## Sunday, December 6, 2015

### 37 | The Illustris Simulation

In this post, we will explore the Illustris Simulation, a numerical cosmological simulation developed by a collaboration of scientists working to better understand the formation of large-scale structure in the universe. The online tools associated with Illustris allow us to explore the data generated from this simulation.

We will begin by looking closely at an over-dense region. By choosing such a region on the simulation explorer, the program will automatically select nearby subhalos, which are essentially regions with very high relative density contrast.

 a selection within the Illustris Explorer

The simulation explorer allows us to learn more about the nature of these halos. Let's start by generating a histogram of the masses of these halos. To make the data easier to visualize, we will consider the logarithm of the mass:

From this plot, it is clear that the mass distribution seems to be shifted slightly more towards the low-mass halos, meaning that low-mass halos are more numerous. From the data provided, we can also learn about the total stellar mass within each halo. From the data set we collected, we find that about 93% of the halo mass comes from the stars contained within it.

At a large scale, the density of dark matter and gas seem to be quite similar, and we can see similar structural features in spatial plots of each component. The following plots show the density of dark matter and gas, respectively.

Note how both appear to peak in similar regions, and exhibit "filamentary" structure elsewhere. However, it's interesting to note that the dark matter appears to be more closely defined to the filaments, whereas the gas is a bit more diffuse.

On a small scale, however, the gas density appears to be more defined than the dark matter density. Again these plots show dark matter and gas density, respectively--this time at a much smaller scale:

In some of the larger galaxies, we can see that the gas density is actually greatest in the disk of the galaxy and note in the nucleus. This is clearly visible in the image below where white represents the highest gas density. Note how the white appears in a ring around the center of galaxy, but the center of the galaxy is actually less dense.

The explorer also allows us to see that the largest galaxies tend to exist in clusters rather than on their own in the universe. This is evident in the screenshot below, where the most massive galaxies have been circled.

Next, we can look at an animation which shows the development of dark matter and gas temperature over the evolution of our universe. This animation seems to show that structure formation is lead by dark matter, then baryons (gas) tend to develop along these filaments of dark matter. However, the gas is initially not very warm. In fact, researchers refer to the time before the gas heats up as the "Dark Ages." Based on the animation, gas did not begin to warm up until a redshift of $$z \approx 10$$, which corresponds to about 0.5 billion years after the Big Bang.

The model also provides us with the approximate rate of star formation as the universe forms. More specifically, the model predicts this by the rate of increase in stellar mass. There are several periods throughout the evolution of the universe when star formation greatly increases. In particular, between $$z = 1.5$$ and $$z=1.2$$, the rate of star formation is quite fast.

As time progresses, the animation shows large structures collapsing and generating "outbursts" of gas. This suggests that as time passes, large objects break up and this ultimately results in the formation of new, smaller structures throughout the universe. These small structures grow until they too collapse on themselves.

Finally, it is curious why this structure forms in filaments. We can see that the dark matter occurs in filaments throughout the universe. The gravity of this dark matter pulls baryons towards it. As a result of this matter being pulled together, stars and galaxies can form. Consequently, this gravitational attraction from the dark matter generates structures along the filaments.

## Saturday, December 5, 2015

### 36 | Worksheet 12.1, Problems 1 and 2(d): Large Scale Structure

1. Linear perturbation theory: In this and the next exercise, we study how small fluctuations in the initial conditions 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 $$\vec{x}$$ and time $$t$$ as $$\rho(\vec{x}, t)$$ and the relative density contrast as
$\delta(\vec{x},t) \equiv \frac{\rho(\vec{x},t) - \bar{\rho}(t)}{\bar{\rho}(t)}$
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, only their amplitude changes. Namely this means that we can factorize
$\delta(\vec{x},t) = D(t)\tilde{\delta}$
where $$\tilde{\delta}(\vec{x})$$ is arbitrary and independent of time, and $$D(t)$$ is a function of time and valid for all $$\vec{x}$$. $$D(t)$$ is not arbitrary and must satisfy a differential equation. Derive this differential equation.

To do this, we need to show that the position-dependent term $$\tilde{\delta}$$ is an arbitrary term in determining the density contrast. We show that the density contrast is given by:
$\delta(\vec{x},t) = D(t) \tilde{\delta}(\vec{x})$
The differential equation is:
$\frac{d^2\delta}{dt^2} + \frac{2\dot{a}}{a}\frac{d\delta}{dt} = 4\pi G \bar{\rho} \delta$
Let's plug our expression for $$\delta$$ into this differential equation:
$\frac{d^2}{dt^2}\left(D(t)\tilde{\delta}(\vec{x})\right) + \frac{2\dot{a}}{a} \frac{d}{dt}\left(D(t) \tilde{\delta}(\vec{x})\right) = 4\pi G \bar{\rho} \left(D(t) \tilde{\delta}(\vec{x})\right)$
This gives us
$D''(t) \tilde{\delta}(\vec{x}) + \frac{2\dot{a}}{a} D'(t) \tilde{\delta}(\vec{x}) = 4\pi G \bar{\rho} D(t)\tilde{\delta}(\vec{x})$
We can see that all of the $$\tilde{\delta}(\vec{x})$$ terms cancel, giving us:
$D''(t) + \frac{2\dot{a}}{a} D'(t) = 4\pi G \bar{\rho} D(t)$
Since all of the position-dependent terms cancel out, we can see that the $$\tilde{\delta}(\vec{x})$$ term is arbitrary and the equation is dependent entirely on time.

(b) Now let us consider a matter dominated flat universe, so that $$\bar{\rho}(t) = a^{-3}\rho_{c,0}$$ where $$\rho_{c,0}$$ is the critical density today, $$3H_0^2/8\pi G$$. Recall that the behavior of the scale factor of this universe can be written $$a(t) = (3H_0 t/2)^{2/3}$$. Solve the differential equation for $$D(t)$$. 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 $$D_-(t)$$.

We know that
$a = \left(\frac{3H_0 t}{2}\right)^{\frac{2}{3}}$
So, the time derivative of $$a$$ is:
$\dot{a} = \frac{2}{3} \left(\frac{3H_0}{2}\right)^{\frac{2}{3}} t^{-\frac{1}{3}}$
Furthermore, we know that:
$\bar{\rho} = a^{-3} \frac{3H_0^2}{8\pi G}$
If we plug these values into our differential equation:
$D''(t) + \frac{2\dot{a}}{a}D'(t) = 4\pi G \bar{\rho} D(t)$
and simplify, we get:
$D''(t) + \frac{4}{3t}D'(t) = \frac{2}{3t^2}D(t)$
Now we need to solve this differential equation. We will start by saying:
$D = kt^q$
The first- and second-order time derivatives of $$D$$ are,
$D'(t) = kqt^{q-1}$
$D''(t) = kq(q-1)t^{q-2}$
If we plug this into our differential equation, we get:
$kq(q-1)t^{q-2} + \frac{4}{3t}kqt^{q-1} = \frac{2}{3t^2}kt^q$
Note that by simplifying, we get:
$kq(q-1)t^{q-2} + \frac{4}{3}kqt^{q-2} = \frac{2}{3}kt^{q-2}$
Now we can cancel all $$k$$ and $$t$$, giving:
$q^2 + \frac{1}{3}q - \frac{2}{3} = 0$
If we factor this, we get find that
$q = -1,\frac{2}{3}$
As a result, we find that:
$\boxed{D_+ \propto t^{2/3} \, \, \, \, \, \, \, \, D_- \propto t^{-1}}$

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

In order for structure to form, by definition, the density contrast must increase. Since $$D(t)$$ represents the density contrast, the component that leads to an increased density contrast with time must be the dominant component of structure formation. Since $$D_+$$ represents an increasing density contrast, this is the dominant component of structure formation.

Earlier, we've shown that in a matter-dominated universe,
$a(t) \propto t^{2/3}$
And here, we have shown that
$D_+(t) \propto t^{2/3}$
Therefore,
$\boxed{D_+(t) \propto a(t)}$

2(d) 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.

Consider a spherical patch of uniform over-density. Let's 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 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. It turns out that this differential equation can be solved analytically for constant $$C$$.

The parametric equation that solves this differential equation for the closed case is:
$r = A(1-\cos \eta)$
$t = B(\eta - \sin \eta)$
where $$\eta$$ is a function of $$t$$. $$A$$ and $$B$$ are positive constants satisfying the relationship $$A^3 = GMB^2$$.

For the open case, the solutions to this differential equation are:
$r = A(\cosh \eta - 1)$
$t = B(\sinh \eta - \eta)$

And finally, for the flat case, the solutions are:
$r = \frac{A}{2} \eta^2$
$t = \frac{B}{6} \eta^3$

Plot $$r$$ as a function of $$t$$ for all three cases, 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.

Here are the plots:
As you can see, the particle turns around and collapses with time in the closed case.

In the open case, the radius of the sphere appears to increase forever in what appears to be approximately linear with time.

Finally, the flat case also exhibits and infinite increase in the radius of the sphere, but the rate at which this distance increases steadily decreases over time. In other words, the velocity approaches zero with time.

## Saturday, November 28, 2015

### 35 | Worksheet 11.1, Problem 4: The Flatness Problem of the Big Bang Model

Despite the success of the Big Bang model mentioned in the lecture, there are also problems with it. These problems mostly have to do with the initial conditions for the Big Bang model, and provide motivations for scientists to seek deeper answers for the origin of the Big Bang. In this exercise, we study for one of these problems.

