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
Monday, December 7, 2015
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
\[\frac{d^2\delta}{dt^2}+\frac{2\dot{a}}{a} \frac{d\delta}{dt}=4\pi G\overline{\rho}\delta\]
(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.
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
\[\delta (x,t)=D(t)\tilde{\delta}(x)\]
\[\delta (x,t)=D(t)\tilde{\delta}(x)\]
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 \(D_- (t)\).
(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
\[\frac{d^2\delta}{dt^2}+\frac{2\dot{a}}{a} \frac{d\delta}{dt}=4\pi G\overline{\rho}\delta\]
\[\frac{d^2\delta}{dt^2}+\frac{2\dot{a}}{a} \frac{d\delta}{dt}=4\pi G\overline{\rho}\delta\]
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
\[\delta (x,t)=D(t)\tilde{\delta}(x)\]
\[\delta (x,t)=D(t)\tilde{\delta}(x)\]
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.
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.)
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 \(T'=T(1+z)^{-1}\).
\[u_{\nu}d\nu = \frac{8\pi h_P \nu^3}{c^3} \frac{1}{e^{\frac{h_P\nu}{k_B T}}-1}d\nu\]
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.
Blog Post 32, WS 10.1, Problem 3: The Observable Universe
3. Observable Universe. It is important to realize that in our Big Bang universe, at any given time, the size of the observable universe is finite. The limit of this observable part of the universe is called the horizon (or particle horizon to be precise, since there are other definitions of horizon.) In this problem, we’ll compute the horizon size in a matter dominated universe in co-moving coordinates.
To compute the size of the horizon, let us compute how far the light can travel since the Big Bang.
(a) First of all - Why do we use the light to figure out the horizon size?
(b) Light satisfies the statement that ds2 = 0. Using the FRW metric, write down the differential equation that describes the path light takes. We call this path the geodesic for a photon. Choosing convenient coordinates in which the light travels in the radial direction so that we can set dθ = dφ = 0, find the differential equation in terms of the coordinates t and r only.
(c) Suppose we consider a flat universe. Let’s consider a matter dominated universe so that aptq as a function of time is known (in the last worksheet). Find the radius of the horizon today (at t = t0).
(Hint: move all terms with variable r to the LHS and t to the RHS. Integrate both sides, namely r from 0 to rhorizon and t from 0 to t0.)
(a) First of all - Why do we use the light to figure out the horizon size?
To compute the size of the horizon, let us compute how far the light can travel since the Big Bang.
(a) First of all - Why do we use the light to figure out the horizon size?
(b) Light satisfies the statement that ds2 = 0. Using the FRW metric, write down the differential equation that describes the path light takes. We call this path the geodesic for a photon. Choosing convenient coordinates in which the light travels in the radial direction so that we can set dθ = dφ = 0, find the differential equation in terms of the coordinates t and r only.
(c) Suppose we consider a flat universe. Let’s consider a matter dominated universe so that aptq as a function of time is known (in the last worksheet). Find the radius of the horizon today (at t = t0).
(Hint: move all terms with variable r to the LHS and t to the RHS. Integrate both sides, namely r from 0 to rhorizon and t from 0 to t0.)
https://upload.wikimedia.org/wikipedia/commons/9/98/Observable_Universe_with_Measurements_01.png |
Since nothing travels faster than the speed of light, the observable universe is limited by the point farthest away whose emitted light has reached us.
(b) Light satisfies the statement that ds2 = 0. Using the FRW metric, write down the differential equation that describes the path light takes. We call this path the geodesic for a photon. Choosing convenient coordinates in which the light travels in the radial direction so that we can set dθ = dφ = 0, find the differential equation in terms of the coordinates t and r only.
The metric is:
\[0=-c^2dt^2+a^2(t)\left(\frac{dr^2}{1-kr^2}+r^2(d\theta^2 + sin^2(\theta)d\phi)\right)\]
We are told to set dθ = dφ = 0, so:
\[0=-c^2dt^2+a^2(t)\left(\frac{dr^2}{1-kr^2}\right)\]
Since we are in a flat universe, k=0.
\[0=-c^2dt^2+a^2(t)dr^2\]
The metric is:
\[0=-c^2dt^2+a^2(t)\left(\frac{dr^2}{1-kr^2}+r^2(d\theta^2 + sin^2(\theta)d\phi)\right)\]
We are told to set dθ = dφ = 0, so:
\[0=-c^2dt^2+a^2(t)\left(\frac{dr^2}{1-kr^2}\right)\]
Since we are in a flat universe, k=0.
\[0=-c^2dt^2+a^2(t)dr^2\]
\[c^2dt^2=a^2(t)dr^2\]
Simplifying gives us our differential result:
Simplifying gives us our differential result:
\[\frac{dr}{dt}=\frac{c}{a(t)}\]
(c) Suppose we consider a flat universe. Let’s consider a matter dominated universe so that a(t) as a function of time is known (in the last worksheet). Find the radius of the horizon today (at t = t0).
(Hint: move all terms with variable r to the LHS and t to the RHS. Integrate both sides, namely r from 0 to rhorizon and t from 0 to t0.)
(Hint: move all terms with variable r to the LHS and t to the RHS. Integrate both sides, namely r from 0 to rhorizon and t from 0 to t0.)
\[\frac{dr}{dt}=\frac{c}{a(t)}\]
Last worksheet, we proved that \(a(t)\propto t^{2/3}\). Let's plug this in.
\[\frac{dr}{dt}=\frac{c}{t^{2/3}}\]
Let's separate variables and integrate.
\[\frac{dr}{c}=\frac{dt}{t^{2/3}}\]
\[\int_0^{r_{horizon}} \frac{dr}{c}=\int_0^{t_0} \frac{dt}{t^{2/3}}\]
Last worksheet, we proved that \(a(t)\propto t^{2/3}\). Let's plug this in.
\[\frac{dr}{dt}=\frac{c}{t^{2/3}}\]
Let's separate variables and integrate.
\[\frac{dr}{c}=\frac{dt}{t^{2/3}}\]
\[\int_0^{r_{horizon}} \frac{dr}{c}=\int_0^{t_0} \frac{dt}{t^{2/3}}\]
\[\frac{r_{horizon}}{c}=3t_0^{1/3}\]
\[\boxed{r_{horizon}=3ct_0^{1/3}}\]
Worked with B. Brzycki, G. Grell, and N. James on this problem.
Blog Post 31, WS 10.1, Problem 2: Circumference to Radius Ratio
Ratio of circumference to radius. Let’s continue to study the difference between closed, flat and open geometries by computing the ratio between the circumference and radius of a circle.
(a) To compute the radius and circumference of a circle, we look at the spatial part of the metric and concentrate on the two-dimensional part by setting dφ = 0 because a circle encloses a two-dimensional surface. For the flat case, this part is just
\[ds^2_{2d}=dr^2 +r^2 d\theta^2\]
The circumference is found by fixing the radial coordinate (r = R and dr = 0) and integrating both sides of the equation (note that θ is integrated from 0 to 2π).
