2. The Local Sidereal Time (LST) is the right ascension that is at the meridian right now. LST = 0:00 is at noon on the Vernal Equinox (the time when the Sun is on the meridian March 20th, for 2013 and 2014).
a) What is the LST at midnight on the Vernal Equinox?
Sidereal time is based on the time it takes for the Earth to complete exactly one rotation, rather than how long it takes for the same for the same point to realign with the Sun. As a result of the Earth's rotation around the Sun, these times differ slightly--LST picks up 4 minutes per day over "regular" time. Therefore, at midnight (12 hours after noon) on the Vernal Equinox, the LST would be 12:02.
b) What is the LST 24 hours later (after midnight in part 'a')?
An extra 24 hours would mean an extra 4 minutes; 12:06.
c) What is the LST right now (to the nearest hour)?
During our class time (2:30-4pm) on Tuesday, February 2, we calculated that the LST would be approximately 1:00--1 month and 18 days before the Vernal Equinox translates to a loss of about 3 hours. At 4pm, this would make the LST about 1:00.
d) What will the LST be tonight at midnight (to the nearest hour)?
Approximately 8 hours after class time would make the LST about 9:00.
e) What LST will it be at sunset on your birthday?
My birthday is July 27. If sunset is at 8pm, that makes this 4 months, 7 days, and 8 hours after noon on March 20th. This translates to a gain of 8:29 and an LST of 16:29.
Sunday, February 7, 2016
Week 2: The Pillars of Creation
"Pillars of Creation" is one of my favorite photographs of all time--in fact, it's actually my desktop background, which hopefully contributes to my astro street cred without being too nerdy. First of all, its colors and form are beautiful. The picture was taken in both visible and infrared spectra, which allows the capture of the densest regions of the structure. The glow around the pillars (which might be my favorite part of the picture) is the vaporization of the material on the periphery. Unfortunately, this means that this incredible structure is gradually disappearing.
Second, this thing is huge--the pillar on the left is about four light years long. It's composed of molecular hydrogen and is an active star forming region. Thinking about the sheer size and the vast amount of hydrogen, the lightest element, that must be present to form structures as large and dense as stars is mind-boggling. In fact, the stars that this thing has created provide a large part of the energy that is destroying the pillars. Kind of sad, isn't it?
A final interesting fact about this photo is that it was taken in 2014 as kind of a tribute to a famous picture taken by the Hubble telescope in 1995--an awesome image in its own right:
Sources
http://www.astronomy.com/-/media/Images/News%20and%20Observing/News/2015/01/Pillars2014.jpg?mw=600
https://en.wikipedia.org/wiki/Pillars_of_Creation
https://upload.wikimedia.org/wikipedia/commons/b/b2/Eagle_nebula_pillars.jpg
Thursday, February 4, 2016
Week 2: Planet X
As probably everyone who is remotely interested in astronomy has heard, a new planet was recently discovered. What's notable about this planet is that it orbits our own Sun, which makes it a candidate for inclusion in The Official Planet Roster Of Our Solar System--a list from which Pluto was crossed off in 2006.
While Planet X (whose name for some reason just reminds me of Chemical X from the Powerpuff Girls) has been detected due to its gravitational effects on other objects in our solar system, including the minor planet Sedna, it hasn't been visually detected. Part of the reason for this is because of how far away it is. Estimates for its closest approach are around 200 AU--almost 7 times that of Neptune, the most distal planet in our solar system (since Pluto's demotion at least)--and its farthest distance could be as large as 1,200 AU. At that distance, it would take light from the Sun almost a week to reach the planet. This makes it pretty difficult to find with a telescope.
In an ironic twist, one of the scientists who published results supporting the existence of Planet X was Mike Brown, a Caltech scientist whose previous work directly led to Pluto's demotion. On discovering Planet X, he said, "Killing Pluto was fun, but this is head and shoulders above everything else." Harsh.