(a) Recall the Friedmann equation that we learned in a previous lecture:
$H^2 = \frac{8\pi G}{3}\rho - \frac{kc^2}{a^2}$
The critical density of the universe $$\rho_c$$ is defined to be the value of $$\rho$$ at which the curvature parameter $$k=0$$. Namely,
$\rho_c (t) = \frac{3H(t)^2}{8\pi G}$
Rewrite the Friedmann equation in terms of the density parameter $$\Omega$$ defined as:
$\Omega (t) \equiv \frac{\rho (t)}{\rho_c (t)}$
and show that:
$1+ \frac{kc^2}{a^2 H^2} = \Omega$

Starting with the Friedmann equation given:
$H^2 = \frac{8\pi G}{3} \rho - \frac{kc^2}{a^2}$
We notice the first term on the right-hand side looks similar to $$\rho_c$$, which is:
$\rho_c = \frac{3H^2}{8\pi G}$
We can rewrite the Friedmann equation in terms of $$\rho_c and \rho$$:
$H^2 = \frac{\rho}{\rho_c}H^2 - \frac{kc^2}{a^2}$
Since the ratio of $$\rho$$ and $$\rho_c$$ is defined to be the density parameter $$\Omega$$, we can rewrite this as:
$H^2 = \Omega H^2 - \frac{kc^2}{a^2}$
Solving for $$\Omega$$, we get:
$\boxed{\Omega = 1 + \frac{kc^2}{H^2 a^2}}$

(b) Consider the epoch of our universe that is dominated by matter. We have already computed the time dependence of $$a$$ and $$H$$ in such a universe. Use this result and (a) to show the time dependence of $$\Omega$$.

In Worksheet 9.1, we found that the time dependence of the scale factor $$a$$ in such a universe is given by:
$a \propto t^{\frac{2}{3}}$
We also know that the Hubble constant is defined as:
$H = \frac{\dot{a}}{a}$
Since we've already found the time dependence of $$a$$, we can easily find the time dependence of $$H$$. Taking the derivative of $$a$$, we find that $$\dot{a} \propto t^{-1/3}$$, so $$H \propto t^{-1}$$. Above, we showed that:
$\Omega - 1 \propto a^{-2}H^{-2}$
Plugging in the time dependences found above, we find that
$\boxed{\Omega -1 \propto t^{2/3}}$

(c) Today, experiments have measured that the density parameter of our universe is within -0.005<$$\Omega -1$$<0.005, namely $$\Omega$$ is close to 1 today. Use your above result to determine the range of $$\Omega -1$$ at the time of CMB formation, using the fact that today the age of universe is about 13.7 billion years and the age of the universe when the CMB was formed was about 380,000 years.

We've shown how the density parameter is related to time, namely:
$\Omega -1 \propto t^{2/3}$
In order to solve for $$\Omega$$ we will say that $$\Omega - 1 = C t^{2/3}$$ where $$C$$ is a proportionality constant.

We know both the time and density parameter today, so we can use these values to find the proportionality constant. Since we know a range for the density parameter, we cannot use equal signs, but instead I will use arrows below:
$C t^{2/3} \Rightarrow \Omega - 1$
$C(13.7\times 10^9 \text{ yr})^{2/3} \Rightarrow 0.005$
$C \Rightarrow 8.73\times 10^{-10}$
Now we can use this constant to find the density parameter range during the time the CMB was formed:
$\Omega - 1 \Rightarrow (8.73\times 10^{-1-})(3.8\times 10^5 \text{ yr}) = 4.58\times 10^{-6}$
So, the range for the density parameter at the time of CMB formation was:
$\boxed{-4.6\times 10^{-6} < \Omega - 1< 4.6\times 10^{-6}}$
This means that $$\Omega$$ is very close to 1.

### 34 | Worksheet 11.1, Problem 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.

 CMB (image: cosmology.berkeley.edu)

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 a photon) satisfies the Planck spectrum. At about the redshift $$z \approx 1100$$ when the universe had the temperature $$T \approx 3000 \text{ K}$$, 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 a redshift $$z$$ with frequency $$\nu$$, what is the its frequency $$\nu'$$ today?

We have shown before that the frequency observed today $$\nu'$$ is equal to the emitted frequency $$\nu$$ multiplied by the scale factor $$a(t)$$ at the time the light was emitted. This gives:
$\nu' = a(t)\nu$
In a previous post, we've also shown that the scale factor is related to the redshift (\z\) corresponding to light emitted at time $$t$$:
$a(t) = \frac{1}{1+z}$
Plugging this into our frequency expression, we get:
$\boxed{\nu' = \frac{\nu}{1+z}}$

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

To answer this question, we need to consider two components of the energy density term. In a most basic sense, energy density is defined as $$\frac{\text{energy}}{\text{volume}}$$. As the universe expands, both of these values change. Let's start with the energy.

As the photon travels, it is redshifted. Since it's frequency changes, its energy is also "redshifted." The energy of a photon is given as $$E = h\nu$$. In (a), we showed that $$\nu$$ changes by a factor of $$(1+z)^{-1}$$. Since frequency and energy are directly proportional, we know that the change in energy must also be proportional to $$(1+z)^{-1}$$.

As the universe expands, the volume gets larger. This essentially "dilutes" photons in the universe. We know that volume is proportional to $$d^3$$, where $$d$$ represents some distance. In a previous worksheet, we showed that the proper distance $$d(t) = a(t)d$$. In terms of volume, this gives $$V(t) = a^3(t)V$$. In terms of redshift, we know have $$V(t) = (1+z)^3 V$$. This changes our energy density term by a factor of $$(1+z)^{-3}$$.

Between the change in energy and change in volume, we know that the energy density must overall change by a factor of $$(1+z)^{-4}$$. So,
$\boxed{u_{\nu'} \mathrm{d}\nu' = (1+z)^{-4} u_{\nu}\mathrm{d}\nu}$

(c) Plug the relation between $$\nu$$ and $$\nu'$$ into the Planck spectrum:
$u_{\nu} \mathrm{d}\nu = \frac{8\pi h \nu^3}{c^3} \frac{1}{\mathrm{e}^{\frac{h\nu}{kT}}-1} \mathrm{d}\nu$
and also multiply it with the overall energy dilution factor that you have just figured out to get the energy density today. Write a final expression as the form $$u_{\nu'}\mathrm{d}\nu$$. Show that it is exactly the same Plank spectrum except that the temperature is now $$T' = T(1+z)^{-1}$$.

If we rewrite the Planck spectrum in terms of the energy density today, we get:
$u_{\nu'} \mathrm{d}\nu' (1+z)^{-4} = \frac{8\pi h (\nu' (1+z)^{-1})^3}{c^3} \frac{1}{\mathrm{e}^{\frac{h\nu' (1+z)^{-1}}{kT}}-1} \mathrm{d}\nu'(1+z)^{-1}$
When we solve for the energy density today $$u_{\nu'} \mathrm{d}\nu'$$, we get:
$u_{\nu'} \mathrm{d}\nu' = \frac{8\pi h \nu'^3}{c^3} \frac{1}{\mathrm{e}^{\frac{h\nu' (1+z)^{-1}}{kT}}-1} \mathrm{d}\nu'$
Note that this equation is the same as above, except for the exponential term which is now:
$\mathrm{e}^{\frac{h\nu' (1+z)^{-1}}{kT}}$
Note that we can simply rewrite this as:
$\mathrm{e}^{\frac{h\nu'}{kT'}}$
where $$T' = T(1+z)^{-1}$$.

(d) As you have just derived, according to the Big Bang model we should observe a black body radiation with temperature $$T'$$ throughout the entire universe. This is the Cosmic Microwave Background. Using the information given at the beginning of this problem, what is this temperature $$T'$$ today?

Above, we found:
$T' = T (1+z)^{-1}$
At the beginning of the problem, we were told that the CMB was formed at a redshift of about $$z \approx 1100$$ when the universe had a temperature $$T = 3000\text{ K}$$. Plugging these values in, we can get an approximate temperature for the CMB today:
$T' = (3000 \text{ K}) (1 + 1100)^{-1} = \boxed{2.72 \text{ K}}$

## Saturday, November 21, 2015

### 33 | Free-form: Meteor Showers

For as long as I can remember, I've headed outside a few times a year to watch meteor showers. A few prominent showers come to mind -- the Perseids, Geminids, and the Leonids. These events are named after the constellations around which they are centered in the nighttime sky.

 Geminid Meteor Shower (image: www.space.com)

Although I've watched these mesmerizing events, I never knew why they occurred. Sure, I knew that meteors or "shooting stars" were caused by chunks of rock falling through Earth's atmosphere and burning up along the way, but why do periodic meteor showers occur?

It turns out that meteor showers occur when Earth passes through debris left by a passing comet. Comets are large collections of rock and ice that have highly elliptical orbits around the Sun.

 (image: cumbriansky.files.wordpress.com)

So-called "short-period" comets travel around the Sun in as few as twenty years, which is on the order of the period of planets within the Solar System. Other comets have much, much longer periods. For example, Comet West is predicted to have an orbital period of as high as 6 million years!

 Comet West (image: WikiCommons | J. Linder)

As these comets approach the Sun, the ice and rock within the comet begin to break apart. This material streams off behind the comet, which has picked up speed as it is pulled towards the Sun. This material is what gives comets the characteristic "tail" that we are used to seeing.

 Comet ISON with its exceptionally long visible tail--20 million kilometers long! (image: http://www.eagleseye.me.uk)

The Earth's orbit occasionally passes through the orbit of a comet. Luckily for the Earth, it would be very unlikely for a comet to actually collide with the planet.

 (image: spaceplace.nasa.gov/)

However, the debris left behind a comet remains long after the comet has passed. When the Earth passes through this collection of rocks left by a comet, some of this material hits the Earth's atmosphere and falls to the ground, thus creating meteors. Because there is a high concentration of this debris in a given spot, we observe thousands and thousands of meteors. This is a meteor shower.

 (image: www.examiner.com)

Because the debris is at a single point along Earth's orbit, we encounter this material once a year. This results in a shower that occurs at the same time every year.