The radius is found by fixing the angular coordinate (θ, dθ = 0) and integrating both sides (note that dr is integrated from 0 to R).
Compute the circumference and radius to reproduce the famous Euclidean ratio 2π.
(b) For a closed geometry, we calculated the analogous two-dimensional part of the metric in Problem (1). This can be written as:
\[ds^2_{2d}=d\xi^2+sin\xi^2 d\theta^2\]
Repeat the same calculation above and derive the ratio for the closed geometry.
The circumference is found by fixing the radial coordinate (r = R and dr = 0) and integrating both sides of the equation (note that θ is integrated from 0 to 2π).
The radius is found by fixing the angular coordinate (θ, dθ = 0) and integrating both sides (note that dr is integrated from 0 to R).
Compute the circumference and radius to reproduce the famous Euclidean ratio 2π.
(b) For a closed geometry, we calculated the analogous two-dimensional part of the metric in Problem (1). This can be written as:
\[ds^2_{2d}=d\xi^2+sin\xi^2 d\theta^2\]
Repeat the same calculation above and derive the ratio for the closed geometry.
Compare your results to the flat (Euclidean) case; which ratio is larger? (You can try some arbitrary values of ξ to get some examples.)
(c) Repeat the same analyses for the open geometry, and comparing to the flat case.
(d) You may have noticed that, except for the flat case, this ratio is not a constant value. However, in both the open and closed case, there is a limit where the ratio approaches the flat case. Which limit is that?
(c) Repeat the same analyses for the open geometry, and comparing to the flat case.
(d) You may have noticed that, except for the flat case, this ratio is not a constant value. However, in both the open and closed case, there is a limit where the ratio approaches the flat case. Which limit is that?
Let's begin.
(a) To compute the radius and circumference of a circle, we look at the spatial part of the metric and concentrate on the two-dimensional part by setting dφ = 0 because a circle encloses a two-dimensional surface. For the flat case, this part is just
\[ds^2_{2d}=dr^2 +r^2 d\theta^2\]
The circumference is found by fixing the radial coordinate (r = R and dr = 0) and integrating both sides of the equation (note that θ is integrated from 0 to 2π).
The radius is found by fixing the angular coordinate (θ, dθ = 0) and integrating both sides (note that dr is integrated from 0 to R).
Compute the circumference and radius to reproduce the famous Euclidean ratio 2π.
The circumference is found by fixing the radial coordinate (r = R and dr = 0) and integrating both sides of the equation (note that θ is integrated from 0 to 2π).
The radius is found by fixing the angular coordinate (θ, dθ = 0) and integrating both sides (note that dr is integrated from 0 to R).
Compute the circumference and radius to reproduce the famous Euclidean ratio 2π.
Let's start with the given equation:
\[ds^2_{2d}=dr^2 +r^2 d\theta^2\]
So, to calculate C, we let r=R and dr=0.
\[ds^2_{2d}=R^2 d\theta^2\]
\[ds_{2d}=R d\theta\]
Integrating, we get:
\[\int ds_{2d}=\int^{2\pi}_0 R d\theta\]
\[C=2\pi R\]
This is as expected.
Now, we solve for radius.
\[ds^2_{2d}=dr^2 +r^2 d\theta^2\]
\[ds^2_{2d}=dr^2\]
Integrating, we get:
\[\int ds_{2d}=\int^R_0 dr\]
\[R=R\]
Now to find our ratio:
\[\frac{C}{R}=\frac{2\pi R}{R}=\boxed{2\pi}\]
This is obvious, but it sets up a useful frame work for the rest of the calculations.
(b) For a closed geometry, we calculated the analogous two-dimensional part of the metric in Problem (1). This can be written as:
\[ds^2_{2d}=d\xi^2+sin\xi^2 d\theta^2\]
Repeat the same calculation above and derive the ratio for the closed geometry.
\[ds^2_{2d}=d\xi^2+sin\xi^2 d\theta^2\]
Repeat the same calculation above and derive the ratio for the closed geometry.
Compare your results to the flat (Euclidean) case; which ratio is larger? (You can try some arbitrary values of ξ to get some examples.)
Let's solve for C.
\[ds^2_{2d}=d\xi^2+sin\xi^2 d\theta^2\]
\[ds^2_{2d}=sin\xi^2 d\theta^2\]
\[ds_{2d}=sin\xi d\theta\]
\[\int ds_{2d}=\int^{2\pi}_{0}sin\xi d\theta\]
\[C=2\pi sin\xi\]
Solving for R:
\[ds^2_{2d}=d\xi^2+sin\xi^2 d\theta^2\]
\[ds^2_{2d}=d\xi^2\]
\[ds_{2d}=d\xi\]
\[\int ds_{2d}=\int^{\xi}_0 d\xi\]
\[R=\xi\]
Now for the ratio:
\[\frac{C}{R}=\boxed{\frac{2\pi sin\xi}{\xi}}\]
We will compare everything is part (c).
(c) Repeat the same analyses for the open geometry, and comparing to the flat case.
\[ds_{2d}^2=\frac{dr^2}{1+r^2}+r^2 d\theta^2\]
Solving for C:
\[ds_{2d}^2=\frac{dr^2}{1+r^2}+r^2 d\theta^2\]
\[ds_{2d}^2=R^2 d\theta^2\]
\[ds_{2d}=R d\theta\]
\[\int ds_{2d}=\int^{2\pi}_0 R d\theta\]
\[C=2\pi R\]
Solving for R:
\[ds_{2d}^2=\frac{dr^2}{1+r^2}\]
\[ds_{2d}=\frac{dr}{(1+r^2)^{1/2}}\]
\[\int ds_{2d}=\int^R_0 \frac{dr}{(1+r^2)^{1/2}}\]
Using fancy hyperbolic trigonometry identities:
\[R=sinh^{-1}(R)\]
Now for the ratio:
\[\frac{C}{R}=\frac{2\pi R}{sinh^{-1}R}=\boxed{\frac{2\pi sinh\xi}{\xi}}\]
(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?
First let's compare our 3 values.
\[\frac{2\pi sin(\xi)}{\xi}\le 2\pi R \le \frac{2\pi sinh(\xi)}{\xi}\]
Now, as we talk the limit as \(\xi\) goes to 0, a remarkable thing happens: they all become \(2\pi R\).
This makes sense mathematically, but more importantly, it makes sense physically. For example, we live on the curved surface of the Earth, but when we are close to it, it looks flat. The same applies for curved 3D space, in small portions it still looks flat.
I worked with B. Brzycki, G. Grell, and N. James on this problem.
Sunday, November 8, 2015
Blog Post 30, WS 9.2, Introduction to the Radio Lab
In the coming weeks, AY17 will be using the millimeter-wave telescope at the Harvard College Observatory to observe the velocities of Giant Molecular Clouds (GMC's) throughout our galaxy.
