## Saturday, November 19, 2011

### Radial Velocities Worksheet Problem 2

We can determine how much the sun is displaced by Jupiter's mass with just a few simple calculations. If we want the center of mass to be at the origin of the system, we have:
where J subscripts denote Jupiter, s subscripts denote the sun, M denotes mass, and a denotes distance from the center of mass. From the image, we see:
We want as, the distance from the center of the sun to the center of mass of the system. We can solve the center of mass equation for this distance and then substitute (a – ap) for as.
In the last step, I've used the approximation MJ << Ms to simplify the expression. The mass of Jupiter is one-one-thousandth of the mass of the sun, and the distance a between the sun and Jupiter is 5 AU. We know that one AU is approximately 200 solar radii, so we can get a in terms of solar radii.
So we see that the sun is 1 solar radii away from the center of mass of the sun-Jupiter system. This means that the mass of Jupiter alone causes the sun to displace by a full solar radius!

Secondary authors: Daniel Lo, Monica He

## Tuesday, November 15, 2011

### What Does it Mean to be a Professional Astronomer? Interview with Graduate Student Kunal Mooley

Joanna and I had the pleasure of sitting down with a Caltech graduate student in the astronomy department this afternoon and discussing the life of a graduate student. We interviewed Kunal Mooley, who was happy to enlighten us about his personal experience as a grad student at Caltech.

The other options are to join industry, financial institutions, or to get a consulting job. The defense industry hires a lot of physics students, and some industries are even affiliated with universities. It's possible to get a degree by working for an industry and getting a scholarship to work toward an upper-level degree while working a typical job. The decision between going into industry and becoming a grad student is mostly driven by whether you prefer making money or exploring research that you find interesting. Also, if you want to be a professor, you need a PhD, so you need to take the grad student track.

How did you decide where to go for grad school?
I looked at the top 10 USA universities for physics. I majored in physics as an undergrad, but I was interested in astronomy since childhood and decided that's what I wanted to do. There's so much physics involved in astronomy that a university with a good physics program was very important. It's also important to look at how good the university is overall--too much specialization in a single field limits the collaboration with other fields, which is important for most recent research.

How did you decide in what part of astronomy to specialize, and when do you decide this?
You start thinking about what kind of research you might want to specialize in as a grad student. It's good to learn a variety of topics within astronomy so that you can better discover what it is that interests you, and also because too much specialization makes it difficult to think outside the box and can limit your competence in all fields. As you work on several projects, you realize what it is that interests you, and then you know how you might want to specialize. If you want to be a professor instead of a research scientist, you have to avoid specializing too much so that you can still teach others in more general fields.

What is your personal ultimate goal as an astronomer?
I like a lot of topics in astronomy, so I'm trying not to specialize too much . I really enjoy teaching, so my goal is to be a professor and do research in a variety of topics. In general, if a grad student discovers they don't like teaching, they work toward being a research scientist instead. Profs usually get better positions, so it's usually a better idea to try to be a prof than a research scientist.

How do you decide which prof you want to work under?
You want to search for professors with the same research interests as you. Also, since your work as a graduate student has a large impact on your scientific standing when applying for jobs as a professor or a research scientist, it's important to find an adviser who is well-known and reputable in their field. You have to be careful about who you originally start working for, because you can end up being biased toward your first research experience, and you want to make sure it's a good experience to begin with. In an environment like Caltech, the department is very small and located in a single place, so it's easy to get to know people and be able to ask around to find out which profs make good mentors. It's also really easy to introduce yourself to profs and get to know them so that you can work on multiple projects and keep from getting too specific in a single topic. At Caltech, you can take 1-2 years before choosing a mentor, but this varies by university. In general, there is time to get to know the department before you have to choose who to work for, and matching up your research interests to a professor's research interests is the easiest way to find a mentor. It's important to realize that your research under your mentor is very important, but it's not the only thing you should be restricted to.

I didn't have many expectations of grad school other than the amount of freedom to pursue any research opportunities I found interesting. I also really looked forward to meeting new people who were eager to collaborate and help in various research topics. Caltech met these expectations, likely because it is such a small university and it's common for professors, grad students, and post-docs to have a helpful personality, which is very important.

How do you know if you want to do research or go into industry?
Before making the decision, make sure you know exactly what's involved in going from grad school to a professorship. There's a lot more freedom in research through a university than through an industry, and you're not forced to do something you many not enjoy as much when you're doing your own research. If you realize partway through grad school that research is not for you and you want to switch to industry, you need to make sure you've taken the appropriate classes to be competent in the field you're switching to. There's not the freedom in industry to learn as you go like there is in research, and working in a real job is much more pressing and restricting. When doing research, you can work with lots of different people who help you through topics you don't understand, and that doesn't happen as much in industry.