 (image: www.space.com)

Because meteor showers are caused by comet remnants, these events are strongest soon after the comet passed near the Sun. This also means that, over time, the meteor showers we are most familiar with will likely fade until the next time the comet passes close to the Sun. Furthermore, it is possible that there are meteor showers that humans have not yet witnessed, as the comets have not passed through Earth's orbit within the last few millennia.

References:

https://stardate.org/nightsky/meteors

http://www.scientificamerican.com/article/what-causes-a-meteor-show/

http://spaceguard.rm.iasf.cnr.it/NScience/neo/neo-what/com-prop.htm

http://www.eagleseye.me.uk/Sky/Wordpress/?p=2959

### 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 universe is called the particle 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) Why do we use the light to figure out the horizon size?

We use the light, as there is no information in the universe that can travel faster than the speed of light. This is to say that if we want to see as far as possible into our universe, we must use light, as this information has traveled the farthest in the finite time since the Big Bang.

 (image: http://frigg.physastro.mnsu.edu)

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

The Friedmann-Robertson-Walker metric is given by:
$\mathrm{d}s^2 = -c^2 \mathrm{d}t^2 + a^2(t) \left[\frac{\mathrm{d}r^2}{1-kr^2} + r^2 (\mathrm{d}\theta^2 + \sin^2 \theta \mathrm{d}\phi^2)\right]$
Since we are dealing with light, we can set $$\mathrm{d}s^2 = 0$$. Also, we are considering light traveling in the radial direction, so $$\mathrm{d}\theta = \mathrm{d}\phi = 0$$. This simplifies the Friedmann-Robertson-Walker metric to:
$c^2 \mathrm{d}t^2 = \frac{a^2(t) \mathrm{d}r^2}{1-kr^2}$
Let's group the $$t$$-dependent terms on the left-hand side and the $$r$$-dependent terms on the right:
$\frac{c^2 \mathrm{d}t^2}{a^2(t)} = \left(\frac{1}{1-kr^2}\right)\mathrm{d}r^2$
Taking the square root of both sides gives:
$\boxed{\frac{c \mathrm{d}t}{a(t)} = \left(\frac{1}{1-kr^2}\right)^{\frac{1}{2}}\mathrm{d}r}$

(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 ($$t = t_0$$).

In (b), we found the following differential equation:
$\frac{c \mathrm{d}t}{a(t)} = \left(\frac{1}{1-kr^2}\right)^{\frac{1}{2}}\mathrm{d}r$
For now, we will consider a flat universe, so $$k =0$$. This means that our expression becomes:
$\frac{c \mathrm{d}t}{a(t)} = \mathrm{d}r$
Now we can integrate both sides. We will integrate over time from $$0 \rightarrow t_0$$ and over the distance $$0 \rightarrow r_h$$, where $$r_h$$ is the radius of the particle horizon:
$\int_0^{t_0} \frac{c}{a(t)}\mathrm{d}t = \int_0^{r_h} \mathrm{d}r$
Our expression includes $$a(t)$$, which we need to consider when taking the integral. In a previous worksheet, we found that $$a \propto t^{2/3}$$ in a matter-dominated universe. Since this is the scenario we are considering in this problem, we will say $$a(t) = \frac{1}{A}t^{2/3}$$ where $$A$$ is a proportionality constant. Our expression is now:
$cA \int_0^{t_0} \frac{1}{t^{\frac{2}{3}}} = \int_0^{r_h}\mathrm{d}r$
Solving this integral gives us the radius of the particle horizon:
$\boxed{r_h = 3cAt_0^{\frac{1}{3}}}$

### 31 | Worksheet 10.1, Problem 2: Ratio of circumference to radius

In this problem, we explored the ratio of the circumference to radius in flat, open, and closed geometries. To do this, we examined 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 $$\mathrm{d}\phi = 0$$ because a circle encloses a two-dimensional surface. For the flat case, this part is just:
$\mathrm{d}s^2 = \mathrm{d}r^2 + r^2 \mathrm{d}\theta^2$
The circumference is found by fixing the radial coordinate ($$r = R$$ and $$\mathrm{d}r =0$$) and integrating both sides of the equation. The radius is found by fixing the angular coordinate ($$\theta$$,$$\mathrm{d}\theta = 0$$) and integrating both sides.

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

We are told that the metric we are working with is given by:
$\mathrm{d}s^2 = \mathrm{d}r^2 + r^2 \mathrm{d}\theta^2$
Let's start by finding the circumference. First, we need to fix the radial coordinate such that $$r = R$$ and $$\mathrm{d}r = 0$$. Doing so gives us;
$\mathrm{d}s^2 = R^2 \mathrm{d}\theta^2$
This simplifies to:
$\mathrm{d}s = R\mathrm{d}\theta$
Integrating over the angles $$0 \rightarrow 2\pi$$,
$\int\! \mathrm{d}s = \int_0^{2\pi} R \mathrm{d}\theta$
This gives us:
$S = 2\pi R$
This means that the circumference of the circle is $$2\pi R$$. This is consistent with what we know about circles.

Now, let's find the radius. To do this, we set $$\theta$$ constant, such that $$\mathrm{d}\theta = 0$$. Our expression now simplifies to:
$\mathrm{d}s = \mathrm{d}r$
If we integrate both sides, we get:
$\int \mathrm{d}s = \int_0^R \mathrm{d}r$
$s = R$
This means that the radius of our circle is $$R$$, which makes perfect sense.

Now, to find the ratio, we simply take the circumference over the radius:
$\frac{\text{circumference}}{\text{radius}} = \frac{2\pi R}{R} = \boxed{2\pi}$

For closed geometry, we calculated the analogous two-dimensional part of the metric in Problem 1. This can be written as:
$\mathrm{d}s^2 = \mathrm{d}\xi^2 + \sin^2 \xi \mathrm{d}\theta^2$
Repeat the same calculation as above to derive the ratio for closed geometry.  Compare your results to the Euclidean case.

Here we have a very similar problem to what we did in (a), so we can use the same technique. Let's start by considering the circumference. To do so, we will set $$\xi$$ constant, such that $$\xi = \Xi, \mathrm{d}\xi = 0$$). Our expression becomes:
$\mathrm{d}s^2 = \sin^2 \Xi \mathrm{d}\theta^2$
$\mathrm{d}s = \sin \Xi \mathrm{d}\theta$
Again, we can integrate both sides:
$\int \mathrm{d}s = \int_0^{2\pi}\sin \Xi\, \mathrm{d}\theta$
Solving gives us that the circumference of the circle in closed geometry is:
$S = 2\pi \sin \Xi$
Now, let's find the radius by setting the angular coordinate constant $$\mathrm{d}\theta =0$$. Simplification yields:
$\mathrm{d}s = \mathrm{d}\xi$
Now we can integrate from 0 to $$\Xi$$:
$\int \mathrm{d}s =\int_0^{\Xi}\mathrm{d}\xi$
Thus, the radius is simply given by $$\Xi$$.

Using these two values we find that the ratio is:
$\frac{\text{circumference}}{\text{radius}} = \boxed{\frac{2\pi \sin \Xi}{\Xi}}$
We can see that this is different from the Euclidean case. In this example, the peak of the ratio is $$2\pi$$ (same as Euclidean), but this only occurs in the limit as $$\Xi \rightarrow 0$$. Otherwise, this ratio is smaller.

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

This is a very similar problem to (b) except that:
$\mathrm{d}s^2 = \mathrm{d}\xi^2 + \sinh^2 \xi\, \mathrm{d} \theta^2$
I won't do the math here, as it is exactly the same as (b) except that we replace the sine function with the hyperbolic sine function. We end up at the conclusion that:
$\frac{\text{circumference}}{\text{radius}} = \boxed{\frac{2\pi \sinh \Xi}{\Xi}}$
Like in the closed geometry case, this ratio is dependent on the radial coordinate. In fact, this ratio has a very similar form to that of the closed 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?

Let's consider the closed case where we found the ratio to be:
$\frac{2\pi \sin \Xi}{\Xi}$
As $$\Xi \rightarrow 0$$, we can use the small angle approximation that says $$\sin x \approx x$$ for small angles. For small values of $$\Xi$$, the ratio we found becomes:
$\frac{2\pi \sin \Xi}{\Xi} \approx \boxed{2\pi}$
Thus, the limit for closed and open geometries as $$\Xi \rightarrow 0$$ is identical to the ratio for the flat geometry.

## Sunday, November 8, 2015

### 30 | Radio Astronomy Lab, Galactic Rotation

In this lab, we will be using the millimeter-wave telescope at the Harvard-Smithsonian Center for Astrophysics to ultimately determine the mass of the Milky Way interior to the Sun's orbital radius. We will accomplish this by measuring the Doppler shifts of the 115 GHz spectral line of CO. This line corresponds to a strong transition in the rotational energy levels of carbon monoxide. At this frequency, we observe photons emitted by the transition from the first energy level back down to the ground state.

Here are some considerations for observing with the radio telescope:

What is going to be the typical integration time per point?

To calculate a reasonable integration time, we need to know a but about noise, as we hope to have an integration time long enough that our signal-to-noise ratio (SNR) is quite high. The root-mean-square error $$\sigma$$ is given by:
$\sigma = \frac{T_{\text{sys}}}{\sqrt{\tau \Delta \nu}}$
Where $$T_{\text{sys}}$$ is the noise of each sample, $$\tau$$ is the integration time, and $$\Delta \nu$$ is the channel width.
We will define the signal strength to be $$T_A$$. Therefore, the SNR is defined as:
$\text{SNR} = \frac{T_A}{\sigma}$
If we solve this for $$\sigma$$ and plug it into the equation we found above, then solve for $$\tau$$, we get:
$\tau = \frac{T_{\text{sys}}^2 (\text{SNR})^2}{T_A^2 \Delta \nu}$
We are told that $$T_{\text{sys}}$$ is about 500 K and that the average signal peak will be about 2 K. Furthermore, we will consider using the 0.5 MHz wide channel. As for the SNR, let's calculate the integration time that corresponds to an average error of 1%, meaning SNR=100.
$\tau = \frac{(500 \text{ K})^2 (100)^2}{(2\text{ K})^2 (0.5\times 10^6 \text{ s}^{-1})} = 1250 \text{ s}$
This means that a good integration time would be about 21 minutes.

