Write about anything?

For my free blog post I thought I would write about something cool I stumbled across that is happening here at Harvard, more specifically at the Harvard Art Museum. Before I even get to my point, I have to say that if you haven't been to our art museum, please go. It's right on campus and has works by amazing artists, including Monet, Picasso, Cezanne, Dali, Polluck, Rubens, and Miro, just to name a few.

Look at how close it is.

I actually have a point, I promise. In the art museum, we have many majestic, ancient, white marble statues such as the one of Hypatia below:

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However, these statues weren't always monochromatic. At some point, each statue was brightly painted (they even had noses at some point). A group of art historians used paint chips left of sculptures along with UV scans to figure out what colors some ancient art pieces were originally painted, and here are the results:

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Although this exhibit is no longer featured at our art museum, research is continuously happening. In the mean time, the recreations are causing quite a stir. Some like seeing the art in its original intention, while others say the bright colors ruin the elegant image of ancient art. What do you prefer?

*Bonus fun fact: some of the ancient roman statues show evidence of having once been painted with bright, colorful hair, suggesting that the ancient romans used colorful dyes in their own hair.

Worksheet 11.2, Problem 1: Flow of Energy

The problem: Stars generate their energy in their cores, where nuclear fusion is taking place. The energy generated is eventually radiated out at the star’s surface. Therefore, there exists a gradient in energy density from the center (high) to the surface (low), but thermodynamic systems tend towards ‘equilibrium.’ In the following sections we will determine how energy flows through the star.

Part A: Inside the star, consider a mass shell of width ∆r, at a radius r. This mass shell has an energy density u + ∆u, and the next mass shell out (at radius r + ∆r) will have an energy density u. Both shells behave as blackbodies.

The net outwards flow of energy, L(r), must equal the total excess energy in the inner shell divided by the amount of time needed to cross the shell’s width ∆r. Use this to derive an expression for L(r) in terms of du/dr , the energy density profile. This is the diffusion equation describing the outward flow of energy.

Once again, an image from the textbook can help us visualize the situation.

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If we keep in mind the answer needs to be in terms of du/dr we can solve: $$L(r) = \frac{dE}{dt} = \frac{dE}{du}\frac{du}{dt} = \frac{dE}{dt}\frac{du}{dr}\frac{dr}{dt}$$ We can simplify this because you know $$\frac{dE}{du} = 4 \pi r^2 dr$$ and $$\frac{dr}{dt} = v$$ and $$v = \frac{c}{\rho \kappa \Delta r}$$ $$L(r) = 4 \pi r^2 \frac{c}{\rho \kappa} \frac{du}{dr}$$

Part B: From the diffusion equation, use the fact that the energy density of a blackbody is u(T(r)) = aT^4 to derive the differential equation: $$\frac{dT(r)}{ dr} \propto - \frac{ L(r) \kappa \rho (r)}{ πr^2acT^3}$$ where a is the radiation constant. You just derived the equation for radiative energy transport!

$$u(T(r)) = aT^4$$ $$du(T(r)) = 4aT^3 d(T(r))$$ From part a: $$du = \frac{L(r) \rho \kappa dr}{4 \pi r^2 c }$$ So $$\frac{L(r) \rho \kappa dr}{4 \pi r^2 c } =  4aT^3 d(T(r))$$ Which simplifies to our answer!   $$\frac{dT(r)}{ dr} \propto - \frac{ L(r) \kappa \rho (r)}{ πr^2acT^3}$$

Acknowledgements: I worked on this worksheet with Sean, April, and Barra.

Worksheet 11.1: Photons Random Walking Out of a Star

 The problem: Consider a photon that has just been created via a nuclear reaction in the center of the Sun. The photon now starts a long and arduous journey to the Earth to be enjoyed by Ay16 students studying on a nice Spring day.

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Part A: The photon does not travel freely from the Sun’s center to the surface. Instead it random walks, one collision at a time. Each step of the random walk traverses an average distance l, also known as the mean free path. On average, how many steps does the photon take to travel a distance ∆r? 

