Twinkle Twinkle Little Star: October 2011

Tuesday, October 25, 2011

Neutron Stars and Their Equations of State

For my summer research with LIGO, I worked a lot with neutron stars. The first portion of my project was making a mass-radius plot of different proposed equations of state. An equation of state provides the relationship between the thermodynamic variables of pressure and energy density.

I worked with/helped develop a TOV solver (which is a code in C) that makes model neutron stars (i.e. given a starting central density, outputs the properties of the resulting star). We're most interested in the masses and radii of these stars.

There were two types of equations of state: analytic and tabular. The analytic equation of state was given by the Tolman-Oppenheimer-Volkoff (TOV) equation:

$\frac{dP(r)}{dr}=-\frac{G}{r^2}\left[\rho(r)+\frac{P(r)}{c^2}\right]\left[M(r)+4\pi r^3 \frac{P(r)}{c^2}\right]\left[1-\frac{2GM(r)}{c^2r}\right]^{-1} \;$

It looks really scary... but luckily the program solved it, so I didn't have to :] The equation basically describes a spherically symmetric, isotropic star in static gravitational equilibrium.

The other equations of state came in the form of density-pressure tables from different researchers in the field. I wrote a function that interpolated the pressure-density relationship so that the code could use it instead of the TOV equation to create a model neutron star.

I also had to look at the constraints for the mass-radius plot.

Minimum Maximal Mass Constraint
The first constraint is based on the observation of a 2 solar-mass neutron star. Any equation of
state that does not predict a mass that reaches two solar masses is disproved by obesrvation.

General Relativity Constraint
The GR constraint requires R > 2GM/c2 because neutron stars must have a radius that
is greater than the Schwarzschild Radius. Any proposed neutron star with a radius below
the threshold would in reality be a black hole.

Finite Pressure Constraint
The finite pressure constraint, according to Lattimer and Prakash (2007) , requires
R >(9/4)GM/c2.

Causality Constraint
To obtain the causal constraint, we recreated the procedures of Koranda et al. (1997) ,
and Lattimer and Prakash (2007). To avoid a supraluminous equation of state in which
the speed of sound is faster than the speed of light, it is necessary that

$\frac{dp}{d\rho}\leq c^2$
.
Rotational Constraint
For a fully relativistic star, the rotational limit can be expressed as
$R_{sph} < 10.4 \bigg(\cfrac{1000Hz}{\nu}\bigg)^{2/3}\bigg(\cfrac{M_{\mathrm{sph}}}{M_{\odot}}\bigg)^{1/3} km$

The final mass-radius plot that includes all of the EOS that I worked with looked like this:

Hydrostatic Equilibrium and the Virial Theorem

Author: Iryna Butsky
Co-authors: Juliette Becker and Monica He

This is a write up of question 1 of the Hydrostatic Equilibrium and the Virial Theorem worksheet.

(a) We're looking for the total potential energy contained in the sun. We know that potential energy is expressed by U = - GMm/r. More specifically, in this case $U = \frac{GM_{shell}M_{int}}{r}$ where M_shell is the mass and M_int is the mass of the sphere inside that shell. We also have $dM_{shell}= \rho(r)4\pi r^2dr$ and $dM_{int}= \rho(r)\frac{4}{3}\pi r^3$ .The change in potential energy, dU: $dU = \frac{GdM_{shell}dM_{int}}{r}$ therefore, after combining the terms we have that $U =\int_{0}^{R_{\odot}} \frac{16}{3}G\rho^2\pi^2 r^4dr = \frac{16}{15}G\rho^2\pi^2R_{\odot}^5$ . Keeping in mind that $\rho = \frac{M}{V} = \frac{M_{\odot}}{\frac{4}{3}\pi R_{\odot}^3}$ we can simplify the expression to $U = \frac{3GM_{\odot}^2}{5R_{\odot}}$

(b) Potential energy is related to force in the following expression: $U = \int F\cdot dr = \int m\frac{d^2r}{dt^2}$ By doing dimensional analysis, we know that U has units of cm^3g^-1s^2 and dr/dt has units of cm/s. Therefore the velocity of a free-falling element would have the following relation to potential energy: $\frac{dr}{dt}=\sqrt{\frac{U}{M}}$

(c) We can replace dr/dt in the equation above with $R_{\odot}/\tau_{ff}$ . We're assuming that the free-fall velocity will be roughly constant. Therefore, if we define tau to be the time it takes a particle to fall from the surface of the star to the center and R_solar as the radius of the Sun, then we can make this substitution to an order of magnitude.

(d) Plugging in our values into this expression $\frac{R_{\odot}}{\tau_{ff}}=\sqrt{\frac{U}{M}}$ we find that $\tau_{ff} = 2065 \;\mathrm{seconds}\simeq 34 \;\mathrm{minutes}$ . This means that if the Sun were truly undergoing free fall collapse, it would collapse in 34 minutes, which is surely a time-scale that we would notice.