Over what range of longitude do you plan to observe?

According to the lab manual, we will be observing at Galactic longitudes of 10° to 70°.

How many positions do you plan to observe?

The manual seems to suggest that we will be observing four giant molecular clouds, so we will be observing at four positions.

At what LST are you going to start observing? At what EST?

We will probably begin our observations 1:00pm EST on (for example) November 10. This corresponds to about 15:34 LST.

Are you going to position switch or frequency switch to "flat field" the spectrum?

We want the flat-field correction to correspond to the sensitivity of the CCD at the specific frequency at which we are observing. As a result, we do not want to switch the frequency to flat field our images. Instead, we can perform a "position switch" to do this.

Above, we started Problem 1 on Worksheet 9.2, so let's finish it:

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

To do this, we can use the Doppler equation, which tells us:
$\Delta v = c \frac{\Delta \nu}{\nu}$
To find the equation in the desired form, we simply need to multiply the speed of light $$c$$ ($$3\times 10^5 \text{ km/s}$$) by $$10^6$$, which corresponds to the MHz term we put in the denominator. This gives us:
$\boxed{\Delta v = 3.8\times 10^{11} \text{km/s}\left(\frac{\Delta \nu}{1 \text{ MHz}}\right)}$

I've shown solutions very similar to (a) and (c) above.

## Saturday, November 7, 2015

### 29 | Worksheet 9.1, Problem 2: General Relativity Modification to Newtonian Friedmann Equation

In the previous post, we derived the Friedmann equation in a matter-only universe using 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 general relativity (GR). Analogous to the equations that we derived previously, 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.

The first Friedmann equation (1):
$\left(\frac{\dot{a}}{a}\right)^2 = \frac{8\pi}{3}G \rho - \frac{kc^2}{a^2} + \frac{\Lambda}{3}$
The second Friedmann equation (2):
$\frac{\ddot{a}}{a} = - \frac{4\pi G}{3c^2} (\rho c^2 + 3P) + \frac{\Lambda}{3}$
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 universes, 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, 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 the time derivative on both sides; plug (2) into your expression to eliminate $$\ddot{a}$$.

The first Friedmann equation is given by:
$\frac{\dot{a}^2}{a^2} = \frac{8\pi}{3} G \rho - \frac{kc^2}{a^2} + \frac{\Lambda}{3}$
If we multiply both sides by $$a^2$$, we get:
$\dot{a}^2 = \frac{8\pi}{3}G \rho a^2 - kc^2 + \frac{\Lambda a^2}{3}$
Now we can take the derivative with respect time time. In this expression, both $$a$$ and $$\rho$$ are functions of time. Therefore, the time derivative is:
$2\dot{a}\ddot{a} = \frac{8\pi}{3} G a^2 \dot{\rho} + \frac{8\pi}{3}\rho (2a\dot{a}) + \frac{\Lambda}{3}2a\dot{a}$
Notice the $$-kc^2$$ term drops out, as this has no time dependence. If we simplify and solve for $$\frac{\ddot{a}}{a}$$, we get:
$\frac{\ddot{a}}{a} = \frac{4\pi G \dot{\rho} a}{3\dot{a}} + \frac{8\pi}{3} G \rho + \frac{\Lambda}{3}$
The second Friedmann equation tells us:
$\frac{\ddot{a}}{a} = -\frac{4\pi G}{3c^2} (\rho c^2 + 3P) + \frac{\Lambda}{3}$
Since we now have two expressions for $$\ddot{a}/a$$, we can set them equal:
$\frac{4\pi G \dot{\rho} a}{3\dot{a}} + \frac{8\pi G \rho}{3} + \frac{\Lambda}{3} = - \frac{4\pi G}{3c^2}(\rho c^2 + 3P) + \frac{\Lambda}{3}$
Algebraic simplifcation and rearrangement to solve for $$\dot{\rho} c^2$$ gives:
$\boxed{\dot{\rho}c^2 = - \frac{3\dot{a}}{a}(\rho c^2 + P)}$

Now we can use these equations to derive some fun consequences of different kinds of universe, some of which describe our own universe.

For simplicity, in the exercise below, let us always set $$k = 0$$, namely consider a flat universe. Luckily for us, state-of-the-art observations suggest that our universe is likely flat.

(b) Cold matter dominated universe. 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 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 third Friedmann equation says:
$\dot{\rho}c^2 = -\frac{3\dot{a}}{a} (\rho c^2 + P)$
In this cold universe, $$P = 0$$. This simplifies our expression:
$\dot{\rho} = -3\rho \frac{\dot{a}}{a}$
Dividing both sides by $$\rho$$ gives:
$\frac{\dot{\rho}}{\rho} = -3 \frac{\dot{a}}{a}$
Now we can integrate both sides. We will use the limits provided in the problem:  $$\rho$$, $$\rho_0$$, $$a$$, and $$a_0$$.
$\int_{\rho_0}^{\rho} \frac{\dot{\rho'}}{\rho'} = -3 \int_{a_0}^{a} \frac{\dot{a'}}{a'}$
The evaluated integral gives:
$\ln{\rho} - \ln{\rho_0} = -3(\ln{a} - \ln{a_0})$
Using log rules, we can write this as:
$\ln{\frac{\rho}{\rho_0}} = -3\ln{\frac{a}{a_0}}$
Exponentiating both sides, we get:
$\boxed{\frac{\rho}{\rho_0} = \left(\frac{a}{a_0}\right)^{-3}}$
We know that $$a_0$$ is the scale factor of the universe in the present day. By definition, $$a_0 =1$$, so we could rewrite this expression as:
$\boxed{\rho = \rho_0 a^{-3}}$

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 proportionally 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 the differential 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. (Hint: When solving this differential equation, recall that at time $$t=0$$ (the Big Bang), $$a=0$$. At time $$t=t_0$$ (present day), $$a = a_0 =1$$.

The first Friedmann equation is:
$\frac{\dot{a}^2}{a^2} = \frac{8\pi}{3} G \rho - \frac{kc^2}{a^2} + \frac{\Lambda}{3}$
However, we will consider the universe flat, so $$k = 0$$. Also, we are considering a cold universe, such that $$\Lambda = 0$$. Therefore, our expression becomes:
$\left(\frac{\dot{a}}{a}\right)^2 = \frac{8\pi}{3}G \rho$
We just showed that $$\rho = \rho_0 a^{-3}$$. If we plug this in for $$\rho$$, we get:
$\left(\frac{\dot{a}}{a}\right)^2 = \frac{8\pi G \rho_0}{3a^3}$
If we solve for $$\dot{a}$$, we arrive at:
$\dot{a} = \left(\frac{8\pi G \rho_0}{3a}\right)^{\frac{1}{2}}$
Let's rewrite $$\dot{a}$$ as $$\mathrm{d}a/\mathrm{d}t$$:
$\frac{\mathrm{d}a}{\mathrm{d}t} = \left(\frac{8\pi G \rho_0}{3a}\right)^{\frac{1}{2}}$
Rearranging a few terms gives:
$a^{\frac{1}{2}}\mathrm{d}a = \left(\frac{8\pi G \rho_0}{3}\right)^{\frac{1}{2}} \mathrm{d}t$
We're only interested in a proportionality between $$a$$ and $$t$$, so we can drop the constants:
$a^{\frac{1}{2}}\mathrm{d}a \propto \mathrm{d}t$
Now we can integrate both sides:
$\int_{0}^{a} a'^{\frac{1}{2}}\mathrm{d}a' \propto \int_{0}^{t} \mathrm{d}t'$
$a^{\frac{3}{2}} \propto t$
So,
$\boxed{a \propto t^{\frac{2}{3}}}$

(c) Radiation dominated universe: 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 the scale factor.

$\dot{\rho}c^2 = -\frac{3\dot{a}}{a} (\rho c^2 + P)$
In this radiation dominated universe, we will still say that $$\Lambda =0$$, but we now are concerned with a pressure term. We are told that for radiation, $$P = \frac{1}{3}\rho c^2$$. The third Friedmann equation becomes:
$\dot{\rho} = -4 \rho \frac{\dot{a}}{a}$
Like we did before, let's divide by $$\rho$$:
$\frac{\dot{\rho}}{\rho} = -4 \frac{\dot{a}}{a}$
Following the same integration steps we used above, this gives us the expression:
$\frac{\rho}{\rho_0} = \left(\frac{a}{a_0}\right)^{-4}$
Or, since $$a_0 = 1$$,
$\boxed{\rho = \rho_0 a^{-4}}$

The result also has a simple interpretation. Imaging the radiation being a collection of photons. Similar to the matter case, the number density of photons is diluted, inversely proportional to the volume. In contrast to the dust particle, each photon can be thought of as a wave. The wavelength of a 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: $$E = h \nu$$, 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 is consistent with the result you got.

Our result is, indeed consistent with this energy consideration. We can see from the above equation that the mass density is inversely proportional to the scale factor. We also previously showed that the scale factor is directly proportional to the wavelength of the radiation. Since wavelength and frequency are inversely proportional by $$\nu = c/\lambda$$, then frequency must be inversely proportional to the scale factor. Furthermore, since energy is directly proportional to frequency, energy must also be inversely proportional to the scale factor.

Again, using the relation between $$\rho$$ and $$a$$ and the first Friedmann equation, show that $$a(t) \propto t^{1/2}$$ for the radiation-only universe.