We know that on average, the displacement for random walk journeys would be zero, which is not very useful to us. Instead, we can look at the displacement squared: $$< \vec{ D }^2> = \sum \vec{ r_i}^2$$ and because the problem tells us $$\Delta r =  \left( < \vec{ D }^2 > \right)^{ \frac{1}{2}}$$ This means $$\Delta r^2 = \sum \vec{r_i}$$ Because r is a vector, the resultant sum would look something like: $$\sum \vec{r_N} = (\vec{r_1} + \vec{r_2} + ... \vec{r_N}) \times ( \vec{r_1} + \vec{r_2} + ... + \vec{r_N}$$ $$\Delta r^2 = \vec{r_1}^2 + \vec{r_2}^2 +...+ \vec{r_N}^2 + 2\vec{r_1}\vec{r_2}cos(\theta)+ 2\vec{r_2}\vec{r_3}cos(\theta) + ... + 2\vec{r_{N-1}}\vec{r_N}cos(\theta)$$ which is equal to: $$\Delta r^2 = nl^2 + 2l^2 \sum cos(\theta)$$ Where l is the average distance traveled defined in the problem. Due to the random nature of the photon's path of travel, $$\sum cos(\theta) = 0$$ So now, $$\Delta r^2 = Nl^2$$ and finally, $$N = \frac{\Delta r^2}{l^2}$$

Part B: What is the photon’s average velocity over the total displacement after many steps? Call this $\vec{v_{diff}}$, the diffusion velocity.

To find velocity of a photon, we need the distance it travels and the time it takes. We can get distance, or rather, displacement from above:   $$\Delta r = lN^{\frac{1}{2}}$$ to find time we can use the fact that we know photons have a speed c, and each are traveling a distance nl: $$t = \frac{d}{v} = \frac{nl}{c}$$ So $$\vec{v_{diff}} = \frac{\Delta r}{t} = \frac{lN^{\frac{1}{2}}c}{nl} = \frac{c}{N}$$ However, we do not always have N, so the formula in terms of l and r is: $$\vec{v_{diff}}= \frac{cl}{\Delta r}$$

Part C: The “mean free path” l is the characteristic (i.e. average) distance between collisions. Consider a photon moving through a cloud of electrons with a number density n. Each electron presents an effective cross-section σ. Give an analytic expression for “mean free path” relating these parameters.

We can draw the situation described as shown in the diagram below, taken from our textbook because it is a much neater drawing than anything I can make in MS paint. 

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Where the volume of the cylinder is $V= \pi r^2 L$

From the described relationship, we can see that the number density multiplied by area would give us the 2D area that would result in a collision. However, we want the 3D area (well, volume?) that would result in a collision, which would give: $$n \sigma = \frac{#_{electrons}}{volume} \cdot \frac{area}{1} = \frac{#_{electrons}}{\pi r^2 L} \cdot \frac{\pi r^2}{1} = \frac{#_{electrons}}{distance}$$ For one electron, we would then have: $$n \sigma = \frac{1}{l}$$ So $$l = {1}{n \sigma}$$

Part D: The mean free path l can also be related to the mass density of absorbers ρ, and the “absorption coefficient” κ (cross-sectional area of absorbers per unit mass). How is κ related to σ? Express vdiff in terms of κ and ρ using dimensional analysis. You have now developed the tools to attack the problem of radiative diffusion.

Part 1: How is κ related to σ? 

The units for kappa and sigma would be:

• $\kappa = \frac{cm^2}{g}$
• $\sigma = \frac{cm^2}{n}$

From this we can deduce: $$\rho = n \bar{m}$$ and $$\kappa = \frac{\sigma}{\bar{m}}$$ Which allows us to find: $$n \ \frac{\rho}{\bar{m}}$$ and $$\sigma = \kappa \bar{m}$$ So plugging those last two equations into    $$l = \frac{1}{n \sigma}$$ gives   $$l = \frac{1}{\rho \kappa}$$ Now we can find velocity using: $$\vec{v_{diff}}= \frac{cl}{\Delta r}$$ $$\vec{v_{diff}}= \frac{c \left( \frac{1}{\rho \kappa} \right)}{\Delta r}$$ $$\vec{v_{diff}}= \frac{c}{ \rho \kappa \Delta r}$$