Saturday, October 22, 2011

Finding the Astronomical Unit

Given this image of Mercury's orbit around the sun, we need to calculate the Astronomical Unit (The distance between the Earth an the Sun). We are given that Mercury's period is 87 days and that the Sun has an angular diameter of about .5 degrees.

The diagram of this problem is represented below. The Earth is the sphere on the left, Mercury is the point M, and the sun is the right edge.

We can start off by calculating the angular amplitude of Mercury's orbit, alpha. Using a ruler and print out of the first image, we can see that the amplitude of Mercury's orbit is about 4mm and that the diameter of the sun is roughly 11cm. Since we know that the sun's angular diameter is 0.5 degrees, we can set up a proportion to find the angular amplitude.
$\frac{\alpha}{0.5}=\frac{0.4}{11}\mathrm{, }\;\;\;\;\; \alpha = 1.8\times10^{-3}\mathrm{degrees}=3.2\times10^{-5}\mathrm{radians}$

Doing a little bit of triangle geometry (the pink triangle), we can see that
$\pi = \alpha + \beta + (\pi-\frac{\theta}{2})$ therefore, $\alpha +\beta =\frac{\theta}{2}$

Now, let's look at the blue triangle. We know that
$tan\Big(\frac{\theta}{2}\Big)=\frac{R_{\oplus}}{x}$ where R is the radius of the Earth and x is the distance from the Earth to Mercury.
Since theta is such a small angle, we can use the sin small angle approximation to find an expression for theta
$\theta = \frac{2R_{\oplus}}{x}$ .

Now let's consider the superposition of the two triangles. From this we f ind the expression:
$tan(\beta)= \frac{R_{\odot}}{AU}$ Once again, using the small angle approximation, we can rewrite this as: $\beta= \frac{R_{\odot}}{AU}$

We can plug this result back in to our previous equation to show that
$\alpha + \frac{R_{\oplus}}{AU} = \frac{R_{\oplus}}{x}$ or $x=\frac{2R_{\oplus}}{\alpha+\frac{R_{\oplus}}{AU}}$

Because we know the period of Mercury (87 days) and the period of the Earth (365 days), we can use that information to find the distance from the sun to Mercury, a_m
$\Bigg(\frac{P_m}{P_{\oplus}}\Bigg)^2=\Bigg(\frac{a_m}{AU}\Bigg)^3$
we know that a_m is just AU - x, so we can rewrite this proportion as:
$\Bigg(\frac{P_m}{P_{\oplus}}\Bigg)^2=\Bigg(\frac{AU-\frac{R_{\oplus}}{\alpha+\frac{R_{\oplus}}{AU}{}}}{AU}\Bigg)^3 = \Bigg(1-\frac{R_{\oplus}}{AU\alpha+R_{\oplus}{}}\Bigg)^3$

we know the radius of the Earth to be 6.38*10^8 cm and alpha = 3.2*10^-5. Plugging in this messy expression into Wolfram Alpha, we find that the astronomical unit,

${\color{Red}AU = 1.25 \times10^{13}cm }$

The real value, however, is known to be about 1.5*10^13 cm. The errors in my calculations come from my estimates in measuring alpha.

We can also use this information to find the mass of the sun.
Using Kepler's third law we have:
$\frac{P_{\oplus}^2}{AU^3}=\frac{4\pi^2}{M_{\odot}G}$ where G is the gravitational constant, which is 6.7*10^-8 in cgs units.

Using WolframAlpha once again, we find that the mass of the sun is:

${\color{Red}M_{\odot} = 1.16 \times10^{33}g}$
This is a little off from the real value of 1.98*10^33g, but this is because our value for the AU was a little off.

Thank you Juliette and Monica for all the help!

Friday, October 21, 2011

Stellar Properties From Afar Worksheet

Author: Iryna Butsky Co-Author: Juliette Becker

We know that the angular diameter of the sun d = 0.5 degrees. We also know that the size of the semi-major axis of the Earth's orbit around the sun, a = 1.5 x 10^13cm.

We can now set up a simple proportion to find the radius of the sun.

$\frac{0.5}{360}=\frac{R_{\odot}}{\pi a_{\oplus}}$ $\frac{0.5}{360}=\frac{R_{\odot}}{\pi (1.5\times10^{13})}$ From this, we find that the radius of the sun is about 6.5 * 10^10 cm.