GMC's are large clusters of Gas and Dust in our galaxy that bit the galactic center (just like everything else). They are uniquely suited to our lab objective because they emit light in the Radio spectrum due to emissions from Carbon Monoxide molecules. This frequency (115.271 GHz) is a very convenient frequency because it is much higher than radio stations and telecommunications, but not high enough to be affected by infrared radiation or visible light. Thus, we can observe during the day.
Our ultimate goal with the lab is to determine a rotation curve for the Milky Way galaxy. We we do this using the following method:
A few quick questions and answers for the lab:
- What is going to be the typical integration time per point?
- Over what range of longitude do you plan to observe? The Lab Handout claims that we will observe in the Gallatic longitude range of 10 degrees to 70 degrees, although most of our targets will be in a similar line of sight.
- How many (l,b) positions do you plan to observe? We have 4 targets, so we will observe at 4 positions.
- Will all your target positions be above 30° when you plan to observe them? Since we are observing around noon, our targets will be visible.
GMC's are large clusters of Gas and Dust in our galaxy that bit the galactic center (just like everything else). They are uniquely suited to our lab objective because they emit light in the Radio spectrum due to emissions from Carbon Monoxide molecules. This frequency (115.271 GHz) is a very convenient frequency because it is much higher than radio stations and telecommunications, but not high enough to be affected by infrared radiation or visible light. Thus, we can observe during the day.
Our ultimate goal with the lab is to determine a rotation curve for the Milky Way galaxy. We we do this using the following method:
We will observe a series of clusters along a similar line of sight. By measuring the doppler shift in their emission lines, we will determine their velocities. In doing so, we will be able to have data for orbital velocities of a range of objects in the Milky Way, allowing us to create a rotation curve.
A few quick questions and answers for the lab:
- What is going to be the typical integration time per point?
To achieve a SNR of 10 or more, we will need o do some math.
\[\tau=\frac{SNR^2 T_{sys}^2}{T_A^2 \Delta \nu}=\frac{10^2 \times (500K)^2}{(3K)^2 \times 0.5 MHz}=\boxed{5.6 sec}\]
- Over what range of longitude do you plan to observe? The Lab Handout claims that we will observe in the Gallatic longitude range of 10 degrees to 70 degrees, although most of our targets will be in a similar line of sight.
- How many (l,b) positions do you plan to observe? We have 4 targets, so we will observe at 4 positions.
- Will all your target positions be above 30° when you plan to observe them? Since we are observing around noon, our targets will be visible.
- At what LST are you going to start observing? At what EST? We are going to be observing around 12:00 Noon EST, and about 15:30 LST.
- Are you going to position switch or frequency switch to “flat field” the spectrum?
To flat field, we will position switch the spectrum because a frequency switch would cause us to get a flat field for the wrong observational wavelength.
Blog Post 29, WS 9.1, Problem 2: Friedmann Applications
In Question 1, you have derived the Friedmann Equation in a matter-only universe in the Newtonian approach. That is, you now have an equation that describes the rate of change of the size of the universe, should the universe be made of matter (this includes stars, gas, and dark matter) and nothing else. Of course, the universe is not quite so simple. In this question we’ll introduce the full Friedmann equation which describes a universe that contains matter, radiation and/or dark energy. We will also see some correction terms to the Newtonian derivation.
(a) The full Friedmann equations follow from Einstein’s GR, which we will not go through in this course. Analogous to the equations that we derived in Question 1, the full Friedmann equations express the expansion/contraction rate of the scale factor of the universe in terms of the properties of the content in the universe, such as the density, pressure and cosmological constant. We will directly quote the equations below and study some important consequences.
The first Friedmann equation:
\[\left(\frac{\dot{a}}{a}\right)^2=\frac{8\pi}{3}G\rho - \frac{kc^2}{a^2}+\frac{\Lambda}{3}\]
The second Friedmann equation:
\[\frac{\ddot{a}}{a}=-\frac{4\pi G}{3c^2}(\rho c^2+3P)+\frac{\Lambda}{3}\]
In these equations, ρ and P are the density and pressure of the content, respectively. k is the curvature parameter; k = -1, 0, 1 for open, flat and closed universe, respectively. Λ 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.
WARNING: MATH AHEAD
First let's multiply both sides of the first equation by \(a^2\) to get:
\[\dot{a}^2=\frac{8\pi}{3}G\rho a^2 - {kc^2}+\frac{\Lambda}{3}a^2\]
Now let's take the derivative with respect to t.
\[2\dot{a}\ddot{a}=\frac{8\pi}{3}G(\dot{\rho}a^2 +2\rho a\dot{a}) - {kc^2}+\frac{2\Lambda a \dot{a}}{3}\]
The second equation solved for \(\ddot{a}\) can be expressed as:
\[\ddot{a}=-\frac{4\pi G}{3c^2}(\rho c^2+3P)a+\frac{\Lambda a}{3}\]
Plugging this in for \(\ddot{a}\) and simplifying gives us:
\[2\dot{a}(-\frac{4\pi G}{3c^2}(\rho c^2+3P)a+\frac{\Lambda a}{3})=\frac{8\pi}{3}G(\dot{\rho}a^2 +2\rho a\dot{a}) - {kc^2}+\frac{2\Lambda a \dot{a}}{3}\]
\[-\frac{8\pi G}{3c^2}(\rho c^2+3P)a\dot{a}+\frac{2\Lambda a\dot{a}}{3}=\frac{8\pi}{3}G(\dot{\rho}a^2 +2\rho a\dot{a}) - {kc^2}+\frac{2\Lambda a \dot{a}}{3}\]
\[-\frac{8\pi G}{3c^2}(\rho c^2+3P)a\dot{a}=\frac{8\pi}{3}G(\dot{\rho}a^2 +2\rho a\dot{a}) - {kc^2}\]
Since our universe is flat, k=0, allowing further simplification.
\[-\frac{1}{c^2}(\rho c^2+3P)a\dot{a}=\dot{\rho}a^2 +2\rho a\dot{a}\]
\[-\rho c^2 a\dot{a} -3Pa\dot{a}=\dot{\rho}a^2 c^2 +2\rho a\dot{a}c^2\]
\[-3Pa\dot{a}=\dot{\rho}a^2 c^2 +3\rho a\dot{a}c^2\]
\[\dot{\rho}a^2 c^2= -3Pa\dot{a}-3\rho a\dot{a}c^2\]
\[\dot{\rho}a^2 c^2= -3a\dot{a}(P+\rho c^2)\]
\[\boxed{\dot{\rho}c^2= -3\frac{\dot{a}}{a}(\rho c^2+P)}\]
This is the Third Friedmann Equation which described the density of the universe as a function of time.
(b) What is the evolution of a cold matter dominated universe? (P=0, \(\Lambda=0\))
Then, solve for a in terms of t using the 1st equation.