Part E: What is the diffusion timescale for a photon moving from the center of the sun to the surface? The cross section for electron scattering is $σ_T = 7 \times 10^{25} cm^2$ and you can assume pure hydrogen for the Sun’s interior. Be careful about the mass of material through which the photon travels, not just the things it scatters off of. Assume a constant density, ρ, set equal to the mean Solar density (N.B.: the subscript T is for Thomson. The scattering of photons by free (i.e. ionized) electrons where both the K.E. of the electron and λ of the photon remain constant—i.e. an elastic collision—is called Thomson scattering, and is a low-energy process appropriate if the electrons aren’t moving too fast, which is the case in the Sun.)

We know: $$t = \frac{d}{v}$$ $$t = \frac{R_{\odot}}{ \frac{c}{ \rho \kappa R_{\odot}}} = \frac{ \rho \kappa R_{\odot}^2}{c}= \frac{ \rho \left( \frac{\sigma}{\bar{m}} \right) R_{\odot}^2}{c}$$ In this case, mass would be mass of a proton, which we can look up along with the density and radius of the sun. Plugging everything in: $$t = \frac{ 1.41 \left( \frac{7 \times 10^{-25} }{1.7 \times 10^{-24}} \right) (7 \times 10^{10})^2}{3 \times 10^{10}} = 9.5 \times 10^{10} s$$

Day Lab Conclusion: Calculating AU

Finally, we can put everything we have done together. We have found (to a degree of accuracy) the angular size, rotational speed, and rotational period of the Sun. Now we can put it all together to find the distance between the Earth and the Sun, a foundational unit of measurement in astronomy, also known as the Astronomical Unit or AU.

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The first step is to find the radius of the Sun from the information we found. We can use the formula: $$v = \frac{2 \pi R}{P}$$ to get $$R = \frac{v \cdot P }{2 \pi }$$ Plugging in our data gives: $$R = \frac{1.442 \times 10^5 \cdot 2.29 \times 10^6}{2 \pi } =  5.256 \times 10^{10} cm$$  

Now that we have the radius, we can set up some simple trigonometry to find AU.

$$tan(\theta/2) = \frac{R}{d}$$ We can use the small angle approximation, but only if our angle is in radians. Converting angular diameter to radians gives: $$\theta = .5750^{\circ} \pm 0.01192^{\circ} \times \frac{\pi}{180^{\circ}} =  1.004 \times 10^{-2} \pm 0.00208 \times 10^{-2} \: rad$$ So we can now isolate d and plug in what we know. $$d = \frac{2R}{\theta}$$ $$d = \frac{2 \times 5.256 \times 10^{10} }{1.004 \times 10^{-2} \pm 0.00208 }$$ $$AU =  1.047 \times 10^{13} \pm 5.05 \times 10^{15} cm$$  In reality, an AU is $1.496 \times 10^{13} cm$ giving us a 42% error (incidentally, the answer to life, the universe, and everything)

Nerdy references aside, 42% error is not amazing. On one level it is pretty cool that we can get a measurement for AU within the correct order of magnitude, but it also means we have significant sources of error. This can come from various steps in our lab:

• Step 1 - Angular Size: Our calculation of the angular size of the Sun was decently accurate, but not perfect. Error in this step could have come from slow reaction times when using the timer, ambiguity with the thickness of the line we used to mark the edge of the Sun. 
• Step 2 - Rotational Speed: When taking data, our images were not as clear as they could have been, which could be seen in the relatively thick bands imaged for the NaD lines. This can also be seen in the shift of the Telluric lines, which we tried to correct for.
• Step 3 - Rotational Period: There are a few possible sources of errors. First, we are averaging data that has already been averaged, each time cutting off significant figures and decreasing the accuracy of the result. Additionally, the data shows that the third measurement (the bottom sunspot) was significantly different from the other two, which indicates something may have gone wrong when taking the data. this could result from the tendency of sunspots to appear to group together at the edges of the Sun, making them hard to tell apart and track.
  
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• Finally, we are also assuming that the sunspots are rotating at exactly the speed of the Sun. While they are a good estimate, they are not going at exactly the same period of the Sun.

All done! Thanks again to my lab group.