Sources:
https://www.centralmaine.com/2016/01/27/the-discovery-of-planet-x/
http://ih0.redbubble.net/image.77569687.6291/raf,220x200,075,f,black.u5.jpg
https://en.wikipedia.org/wiki/Pluto
https://en.wikipedia.org/wiki/90377_Sedna
http://www.sciencemag.org/news/2016/01/feature-astronomers-say-neptune-sized-planet-lurks-unseen-solar-system
https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEim9temp3WPCmNNxpr03SbtAAKjeBqRM6ZpyowVuvYbQtxMNYOsXCcvExzwsHez5aPwTM_fhOsjAQ1ntIIVV9SgRLKBPnRRfmYVF6P7_FzefLwGhGpxMVaImToYCHMhl6sUErQ3MsEfNv81/s1600/sad+pluto4.jpg
While Planet X (whose name for some reason just reminds me of Chemical X from the Powerpuff Girls) has been detected due to its gravitational effects on other objects in our solar system, including the minor planet Sedna, it hasn't been visually detected. Part of the reason for this is because of how far away it is. Estimates for its closest approach are around 200 AU--almost 7 times that of Neptune, the most distal planet in our solar system (since Pluto's demotion at least)--and its farthest distance could be as large as 1,200 AU. At that distance, it would take light from the Sun almost a week to reach the planet. This makes it pretty difficult to find with a telescope.
In an ironic twist, one of the scientists who published results supporting the existence of Planet X was Mike Brown, a Caltech scientist whose previous work directly led to Pluto's demotion. On discovering Planet X, he said, "Killing Pluto was fun, but this is head and shoulders above everything else." Harsh.
Sources:
https://www.centralmaine.com/2016/01/27/the-discovery-of-planet-x/
http://ih0.redbubble.net/image.77569687.6291/raf,220x200,075,f,black.u5.jpg
https://en.wikipedia.org/wiki/Pluto
https://en.wikipedia.org/wiki/90377_Sedna
http://www.sciencemag.org/news/2016/01/feature-astronomers-say-neptune-sized-planet-lurks-unseen-solar-system
https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEim9temp3WPCmNNxpr03SbtAAKjeBqRM6ZpyowVuvYbQtxMNYOsXCcvExzwsHez5aPwTM_fhOsjAQ1ntIIVV9SgRLKBPnRRfmYVF6P7_FzefLwGhGpxMVaImToYCHMhl6sUErQ3MsEfNv81/s1600/sad+pluto4.jpg
Friday, January 29, 2016
Week 1: The Henderson-Hasselbalch Equation
Acidity and basicity are ways of measuring the concentration of protons in a solution--protons are acidic, so an acidic solution is characterized by having a high [H+]. One equation that is commonly used in biology and chemistry is the Henderson-Hasselbalch equation (which I primarily chose to write on because of its cool name). This equation relates the pH of a solution (the negative log of [H+]) with the chemical's acidity (pKa--an inherent property of the chemical) and its concentrations in both its acidic ([HA]) and basic ([A-]) forms: \[pH=pK_a+log(\frac{[A^-]}{[HA]})\]If we imagine a solution of a chemical with a defined pKa and a known ratio of \(\frac{[A^-]}{[HA]}\), we can easily calculate the pH of that solution using this equation. This equation is especially useful in buffers--solutions whose pH's change very little even when strong acids or strong bases are added.
Sources:
https://en.wikipedia.org/wiki/Henderson%E2%80%93Hasselbalch_equation
https://en.wikipedia.org/wiki/Buffer_solution
Sources:
https://en.wikipedia.org/wiki/Henderson%E2%80%93Hasselbalch_equation
https://en.wikipedia.org/wiki/Buffer_solution
Week 1: Introduction
Hi! My name is Kaelyn, and I'm a junior concentrating in Molecular and Cellular Biology (aka the best concentration). On top of that, I'm interested in pursuing Astrophysics as a secondary, which is why I'll be taking Astro 16. I had a great time in Astro 17 last semester, but am even more excited for this semester. As I have much more experience with the life sciences than I do physics, I'm especially excited to learn a bit more about astrochemistry and astrobiology. My ultimate goal is to go to medical school and become a surgeon, but after that I wouldn't turn down an astronaut job!
Wednesday, December 9, 2015
Blog #37: Illustris Simulation
The Halo Mass Function
When I zoomed in on a random region of high density, this is what I found:
Because I am technologically inept, here is a hand-drawn histogram of the masses of the subhaloes that I zoomed in. It appears that low-mass haloes are more common than high-mass haloes.
Because I'm only mostly technologically inept, I was able to use Excel to calculate that approximately 91% on average of the mass of the subhaloes (or at least the subhaloes that Illustris provided me with) was stellar mass.