What are the best and worst parts of being a grad student?
I can't think of anything bad about it! The best part is the flexibility of work--you can work at your own pace with who you want, you're not pressured the same way you would be if you had a real job in industry. The flexible and helpful learning environment is necessary to be able to survive the pressures of working toward a career, and Caltech does a great job of providing such an environment. You also get to know a lot of interesting people and can work on whatever projects you want, so it's really easy to keep your knowledge broad across lots of topics. You also get lots of chances to hone your presenting and teaching skills and become a more well-rounded person, which you might not get in industry. In general, universities offer many more opportunities and a much more pleasant environment than getting a job in industry.

Do grad students usually take advantage of the opportunities presented by universities?
If they want to, they definitely do. It's not difficult to go out and find a research project you want to work on, or find a way to do any extracurricular activities. As long as a student is interested in something, they can find something interesting to do.

Kunal is a third-year grad student whose main interest is radio astronomy. He wants to become a professor, and he's very happy with Caltech's grad program.

## Saturday, November 12, 2011

### Class Activity: Virializing the Night Away Problems 5-9

If we have a star that has ceased fusion in the core, as we add more mass to the star, the radius should increase until the star is massive enough that it begins to shrink under its own weight. Then, when the star is small enough, it becomes so dense that it cannot get any denser due to the uncertainty principle--electrons can't be in the same state as other electrons, so they must occupy other energy levels instead and this means there is an upper bound to the density of a star. Once this bound is reached, it cannot get more dense, and increasing mass must increase the radius again.

By the virial theorem (twice the kinetic energy plus the gravitational energy of a system is 0), we can determine the relationship between the kinetic energy of the electrons in a white dwarf and the gravitational energy of the dwarf.
where U is the gravitational energy, which is found by integrating the gravitational energy over the mass of the entire star, and T is the kinetic energy, which is the number of electrons ne times the thermal energy per particle.

We know that degeneracy pressure becomes important in a white dwarf when the inter-particle spacing is of order of the de Broglie wavelength, which is given by:
where h is Planck's constant and p is the momentum of a particle. The uncertainty in the electron's position is also of order of the de Broglie wavelength, since uncertainty in position is on the order of the position. Since the inter-particle spacing is dependent on the number density of electrons, Ne, so is the uncertainty in the position of the electrons, and we have:
The inter-particle spacing cubed is essentially the volume per particle, which is the inverse of the number density, which is the particles per volume. The kinetic energy of a particle is related to the momentum of the particle by:
The uncertainty in momentum is on the order of the momentum of the particle, and by the Heisenberg uncertainty principle in 3 dimensions,
we can relate the uncertainty in position and the uncertainty in momentum.
This is the relationship between the kinetic energy per particle in the white dwarf and the dwarf's electron number density, where h is Planck's constant and m is the mass of an electron. To relate the electron number density to the total mass and radius of the white dwarf, we realize that the total mass will be the number of electrons times the mass of an electron plus the number of protons times the mass of a proton plus the number of neutrons times the mass of a neutron. We can take the mass of an electron to be negligible when compared to the mass of a proton. We can take the number of electrons to be some multiple of the number of neutrons, and let's call this factor A. The mass of a proton is approximately equal to the mass of a neutron. Then we have:
where ne is the number of electrons, and V is the volume of the star. This is the relationship between the number density of electrons and the mass and radius of the white dwarf. We substitute this back into the virial theorem (see above), using the previously found expression for the kinetic energy per particle and for the electron number density, keeping in mind that the total kinetic energy is the number density times the volume times the kinetic energy per particle:
In the last couple steps we've dropped all constants to obtain the relationship between the radius and mass of a white dwarf.