Day Lab - Step 3: Determine the Rotational Period of the Sun

Now we have all the information needed except for the Sun's rotation around its own axis. Following Galileo's example, we can use sunspots as tracers of the Sun's rotation

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Unfortunately, whenever my lab group checked, there were no large, obvious sunspots to track. Instead, we used previously recorded video data (much like the gif above) to track sunspots over time. I tracked three sunspots from 2001:

The top sunspot traveled:

• 51 degrees in 3:12:48 days = 305280 seconds
• 53 degrees in 3:17:36 days = 322560 seconds
• The average is 52 degrees in $3.14 \times 10^5$ seconds (pi!), giving a period of: $$\frac{P}{360^{\circ}} =  \frac{seconds \: traveled }{degrees \: traveled}$$   $$P =  \frac{3.14 \times 10^5 \times 360^{\circ}}{52^{\circ}} = 2.17 \times 10^6 s$$

The middle sunspot traveled:

• 50 degrees in 3:12:48 days = 305280 seconds
• 40 degrees in 2:17:36 days = 236160 seconds
• The average is 45 degrees in $2.71 \times 10^5$ seconds, giving a period of: $$\frac{P}{360^{\circ}} =  \frac{seconds \: traveled }{degrees \: traveled}$$   $$P =  \frac{2.71 \times 10^5 \times 360^{\circ}}{45^{\circ}} = 2.17 \times 10^6 s$$

The bottom sunspot traveled:

• 83 degrees in 6:19:12 days = 587520 seconds
• 30 degrees in 2:11:12 days = 213120 seconds
• The average is 56.5 degrees in $4.00 \times 10^5$ seconds, giving a period of: $$\frac{P}{360^{\circ}} =  \frac{seconds \: traveled}{degrees \: traveled}$$   $$P =  \frac{4.00 \times 10^5 \times 360^{\circ}}{56.5^{\circ}} = 2.52 \times 10^6 s$$

The average is $2.29 \times 10^6 s$ or 26.5 days. Interestingly, the first two spots were very consistent and decently close to the actual rotational period of the Sun (24.47 days) and the bottom sunspot was significantly farther than the other two data points. Alas, we cannot throw out data just because we know it is off and get a more accurate answer. However, we can pay attention to this discrepancy as a potential source of error.

Due to a lab group of uneven numbers, I worked on this section alone.

Day Lab - Step 2: Determine the Rotational Speed of the Sun

Now we can use measurements of doppler shift to measure the radial velocity of the Sun. Because the Sun is rotating, part of it is rotating toward us and part of it is rotating away, causing the light we receive to be shifted. If we can account for this shift, we can find the rotational speed of the Sun.

However, we do not know where the axis of the Sun lies, so we took 8 measurements around the circumference of the sun and chose the two that were the most different: the bottom left and top right.

So what exactly were we measuring? We took spectral measures of NaD (sodium) lines by carefully aligning a sodium lamp to find the sodium lines on a spectrum and then filtering in sunlight and taking data on those same spectral lines. Numerically, the data is not very exciting:

However, If we graph it, we can see the two sodium lines and to the left of them, Telluric absorption lines which come from water in the atmosphere.

Next we isolate the bottom left line and top right line as discussed above. Zooming in to each of the NaD lines allows us to find the shift between each line by using a best fit curve.

 

However, our measurements are never perfect (it is really hard not to bump the table where the experiment is taking place). We can correct for errors by looking at the Telluric lines (they should not be shifted as they come from Earth's atmosphere).

The offset between each line (in this case 1.089 pixels) tells us how much we need to shift each of the NaD lines. Taking the shift into account, we can use the corrected shift between each NaD line (about 3 pixels) to find the velocity of the sun. (Actually, excel does this for us by converting the data to angstroms and using to formula $\frac{\delta v}{c} = \frac{\delta \lambda }{\lambda}$ where lambda is the wavelength of each NaD line). 

We get velocities of 1.473 km/s  and 1.411 km/s. Our average velocity is then:

$$\bar{v}= \frac{1.473 + 1.411}{2} = 1.442 \frac{km}{s} \times \frac{1 \times 10^5 cm}{1km} = 1.442 \times 10^5 \frac{cm}{s}$$ .