Therefore, the AU, in solar diameters is $\frac{1.5\times10^{13}}{2\times6.5\times10^{10}}\simeq 100$
From Newton's version of Kepler's third law we have the following equation $\frac {P^2} {a^3} = \frac {4 \pi^2} {M G},$ we know that P =31,556,926 secons , a = 1.5*10^13cm, and G = 6.674*10^-8 cm^3 g^-1s^-2. Plugging the numbers in we find that the mass of the Sun is about 2*10^33g.

Using only a lightbulb and our skin, we were able to figure out the luminosity of the Sun. We are given that the power output of the light bulb is 100 W with an efficiency of .1. Therefore, the light bulb emits the equivalence of 10 W. We know that $L=F\times4\pi R^2$ . Therefore, we can set up a simple proportion to find the Luminosity of the sun. We decided that the energy given off from the light bulb feels roughly the same as the sun on a warm day at about 5cm distance. Therefore, $\frac{L_{\odot}}{10W}=\frac{(1.5\times10^{13})cm}{5cm}$ the Luminosity of the sun is about 10^26 Watts or about 10^33erg/second
We can find the effective temperature of the sun using the equation: $\lambda_{max}T=0.002897755 \;m\; K$ We know that lambda max is equal to 500 nanometers, we find that the effective temperature of the sun is 5.8*10^3 degrees Kelvin.
Once again, we know that $L=F\times4\pi R^2$ , and we found the solar luminosity to be 10^33 ergs/second. We can set up a proportion relation to find the bolometric flux at the Earth (the Solar Unit). $F_{bol}=\frac{L_{\odot}}{4\pi R_{Au}^2}=\frac{10^{33}}{4\pi(1.5\times 10^{13})^2}$ Therefore, the Solar Unit is 3.5*10^5 ergs/cm^2/s

Tuesday, October 18, 2011

Drops of Jupiter

Last week, a group of us were walking back together from a TALC session. I asked the group about about a star/planet that had been catching my attention for the past couple weeks. Nathan was first to reply and say that it was Jupiter. I was really excited that it was so easily visible. Nathan then let us use his binoculars to observe Jupiter. Much to my surprise, I could see two of Jupiter's moons. I'm not really sure which moons I saw... my guess would be Ganymede and/or Callisto, because they are the largest. I haven't had very much experience observing the night sky with binoculars or telescopes, so this was all super exciting to me! Thanks Nathan! :]

Sunday, October 16, 2011

Horsehead Nebula

I've always seen pictures of the Horsehead Nebula, but I just realized that I know nothing about it. So I thought this would be a good opportunity to figure that out.

The Horsehead Nebula, also known as Barnard 33, is located on the east side of Orion's Belt, about 1500 light years away from the Earth. It gets its name because the shape of the dust and gas clouds resembles that of a horse's head when views from the Earth. The Horsehead Nebula is a dark nebula, which means that it is a very dense interstellar cloud. In this case, the Horsehead Nebula obscures the light of the emission nebula IC 434, in which it is located.

The red glow around the nebula comes from hydrogen gas that is ionized by the star system Sigma Orionis. Sigma Orionis is system of five stars in Orion. The main components of the system are two hydrogen-fusing dwarf stars in binary, Sigma Orionis AB. The bright spots just inside the base of the nebula are young stars being born.

Monday, October 10, 2011

Too Bright to See

I love being a student in Pasadena in L.A. County. I love that everything is within driving distance and that there's always something open at obscure hours of the night (because let's face it, Caltech students, as well as Astronomers, are predominately nocturnal). Ironically, one of my favorite features about L.A. is also what I like least about it. The city never sleeps. Light pollution (among many other forms of pollution) make star-gazing a rare opportunity. Whenever I find myself outside after sunset, I always look up at the stars, and am generally disappointed with the limited view.

Lately, I've been extra-reminiscent of the times that I was lucky enough to star-gaze in complete darkness. I used to go camping all the time with my family. One of my absolute favorite memories was just sitting on the edge of a lake, in the middle of a moon-less night, staring at the sky. The view was absolutely breathtaking and awe-inspiring.

I think it would be a really cool idea to organize a class star-gazing trip sometime in the following weeks. Theory is awesome, and in my opinion greatly contributes to the beauty of astronomy; but there's nothing like looking up at a clear, starry sky.

This picture was taken at the Jackson Meadows Reservoir in Lake Tahoe National Forest.

This picture was taken in a small village by the name of Shevchenky in the Ukraine. The bright stripe on the left hand side is a small two-lane road that is the only "major" road for miles and miles. The long time-exposure on the camera managed to capture a car driving by :]

Sunday, October 9, 2011

The Celestial Sphere and Observation Planning

Author: Iryna Butsky
Co-Authors: Juliette Becker and Tommy Heavey

This is a write-up of question #1 on worksheet #2.