Let's start with the Third Friedmann Equation.
\[\dot{\rho}c^2= -3\frac{\dot{a}}{a}(\rho c^2+P)\]
We are told that \(P=0\) and \(\Lambda=0\), so:
\[\dot{\rho}= -3\frac{\dot{a}}{a}\rho \]
We now need to solve this differential equation, so:
\[\frac{\dot{\rho}}{\rho}= -3\frac{\dot{a}}{a}\]
\[\int^{\rho}_{\rho_0}\frac{d\rho}{\rho}= -3\int^a_{a_0}\frac{da}{a}\]
\[ln\left(\frac{\rho}{\rho_0}\right)=-3ln\left(\frac{a}{a_0}\right)\]
\[\boxed{\frac{\rho}{\rho_0}=\left(\frac{a}{a_0}\right)^{-3}}\]
Now we can use this result along with the 1st equation to find how a and t scale. This means we can drop all constants.
\[\left(\frac{\dot{a}}{a}\right)^2=\frac{8\pi}{3}G\rho - \frac{kc^2}{a^2}+\frac{\Lambda}{3}\]
Earlier we were told that \(P=0\), \(\Lambda=0\), and \(k=0\) so:
\[\left(\frac{\dot{a}}{a}\right)^2=\frac{8\pi}{3}G\rho\]
Let's drop all constants.
\[\left(\frac{\dot{a}}{a}\right)^2 \propto \rho\]
Let's apply our previous result and integrate:
\[\left(\frac{\dot{a}}{a}\right)^2 \propto a^{-3}\]
\[\dot{a}^2 \propto a^{-1}\]
\[\dot{a} \propto a^{-1/2}\]
\[\frac{da}{dt} \propto a^{-1/2}\]
\[a^{1/2}da \propto dt\]
\[\int a^{1/2}da \propto \int dt\]
\[a^{3/2} \propto t\]
\[\boxed{a \propto t^{2/3}}\]
(c) Radiation dominated universe: \(P=\rho c^2 /3\), \(\Lambda =0\).
I'll omit the math from here on out and just give answers.
\[\boxed{\frac{\rho}{\rho_0}=\left(\frac{a}{a_0}\right)^{-4}}\]
\[\boxed{a \propto t^{1/2}}\]
(d) Dark Energy dominated universe \(P=0\) \(\rho=0\)
\[\boxed{a \propto e^t}\]
(e) Suppose the energy density of a universe at its very early time is dominated by half matter and half radiation. (This is in fact the case for our universe 13.7 billion years ago and only 60 thousand years after the Big Bang.) As the universe keeps expanding, which content, radiation or matter, will become the dominant component? Why?
(e) Suppose the energy density of a universe at its very early time is dominated by half matter and half radiation. (This is in fact the case for our universe 13.7 billion years ago and only 60 thousand years after the Big Bang.) As the universe keeps expanding, which content, radiation or matter, will become the dominant component? Why?
(f) Suppose the energy density of a universe is dominated by similar amount of matter and dark energy. (This is the case for our universe today. Today our universe is roughly 68% in dark energy and 32% in matter, including 28% dark matter and 5% usual matter, which is why it is acceleratedly expanding today.) As the universe keeps expanding, which content, matter or the dark energy, will become the dominant component? Why? What is the fate of our universe?
So far we have determined that radiation disperses faster than matter. Dark Energy on the other hand, disperses slower than matter, making it the dominant component as time passes. Thus the fate of our universe, is for matter and radiation to be spread thin, while dark energy becomes the primary component of the universe.
I worked on this problem with B. Brzycki, G. Grell, and N. James.
Blog Post 28, WS 9.1, Problem 1: Deriving Friedmann
The Friedmann equations describe the dynamics of a universe filled with mass. We will derive them Newtonian Mechanics. Consider a universe filled with matter which has a mass density \(rho(t)\). Note that as the universe expands or contracts, the density of the matter changes with time, which is why it is a function of time t.
Now consider a mass shell of radius R within this universe. The total mass of the matter enclosed by this shell is M. In the case we consider (homogeneous and isotropic universe), there is no shell crossing, so M is a constant.
Now consider a mass shell of radius R within this universe. The total mass of the matter enclosed by this shell is M. In the case we consider (homogeneous and isotropic universe), there is no shell crossing, so M is a constant.
WARNING: MATH AHEAD
(a) Find the shell's acceleration.
This is pretty simple. We will use Newton's gravity equation.
\[F_g=\frac{GMm}{R^2}\]
We know that \(\dot(v)=F/m\), so:
\[\dot{v}=\frac{GM}{R^2}\]
(b) Convert to energy.
We will use the quick trick of multiplying both sides by velocity and integrating.
\[\dot{v}v=\frac{GMv}{R^2}\]
\[vdv=\frac{GMdR}{R^2}\]
\[\frac{1}{2}v^2=\frac{GM}{R}+C\]
\[\frac{1}{2}\dot{R}^2-\frac{GM}{R}=C\]
(c) Express using mass density.
\[M=\frac{4}{3}\pi R^3 \rho\]
\[\frac{1}{2}\dot{R}^2-\frac{G(\frac{4}{3}\pi R^3 \rho)}{R}=C\]
\[\frac{1}{2}\dot{R}^2-\frac{G4\pi R^2 \rho}{3}=C\]
\[\left(\frac{\dot{R}}{R}\right)^2=\frac{8\pi G \rho}{3}+\frac{2C}{R^2}\]
(d,e) Express using R=a(t)r where r is the coming radius of the sphere.
\[\left(\frac{\dot{R}}{R}\right)^2=\frac{8\pi G \rho}{3}+\frac{2C}{R^2}\]
\[\left(\frac{\dot{a}}{a}\right)^2=\frac{8\pi G \rho}{3}+\frac{2C}{a^2 r^2}\]
(f) In the last worksheet we showed that \(H(t) = \frac{\dot{a}}{a}\). Express the First Friedmann Equation using this term. We can also turn \(2c/r^2\) into \(-kc^2\), a curvature term.
\[\left(\frac{\dot{a}}{a}\right)^2=\frac{8\pi G \rho}{3}+\frac{2C}{a^2 r^2}\]
\[\boxed{H^2=\frac{kc^2}{a^2}+\frac{8\pi G \rho}{3}}\]
(g) Derive the Second Friedmann Equation by expressing the acceleration of our initial shell in terms of the universal density.
From before we established:
\[\dot{v}=\frac{GM}{R^2}\]
\[\ddot{R}=\frac{4}{3}GM\pi R \rho\]
\[\boxed{\frac{\ddot{a}}{a}=-\frac{4}{3}\pi G \rho}\]
These Friedmann equations only apply to a universe that contains only matter without any pressure. The more extensive Friedmann equations come from General Relativity.
I worked with B. Brzycki, G. Grell, and N. James on this problem.