Thanks to my lab group: Richard, Daniel, Jonathan, Corey, and Sean and thanks to John and Allison for walking us through this process. 

Day Lab - Step 1: Determine the Angular Size of the Sun

From the lab: The sun appears to move all the way around the Earth (360 degrees) in 24 hours. If we measure how long (minutes and seconds) the Sun takes to move its own diameter along the sundial in our lab, we can measure its angular diameter, in degrees.

 
In our case, the "sundial" is created from focussing an image of the sun through a lens in a window onto an easel. To measure the angular diameter of the Sun, we marked the right edge of the Sun's image and timed how long it took for that image to cross that marking (basically, how long it takes the sun to traverse its diameter).

Our results were:

• 2:17.59 minutes  = 137.59 seconds
• 2:14.46 minutes = 134.46 seconds
• 2:21.45 minutes = 141.45 seconds
• 2:17.51 minutes = 137.51 seconds

This data set has a mean of 137.75 seconds and a standard deviation of 2.86 so average time is: $$\bar{t} = 137.75 \pm 2.86$$

So what can we do with this information? We can find the angular diameter of the sun because we know the sun traverses about 360 degrees in our sky each day. 

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We examined a fraction of that day and can therefore set up the proportion below: $$\frac{\theta}{\bar{t}} = \frac{360^{\circ}}{1 \: day}$$  Rearranging the equation and converting days to seconds gives: $$\theta = \frac{\bar{t} \times 360^{\circ}}{8.64 \times 10^4 s}$$ which with our result is:  $$\theta = \frac{137.75 \pm 2.86 \times 360^{\circ}}{8.64 \times 10^4 s}$$   $$\theta = \frac{137.75 \pm 2.86 \times 360^{\circ}}{8.64 \times 10^4 s}$$    $$\theta = .5750^{\circ} \pm 0.01192^{\circ}$$ The most accurate measurement of the angular size of the sun I could find online was .573 degrees. Our results did not quite match up, but they came close, only a 0.35% error.

Acknowledgements: I worked with Corey, Daniel, Richard, Sean, and Jonathan.

Fritz Zwicky - Dark Matter in the Coma Cluster

This week we will be walking through Fritz Zwicky's 1933 paper "Dark Matter in the Coma Cluster."

Coma Cluster

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The Coma Cluster, which is 321 million light years away from Earth, is home to over 1000 identified galaxies (1). In the middle shines two supergiant elliptical galaxies, on of which (NGC 4839) is pictured above. Together with the Leo Cluster, it forms the Coma Supercluster.

In 1933, Zwicky examined the Coma Cluster. Based on the assumption that the cluster has reached a stationary state, he uses the property $K= - \frac{1}{2}U$ from the virial theorem and applies it to the cluster. Yet before he can use this formula, important information must be set up. The cluster has a radius of $10^{24}$ cm with about 800 nebulae, each with an average mass of $10^9$ solar masses. Because the sun's mass is about $2 \times 10^{33}$ g he can find the mass of the cluster:

$$M  \approx M_{\odot} \times 10^9 \times Number_{nebulae}$$ $$M  \approx 2 \times 10^{33} \times 10^9 \times 800 \approx 1.6 \times 10^{45} g$$

Next, he sets up a formula we found this week and relates it to the virial theorem to find veloctity: $$U = -\frac{3GM^2}{5R}$$ $$K= \frac{1}{2}Mv^2= - \frac{1}{2} U =  \frac{3GM^2}{10R}$$ Solving for v gives: $$v = \sqrt{ \frac{3GM}{5R}} = \sqrt{ \frac{3 \times 6.7 \times 10^{-8} \times 1.6 \times 10^{45}}{5 \times 10^{24}}} = 8.0 \times 10^6 \frac{cm}{s}$$

However, there is something off about this calculated velocity, it is very different from the actual velocity. The density of the Coma Cluster would have to be at least 400 times greater than the density calculated based on the matter we can see. This must mean there is something we cannot see adding density to the cluster, which Zwicky names dark matter.