Exploring Structure and Reionization in the Illustris Simulation
Large scale:
Small scale:
The dark matter appears to be more confined to the filamentary structure than the gas is. This could be because it is denser than the gas and therefore it is more favorable for it to aggregate along the filaments, whereas gas is less affected by gravity and therefore more free to diffuse throughout space. On the smaller scale, however, the gas appears to aggregate more tightly than the dark matter does.
In medium to large galaxies, the gas appears to be densest near the center of the disk. This makes sense given what we have learned about the structure of the Milky Way, that a higher density of matter, including gas, towards its center.
The most massive galaxies tend to cluster, which makes sense given their gravitational attraction to each other.
When I zoomed in on a random region of high density, this is what I found:
Because I am technologically inept, here is a hand-drawn histogram of the masses of the subhaloes that I zoomed in. It appears that low-mass haloes are more common than high-mass haloes.
Exploring Structure and Reionization in the Illustris Simulation
Large scale:
 |
| Dark matter density |
 |
| Gas density |
Small scale:
 |
| Dark matter density |
 |
| Gas density |
In medium to large galaxies, the gas appears to be densest near the center of the disk. This makes sense given what we have learned about the structure of the Milky Way, that a higher density of matter, including gas, towards its center.
 |
| Gas density (blue is denser) relative to galaxies (circles) |
 |
| The clustering of galaxies (circles) |
After watching the video, it seems clear that the dark matter is leading the structure formation. The formation of filaments is apparent much earlier in the dark energy box.
This is only a bit after the beginning of the "Epoch of Reionization," when hydrogen atoms become ionized (indicated by the color that is starting to appear in the gas box). According to the video, it appeared to start around a redshift of 7.7, or 0.7 billion years.
One of the fastest rates of star formation appears to be at redshift of between 3.50-2.90 or so, although it seems that star formation goes through cycles of speeding up and slowing down. It also appears that new objects are formed when large objects break up, as there are explosions occurring in the gas box, which will eject matter away to form smaller objects later.
Structures probably form along these filaments because that is where the dark matter is. The dark matter gravitationally attracts normal matter, which makes it more favorable for structures to form there instead of in random regions of space.
Blog #36: WS12.1, #1&2d
1. Linear perturbation theory. In the early universe, the matter/radiation distribution of the universe is very homogenous and isotropic. At any given time, let us denote the average density of the universe as \(\bar{\rho}(t)\). Nonetheless, there are some tiny fluctuations and not everywhere exactly the same. So let us define the density at comoving position r and time t as \(\rho(x,t)\) and the relative density contrast as \[\delta(r,t)\equiv\frac{\rho(r,t)-\bar{\rho}(t)}{\bar{\rho}(t)}.\]In this exercise we focus on the linear theory, namely, the density contrast in the problem remains small enough so we only need to consider terms linear in \(\delta\). We assume that cold dark matter, which behaves like dust (that is, it is pressureless) dominates the content of the universe at the early epoch. The absence of pressure simplifies the fluid dynamics equations used to characterize the problem.
a) In the linear theory, it turns out that the fluid equations simplify such that the density contrast \(\delta\) satisfies the following second-order differential equation: \[\frac{d^2\delta}{dt^2}+\frac{2\dot{a}}{a}\frac{d\delta}{dt}=4\pi G\bar{\rho}\delta,\]where a(t) is the scale factor of the universe. Notice that remarkably in the linear theory this equation does not contain spatial derivatives. Show that this means that the spatial shape of the density fluctuations is frozen in comoving coordinates and only their amplitude changes. Namely this means that we can factorize \[\delta(x,t)=D(t)\tilde{\delta}(x),\]where \(\tilde{\delta}(x)\) is arbitrary and independent of time, and D(t) is a function of time and valid for all x. D(t) is not arbitrary and must satisfy a differential equation. Derive this differential equation.
We can check that this is a solution by plugging the third equation into the second.
