Monday, December 7, 2015

Blog Post 37, ILLUSTRIS Simulation

The Illustis Simulation is a computer simulation of an entire universe using numerical calculations.  This project simulates most if not all relevant physics to high accuracy and results in a statistically accurate model of the universe in which one can study the distributions of gas, heat, dark matter, strs, etc.

First we will examine a random overdense region for halo data.  We will then plot this data in a histogram and interpret it.
This plot shows \(log(M)\) of the galactic halos on the x axis, and how many of these occurred in our sample location.  The histogram shows a clear trend in halo size, in that low mass halos are much more common than high mass halos.  Remember, we are using a log scale, so a halo in the 14-14.5 bin is 10,000 times more massive than one in the 10-10.5 bin.
Quickly analyzing the full dataset for our selected halos, we find that on average, 15% of the halo mass is stellar mass.  That implies that the rest is primarily dark matter.  Cool!

Using the simulation, we can make observations about the structure of a universe like our own.

Gas and Dark Matter Densities:

(Gas density (left) and Dark Matter density (right))

On a large scale, Gas and Dark Matter densities seem to correlate with one another, with the gas density being more spread out than the dark matter.  On a small scale (see below) this same trend is visible, but with the gas being very poorly defined compared to the stark definition of the dark matter filamentary structure.  This is likely due to baryonic interaction of the gas matter with itself, causing a counter-force to simply gravitation.  This greater degree of randomness could serve as a viable explanation for the greater dispersion of normal matter.

(Gas Density and Dark Matter density of a single cluster)

In a single galaxy (see below), the Gas is highly concentrated at the nucleus, while the dark matter is more dispersed in the halo.
(Gas density and dark matter density above the stellar material of the galaxy)

The most massive galaxies tend to be found in clusters, not in the background field.

Gas Temperature Evolution:
The following observations are based upon a video derived from the simulation.  This video can be found at: http://www.illustris-project.org/movies/illustris_movie_cube_sub_frame.mp4

Stars first begin to form win the early universe along the dark matter filaments as shown by the increase in gas temperature.  This starts slowly, then accelerates, reaching maximum star formation rate around redshift 1.5 to 1.0, where massive amounts of stars form.  The first stars begin to form at about 1 billion years after the Big Bang (Redshift 5.75), although they do not widely populate the universe until about 2.5 billion years after the Big Bang (Redshift: 2.75).  This marks the end fate "Dark Ages."  In the simulation, structure formation usually occurs through parts of very large structures collapsing due to gravity, breaking the very large structures into smaller more compact ones.  However, over time, these smaller structures tend to consume their neighbors, forming larger high density structures throughout the universe.  These structures form along filaments because gravity from dark matter is strongest there.  This gravity will pull both dark matter and gas together to form structures.

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

Sunday, December 6, 2015

Blog Post 36, WS 12.1, Problem 1, 2(d): Large Scale Structures

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

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

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

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

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

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

Let's begin.

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

Let's start with our equation:
\[\frac{d^2\delta}{dt^2}+\frac{2\dot{a}}{a} \frac{d\delta}{dt}=4\pi G\overline{\rho}\delta\]
Now, let's use the second equate to substitute in for delta.
\[\frac{d^2D(t)\tilde{\delta}(x)}{dt^2}+\frac{2\dot{a}}{a} \frac{dD(t)\tilde{\delta}(x)}{dt}=4\pi G\overline{\rho}D(t)\tilde{\delta}(x)\]
Now \(\tilde{\delta}\) cancels.
\[\frac{d^2D(t)}{dt^2}+\frac{2\dot{a}}{a} \frac{dD(t)}{dt}=4\pi G\overline{\rho}D(t)\]
This resultant equation is entirely dependent upon t, as x does not appear anywhere in it, thus it is time independent, and is our differential equation of interest.