Sunday, November 1, 2015
Blog Post 27, WS 8.1, Problem 3: Hubble Flow and Constant
It is not strictly correct to associate this ubiquitous distance-dependent redshift we observe with the velocity of the galaxies (at very large separations, Hubble’s Law gives ‘velocities’ that exceeds the speed of light and becomes poorly defined). What we have measured is the cosmological redshift, which is actually due to the overall expansion of the universe itself. This phenomenon is dubbed the Hubble Flow, and it is due to space itself being stretched in an expanding universe.
Since everything seems to be getting away from us, you might be tempted to imagine we are located at the centre of this expansion. But, as you explored in the opening thought experiment, in actuality, everything is rushing away from everything else, everywhere in the universe, in the same way. So, an alien astronomer observing the motion of galaxies in its locality would arrive at the same conclusions we do.
In cosmology, the scale factor, a(t), is a dimensionless parameter that characterizes the size of the universe and the amount of space in between grid points in the universe at time t. In the current epoch, t = t0 and a(t0) = 1. a(t) is a function of time. It changes over time, and it was smaller in the past (since the universe is expanding). This means that two galaxies in the Hubble Flow separated by distance d0 = d(t0) in the present were d(t)= a(t)d0 apart at time t.
The Hubble Constant is also a function of time, and is defined so as to characterize the fractional rate of change of the scale factor:
\[H(t)=\frac{1}{a(t)} \frac{da}{dt}\rvert_t\]
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, \(\dot{a}= da/dt\), has been constant for all time. How long ago was the Big Bang (i.e. when a(t=0)=0)? How does this compare with the age of the oldest globular clusters (~12 Gyr)? What you will calculate is known as the Hubble Time.
(b) What is the size of the observable universe? What you will calculate is known as the Hubble Length.
https://upload.wikimedia.org/wikipedia/commons/c/c2/Lambda-Cold_Dark_Matter,_Accelerated_Expansion_of_the_Universe,_Big_Bang-Inflation.jpg |
(a) Assume the rate of expansion, \(\dot{a}= da/dt\), has been constant for all time. How long ago was the Big Bang (i.e. when a(t=0)=0)? How does this compare with the age of the oldest globular clusters (~12 Gyr)? What you will calculate is known as the Hubble Time.
It's time for some math.
We want to solve for \(t_0\).
Let's start with the given equation:
\[H(t)=\frac{1}{a(t)} \frac{da}{dt}\]
We can define \(da/dt=\dot{a}\).
\[H(t)=\frac{1}{a(t)} \dot{a}\]
\[H_0=\frac{\dot{a}}{a(t_0)}\]
We are also told that \(a(t_0)=1\), thus:
\[H_0=\frac{\dot{a}}{1}\]
\[H_0=\dot{a}\]
We will now change \(\dot{a}\) back to \(da/dt\).
\[H_0=\frac{da}{dt}\]
Let's integrate.
\[H_0 dt=da\]
\[\int^{t_0}_0 H_0 dt=\int^{a(t_0)=1}_0 da\]
\[H_0 t_0=1\]
\[t_0=H_0^{-1}\]
We found in a previous problem that \(H_0=68 km/s/Mpc\), thus:
\[\boxed{t_H=4.6\times 10^{17} sec=14.4 billion years}\]
This is the Hubble time.
(b) What is the size of the observable universe? What you will calculate is known as the Hubble Length.
The fastest traveling entity in the universe is light, thus the Hubble distance can be expressed as:
\[D_H=t_H c\]
\[D_H=(4.6\times 10^{17} sec)\times (3\times 10^10 cm\sec)\]
\[\boxed{D_H=1.36\times 10^{28}cm=4400Mpc=4.4Gpc}\]
Lengths of this scale are certainly, astronomical.
(Pro-tip: click the link).
I worked on this problem with B. Brzycki, G. Grell, N. James.
(Pro-tip: click the link).
I worked on this problem with B. Brzycki, G. Grell, N. James.
Blog Post 26, WS 8.1, Problem 1: Spatial Expansion
Before we dive into the Hubble Flow, let’s do a thought experiment. Pretend that there is an infinitely long series of balls sitting in a row. Imagine that during a time interval ∆t the space between each ball increases by ∆x.
(a) Look at the shaded ball, Ball C, in the figure above. Imagine that Ball C is sitting still (so we are in the reference frame of Ball C). What is the distance to Ball D after time ∆t? What about Ball B?
(b) What are the distances from Ball C to Ball A and Ball E?
(c) Write a general expression for the distance to a ball N balls away from Ball C after time ∆t. Interpret your finding.
(d) Write the velocity of a ball N balls away from Ball C during ∆t. Interpret your finding.
(b) What are the distances from Ball C to Ball A and Ball E?
This requires a little bit more thought. Since A and E are two balls away, each will move a distance 2\(\Delta\)x. Thus:
(a) Look at the shaded ball, Ball C, in the figure above. Imagine that Ball C is sitting still (so we are in the reference frame of Ball C). What is the distance to Ball D after time ∆t? What about Ball B?
(b) What are the distances from Ball C to Ball A and Ball E?
(c) Write a general expression for the distance to a ball N balls away from Ball C after time ∆t. Interpret your finding.
(d) Write the velocity of a ball N balls away from Ball C during ∆t. Interpret your finding.
This problem is very hypothetical but its a great introduction into spatial expansion of the universe.
(a) Look at the shaded ball, Ball C, in the figure above. Imagine that Ball C is sitting still (so we are in the reference frame of Ball C). What is the distance to Ball D after time ∆t? What about Ball B?
This pretty easy, the problem says that the distance increases by \(\Delta\)x every \(\Delta\)t, also D and B are both 1 ball away from C. Thus:
\[d_{D,C}(\Delta t)=d_{B,C}(\Delta t)=\Delta x\]
(b) What are the distances from Ball C to Ball A and Ball E?
This requires a little bit more thought. Since A and E are two balls away, each will move a distance 2\(\Delta\)x. Thus:
\[d_{E,C}(\Delta t)=d_{A,C}(\Delta t)=2\Delta x\]
(c) Write a general expression for the distance to a ball N balls away from Ball C after time ∆t. Interpret your finding.
Let's look at our last two answers. The distance travelled by a ball seems to be proportionate to the number of balls between the two endpoints. Thus, we can write:
\[d_N(\Delta t)=N\Delta x\]
This implies that the further away an object is, the more it moves away in a time-step.
This implies that the further away an object is, the more it moves away in a time-step.
(d) Write the velocity of a ball N balls away from Ball C during ∆t. Interpret your finding.
Velocity can be written as \(\frac{\Delta x}{\Delta t}\). Thus we can modify our previous expression:
\[v_N(\Delta t)=N\frac{\Delta x}{\Delta t}=N\Delta v\]
Let's interpret our findings. We can come up with 2 general rules for our forceless expanding space:
1. Objects always move away for each other
2. The farther away an object is, he faster it moves further away.
These rules hold true in our real universe provided that no other forces act upon the objects. For instance, distant galaxies and quasars are moving away from us faster and faster every day. However,
nearby galaxies such as Andromeda are close enough to be affected by the force of gravity, thus Andromeda is ready to collide with us in the Milkomeda collision. This spatial expansion is even occurring on the space between our cells and between our very atoms, but the fundamental forces are more than strong enough to make spatial expansion not even register of such small scales.