Recall that our modified version of the first Friedmann equation is:
$\left(\frac{\dot{a}}{a}\right)^2 = \frac{8\pi}{3}G \rho$
As we did before, we can plug in the expression we found for $$\rho$$, which gives:
$\left(\frac{\dot{a}}{a}\right)^2 = \frac{8\pi G \rho_0}{3a^4}$
Following the exact same steps as we did for (b), we arrive at the following integral:
$\int a \mathrm{d}a \propto \int \mathrm{d}t$
So,
$a^2 \propto t$
Therefore, we conclude that:
$\boxed{a \propto t^{\frac{1}{2}}}$

(d) Cosmological constant/dark energy dominated universe: Imagine a universe dominated by the cosmological-constant-like term. Namiely 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 righ-hand side of the 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?

The third Friedmann equation tells us:
$\dot{\rho} c^2 = -3 \frac{\dot{a}}{a} (\rho c^2 + P)$
In this model of the universe, $$\rho = 0$$ and $$P=0$$. Therefore,
$\dot{\rho} = 0$
This suggest that the energy density of the dark energy does not 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?

When we allow $$P =0$$ and $$\rho=0$$, the first Friedmann equation is simply:
$\left(\frac{\dot{a}}{a}\right)^2 = \frac{\Lambda}{3}$
This means that the Hubble parameter of this universe is:
$\boxed{H = \left(\frac{\Lambda}{3}\right)^{\frac{1}{2}}}$
We can rewrite the first Friedmann equation as:
$\frac{\mathrm{d}a}{\mathrm{d}t} = \left(\frac{\Lambda}{3}\right)^{\frac{1}{2}}a$
Rearranging and dropping constants, we are left with:
$\frac{\mathrm{d}a}{a} \propto \mathrm{d}t$
Integrating both sides gives:
$\int \frac{\mathrm{d}a}{a} \propto \int \mathrm{d}t$
$\ln{a} \propto t$
Therefore, we arrive at the relationship that:
$\boxed{a \propto \mathrm{e}^t}$

(e) Suppose the energy density of a universe at its very early time is dominated by half matter and half radiation. (This is 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?

In the questions, above we derived relationships between the scale factor of the universe and the denisty. In a matter-dominated universe:
$\rho_m \propto a^{-3}$
$\rho_r \propto a^{-4}$
By these relations, we can see that the density of radiation decays quicker than the density of matter. Therefore, radiation will be "diluted" more rapidly. As a result, in this half matter/half radiation universe, matter will be the dominant component.

(f) Suppose the energy density of a universe is dominated by similar amounts 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?

As mentioned above, a matter-dominated universe can be described by:
$\rho_m \propto a^{-3}$
However, we showed that the density of dark energy did not have a time dependence. Therefore, as the universe continues to expand, dark energy will become the dominant component. With this knowledge, we can consider the scale factor-time relation we found for a dark energy universe:
$a \propto \mathrm{e}^t$
This suggests that as we become a dark energy-dominated universe, the size of the universe will expand exponentially!

### 28 | Worksheet 9.1, Problem 1: Matter-only Model of the Universe

Consider a universe filled with matter which has a mass density of $$\rho(t)$$. 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 constant.

(a) What is the acceleration of this shell? Express the acceleration as the time derivative of the velocity, $$\dot{v}$$.

Acceleration is caused by a net force acting on a mass. In this matter-only model, the only force to consider is the force due gravity. This is given by:
$F_g = -\frac{GMm}{R^2}$
Newton's second law tells us:
$F_{\text{net}} = ma$
Or, using our notation:
$F = m\dot{v}$
Setting these two expressions equal:
$m\dot{v} = -\frac{GMm}{R^2}$
Solving for acceleration gives us:
$\boxed{\dot{v} = -\frac{GM}{R^2}}$

(b) To derive an energy equation, it is a common trick to multiply both sides of your acceleration equation by $$v$$. Turn your velocity $$v$$ into $$\frac{\mathrm{d} R}{\mathrm{d}t}$$, cancel $$\mathrm{d} t$$, and integrate both sides of the equation. This is an indefinite integral, so you will have a constant of integration; combine these integration constants and call their sum $$C$$. You should arrive at the following equation:
$\frac{1}{2}\dot{R}^2 - \frac{GM}{R} = C$
Convince yourself the equation you've written down as units of energy per unit mass

The expression we derived above was:
$\dot{v} = -\frac{GM}{R^2}$
Let's multiply both sides by $$v$$:
$v\dot{v} = -\frac{GM}{R^2}v$
We can write $$\dot{v}$$ as $$\mathrm{d} v /\mathrm{d}t$$. On the other side of the equation, we can write $$v$$ as $$\mathrm{d}R/\mathrm{d}t$$. This gives us:
$v \frac{\mathrm{d}v}{\mathrm{d}t} = -\frac{GM}{R^2} \frac{\mathrm{d}R}{\mathrm{d}t}$
Using the trick of canceling out the $$\mathrm{d}t$$ on each side of the equation, we get:
$v \mathrm{d}v = -\frac{GM}{R^2}\mathrm{d}R$
Now we can integrate both sides:
$\int v \mathrm{d}v = -GM \int \frac{1}{R^2}\mathrm{d}R$
$\frac{1}{2} v^2 + C_1 = \frac{GM}{R} + C_2$
Where $$C_1$$ and $$C_2$$ are constants of integration. We can rearrange this equation and combine the constants of integration, which gives us:
$\frac{1}{2}v^2 - \frac{GM}{R} = C$
Finally, let's write $$v$$ as $$\dot{R}$$:
$\boxed{\frac{1}{2}\dot{R}^2 - \frac{GM}{R} = C}$
Now, let's check out the units on this equation are energy per unit mass.

We can see from the $$\dot{R}$$ that this equation has units $$\frac{[\text{distance}]^2}{[\text{time}]^2}$$. Energy can be expressed as $$\frac{[\text{mass}][\text{distance}]^2}{[\text{time}]^2}$$. Therefore, our equation does indeed have units of energy per unit mass.

(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 $$\left(\frac{\dot{R}}{R}\right)^2$$.

The expression we found was:
$\frac{1}{2}\dot{R}^2 - \frac{GM}{R} = C$
We can replace $$M$$ in terms of the mass density $$\rho$$. We know that $$M = \frac{4}{3}\pi \rho R^3$$. So,
$\frac{1}{2}\dot{R}^2 - \frac{G(\tfrac{4}{3}\pi \rho R^3)}{R} = C$
Simplifying and rearranging gives:
$\dot{R}^2 = 2C + \frac{8}{3}\pi G \rho R^2$
To get the expression in the desired form, all we have to do is divide both sides by $$R^2$$. This gives:
$\boxed{\left(\frac{\dot{R}}{R}\right)^2 = \frac{2C}{R^2} + \frac{8}{3}\pi G \rho}$

(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 $$a(t)$$.

Our equation is:
$\left(\frac{\dot{R}}{R}\right)^2 = \frac{2C}{R^2} + \frac{8}{3}\pi G \rho$
We need to write this in terms of $$a$$ and $$r$$ instead of $$R$$. Let's begin with the left side of the equation. We will consider $$\frac{\dot{R}}{R}$$. We can express this as:
$\frac{\dot{R}}{R} = \frac{\mathrm{d}R/\mathrm{d}t}{R}$
But what is $$\frac{\mathrm{d}R}{\mathrm{d}t}$$? Let's take the derivative of $$R$$ incorporating the comoving radius and scale factor:
$\frac{\mathrm{d}R}{\mathrm{d}t} = \frac{\mathrm{d}}{\mathrm{d}t}(ar) = a\dot{r} + \dot{a}r$
However, the comoving radius does not change with time, so $$\dot{r} = 0$$. This simplifies our expression to:
$\frac{\mathrm{d}R}{\mathrm{d}t} = \dot{a}r$
Since $$R = ar$$, the left side of our equation can simply be written as:
$\left(\frac{\dot{a}r}{ar}\right)^2$
This simplifies to:
$\left(\frac{\dot{a}}{a}\right)^2$
Now, let's deal with the right side of the equation. Here, the only change that needs to be made is to change $$R$$. So, our final expression is:
$\boxed{\left(\frac{\dot{a}}{a}\right)^2 = \frac{8}{3}\pi G \rho + \frac{2C}{(ar)^2}}$

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

I showed this in (d).

(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 $$\frac{2C}{r^2} = -kc^2$$ where $$k$$ is the "curvature" parameter, you will get the first Friedmann equation. This equation tells us about how the shell expands or contracts; in other words, it tells about the Hubble expansion (or contraction) rate of the universe.

We know that $$H(t) = \frac{\dot{a}}{a}$$, which means that the left side of our equation is simply $$H^2$$. We are told that the constant $$\frac{2C}{r^2} = -kc^2$$. The second term of the right side of the equation is $$\frac{2C}{(ar)^2}$$. Substituting $$-kc^2$$ our final expression becomes:
$\boxed{H^2 = \frac{8}{3}\pi G \rho - \frac{kc^2}{a^2}}$

(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 $$a(t)r$$. You should see that $$\frac{\dot{a}}{a} = -\frac{4\pi}{3}G \rho$$, 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, which is not captured in this 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.

We can accomplish this through a similar process of what we did above to derive the first Friedmann equation. Recall the expression we derived by equating Newton's second law with the expression of force due to gravity:
$m\ddot{R} = -\frac{GMm}{R^2}$
This becomes:
$\ddot{R} = -\frac{GM}{R^2}$
Now let's use the mass density to replace $$M$$.
$\ddot{R} = -\frac{GM}{R^2} = \frac{4}{3}\pi G \rho R$
Dividing by $$R$$:
$\frac{\ddot{R}}{R} = \frac{4}{3}\pi G \rho$
Now, we need to write this in terms of the scale factor $$a$$. Before we showed that $$\dot{R} = \dot{a}r$$. Now we need to take the derivative of $$\dot{R}$$:
$\ddot{R} = \frac{\mathrm{d}}{\mathrm{d}t}(\dot{R}) = \frac{\mathrm{d}}{\mathrm{d}t} (\dot{a}r) = \ddot{a}r + \dot{a}\dot{r}$
As we said before, $$\dot{r} = 0$$ since the comoving radius does not change. Therefore,
$\ddot{R} = \ddot{a}r$
Plugging this into our equation gives:
$\boxed{\frac{\ddot{a}}{a} = - \frac{4}{3}\pi G \rho}$

## Friday, October 30, 2015

### 27 | Worksheet 8.1, Problem 4: Scale factor-comological redshift relation

There is a direct relationship between the scale factor of the universe during the epoch in which the light we receive from an object was originally emitted $$a$$ and the cosmological redshift $$z$$. One way to derive this is to consider a light ray with wavelength $$\lambda_e$$ that was emitted at time $$t_e$$ and traveling towards us, and track the change in its wavelength as it travels across space and time while the universe expands.