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

We will start with the equation from before.
\[\frac{d^2D(t)}{dt^2}+\frac{2\dot{a}}{a} \frac{dD(t)}{dt}=4\pi G\overline{\rho}D(t)\]
We now substitute in for D(t) and \(\overline{\rho}\).
\[\frac{d^2t^q}{dt^2}+\frac{2\dot{a}}{a} \frac{dt^q}{dt}=4\pi G\rho_{c,0}a^{-3}t^q\]
Now, let's focus on the \(\frac{\dot{a}}{a}\) term.
We are told that \(a(t)=(3H_0 t/2)^{2/3}\), so:
\[\frac{\dot{a}}{a}=\frac{2}{3t}\]
Plugging this in, we get:
\[\frac{d^2t^q}{dt^2}+\frac{4}{3t} \frac{dt^q}{dt}=\frac{16\pi G\rho_{c,0}}{9H_0^2 t^2}t^q\]
Taking our derivatives, we get:
\[q(q-1)t^{q-2}+\frac{4qt^{q-2}}{3}=\frac{16\pi G\rho_{c,0}}{9H_0^2}t^{q-2}\]
\(t^{q-2}\) cancels, leaving:
\[q^2+\frac{q}{3}=\frac{16\pi G\rho_{c,0}}{9H_0^2}\]
We now substitute in for \(\rho_{c,0}\).
\[q^2+\frac{q}{3}=\frac{16\pi G3H_0^2}{9\pi 8 G H_0^2}\]
This simplifies to:
\[q^2+\frac{q}{3}=\frac{2}{3}\]
Solving for q we get:
\[q=-1,2/3\]
Finalizing this, we get:
\[\boxed{D_+(t)\propto t^{2/3} \text{     } D_-(t)\propto t^{-1}}\]

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

The D+ component will always be dominant, because as t increases, the D- term will go to 0 since \(\infty^{-1}=0\).
Time for just a little more math:
\[D_+(t)\propto t^{2/3}\]
\[a(t)=(3H_0 t/2)^{2/3}\propto t^{2/3}\]
Thus:
\[D_+(t)\propto t^{2/3} \propto a(t)\]
\[\boxed{D_+(t)\propto  a(t)}\]

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

The three cases are an Open Universe (Blue), a Flat Universe (Red), and an a Closed Universe (Black).


I worked with B. Brzycki and N. James on this (last) problem.

Monday, November 30, 2015

Blog Post 35, WS 11.1, Problem 2: Cosmic Microwave Background

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 equi- librium. In particular, the spectrum of the electromagnetic radiation (the particle content of the electromagnetic radiation is called photon) satisfies (1). At about the redshift z = 1100 when the universe had the temperature T = 3000K, almost all the electrons and protons in our universe are combined and the universe becomes electromagnetically neutral. So the electromagnetic waves (photons) no longer get absorbed or scattered by the rest of the contents of the universe. They started to propagate freely in the universe until reaching our detectors. Interestingly, even though the photons are no longer in equilibrium with its environment, as we will see, the spectrum still maintains an identical form to the Planck spectrum, albeit characterized by a different temperature.

(a) If the photons was emitted at redshift z with frequency ν, what is its frequency ν' today?

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

(c) Plug in the relation between ν and ν' into the Planck spectrum:
\[u_{\nu}d\nu = \frac{8\pi h_P \nu^3}{c^3} \frac{1}{e^{\frac{h_P\nu}{k_B T}}-1}d\nu\]
and also multiply it with the overall energy density dilution factor that you have just figured out above to get the energy density today. Write the final expression as the form uν1dν1. What is uν 1 ? This is the spectrum we observe today. Show that it is exactly the same Planck spectrum, except that the temperature is now \(T'=T(1+z)^{-1}\).

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


(a) If the photons was emitted at redshift z with frequency ν, what is its frequency ν' today? 
Let's use the Doppler equation.
\[z=\frac{\lambda'-\lambda}{\lambda}\]
Converting wavelength to frequency, we find that:
\[z=\frac{\nu-\nu'}{\nu'}\]
Solving for \(\nu'\), we get:
\[\boxed{\nu'=\frac{\nu}{z+1}}\]

(b) If a photon at redshift z had the energy density uνdν, what is its energy density uν'dν' today? (Hint, consider two effects: 1) the number density of photons is diluted; 2) each photon is redshifted so its energy, E = hν, is also redshifted.)
For this problem we will use our diagram above.  Firstly, energy density decreases by a factor of 3 because the universe is expanding in 3 spatial dimensions.  Then, the energy of each photon decease by an additional factor because the expansion increases the photon's wavelength.  This sums to a factor of 4.  Mathematically, it looks like this, where \(a\) is the expansion factor, and \(a'\) is the expansion factor today.
\[\frac{u_{\nu'}d\nu'}{u_{\nu}d\nu}=\left(\frac{a}{a'}\right)^4=(z+1)^{-4}\]