I worked with B. Brzycki, G. Grell, and N. James on this problem.
Blog Post 25, WS 7.2, Problem 5: More Quasar Redshift
You may also have noticed some weak “dips” (or absorption features) in the spectrum:
a) Suggest some plausible origins for these features. By way of inspiration, you may want to consider what might occur if the bright light from this quasar’s accretion disk encounters some gaseous material on its way to Earth. That gaseous material will definitely contain hydrogen, and those hydrogen atoms will probably have electrons occupying the lowest allowed energy state.
(b) A spectrum of a different quasar is shown below. Assuming the strongest emission line you see here is due to Lyα, what is the approximate redshift of this object?
c) What is the most noticeable difference between this spectrum and the spectrum of 3C 273? What conclusion might we draw regarding the incidence of gas in the early Universe as compared to the nearby Universe?
a) Suggest some plausible origins for these features. By way of inspiration, you may want to consider what might occur if the bright light from this quasar’s accretion disk encounters some gaseous material on its way to Earth. That gaseous material will definitely contain hydrogen, and those hydrogen atoms will probably have electrons occupying the lowest allowed energy state.
SPOILER ALERT: The answer is in the question! Gas accounts for the spikes perfectly. Hydrogen gas in the ground state that is bombarded by high-energy radiation such as quasar emissions will absorb wavelengths of radiation corresponding to jumps in electron energy levels. This will strip the spectrum of such wavelengths creating the jagged spectrum that we see.
(b) A spectrum of a different quasar is shown below. Assuming the strongest emission line you see here is due to Lyα, what is the approximate redshift of this object?
Let's use the Doppler equation.
\[\frac{\lambda_{observed}-\lambda_{emitted}}{\lambda_{emitted}}=z\]
\[\frac{5600-1215.67}{1215.67}=z\]
\[\boxed{z=3.6}\]
c) What is the most noticeable difference between this spectrum and the spectrum of 3C 273? What conclusion might we draw regarding the incidence of gas in the early Universe as compared to the nearby Universe?
Because this object is very far away and very old, the universe was much younger when its light was first emitted, thus it passed through gas and dust from the young universe on its way to us. The spectrum picture is a little pixelated, but the dark part on the right half is actually a series of tons of absorption lines caused by the light passing through cosmic gas and dust. Since these lines are much more numerous than those in the spectra of closer objects, the gas in the earlier universe must have been much more dense in order to produce the lines. Neat!
I worked with B. Brzycki, G. Grell, and N. James on this problem.
Blog Post 24, WS 7.2, Problem 4: Quasar Mass
4. One feature you surely noticed was the strong, broad emission lines. Here is a closer look at the strongest emission line in the spectrum:
This feature arises from hydrogen gas in the accretion disk. The photons radiated during the accretion process are constantly ionizing nearby hydrogen atoms. So there are many free protons and electrons in the disk. When one of these protons comes close enough to an electron, they recombine into a new hydrogen atom, and the electron will lose energy until it reaches the lowest allowed energy state, labeled n = 1 in the model of the hydrogen atom shown below (and called the ground state): It turns out that that strongest emission feature you observed in the quasar spectrum above arises from Lyα emission from material orbiting around the central black hole.
On its way to the ground state, the electron passes through other allowed energy states (called excited states). Technically speaking, atoms have an infinite number of allowed energy states, but electrons spend most of their time occupying those of lowest energies, and so only the n = 2 and n = 3 excited states are shown above for simplicity.
Because the difference in energy between, e.g., the n = 2 and n = 1 states are always the same, the electron always loses the same amount of energy when it passes between them. Thus, the photon it emits during this process will always have the same wavelength. For the hydrogen atom, the energy difference between the n = 2 and n = 1 energy levels is 10.19 eV, corresponding to a photon wavelength of λ = 1215.67 Angstroms. This is the most commonly-observed atomic transition in all of astronomy, as hydrogen is by far the most abundant element in the Universe. It is referred to as the Lyman α transition (or Lyα for short).
(a) Recall the Doppler equation:
\[\frac{\lambda_{observed}-\lambda_{emitted}}{\lambda_{emitted}}=z \approx \frac{v}{c}\]
Using the data provided, calculate the redshift of this quasar.
(b) Again using the data provided, along with the Virial Theorem, estimate the mass of the black hole in this quasar. It will help to know that the typical accretion disk around a 108 M⊙ black hole extends to a radius of r = 1015 m.
This feature arises from hydrogen gas in the accretion disk. The photons radiated during the accretion process are constantly ionizing nearby hydrogen atoms. So there are many free protons and electrons in the disk. When one of these protons comes close enough to an electron, they recombine into a new hydrogen atom, and the electron will lose energy until it reaches the lowest allowed energy state, labeled n = 1 in the model of the hydrogen atom shown below (and called the ground state): It turns out that that strongest emission feature you observed in the quasar spectrum above arises from Lyα emission from material orbiting around the central black hole.
On its way to the ground state, the electron passes through other allowed energy states (called excited states). Technically speaking, atoms have an infinite number of allowed energy states, but electrons spend most of their time occupying those of lowest energies, and so only the n = 2 and n = 3 excited states are shown above for simplicity.
Because the difference in energy between, e.g., the n = 2 and n = 1 states are always the same, the electron always loses the same amount of energy when it passes between them. Thus, the photon it emits during this process will always have the same wavelength. For the hydrogen atom, the energy difference between the n = 2 and n = 1 energy levels is 10.19 eV, corresponding to a photon wavelength of λ = 1215.67 Angstroms. This is the most commonly-observed atomic transition in all of astronomy, as hydrogen is by far the most abundant element in the Universe. It is referred to as the Lyman α transition (or Lyα for short).
\[\frac{\lambda_{observed}-\lambda_{emitted}}{\lambda_{emitted}}=z \approx \frac{v}{c}\]
Using the data provided, calculate the redshift of this quasar.
(b) Again using the data provided, along with the Virial Theorem, estimate the mass of the black hole in this quasar. It will help to know that the typical accretion disk around a 108 M⊙ black hole extends to a radius of r = 1015 m.
OK, let's begin.
(a) Recall the Doppler equation:
\[\frac{\lambda_{observed}-\lambda_{emitted}}{\lambda_{emitted}}=z \approx \frac{v}{c}\]
Using the data provided, calculate the redshift of this quasar.
\[\frac{\lambda_{observed}-\lambda_{emitted}}{\lambda_{emitted}}=z \approx \frac{v}{c}\]
Using the data provided, calculate the redshift of this quasar.