\(\frac{d\delta}{dt}=\dot{D}(t)\tilde{\delta}(x)\)
\(\frac{d^2\delta}{dt^2}=\ddot{D}(t)\tilde{\delta}(x)\)
\(\ddot{D}(t)\tilde{\delta}(x)+\frac{2\dot{a}}{a}\dot{D}(t)\tilde{\delta}(x)=4\pi G\bar{\rho}D(t)\tilde{\delta}(x)\)
\(\ddot{D}(t)+\frac{2\dot{a}}{a}\dot{D}(t)=4\pi G\bar{\rho}D(t)\)
This is a solution to the differential equation.
b) Now let us consider a matter-dominated flat universe, so that \(\bar{\rho}(t)=a^{-3}\rho_{c,0}\) where \(\rho_{c,0}\) is the critical density today, \(3H^2_0/8\pi G\) as in WS11.1. Recall that the behavior of the scale factor of this universe can be written \(a(t)=(3H_0t/2)^{2/3}\), which you learned in previous worksheets, and solve the differential equation for D(t). Hint: you can use the ansatz \(D(t)\propto t^q\) and plug it into the equation that you derived above; you will end up with a quadratic equation for q, There are two solutions, and the general solution for D is a linear combination of two components: one gives you a growing function in t, denoting it as \(D_+(t)\); another decreasing function in t, denoting it as \(D_-(t)\).
By this ansatz, we can say:
\(D(t)=t^q\)
\(\dot{D}(t)=qt^{q-1}\)
\(\ddot{D}(t)=q(q-1)t^{q-2}\)
We can then plug into the differential equation.
\((q)(q-1)t^{q-2}+\frac{2\dot{a}}{a}qt^{q-1}=4\pi G\bar{\rho}t^q\)
We can multiply by \(t^{2-q}\) and substitute in today's critical density.
\((q)(q-1)+\frac{2\dot{a}}{a}qt-4\pi Ga^{-3}\rho_{c,0}t^2=0\)
We plug in the critical density and \(\frac{2}{3t}\) for \(\frac{\dot{a}}{a}\).
\(q^2+\frac{1}{3}q-\frac{2}{3}=0\)
\(q=\frac{2}{3}, -1\)
c) Explain why the \(D_+\) component is generically the dominant one in structure formation, and show that in the Einstein-de Sitter model, \(D_+(t)\propto a(t)\).
Since this is a linear combination, it means that \(D(t)=D_+(t)+D_-(t)=At^{2/3}+Bt^{-1}\).
\(D_+(t)\propto t^{2/3}\propto a(t)\). This term will grow as time increases, since t's exponent is positive.
\(D_-(t)\propto t^{-1}\). This term will decrease as time increases, since t's exponent is negative.
As a result, \(D_+(t)\) will come to dominate as time moves forward.
2. Spherical collapse. Gravitational instability makes initial small density contrasts grow in time. When the density perturbation grows large enough, the linear theory, such as the one presented in the above exercise, breaks down. Generically speaking, non-linear and non-perturbative evolution of the density contrast have to be dealt with in numerical calculations. However, in some very special situations, analytical treatment is possible and provides some insights to some important natures of gravitational collapse. In this exercise we study such an example.
d) Plot r as a function of t for all three cases (open, closed, and flat universe), and show that in the closed case, the particle turns around and collapses; in the open case, the particle keeps expanding with some asymptotically positive velocity; and in the flat case, the particle reaches an infinite radius but with a velocity that approaches zero.
Closed case: blue
\(r=A(1-cos\eta)\)
\(t=B(\eta-sin\eta)\)
The particle's radius increases, but then decreases until the particle collapses to zero again.
Open case: red
\(r=A(cosh\eta-1)\)
\(t=B(sinh\eta-\eta)\)
The particle's radius appears to continue to increase.
Flat case: green
\(r=A\eta^2/2\)
\(t=B\eta^3/6\)
The particle's radius appears to increase with a decreasing velocity.
The x-axis is time and the y-axis is radius. A and B were arbitrarily assigned values of 1, but in practice these values depend on the mass of the patch of over-density.
a) In the linear theory, it turns out that the fluid equations simplify such that the density contrast \(\delta\) satisfies the following second-order differential equation: \[\frac{d^2\delta}{dt^2}+\frac{2\dot{a}}{a}\frac{d\delta}{dt}=4\pi G\bar{\rho}\delta,\]where a(t) is the scale factor of the universe. Notice that remarkably in the linear theory this equation does not contain spatial derivatives. Show that this means that the spatial shape of the density fluctuations is frozen in comoving coordinates and only their amplitude changes. Namely this means that we can factorize \[\delta(x,t)=D(t)\tilde{\delta}(x),\]where \(\tilde{\delta}(x)\) is arbitrary and independent of time, and D(t) is a function of time and valid for all x. D(t) is not arbitrary and must satisfy a differential equation. Derive this differential equation.