(a) Two observers along the path of this light ray are separated by a small distance $$\mathrm{d}r$$. If the light ray passes by them at time $$t$$, write down their relative velocity $$\mathrm{d}v$$ as a function of $$H(t)$$ and $$\mathrm{d}r$$. Then write their redshifts $$\mathrm{d}z = \mathrm{d}\lambda / \lambda$$ with respect to each other.

We can start by recalling the Hubble law:
$v = H(t) d$
Using this expression, the relative velocity over a small distance $$\mathrm{d}r$$ becomes:
$\mathrm{d}v = H(t) \mathrm{d}r$
Now let's deal with the redshifts. We have two expressions for the redshift. The first is given by:
$z = \frac{v}{c}$
Let's consider the redshift over a small change in velocity $$\mathrm{d}v$$. Our expression becomes:
$\mathrm{d}z = \frac{\mathrm{d}v}{c}$
The Hubble law gave us an expression for $$\mathrm{d}v$$ in terms of the Hubble constant and the distance. If we plug this in, we get:
$\mathrm{d}z = \frac{H(t) \mathrm{d}r}{c}$
The second expression for the redshift is given in the problem:
$\mathrm{d}z = \frac{\mathrm{d}\lambda}{\lambda}$
Now that we have two expressions for $$\mathrm{d}z$$, let's set them equal.
$\boxed{\frac{\mathrm{d}\lambda}{\lambda} = \frac{H(t) \mathrm{d}r}{c}}$

(b) Let's gradually work the time differential into the picture. Express the time it took light to travel between the two observers $$\mathrm{d}t$$ in terms of $$\mathrm{d}r$$ and $$c$$. Also, rearrange the definition of the Hubble constant to isolate for $$\mathrm{d}t$$.

The time it takes the light to travel between the observers can be found using the simple relationship that velocity is equal to distance over time. Using our differentials and the speed of light, this gives:
$\boxed{\mathrm{d}t = \frac{\mathrm{d}r}{c}}$
The Hubble constant is given by:
$H(t) = \frac{1}{a(t)}\frac{\mathrm{d}a}{\mathrm{d}t}$
Solving for the time differential and saying $$a = a(t)$$,
$\boxed{\mathrm{d}t = \frac{\mathrm{d}a}{H(t)a}}$

(c) Combine your answers in (b) with part (a) to show that $$\mathrm{d}\lambda / \lambda = \mathrm{d}a / a$$. This relation means that at time $$t$$ and over time interval $$\mathrm{d}t$$, the fractional change in the light ray's wavelength equals the fractional change in the scale factor.

In (a) we found:
$\frac{\mathrm{d}\lambda}{\lambda} = \frac{H(t) \mathrm{d}r}{c}$
Let's solve this for $$H(t)$$:
$H(t) = \frac{c \mathrm{d}\lambda}{\lambda \mathrm{d}r}$
In (b) we found the following two expressions for the time differential:
$\mathrm{d}t = \frac{\mathrm{d}r}{c}$
$\mathrm{d}t = \frac{\mathrm{d}a}{H(t)a}$
If we set these two equal and solve for $$H(t)$$, we get:
$H(t) = \frac{c \mathrm{d}a}{a\mathrm{d}r}$
Now we can combine our results from (a) and (b):
$\frac{c \mathrm{d}\lambda}{\lambda \mathrm{d}r} = \frac{c \mathrm{d}a}{a\mathrm{d}r}$
Sure enough, this simplifies to:
$\boxed{\frac{\mathrm{d}\lambda}{\lambda} = \frac{\mathrm{d}a}{a}}$

(d) Solve this differential equation for $$\lambda(a)$$ up to a constant $$C$$. This is the wavelength of the light ray as a function of the scale factor.

We found that:
$\frac{\mathrm{d}\lambda}{\lambda} = \frac{\mathrm{d}a}{a}$
To solve this differential equation, we can integrate both sides of our expression.
$\int \frac{\mathrm{d}\lambda}{\lambda} = \int \frac{\mathrm{d}a}{a}$
$\ln{\lambda} = \ln{a} + C'$
Now we can exponentiate both sides to get rid of the logarithms. This gives us:
$\lambda = \mathrm{e}^{\ln{a} + C'} = a\cdot \mathrm{e}^{C'}$
We can write this same expression as:
$\boxed{\lambda(a) = Ca}$

(e) Use appropriate boundary conditions on your solution above to show that $$C = \lambda_{\text{observed,today}}$$.

Let's say $$\lambda_0 = \lambda_{\text{observed,today}}$$. In the present, the value of the scale factor $$a$$ is, by definition, equal to 1. Since $$\lambda_0$$ is the observed wavelength at the present time,
$\lambda(a=1) = \lambda_0 = C(1) = C$
Therefore, $$C = \lambda{\text{observed,today}}$$.

(f) What is the original wavelength of the light $$\lambda_e$$ (i.e. what is the wavelength of the light ray at time of emission $$t_e$$ )?

Recall that the scale factor $$a$$ is a function of time $$t$$. As a result, we can use the expression we derived above to say:
$\lambda_e = Ca(t_e)$
However, we just showed that $$C = \lambda_0$$, where $$\lambda_0$$ is the observed wavelength in the present. Using this, our expression for the emitted wavelength becomes:
$\boxed{\lambda_e = \lambda_0 a(t_e)}$

(g) What is the measured redshift for the source of this light ray in terms of the scale factor of the universe at the time of emission?

The redshift is defined as:
$z = \frac{\lambda_o - \lambda_e}{\lambda_e}$
Where $$\lambda_o$$ is the observed wavelength and $$\lambda_e$$ is the emitted wavelength. In terms of our variables, $$\lambda_o = \lambda_0$$ and $$\lambda_e$$ is the same. So, our expression is:
$z = \frac{\lambda_0 - \lambda_e}{\lambda_e}$
In (f), we showed that $$\lambda_e = \lambda_0 a(t_e)$$. Using this, our redshift expression becomes:
$z = \frac{\lambda_0 - \lambda_0 a(t_e)}{\lambda_0 a(t_e)} = \frac{1 - a(t_e)}{a(t_e)}$
If we say represent the scale factor at time of emission as $$a_e = a(t_e)$$, we can express the measured redshift for the source of the light ray as:
$\boxed{z = \frac{1 - a_e}{a_e}}$

(h) You should have noticed that the redshift is a unique function of the scale factor. Alternatively, you can express the scale factor unambiguously in terms of the redshift. Interpret this finding. How large was the universe at the time the light from a galaxy at $$z=3$$ was emitted compared to the present-day?

To find this expression, we can simply rearrange the expression we just derived:
$z = \frac{1 - a_e}{a_e}$
Rearranging, we get:
$\boxed{a_e = \frac{1}{1+z}}$
This result means that a source with a larger observed redshift emitted it's light longer ago. Based on other relations we've shown, this also directly shows that a larger redshift represents light from an object that is farther away.

Let's use this expression to determine the scale factor of the universe at the time light was emitted from a source with an observed redshift of $$z = 3$$.
$a_e = \frac{1}{1+z} = \frac{1}{1+3} =\boxed{\frac{1}{4}}$
This means that at the time the light from this source was emitted, the observable universe had a Hubble length 1/4 the size it is today.

### 26 | Worksheet 8.1, Problem 3: Size of the Universe

It is not strictly correct to associate the distance-dependent redshift with the velocity of galaxies. At very large separations, Hubble's law gives "velocities" that exceed the speed of light and the redshift 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 form 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 between grid points in the universe at time $$t$$. In the current epoch, $$t = t_0$$ and $$a(t_0) \equiv 1$$. The scale factor 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:
$H(t) = \frac{1}{a(t)} \frac{\mathrm{d}a}{\mathrm{d}t}\bigg|_{t}$
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 $$\dot{a} \equiv \mathrm{d}a/\mathrm{d}t$$ has been constant for all time. How long ago was the Big Bang (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.

We will start with the basic physical idea that velocity is equal to distance divided by time, or:
$v = \frac{d}{t}$
We are interested in a time, so let's solve for $$t$$:
$t = \frac{d}{v}$
The Hubble law tells us $$v = H(t)d$$. The Hubble constant $$H$$ is related to the amount of time that has passed since the beginning of the universe. That said, we should evaluate $$H$$ at the present ($$t = t_0$$) to account for all time that has passed. Based on what we did earlier in this worksheet, we will say that $$H(t=t_0) = H_0$$. Our expression is now:
$v = H_0 d$
Let's plug this $$v$$ into the expression we derived earlier. We will call this particular time $$t_H$$, which is the Hubble time.
$t_H =\frac{d}{H_0 d}$
So,
$\boxed{t_H = \frac{1}{H_0}}$
The most recently calculated value for $$H_0$$ was by the Planck Satellite in 2013. These measurements found $$H_0 = 67.8 \text{ km} \text{ s}^{-1} \text{ Mpc}^{-1}$$. In cgs, this is $$2.26\times 10^{-18} \text{ s}^{-1}$$. Plugging this into our equation,
$t_H = \frac{1}{ 2.26\times 10^{-18} \text{ s}^{-1}} = \boxed{4.42\times 10^{17} \text{ s}}$
This suggests that the Big Bang occurred about $$4.4 \times 10^{17} \text{ s}$$ ago.