So, the rest wavelength that we are observing is 1215.67\(\unicode[serif]{xC5}\). However, the intensity in the data peaks at the wavelength 1407 \(\unicode[serif]{xC5}\). Let's use the equation:
\[\frac{\lambda_{observed}-\lambda_{emitted}}{\lambda_{emitted}}=z \approx \frac{v}{c}\]
\[\frac{1407-1215.67}{1215.67}=z \approx \frac{v}{c}\]
\[\boxed{z=0.157\approx \frac{v}{c}}\]
(b) Again using the data provided, along with the Virial Theorem, estimate the mass of the black hole in this quasar. It will help to know that the typical accretion disk around a 108 M⊙ black hole extends to a radius of r = \(10^{15}\) m.
This relies on a few different equations. We need our Doppler that was used in part (a), but we also need Virial Theorem solved for mass.
\[K=-\frac{1}{2}U\]
\[Mv^2=\frac{3GM^2}{5R}\]
\[M=\frac{5v^2 R}{3G}\]
We are given R, but we now need V. For this, we will use the Doppler data.
We know the approximate speed that the black hole is traveling away from us: 0.157c. If we repeat this calculate for the matter at the edge of the spectrum spike, then we can get the speed of the outer edge matter. By subtracting the two, we can get the relative speed of the outside matter.
\[\frac{\lambda_{observed}-\lambda_{emitted}}{\lambda_{emitted}}=z \approx \frac{v}{c}\]
\[\frac{1395-1215.67}{1215.67}=z \approx \frac{v}{c}\]
\[\frac{v}{c}=0.148\]
Now let's compare the two:
\[\Delta v=v_{BH}-v_1=0.157c-0.148c=0.009c=2.7\times 10^8cm/s\]
Let's plug this into our final equation:
\[M=\frac{5v^2 R}{3G}\]
\[M=\frac{5(2.7\times 10^8cm/s)^2 (10^{17 cm})}{3(6.7\times 10^{-8}cm^3 g^{-1}s^{-2}}\]
\[\boxed{M_{BH}=1.8\times 10^{41}g=9.1\times 10^7 M_{\odot}}\]
I worked with B. Brzycki, G. Grell, and N. James.
Saturday, October 24, 2015
Blog Post 23, Free Form: Astronomy and Politics: Interplanetary Law
Who owns space? For that matter, who owns the moon, or the orbit of a satellite? While astronomy is generally a very pure science, it has allowed mankind to leave our planet's surface and go to new unclaimed territory. The real question here is, if you go to the moon and build a moon-house there, is that your property? Can you own part of the moon? If you happen to blow up your moon-neighbor's house, under what law code is this illegal? This may seem very futuristic, but the nations of Earth actually put forth legal proceeding on these very question (okay fine, not these exact questions, but close enough).
On October 10th, 1967, the United Nations enacted the "Outer Space Treaty," also known as the "Treaty on Principles Governing the Activities of States in the Exploration and Use of Outer Space, including the Moon and Other Celestial Bodies." That's quite a mouthful, so we will call it the OST (an initialism that I invented about 15 seconds ago).
On October 10th, 1967, the United Nations enacted the "Outer Space Treaty," also known as the "Treaty on Principles Governing the Activities of States in the Exploration and Use of Outer Space, including the Moon and Other Celestial Bodies." That's quite a mouthful, so we will call it the OST (an initialism that I invented about 15 seconds ago).
The OST was originally put forth as a military treaty which bans the placement of usage of weapons of mass destruction in space, in orbit around Earth, or on the moon or other celestial bodies. So basically, no satellite nukes. Interestingly, it does not address conventional weapons, so you can put as many missiles in orbit as you want, but they can't be nuclear.
The next part of the OST bans individual or national ownership of the any celestial body or resource, leaving them as the "common heritage of mankind." Unfortunately, this means that you can't own a lunar golf course. What a shame. This doesn't mean that all satellites and equipment are public property, because all space equipment still belongs to the state that launched it. This all means that the owner of the object is liable for any damage caused by said object. This means that in the hit film Gravity, Russia has to pay for all of the equipment the its defunct satellite broke.
The OST also addresses individual activities in space. If a non-governmental party wishes to perform actions in space, they need to work with the jurisdiction of a signing nation of the OST. For instance, SpaceX works under the United States' jurisdiction.
IN 1979, a second treaty called the "Moon Treaty" (MT (also my own initialism)) was created. This treaty tightened restrictions on space resource allocation, military functions, and generally attempted to sum up the OST and a few other treaties that hd been put in place in the years between the OST and MT. Unfortunately, No space capable nations signed it. This treaty has only been ratified by 16 nations, including Australia, Chile, Mexico, Pakistan and Turkey, none of which have space programs. The US has not ratified it, neither has any space capable country. Thus, it has no real power at all.
So, let's answer our initial questions:
Who owns space? All Humans
Who owns the moon, or the orbit of a satellite? Mankind owns the moon, the launching nation owns the satellite.
If you go to the moon and build a moon-house there, is that your property? The house is your property, but the land you placed it on is not.
Can you own part of the moon? No
If you happen to blow up your moon-neighbor's house, under what law code is this illegal? This is illegal under the OST as it is a "harmful interference with activities in the peaceful exploration and use of outer space."
So, there's a bit of space law for you. One cool closing point: the OST claims that outer space is "common heritage of mankind." I guess we just declared ownership of all alien worlds. No wonder that always try to wipe us out in movies.
Sources:
https://en.wikipedia.org/wiki/Outer_Space_Treaty
https://en.wikipedia.org/wiki/Space_law
https://en.wikipedia.org/wiki/Moon_Treaty
Sources:
https://en.wikipedia.org/wiki/Outer_Space_Treaty
https://en.wikipedia.org/wiki/Space_law
https://en.wikipedia.org/wiki/Moon_Treaty
Blog Post 22, WS 7.1, Problem 4: Type Ia Supernova Energy Conservation
Calculate the total energy output, in ergs of the explosion assuming that the white dwarf's mass is converted to energy via fusion of carbon into nickel. Note that the process of carbon fusion is not entirely efficient, and only about 0.1% of this mass will be radiated away as electromagnetic radiation.
How does this compare to the total binding energy, in ergs, of the original white draw? Does the white dwarf explode completely, or is some mass left over in the form of a highly concentrated remnant?
Let's begin.
First let's find the energy of the explosion. It's time for a classic equation:
\[E=Mc^2\]
We are told that only 0.1% of the mass is converted to energy, so:
\[E=0.001Mc^2\]
Let's plug and chug.
\[E=0.001(1.4M_{\odot})(3\times 10^{10}cm/s)^2\]
How does this compare to the total binding energy, in ergs, of the original white draw? Does the white dwarf explode completely, or is some mass left over in the form of a highly concentrated remnant?
Let's begin.