(c) Plug in the relation between ν and ν' into the Planck spectrum:
\[u_{\nu}d\nu = \frac{8\pi h_P \nu^3}{c^3} \frac{1}{e^{\frac{h_P\nu}{k_B T}}-1}d\nu\]
and also multiply it with the overall energy density dilution factor that you have just figured out above to get the energy density today. Write the final expression as the form uν1dν1. What is uν 1 ? This is the spectrum we observe today. Show that it is exactly the same Planck spectrum, except that the temperature is now \(T'=T(1+z)^{-1}\). 
It's time for more algebra.
Firstly we recall that \(\nu=\nu'(z+1)\). Taking the derivative, we find that \(d\nu=d\nu'(z+1)\).
Pluggin in we get:
\[u_{\nu'}d\nu' =\frac{1}{(z+1)^4} \frac{8\pi h_P (\nu'(z+1))^3}{c^3} \frac{1}{e^{\frac{h_P(\nu'(z+1))}{k_B T}}-1}d\nu'(z+1)\]
Simplifying, we get:
\[u_{\nu'} =\frac{8\pi h_P \nu'}{c^3} \frac{1}{e^{\frac{h_P(\nu'(z+1))}{k_B T}}-1}\]
This is identical to the original expression except for temperature, which is now scaled by a factor of 1/(z+1).
Thus:
\[\boxed{T'=T(z+1)^{-1}}\]

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

Let's plug and chug.
\[T'=T(z+1)^{-1}\]
\[T'=3000K(1100+1)^{-1}\]
\[\boxed{T'=2.72K}\]

I worked with B. Brzycki, G. Grell, and N. James on this problem.

Blog Post 34, WS 11.1, Problem 1: Temperature of the Universe


1. Temperature of the Universe. Remember that, although the universe today is dominated by dark energy and matter (including ordinary matter and dark matter), much earlier on it was dominated by radiation. In this exercise we study the temperature evolution of a radiation dominated universe.
When the electromagnetic wave is in equilibrium with the environment, its spectrum is uniquely determined by the temperature of the equilibrium. This state is called the blackbody radiation. The spectrum is called the Planck spectrum, named after the physicist discovered it. The energy density per frequency interval dν of the black body radiation is given by
\[u_{\nu}d\nu = \frac{8\pi h_P \nu^3}{c^3} \frac{1}{e^{\frac{h_P\nu}{k_B T}}-1}d\nu\]
where hP is the Planck constant, kB is the Boltzmann constant, ν is the frequency, and T is the temperature.

(a) How is the equation for uνdν different from the equation for flux in previous worksheets?

(b) Integrate the Planck spectrum over the frequency and figure out how the energy density u of the black body radiation depends on temperature T. Namely, figure out the power n in \(u \propto T^n\). (Since only the functional form of T is important here, in this exercise you do not have to figure out the exact value of the T-independent coefficient a.)

(c) Remind yourself how the energy density of the radiation dominated universe depends on the scale factor a.

(d) Combine the two results and see how the temperature T of the universe depends on the scale factor a. Explain why this result implies that the early universe is very hot.


(a) How is the equation for uνdν different from the equation for flux given in our previous work- sheets? 
In previous worksheets, we dealt with energy per unit of time per unit of area.  This equation expresses total energy density in a volume of space.

(b) Integrate the Planck spectrum over the frequency and figure out how the energy density u of the black body radiation depends on temperature T. Namely, figure out the power n in \(u \propto T^n\). (Since only the functional form of T is important here, in this exercise you do not have to figure out the exact value of the T-independent coefficient a.) 
\[u_{\nu}d\nu = \frac{8\pi h_P \nu^3}{c^3} \frac{1}{e^{\frac{h_P\nu}{k_B T}}-1}d\nu\]
We want to integrate this over all frequencies to find total energy density's relation to temperature. 
\[\int_0^{\infty} u_{\nu}d\nu = \int_0^{\infty}\frac{8\pi h_P \nu^3}{c^3} \frac{1}{e^{\frac{h_P\nu}{k_B T}}-1}d\nu\]
This may be difficult to integrate, and we only care about our variable of integration, T, and u, so we can drop all other constants.  
\[\int_0^{\infty} u_{\nu}d\nu \propto \int_0^{\infty}\frac{\nu^3}{e^{\frac{\nu}{T}}-1}d\nu\]
Now we will ignore the -1 in the denominator.  This may seem sketchy, but the use of a Taylor Series approximation allows it.
\[u \propto \int_0^{\infty}\frac{\nu^3}{e^{\frac{\nu}{T}}}d\nu\]
This integral is solvable through integration by parts.  Luckily, we have WolframAlpha to take care of that for us.
Our final result is:
\[\boxed{u \propto T^4}\]

