)}{4.302*10^{-3}pc(km/s)^2/M_\odot}=3.4*10^7M_\odot\)

ii. Using the velocity dispersion in the inner 1", estimate the mass enclosed. 

It looks like the dispersion at 1" is about 100km/s. We can use the Virial Theorem again.
\(M_{tot}=\frac{(100km/s)^2(3.7pc)}{4.302*10^{-3}pc(km/s)^2/M_\odot}=8.6*10^6M_\odot\)

iii. Using the average velocity dispersion over the entire range of the plot, deduce the mass of the central supermassive black hole with the \(M-\sigma\) relation. 

Over the plot, the average dispersion looks to be about 200km/s.
\(M_{SMBH}=1.35*10^8M_\odot[\frac{200km/s}{200km/s}]^4=1.35*10^8M_\odot\)

iv. Estimate the mass of the SMBH using the velocity dispersion in the inner 0.1". 

If we assume the dispersion is 200km/s, we can use the Virial Theorem again, after adjusting the radius.
\(M_{SMBH}=\frac{(200km/s)^2(0.37pc)}{4.302*10^{-3}pc(km/s)^2/M_\odot}=3.4*10^6M_\odot\)

Blog #17: WS6.1, #2

2. The Hubble Classes have characteristic surface light distribution profiles. They are fairly well-described by parametrized equations in the form I(R) where I = intensity and R = distance from the center. Often each is scaled to their value at the effective radius, such that \(I_e=I(R_e)\). The most general profile is the Sérsic Profile, given by 
\(I(R)=I_eexp(-b_n[(\frac{R}{R_e})^{1/n}-1])\)
The constant \(b_n\) depends on the shape parameter n. n = 4 gives rise to the famous \(r^{1/4}\)-law, or the de Vaucoleur Profile, which approximates elliptical and the bulge of spiral galaxies. n = 1, on the other hand, is equivalent to a simple exponential profile, which often corresponds with the outskirts of spiral galaxies. The best fit is often obtained by a combination of the functional forms. 
a) Another way to write the exponential profile is \(I(R)=I_0exp(-R/b)\), where \(I_0\) is the central surface brightness and b is a characteristic lengthscale, a constant. 
i. Describe what b is, and suggest how one might measure it for a given spiral galaxy. 

b is the scale radius of the galaxy, which is defined as the radius that contains a specific proportion of the matter in the galaxy. To measure the scale radius, you could measure the fluxes in "bins" across the galaxy, then fit those measurements to the function \(I(R)=I_0exp(-R/b)\). Since you know how far each bin is from the center of the galaxy, you should be able to calculate b.

ii. The Milky Way has an estimated b = 3.5kpc. Plot the disk surface brightness profile relative to the central brightness and indicate the location of the characteristic lengthscale from the center. 

At the scale radius, b = R, so we can solve \(I(R)=I_0exp(-b/b)\). 

\(I(b)=I_0e^{-1}\)

\(I(b)=\frac{I_0}{e}\) 

iii. Assuming the entire Milky Way (including the central bulge) can be described by this profile, and that it is circularly symmetric. Approximately what fraction of all its stars are interior to the Sun (at 8kpc)? 

To find the fraction of the stars that are interior to the sun, we want to divide the total intensity in the sphere with radius = 8kpc by the total intensity of the Milky Way, regardless of radius. To find these total intensities, we can integrate the intensity function over the desired range. 

The intensity interior to the Sun: 

\(\int_{0}^{8}I(r)dr=I_0\int_{0}^{8}e^{-r/b}dr=-I_0be^{-r/b}|^8_0\)

The intensity of the entire Milky Way: 

\(\int_{0}^{\infty}I(r)dr=I_0\int_{0}^{\infty}e^{-r/b}dr=-I_0be^{-r/b}|^{\infty}_0\)

We can then take a ratio (thus canceling out the constants): 

\(\frac{e^{-r/b}|^8_0}{e^{-r/b}|^{\infty}_0}\)

Finally, we can evaluate: 

\(\frac{e^{-8/b}-e^0}{e^{-\infty/b}-e^0}=\frac{e^{-8/b}-1}{-1}=1-e^{-8/b}\) of the stars in the Milky Way are interior to the Sun.