We can check that this is a solution by plugging the third equation into the second.
\(\frac{d\delta}{dt}=\dot{D}(t)\tilde{\delta}(x)\)
\(\frac{d^2\delta}{dt^2}=\ddot{D}(t)\tilde{\delta}(x)\)
\(\ddot{D}(t)\tilde{\delta}(x)+\frac{2\dot{a}}{a}\dot{D}(t)\tilde{\delta}(x)=4\pi G\bar{\rho}D(t)\tilde{\delta}(x)\)
\(\ddot{D}(t)+\frac{2\dot{a}}{a}\dot{D}(t)=4\pi G\bar{\rho}D(t)\)
This is a solution to the differential equation.
b) Now let us consider a matter-dominated flat universe, so that \(\bar{\rho}(t)=a^{-3}\rho_{c,0}\) where \(\rho_{c,0}\) is the critical density today, \(3H^2_0/8\pi G\) as in WS11.1. Recall that the behavior of the scale factor of this universe can be written \(a(t)=(3H_0t/2)^{2/3}\), which you learned in previous worksheets, and solve the differential equation for D(t). Hint: you can use the ansatz \(D(t)\propto t^q\) and plug it into the equation that you derived above; you will end up with a quadratic equation for q, There are two solutions, and the general solution for D is a linear combination of two components: one gives you a growing function in t, denoting it as \(D_+(t)\); another decreasing function in t, denoting it as \(D_-(t)\).
By this ansatz, we can say:
\(D(t)=t^q\)
\(\dot{D}(t)=qt^{q-1}\)
\(\ddot{D}(t)=q(q-1)t^{q-2}\)
We can then plug into the differential equation.
\((q)(q-1)t^{q-2}+\frac{2\dot{a}}{a}qt^{q-1}=4\pi G\bar{\rho}t^q\)
We can multiply by \(t^{2-q}\) and substitute in today's critical density.
\((q)(q-1)+\frac{2\dot{a}}{a}qt-4\pi Ga^{-3}\rho_{c,0}t^2=0\)
We plug in the critical density and \(\frac{2}{3t}\) for \(\frac{\dot{a}}{a}\).
\(q^2+\frac{1}{3}q-\frac{2}{3}=0\)
\(q=\frac{2}{3}, -1\)
c) Explain why the \(D_+\) component is generically the dominant one in structure formation, and show that in the Einstein-de Sitter model, \(D_+(t)\propto a(t)\).
Since this is a linear combination, it means that \(D(t)=D_+(t)+D_-(t)=At^{2/3}+Bt^{-1}\).
\(D_+(t)\propto t^{2/3}\propto a(t)\). This term will grow as time increases, since t's exponent is positive.
\(D_-(t)\propto t^{-1}\). This term will decrease as time increases, since t's exponent is negative.
As a result, \(D_+(t)\) will come to dominate as time moves forward.
2. Spherical collapse. Gravitational instability makes initial small density contrasts grow in time. When the density perturbation grows large enough, the linear theory, such as the one presented in the above exercise, breaks down. Generically speaking, non-linear and non-perturbative evolution of the density contrast have to be dealt with in numerical calculations. However, in some very special situations, analytical treatment is possible and provides some insights to some important natures of gravitational collapse. In this exercise we study such an example.
d) Plot r as a function of t for all three cases (open, closed, and flat universe), and show that in the closed case, the particle turns around and collapses; in the open case, the particle keeps expanding with some asymptotically positive velocity; and in the flat case, the particle reaches an infinite radius but with a velocity that approaches zero.
Closed case: blue
\(r=A(1-cos\eta)\)
\(t=B(\eta-sin\eta)\)
The particle's radius increases, but then decreases until the particle collapses to zero again.
Open case: red
\(r=A(cosh\eta-1)\)
\(t=B(sinh\eta-\eta)\)
The particle's radius appears to continue to increase.
Flat case: green
\(r=A\eta^2/2\)
\(t=B\eta^3/6\)
The particle's radius appears to increase with a decreasing velocity.
 |
| Zoomed in |
 |
| Zoomed out |
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