Secondary author: Melodie Kao

### Class Activity: Formation of Stars Problem 2

If we have a small particle in an orbit of eccentricity 1 (essentially a straight line with the mass around which the particle is orbiting at one end of the line--see picture), then half of the period of this orbit is the time it takes the particle to "fall" in toward the center, where the mass is. This is a decent approximation of how long it takes a cloud of interstellar gas to collapse into a star. The mass about which the particle is orbiting is actually the point mass that comes from considering the mass of the cloud as a point mass, which is valid because as the particle falls in, all the mass inside of it falls as well, so the total mass does not change. We can represent this mass as the volume times the initial average density, then we just need Newton's version of Kepler's third law to obtain the free fall time.

where P is the period of the "orbit," a is the semi-major axis of the orbit, and ρ is the average density of the mass M. The free fall time is half the period, and the radius R is two times the semi-major axis a.
The assumption used to obtain this final step is that the cloud that is collapsing is spherical, which is a valid assumption because oftentimes only part of a cloud will collapse to a star. The part that collapses is a spherical part of the whole, even if the whole cloud isn't spherical.

We know that if the free fall time is longer than the time it takes for a pressure wave (like sound) to traverse the cloud, then the cloud won't be unstable enough to collapse. This is because the beginning of collapse will create a pressure wave, and if that moves through the cloud quickly enough, it will push out the other side of the cloud when it reaches it. This won't cause a collapse, since the cloud isn't pushing inward from all directions. However, if the pressure wave can't make it to the other side of the cloud before that side of the cloud has already collapsed inward somewhat, the cloud will be unstable enough to collapse fully.

The time it takes for a pressure wave to traverse the cloud is the width of the cloud, 2R, over the average speed of sound, cs. Equating this to the free fall time and solving for the radius R will give the minimum possible size of a cloud in order for it to be unstable enough to collapse. If R were any smaller, the sound wave would pass through the cloud in less time, and it will not be unstable. For an isothermal gas of constant density, the speed of sound is constant.
This is an order-of-magnitude estimate of the Jean's Length, which, again, is the minimum size for a cloud to be unstable enough to collapse.
If we consider a spherical cloud collapsing isothermally with an initial radius of the Jean's Length, we can discover what the new Jean's Length would be after the cloud has collapsed to half of its original radius. At half of the original radius, the average density will be 8 times greater than what it was before, since density depends on the inverse of radius cubed. The speed of sound squared is dependent on the pressure over the density, and since pressure is dependent on inverse surface area of the cloud, the pressure will increase by a factor of 4 when the radius decreases by a factor of 2.
So we see that the radius of the cloud changing by a factor of 1/2 makes the Jean's Length change by a factor of 1/4. This means that once the cloud contracts to half its size, the new-sized cloud can fragment into pieces and still be unstable enough to collapse. This means that a single cloud that is just barely large enough to collapse can form more than one star when it collapses, since parts of it will collapse independently once it begins to contract.

Secondary authors: Joanna Robaszewski, Daniel Lo, Melodie Kao

## Sunday, October 30, 2011

### Class Activity: Stellar Properties from Afar Problem 1

Given that the angular width of the sun is .5° and that 1 AU is 1.5 * 1013 cm, it's possible to find the radius of the Sun in centimeters. If we consider the angle subtended by the Sun from a certain point on Earth, we have a right triangle with one leg equal to 1 AU and one angle equal to half of the angular width of the sun, .25° or 15 arcminutes.
By using a trigonometric relation, we can find a value for Rs, the radius of the Sun.
So the radius of the Sun is 6.55 * 1010 cm. Next, we can get 1 AU in solar diameters by simply dividing 1 AU in centimeters by the radius of the Sun in centimeters:
So 1 AU is approximately 100 solar diameters. Now, using Newton's version of Kepler's third law, we can get the mass of the Sun.
This is Newton's version of Kepler's third law, if we assume that the mass of the Sun is significantly greater than the mass of the Earth, which it is. Then P is the period of the Earth, a is 1 AU, G is the gravitational constant, and M is the mass of the sun. Rearranging and plugging in the correct numbers:
We see that the mass of the sun from Kepler's third law is 6.4*1032 g.

Secondary author: Joanna Robaszewski

## Wednesday, October 19, 2011

### Lab Activity 2: The Astronomical Unit

Take a look at this image of Mercury passing in front of the Sun, as taken from the polar Earth-orbiting TRACE satellite.

From this image and a few other facts, it's possible to discover the distance from the Earth to the Sun, which is defined as one AU (astronomical unit). The other information necessary is that the Sun's angular width is 0.5°, Mercury's orbital period is 87 days, and the radius of the Earth is 6.738 * 108 cm.