(c) Remind yourself how the energy density of the radiation dominated universe depends on the scale factor a. 
We solved this on a previous worksheet dealing with Friedmann Applications.  
Our result was: 
\[u\propto a^{-4}\]

(d) Combine the two results and see how the temperature T of the universe depends on the scale factor a. Explain why this result implies that the early universe is very hot. 
This is simple algebra.
\[u\propto a^{-4}\text{,  } u \propto T^4\]
\[a^{-4}\propto T^4\]
\[T\propto a^{-1}\]
\[\boxed{T\propto\frac{1}{a}}\]

This makes sense because In the early universe when \(a\) was very small, \(T\) would be very large.

I worked with B. Brzycki, G. Grell, and N. James on this problem.

Saturday, November 21, 2015

Blog Post 33, Free Form: Lasers

Everybody loves lasers. They are just that cool. Let's talk about what they are, what they do, and how they work.

LASER is actually an acronym for Light Amplification by the Stimulated Emission of Radiation.  This is precisely how a laser works.

Here is a laser diagram from Wikipedia.  
1) The Gain Medium
2) Laser Pumping Energy
3) High Reflector
4) Output Coupler
5) Laser Beam

So, how does this all work?  First, the pumping energy is applied to the gain medium.  This causes the medium's electrons to enter excited states, fall back to ground state, and release photons.  These photons will all be of the same wavelength.  The photons will bounce back and forth off of the high reflector and output coupler until they pass through the coupler in the form of a laser beam.  Unlike a flashlight element which emits light in all directions equally, a laser emits light in a single focussed beam.  This allows lasers to achieve very high energy densities at very low powers.  

For example, the flux/area of the sun on a sunny day is about \(1.4kW/m^2\).  Let's assume one uses a 5mW red presentation laser pointer.  This is the cheap kind that you can buy anywhere.  If we assume that the dot from the pointer is about 0.5cm x 0.5cm, it ha an area of \(0.25cm^2=0.000025m^2\).  Thus, the flux/area of our cheap laser is \(0.2kW/m^2\).  That's about 1/7 of the sun's output.  This is why such lasers of 5mW and below are considered pointers, they can accidentally hit a person's eye, and the blink reflex will prevent any sustained damage forth exposure.  It's not good for the ye, but it won't do any permanent damage.  

Of course, 5mW pointers are old school.  Companies sell handheld lasers with powers up to around 3.5W.  For reference, a 1W laser has an output of \(40kW/m^2\), or almost 30 times the energy density of sunlight.  Retinal exposure to a 1W laser almost always results in blindness.  

Laser color has a large impact on its visibility.  Red lasers are very simple in structure, and can be easily made at high powers, but the human eye is not very sensitive to red light.  Green lasers are the exact opposite.  We have not yet developed a direct green laser diode.  This means that we have to use infrared light, and pas it through a crystal to halve its wavelength of 1064nm infrared to 532nm green.  This makes green lasers not very power efficient, and very complex in structure.  However, the human eye is very sensitive to green light, making green lasers beams of a given power far outshine their equal red counterparts.  In recent years, blue 445nm diodes have become available.  These diodes share the best feature of red and green lasers.  They are direct diodes, so it is easy to create high power outputs, and the human eye is sensitive to blue light, making blue lasers show up easily even in daylight.  

Due to user irresponsibility and plane accidents caused by careless pointing of powerful lasers, the government has cracked down on handheld devices, limiting imports, and setting power limitations on what is considered a pointer, and what is not.  Many foreign laser manufacturers will no longer ship to the US due to trouble getting their products through customs.  Amusingly, there has been no such limitations placed on laser components, so the best way to get a high power laser pointer these days is to buy the parts on eBay, and build it yourself.

Note: I do not condone the use or construction of lasers without proper expertise and safety.  Appropriate wavelength safety goggles should be worn at all times when using lasers of output greater than 5mW.  Lasers should never be pointed at people, animals, or highly reflective surfaces.