Next, we want to compare this value to the age of the oldest known globular clusters, which is about 12 Gyr. Let's convert $$t_H$$ into Gyr:
$4.4 \times 10^{17} \text{ s} \times \frac{1 \text{ yr}}{\pi \times 10^7 \text{ s}} \times \frac{1 \text{ Gyr}}{10^9 \text{ yr}} = \boxed{14 \text{ Gyr}}$
This means that the universe is about 2 Gyr older than the oldest known globular clusters.

(b) What is the size of the observable universe? This is known as the Hubble length.

The observable universe consists of the matter in the universe that emitted photons that have had sufficient time to reach Earth. We can use the Hubble law to evaluate this. The Hubble law is given by:
$v = H(t) d$
We are interested in a distance, so let's solve for $$d$$:
$d = \frac{v}{H(t)}$
We are interested in moving photons, which travel at the speed of light $$c$$,
$d = \frac{c}{H(t)}$
As in the last problem, we want to consider the total amount of time from the Big Bang to the present ($$t = t_0$$). Again, we will say $$H(t = t_0) = H_0$$. Let's call this unique distance $$d_H$$:
$\boxed{d_H = \frac{c}{H_0}}$
This is the Hubble length.

Now let's plug in some numbers. We will again us the value for the Hubble constant as determined by the Planck satellite data.
$d_H = \frac{3\times 10^{10}\text{ cm} \text{ s}^{-1} }{2.26 \times 10^{-18} \text{ s}^{-1}} = \boxed{1.3 \times 10^{28} \text{ cm}}$
This is the size of the observable universe, known as the Hubble length. This value is often reported in lightyears. Converting our value, we get:
$d_h = 1.38\times 10^{10} \text{ ly}$
This means that the Hubble length is about 13.8 billion light years.

### 25 | Worksheet 7.2, Probelm 2: Eddington Limit

There is a hard upper limit to the luminosity of a accretion disk/black hole system -- and to the luminosity of any accreting compact object. Consider that the photons being emitted in this scenario will interact with the surrounding material (which has yet to accrete onto the black hole). These photons will undergo Thomson scattering off of electrons in this material. In detail, the electric field of the incident lightwave (i.e. the photon) will accelerate an electron, causing it to then re-emit radiation. The photons are therefore able to transfer some of their momentum to the infalling gas.

The energy flux of these photons at a distance $$r$$ from the black hole is:
$F = \frac{L}{4\pi r^2}$
Then, recall that the momentum of a photon of energy $$E$$ is simply $$p = E/c$$. Therefore, the momentum flux at $$r$$ from the black hole is $$\frac{L}{4\pi c r^2}$$.

Finally, the rate of momentum transfer to the surrounding electrons (or the force due to photons $$f_{\text{rad}}$$) is modulated by the Thomson cross section, $$\sigma_t = 6.6524 \times 10^{-25} \text{ cm}^2$$ (i.e. the effective area of an electron interacting with a photon):
$f_{\text{rad}} = \sigma_t \frac{L}{4\pi c r^2}$
(a) When this force from radiation pressure exceeds the force of gravity, accretion is halted and all the gas is blown away. For a black hole mass $$M_{\text{BH}}$$, derive the maximum possible luminosity due to accretion. This is called the Eddington limit or Eddington luminosity.

We are told that the rate of momentum transfer to the surrounding electrons, a force, is given by:
$f_{\text{rad}} = \sigma_t \frac{L}{4\pi c r^2}$
We are interested in the point at which this radiation force (outward) exceeds the force of gravity holding everything together. The gravitational force is generally given as:
$f_{\text{g}} = \frac{GMm}{r^2}$
For this problem, $$M$$ represents the central mass of the system, so $$M = M_{\text{BH}}$$. The gas is held near the black hole by the gravitational force between the black hole and the gas, which is made up of atoms. The force $$f_{\text{rad}}$$ is a force on a single electron--part of an atom. Most of the mass of the atom being considered involves the nucleus made up of neutrons and protons, which have mass $$m_p$$. So, we will say that $$m = m_p$$. Our expression for the force due to gravity is now:
$f_{\text{g}} = \frac{G M_{\text{BH}} m_p}{r^2}$
We want to know about the situation when $$f_{\text{g}} = f_{\text{rad}}$$, so:
$\sigma_t \frac{L_{\text{Edd}}}{4\pi c r^2} = \frac{G M_{\text{BH}} m_p}{r^2}$
If we solve for $$L_{\text{Edd}}$$, we get:
$\boxed{L_{\text{Edd}} = \frac{4\pi c G M_{\text{BH}} m_p}{\sigma_t}}$
This is the Eddington limit.

BONUS: Express the Eddington luminosity as a number of solar luminosities $$L_{\odot}$$ and the black hole mass in solar masses $$M_{\text{BH}} / M_{\odot}$$.

We should start by gathering all constants and finding the value that these equal. Our expression for the Eddington luminosity is:
$L_{\text{Edd}} = \frac{4\pi c G M_{\text{BH}} m_p}{\sigma_t}$
Grouping the constants, we get:
$L_{\text{Edd}} = \frac{4\pi c G m_p}{\sigma_t} M_{\text{BH}}$
Plugging in these values, we get:
$L_{\text{Edd}} = \frac{4\pi (3\times 10^{10} \text{ cm} \text{ s}^{-1})(6.67\times 10^{-8} \text{ cm}^3 \text{ g}^{-1} \text{ s}^{-2})(1.67 \times 10^{24} \text{ g})}{6.65 \times 10^{-25} \text{ cm}^2} M_{\text{BH}} = 6.31 \times 10^4 \text{ cm}^2 \text{ s}^{-3} M_{\text{BH}}$
Now we can write this in terms of solar masses $$M_{\odot} = 2\times 10^{33} \text{ g}$$:
$L_{\text{Edd}} = 6.31 \times 10^4 \text{ cm}^2 \text{ s}^{-3} M_{\text{BH}} \times \frac{M_{\odot}}{M_{\odot}} = 1.26 \times 10^{38} \text{ erg} \text{ s}^{-1} \left(\frac{M_{\text{BH}}}{M_{\odot}}\right)$
The last step is to write this in terms of the solar luminosity $$L_{\odot} = 3.8\times 10^{33} \text{ erg} \text{ s}^{-1}$$:
$L_{\text{Edd}} = \frac{1.26 \times 10^{38} \text{ erg} \text{ s}^{-1}}{3.8\times 10^{33} \text{ erg} \text{ s}^{-1}}\left(\frac{M_{\text{BH}}}{M_{\odot}}\right)$
So,
$\boxed{L_{\text{Edd}} = 3.3\times 10^{4}\left(\frac{M_{\text{BH}}}{M_{\odot}}\right) L_{\odot}}$
(b) If the SMBH in Andromeda were accreting at 20% of its Eddington luminosity, how bright would it be? How does this value compare with the $$L_{\text{disk}}$$ you calculated in 1(e)?

We just derived a useful expression for solving this problem:
$L_{\text{Edd}} = 3.3\times 10^{4}\left(\frac{M_{\text{BH}}}{M_{\odot}}\right) L_{\odot}$
Let's begin by finding the Eddington luminosity of Andromeda's black hole. We know that the mass of Andromeda's black hole is around $$M_{\text{BH}} = 10^8 M_{\odot}$$.
So,
$L_{\text{Edd}} = 3.3\times 10^{4}\left(\frac{10^8 M_{\odot}}{M_{\odot}}\right) (3.8 \times 10^{33} \text{ erg} \text{ s}^{-1} ) =3.3 \times 10^{12} L_{\odot} = 1.3 \times 10^{46} \text{ erg} \text{ s}^{-1}$
We are interested in the value of 20% of the Eddington luminosity, so:
$L_{20\%} = 0.2 \times 1.3 \times 10^{46} \text{ erg} \text{ s}^{-1} = \boxed{2.5 \times 10^{45} \text{ erg} \text{ s}^{-1}}$
The value we found for the approximate luminosity of this accretion disk was $$5.7 \times 10^{45} \text{ erg} \text{ s}^{-1}$$. These two values are within the same order of magnitude, the approximate luminosity of Andromeda's accretion disk being about half of the luminosity at 20% of the Eddington luminosity.

## Thursday, October 29, 2015

### 24 | Worksheet 7.2, Problem 1: Black Hole Accretion Disk

Many galaxies host a central supermassive black hole. In galaxies that contain large amounts of gas rotating along with the stars in their disks, some of the gas will end up surrounding the black hole. As the gas falls inward, centrifugal forces will cause it to settle into a much smaller, denser disk which rotates around the central black hole. This is known as an accretion disk.

The orbital period of the gas in the accretion disk changes with radius. This means that adjacent packets of gas in the disk will rub against each other, generating frictional heat which is released as radiation. The energy loss results in the material moving closer to the black hole, eventually falling onto it (hence, accretion).

Let's figure out how hot and bright this material ends up being around a black hole of mass $$M_{\text{BH}}$$.

 artistic rendering of an accretion disk around a black hole (image: astrobites.org)

(a) First, imagine a gas packet of mass $$\mathrm{d}M$$ in the accretion disk. The gas packet falls from a radius of $$r + \mathrm{d}r$$ to a radius of $$r$$ during the accretion process, losing potential energy. With the help of the Virial Theorem, write down the thermal energy the packet gains as a result ($$\mathrm{d}E$$. Then, simply divide this $$\mathrm{d}E$$ by $$\mathrm{d}t$$ to express the luminosity of the packet $$\mathrm{d}L$$ in terms of $$M_{\text{BH}}$$, $$r$$, and the mass accretion rate $$\mathrm{d}M/\mathrm{d}t = \dot{M}$$.