The slight sinusoidal "wiggle" comes from the fact that the TRACE satellite is orbiting the Earth as its taking images of Mercury, so the image of Mercury projected against the non-moving background of the sun moves up as the satellite moves down to the south pole, and down again as the satellite moves back up to the north pole. By extending the disk of the Sun into a full circle (using a printout of the image and a compass), it's possible to estimate how many "wiggles" would fit into the full height of the Sun, which we know to be 0.5°. From this, we can find the angular width of the "wiggle," which, through some tricky geometry, will give us the distance from the Earth to the Sun. We determined that about 50 "wiggles" would fit into the width of the sun, so the angular width of the sinusoid is .5/50 = 1/100th of a degree.

In the above, ae is the Earth's orbital semi-major axis (1 AU), am is Mercury's orbital semi-major axis, and Re is the radius of the Earth. The angles x, y, and z are also shown on the image above, and the angle y is the angular width of the sinusoid, which we found to be .001°. From considering the triangle formed from the center of the earth to the satellite to mercury and by using the small angle approximation, we find:
We know the radius of the radius of the Earth and the angle y, so we can find ae – am. The small angle approximation requires the angle to be in radians, so y/2 = 8.725 * 10-6 radians. We find that ae – am = 7.723 * 1013 cm. Now we need Kepler's laws to obtain the astronomical unit. Kepler's Third Law states:
This means that the ratio of the periods of Mercury's and Earth's orbits squared is equal to the ratio of the distances from the Sun of Mercury's and Earth's orbits cubed. We don't have ae or am, but we do have ae - am. Plugging this in:
and rearranging for ae:
Solving this, we finally find a value for the astronomical unit! 1 AU = 1.25 * 1014 cm. The actual value for the astronomical unit is 1.49 * 1013 cm, so this is a percent error of:
This is obviously a very high percent error, which is due to the not-very-exact method of estimating the number of "wiggle" widths in the full width of the sun, since that was the only estimated value. Since the calculated value for the astronomical unit is too big, this means that the value for ae - am is too big, which means the estimated width y is too small.

Now we can obtain the distance Mercury is from the Sun from our value for ae - am and our value for ae. We find that am = 4.78* 1013 cm, and the actual distance between Mercury and the Sun is 5.79 * 1012 cm. This percent error is 725%, which is to be expected since our percent error on the astronomical unit is so high.

Finally, we can obtain the mass of the Earth from Kepler's Third Law and the mass of the Sun.
Note that it is also possible to obtain the mass of the sun by neglecting the mass of Mercury in its orbital calculation in Kepler's Third Law,
but since it requires the distance from Mercury to the Sun and that distance has such a large percent error, I will just use the actual value for the mass of the Sun. We find, by rearranging Kepler's Third Law and plugging in the values for Pe, ae, and Ms:
where the mass is measured in grams. The actual value for the mass of the Earth is 5.974 * 1027 g, so my found value has a percent error of 2.5 * 1017 %. This is to be expected, since the value of the astronomical unit is used in the calculation, and if that has a percent error of 742%, then the value for the mass of the Earth will have a percent error of at least that cubed, since the astronomical unit is cubed in the equation for mass of the Earth. The actual percent error is higher because there are other factors in the equation that help in determining the value of the Earth's mass.

From just a few important facts, we've managed to find the astronomical unit, the distance from Mercury to the Sun, and the mass of the Earth. Our values are not very good, but they all stemmed from an estimate using a compass on a piece of paper, so this is to be expected.

## Monday, October 17, 2011

### Class Activity: Black Body Radiation Problem 2 b and c

In order to define a relationship between the wavelength corresponding to the peak intensity of radiation emitted from a black body (λmax) and that black body's temperature (T), we must begin with the equation for specific intensity, Bλ(T):
The maximum of this function will be the λ such that
where the derivative is evaluated with T held constant (since we are finding λmax for a given T). If we make a simple variable substitution, the calculation becomes much easier:
Then we need:
This is a difficult equation to solve, but if we approximate ex to second order, then we have:
Plugging in the values for h, c, and k, we find:
where the units of λmaxT are cm · K. Wien's law is:
For a second-order approximation of ex, this is quite close.

Now let's consider very small photon energies, so that hv << kT. Then we can use a second-order approximation for ex again to simplify Bv(T) for small photon energies.
Inserting this into the equation for Bv(T), we obtain:
This is the simplified form of Bv(T) for hv << kT.

Radio astronomers like to talk about "brightness temperature" instead of actual brightness because the brightness is related to the temperature, but the temperature is more helpful in discovering the black body spectrum of an object than its brightness. The black body spectrum, if the temperature is known, gives the most prominent wavelength of a black body's radiation, which is a useful value in classifying an object.

Secondary authors: Joanna Robaszewski, Lauren Gilbert