First let's find the energy of the explosion. It's time for a classic equation:
\[E=Mc^2\]
We are told that only 0.1% of the mass is converted to energy, so:
\[E=0.001Mc^2\]
Let's plug and chug.
\[E=0.001(1.4M_{\odot})(3\times 10^{10}cm/s)^2\]
\[E=0.001(2.8\times 10^{33}g)(3\times 10^{10}cm/s)^2\]
\[\boxed{E=2.52\times 10^{51} ergs}\]
Now lets find the potential energy of the original system.
\[U=\frac{3GM^2}{5R}\]
\[U=\frac{3(6.7\times 10^{-8}cm^3 g^{-1}s^{-2})(2.8\times 10^{33}g)^2}{5(12\times 10^8 cm)}\]
\[\boxed{E=2.52\times 10^{51} ergs}\]
Now lets find the potential energy of the original system.
\[U=\frac{3GM^2}{5R}\]
\[U=\frac{3(6.7\times 10^{-8}cm^3 g^{-1}s^{-2})(2.8\times 10^{33}g)^2}{5(12\times 10^8 cm)}\]
\[\boxed{U=2.6\times 10^{50} ergs}\]
Thanks to approximations and estimations of mass consumption etc, these values are not equal, however, in the grand scheme of things, they are the same. This implies that Type Ia SNe leave nothing behind and the white draw is completely blown apart!
I worked on the problem with B. Brzycki, G. Grell, and N. James.
Blog Post 21, WS 7.1, Problem 1: White Dwarf Pressure
White Dwarfs are supported internally again the force of gravity by "electron degeneracy" pressure. The maximum mass that can be supported by this exotic form of pressure is 1.4 M(sun) (also known as the Chandrasekhar Mass.) The radius of our white dwarf is approximately twice the radius of the Earth, or \(12\times 10^8\) cm.
Given this mass M, and radius R, derive an algebraic expression for the eternal pressure of a white dwarf with these properties. Start with Virial Theorem, and assume that the white dwarf is an ideal gas with uniformly distributed mass.
As the question prompted, lets start with Virial Theorem.
\[K=-\frac{1}{2}U\]
Next we should apply the kinetic energy of a particle in 3D space:
\[K_p=\frac{3}{2}kT\]
Where k is the Boltzmann Constant and T is the temperature.
I few assume that we have N particles, the the total kinetic energy is:
\[K=\frac{3}{2}NkT\]
Now let's add our usual expression for U.
\[K=-\frac{1}{2}U\]
Given this mass M, and radius R, derive an algebraic expression for the eternal pressure of a white dwarf with these properties. Start with Virial Theorem, and assume that the white dwarf is an ideal gas with uniformly distributed mass.
As the question prompted, lets start with Virial Theorem.
\[K=-\frac{1}{2}U\]
Next we should apply the kinetic energy of a particle in 3D space:
\[K_p=\frac{3}{2}kT\]
Where k is the Boltzmann Constant and T is the temperature.
I few assume that we have N particles, the the total kinetic energy is:
\[K=\frac{3}{2}NkT\]
Now let's add our usual expression for U.
\[K=-\frac{1}{2}U\]
\[\frac{3}{2}NkT=\frac{GM^2}{2R}\]
Simplifying we get:
\[3NkT=\frac{GM^2}{R}\]
We want to solve for pressure, so let's look at ideal gas law.
\[PV=NkT\]
We have some similar variables here. This is a very good thing.
Let's solve both of our equations for NkT.
\[NkT=PV\]
\[NkT=\frac{GM^2}{3R}\]
Now we can set them equal to one another.
\[PV=\frac{GM^2}{3R}\]
Solving for P we get:
\[P=\frac{GM^2}{3RV}\]
V isn't in our given variables, but since the star is a sphere, we can express V in terms of R.
\[V=\frac{4}{3}\pi R^3\]
\[P=\frac{GM^2}{3R(\frac{4}{3}\pi R^3)}\]
\[\boxed{P=\frac{GM^2}{4\pi R^4}}\]
Qualitatively this makes sense, because as Mass increases pressure increases, and as Radius decreases, pressure increases. This is the perfect recipe for an explosion. Interestingly, (derived in AY16), the Radius of a White Dwarf Scales with Mass to the 1/3. \(M \sim \frac{1}{R^3}\)
Thus:\[P \sim R^{-10}\]
Or:
\[P \sim M^{3.33}\]
This is cool because we can scale pressure to a single variable, so with a critical mass, we also can find a critical pressure and radius.
I worked with B. Brzycki, G. Grell, and N. James on this problem.
\[\boxed{P=\frac{GM^2}{4\pi R^4}}\]
Qualitatively this makes sense, because as Mass increases pressure increases, and as Radius decreases, pressure increases. This is the perfect recipe for an explosion. Interestingly, (derived in AY16), the Radius of a White Dwarf Scales with Mass to the 1/3. \(M \sim \frac{1}{R^3}\)
Thus:\[P \sim R^{-10}\]
Or:
\[P \sim M^{3.33}\]
This is cool because we can scale pressure to a single variable, so with a critical mass, we also can find a critical pressure and radius.
I worked with B. Brzycki, G. Grell, and N. James on this problem.
Sunday, October 18, 2015
Blog Post 20: Hubble Tuning Fork
E0 Galaxy A near perfectly spherical galaxy with no real defining structure. http://www.noao.edu/image_gallery/images/d5/m89.jpg |
E1 Galaxy An elliptical galaxy with a length to width ratio of 9:10. M87 is pictured. http://www.astronomynotes.com/galaxy/m87.gif |
E4 Galaxy An elliptical galaxy with a l/w of 6:10. Types E2 and E3 have ratio of 8:10 and 7:10. http://cseligman.com/text/atlas/ngc14.htm |
E7 Galaxy An elliptical galaxy with a l/w of 3:10. Types E5 and E6 have ratio of 5:10 and 4:10. http://www.mhhe.com/physsci/astronomy/fix/student/images/23f06.jpg |
S0 Galaxy Lenticular galaxy with a bulge and disk, but no arms or spiral. http://bama.ua.edu/~rbuta/nearirs0/plate014c.jpg |
Sa Galaxy Spiral Galaxy with tightly bound arms. http://www.noao.edu/image_gallery/images/d4/m81y.jpg |
SBa Galaxy Spiral Barred Galaxy with tightly bound arms. https://en.wikipedia.org/wiki/Barred_spiral_galaxy |
SBb Galaxy Spiral Barred Galaxy with loosely bound arms. http://nrumiano.free.fr/Images_gx2/Jalf350.gif |
Sb galaxy Spiral Galaxy with loosely bound arms. http://www.leviathanastronomy.com/image-gallery.html |
SBc Galaxy Spiral Barred Galaxy with very loosely bound arms. http://www.astronomy.ohio-state.edu/~pogge/Ast162/Unit4/Galaxies/ |
Sc Galaxy Spiral galaxy with very loosely bound arms. http://www2.lowell.edu/rsch/LMI/gallery/n6946_2.jpg |
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