The Virial Theorem tells that:
$K = -\frac{1}{2}U$
In this example, we will use $$E$$ to represent the thermal energy released by the system. This becomes,
$E = -\frac{1}{2}U$
Now we need an expression for the potential energy. In reality, we are actually interested in a value for $$\Delta U$$ that represents the change in gravitational potential energy. The general expression for gravitational expression is:
$U = -\frac{GMm}{r}$
For this problem, we will let $$M$$ represent the mass of the black hole such that $$M = M_{\text{BH}}$$. We will then let $$m$$ represent the mass of the gas packet in the accretion disk, $$dM$$. Finally, the distance of the gas packet from the center of the black hole changes from $$r + \mathrm{d}r \rightarrow r$$. Our expression for change in potential energy becomes:
$\mathrm{d}U = -G M_{\text{BH}} \mathrm{d}M \left(\frac{1}{r + \mathrm{d}r} - \frac{1}{r}\right)$
A bit of simplification of this expression gives us:
$\mathrm{d}U = - \frac{G M_{\text{BH}} \mathrm{d}M \mathrm{d}r}{r^2}$
Plugging this into the Virial Theorem expression $$\mathrm{d}E = -\frac{1}{2}\mathrm{d}U$$, we get:
$\boxed{\mathrm{d}E = \frac{G M_{\text{BH}} \mathrm{d}M \mathrm{d}r}{2r^2}}$
If we divide this entire expression by $$\mathrm{d}t$$, we will get a value for the luminosity of the packet $$\mathrm{d}L$$:
$\mathrm{d}L = \frac{\mathrm{d}E}{\mathrm{d}t} = \frac{G M_{\text{BH}} \mathrm{d}r}{2r^2}\frac{\mathrm{d}M}{\mathrm{d}t}$
We can denote $$\mathrm{d}M / \mathrm{d}t$$ as $$\dot{M}$$:
$\boxed{\mathrm{d}L = \frac{GM_{\text{BH}} \dot{M}}{2r^2} \mathrm{d}r}$

(b) Now lets assume the disk gas radiates like a blackbody at the same radius where the potential energy is released. Using the Stefan-Boltzmann law, write down an expression for the luminosity from a given annulus in the disk (between $$r$$ and $$r + \mathrm{d}r$$) in terms of the temperature in that annulus $$T(r)$$.

The Stefan-Boltzmann law relates the flux from a blackbody to the temperature of the blackbody:
$F = \sigma T(r)^4$
Luminosity is given by $$\text{flux} \times \text{area}$$. We can think of the surface area of this annulus by multiplying its circumference $$2 \pi r$$ by the width $$\mathrm{d}r$$. We need to be careful here: we actually need to multiply this value by two, as there is luminosity from both sides of the annulus (so double the surface area). Using these relationships, we can say that the luminosity of the annulus $$mathrm{d}L$$ is given by:
$\boxed{\mathrm{d}L = 4\pi \sigma r T(r)^4 \mathrm{d}r}$

(c) Now, set $$\mathrm{d}L$$ from (a) equal to the expression derived in (b). Solve for $$T(r)$$ and express in terms of the mass accretion rate $$\dot{M}$$.

Our expression in (a) was:
$\mathrm{d}L = \frac{G M_{\text{BH}} \dot{M}}{2r62}\mathrm{d}r$
The expression we just derived in (b) is:
$\mathrm{d}L = 4\pi \sigma r T(r)^4 \mathrm{d}r$
We can now set these two expressions equal and solve for $$T(r)$$. Doing so gives:
$\boxed{T(r) = \left(\frac{G M_{\text{BH}} \dot{M}}{8\pi \sigma r^3}\right)^{\frac{1}{4}}}$

(d) Finally, integrate your expression for $$\mathrm{d}L$$ over $$\mathrm{d}r$$ to find the total disk luminosity $$L_{\text{disk}}$$. Let's say here that the disk extends from inner radius $$r_{\text{in}}$$ to outer radius $$r_{\text{out}}$$. If $$r_{\text{out}} \gg r_{\text{in}}$$, what is $$L_{\text{disk}}$$?

Here, we will integrate the expression we found in (a). Recall that this expression was given by:

$\mathrm{d}L = \frac{G M_{\text{BH}} \dot{M}}{2r62}\mathrm{d}r$
So,
$L_{\text{disk}} = \int_{r_{\text{in}}}^{r_{\text{out}}} \frac{G M_{\text{BH}} \dot{M}}{2r62}\mathrm{d}r = \frac{G M_{\text{BH}} \dot{M}}{2}\left( \frac{1}{r_{\text{in}}} - \frac{1}{r_{\text{out}}}\right)$
However, if we say that $$r_{\text{out}} \gg r_{\text{in}}$$, we can say that $$r_{\text{out}}\rightarrow 0$$. In this case, our expression for $$L_{\text{disk}}$$ becomes:
$\boxed{L_{\text{disk}} = \frac{G M_{\text{BH}} \dot{M}}{2r_{\text{in}}}}$

(e) Astronomers find it useful to express the $$L_{\text{disk}}$$ you found in (d) as a fraction $$\eta$$ of the the total luminosity that would be released if the entire rest mass of the disk were converted to energy, $$\dot{M}c^2$$:
$L_{\text{disk}} = \eta \dot{M} c^2$
Show that $$\eta$$ is given by:
$\eta = \frac{1}{2} \frac{G M_{\text{BH}}}{c^2 r_{\text{in}}}$
We think of this as the radiative efficiency of the disk. This efficiency apparently has a strong dependence on the inner disk radius $$r_{\text{in}}$$.

We previously found that:
$L_{\text{disk}} = \frac{GM_{\text{BH}} \dot{M}}{2 r_{\text{in}}}$
Now we are told to write $$L_{\text{disk}}$$ as:
$L_{\text{disk}} = \eta \dot{M}c^2$
Let's set these two expressions equal and solve for $$\eta$$:
$\eta \dot{M}c^2 = \frac{GM_{\text{BH}} \dot{M}}{2 r_{\text{in}}}$
$\boxed{\eta = \frac{G M_{\text{BH}}}{2c^2 r_{\text{in}}}}$

BONUS: Express $$L_{\text{disk}}$$ in terms of the Schwarzschild Radius of the black hole $$R_s$$.

We've previously found that the Schwarzschile Radius is given by:
$R_s = \frac{2 G M_{\text{BH}}}{c^2}$
Now let's look at our expression for $$\eta$$ that we found above and rewrite this in terms of $$R_s$$:
$\eta = \frac{G M_{\text{BH}}}{2c^2 r_{\text{in}}}$
We can rearrange this to be:
$\eta = \frac{G M_{\text{BH}}}{c^2} \frac{1}{2r_{\text{in}}}$
Note the similarity to our expression for $$R_s$$. At this point it is easy to say that:
$\eta = \frac{G M_{\text{BH}}}{c^2} \frac{1}{2r_{\text{in}}} = \frac{1}{2}R_s \frac{1}{2 r_{\text{in}}} = \frac{R_s}{4 r_{\text{in}}}$
Now recall that $$L_{\text{disk}} = \eta \dot{M} c^2$$, so:
$L_{\text{disk}} = \eta \dot{M} c^2 = \left(\frac{R_s}{4 r_{\text{in}}}\right) \dot{M} c^2$
So,
$\boxed{ L_{\text{disk}} = \frac{R_s \dot{M} c^2}{4 r_{\text{in}}}}$

(f) Accretion disks around massive black holes tend to have low efficiencies of $$\eta \approx 0.1$$ and can accrete up to $$\dot{M} = 1 M_{\odot} \text{yr}^{-1}$$. What is the resulting disk luminosity? How hot would an accretion disk around the SMBH in Andromeda be?

To determine the disk luminosity, we can use the expression we just derived:
$L_{\text{disk}} = \eta \dot{M} c^2$
We know values for each variable, but we need to ensure the units are correct. The accretion rate $$\dot{M}$$ is given in units of solar masses per year. Let's convert this to cgs. First, we know $$M_{\odot} = 2\times 10^{33} \text{ g}$$. We also know that there are approximately $$\pi \times 10^7 \text{ s} \text{ yr}^{-1}$$. So,
$\dot{M} = \frac{1 M_{\odot}}{1 \text{ yr}} \times \frac{2\times 10^{33} \text{ g}}{1 M_{\odot}} \times \frac{1 \text{ yr}}{\pi \times 10^7 \text{ s}} = 6.4 \times 10^{25} \text{ g} \text{ s}^{-1}$
Now we can plug values into our luminosity expression:
$L_{\text{disk}} = \eta \dot{M} c^2 = (0.1)\cdot (6.4 \times 10^{25} \text{ g} \text{ s}^{-1})\cdot (3\times 10^{10} \text{ cm} \text{ s}^{-1})^2 = \boxed{5.7 \times 10^{45} \text{ erg} \text{ s}^{-1}}$
To calculate the temperature of the accretion disk around Andromeda's supermassive black hole, we can use the Stefan-Boltzmann law:
$L = 4 \pi R^2 \sigma T^4$
Solving for $$T$$, we get:
$T = \left(\frac{L}{4\pi R^2 \sigma}\right)^{\frac{1}{4}}$
We just found the value of $$L$$, $$\sigma$$ is a known constant, so all we need to calculate the temperature of Andromeda's accretion disk is the radius of this disk. I do not know how large the accretion disk in Andromeda is, but this source suggests an average radius for the accretion disk is about $$10^{14} \text{ cm}$$. So,
$T = \left(\frac{5.7 \times 10^{45} \text{ erg} \text{ s}^{-1}}{4\pi (10^{14} \text{ cm})^2 (5.7 \times 10^{-5} \text{ erg} \text{ cm}^{-2} \text{ s}^{-1} \text{ K}^{-4}) }\right)^{\frac{1}{4}} = \boxed{1.7 \times 10^{5} \text{ K}}$