Before getting started on the actual question, I'm going to quickly go over the basic idea behind sidereal time. As the Earth rotates around the sun, it actually rotates about 361 degrees for every earth-day. A sidereal day uses a very distant star as reference. In this case, the Earth's orbit around the sun is considered negligible. In a sidereal day, the Earth rotates 360 degrees. Setting up a proportional relationship:

$\frac{1^0}{360^0}=\frac{{x}}{1440}$

we find that the sidereal-day differs from the earth-day by 4 minutes (or 10 seconds/ hour). In other words, 24 hours in earth time will read 24 hours and 4 minutes in LST (Local Sidereal Time).

Given that LST = 0:00 at noon on the Vernal Equinox:

(a) The LST at midnight on the Vernal Equinox will be 0:00 + :02 = 0:02
(b) The LST 24 hours later will be 0:02 + 12:00 + 0:04 = 12:06
(c) The LST right now (October 09, 3:00 pm = 15:00 army time): The Vernal Equinox was on September 3. Right now, it is 16 days and 3 hours later. So the LST is: 15:00 + (16 days)(4 minutes/day) + (3 hours)(10 seconds/hour) = 16:04:30
(d) Tonight at midnight, the LST will be 0:00 + (16 days)(4 minutes/day) + (12 hours)(10 seconds/hour) = 1:06 (October 10)
(e) My birthday is December 18, which is 86 days after the Vernal Equinox. The LST at 16:30 (sunset) on my birthday will be: 16:30 + (86 days)(4 minutes/day) + (4.5 hours)(10 seconds/hour) = 20:14:45

Asteroid (35396) 1997 XF11

I remember the day as if it were yesterday. I was 10 years old and watching some program on the Discovery Channel when I heard horrific news. Scientists predicted that on October 26, 2028, asteroid (35396) 1997 XF11 would come very close to, and possibly collide with the Earth. I freaked out. I was so sure I had just heard the announcement of the end of the world (and I wouldn't even live to see 37!). Although my dad tried to reassure me that there was no need to worry, I couldn't feel safe until I found some factual evidence. After some time on the internet, I learned that the chances of the asteroid actually colliding with the Earth were very slim. However I never forgot that "near-death experience", and to this day, I've never once forgotten the asteroid's name.

I thought now would be a good time to revisit the topic. The picture below shows the asteroid (35396) 1997 XF11. No, I can't really tell which one it is either - I just thought it was really cool that I could even find it!

Asteroid (35396) 1997 XF 11 was discovered on December 6, 1997 by James V. Scotti. In 1998, equations showed that although the chances of a collision were small, the possibility was not out of the question. Prior to fully checking his results with his colleagues, Brian Mardsen released his calculations that the chances of a collision were as high as one in a thousand. Media all over the world made this a front-page story. For any of you that are worried, the current prediction is this: on October 26, 2028, this asteroid will get as close as 930,000 kilometers to the Earth. To put things in perspective, the moon is about 380,000 kilometers away from the Earth. Though (35396) 1997 XF 11 will likely pass even closer to the Earth in successive orbits, there is, so far, no need to worry about collision. It is currently on the PHA (Potentially Hazardous Asteroids) list. I can sleep better at night knowing that there are dedicated scientists keeping their eye on this.

Measuring the Radius of the Earth

In this lab, we're going to calculate the radius of the Earth using nothing but a stopwatch and the horizon. We chose the lovely Santa Monica Beach for our experiment. As the sun began approaching the horizon, I lay down to be eye level with the ocean. As soon as I saw the bottom of the sun touch the horizon, I gave the signal, and my partners started their stopwatches. As soon as they saw the sun touch the horizon, they stopped the watches and recorded the times. This measurement is all we need to calculate the Earth's radius.

Although the picture above isn't quite to scale with our experiment, r represents the radius of the Earth and h represents the height of the standing person with the stopwatch. We know that in 24 hours, the earth makes a rotation of 360 degrees. With this information, we can set up a very simple relationship to find the change in angle of the sun we observed. Let t be the measured time, and note that there are 86,400 seconds in 24 hours.

$\frac{t}{86400}=\frac{\theta}{2\pi}$

Once we have theta, we can use that to find r using the following equation:

$cos\theta=\frac{r}{r+h}$

Using the measurements t = 10 seconds and h = 1.9 meters, I found that r = 6807.225 (roughly 6,800) kilometers. The actual radius of the earth is 6,378.1 kilometers. Our estimation was close (only about 7% off!).

There were a couple factors that contributed to the possible errors of this calculation. The first is human error. With more people trying to take the same measurements, we could have eliminated some of the error by averaging out our times. The atmosphere wasn't very clear on that day, which affected our view of the sun and made it more difficult to see when the sun touched the horizon.

Overall, I feel like the lab was a success. Not to mention that it was the most fun I've had doing a Caltech lab! I may just be biased as a Californian, but there's nothing that compares to the Pacific Ocean.