Yet before Zwicky concludes the presence of dark matter, he checks for any sources of error. First, he examines the fact that the Coma Cluster might not be in equilibrium, meaning that instead of the viriral theorem, he would have to use $$U= -K$$ but that only changes the answer by a factor of 2, not 400.

Next, he examines the possibility that the observed velocities are real, meaning the cluster if flying apart. However, there is not enough proof for this as all other data suggests nebulae do not reach those speeds.

Finally, Zwicky corrects for redshift to see if that could be a source of error. However, this only corresponds to a speed of 10 m/sec and is not large enough to account for our answer.

In conclusion, Zwicky discovered the existence of dark matter, one of the most exciting subjects in astronomy today!

Worksheet 9, Problem 2: Forming Stars

The problem: Giant molecular clouds occasionally collapse under their own gravity (their own “weight”) to form stars. This collapse is temporarily held at bay by the internal gas pressure of the cloud, which can be approximated as an ideal gas such that P “ nkT, where n is the number density (cm´3 ) of gas particles within a cloud of mass M comprising particles of mass $\bar{m}$ (mostly hydrogen molecules, H2), and k is the Boltzmann constant, k = 1.4 x 10^-16 erg K^-1 .

Part A: What is the total thermal energy, K, of all of the gas particles in a molecular cloud of total mass M? (HINT: a particle moving in the ith direction has $E_{thermal} = \frac{1}{ 2}mv_i^2  = \frac{1}{2}$ kT. This fact is a consequence of a useful result called the Equipartition Theorem.)

The problem is basically telling us: $$K= \sum E_{thermal} = \sum^i \frac{3}{2} kT$$ We can rewrite i as: $$i = \frac{M}{\bar{m}}$$ So $$K = \sum^{\frac{M}{\bar{m}}} \frac{3}{2} kT$$ This is really just $$K = \frac{3}{2} kT \frac{M}{\bar{m}}$$

Part B: What is the total gravitational binding energy of the cloud of mass M?

We found this in the last worksheet to be: $$U = - \frac{GM^2}{R}$$

Part C: Relate the total thermal energy to the binding energy using the Virial Theorem, recalling that you used something similar to kinetic energy to get the thermal energy earlier.

We know from the last worksheet that $$K = - \frac{1}{2} U$$ When combined with our answer from part A we get $$\frac{3}{2} kT \frac{M}{\bar{m}} = - \frac{1}{2} U$$ $$U= -3kT \frac{M}{\bar{m}}$$ To put this in an easier form for the next problem, we can combine our answer with the formula in part b to give us: $$\frac{GM^2}{R}= 3kT \frac{M}{\bar{m}}$$

Part D: If the cloud is stable, then the Viriral Theorem will hold. What happens when the gravitational binding energy is greater than the thermal (kinetic) energy of the cloud? Assume a cloud of constant density ρ.

If the gravitational binding energy is greater than the thermal energy of the cloud, the could will begin to collapse in on itself (mathematically, radius will decrease). This will cause the pressure, density, and temperature of the could to increase - the basic steps of star formation.

Part E: What is the critical mass, MJ , beyond which the cloud collapses? This is known as the “Jeans Mass.”

To solve, we can set up the inequality described in part D: $$\frac{GM^2}{R} > 3kT \frac{M}{\bar{m}}$$ Which can be simplified to $$M > \frac{3kTR}{G \bar{m}}$$ This would actually be our expression for the Jeans Mass $$M_J = \frac{3kTR}{G \bar{m}}$$ However, it might be more useful to have this expression in terms of density $$\rho = \frac{M}{\frac{4}{3} \pi R^3}$$ $$M_J = \frac{3kT}{G \bar{m}} \sqrt[3]{\frac{3M_J}{4 \pi \rho}}$$ This simplifies to $$M_J = \left( \frac{3kT}{G \bar{m}} \right)^{\frac{3}{2}} \left( \frac{3}{4 \pi \rho} \right)^{\frac{1}{2}}$$

Part F: What is the critical radius, RJ , that the cloud can have before it collapses? This is known as the “Jeans Length.”

Using the previous inequality $$M > \frac{3kTR}{G \bar{m}}$$ we can solve for R $$R > \frac{G M \bar{m}}{3kT}$$ $$R_J = \frac{G M \bar{m}}{3kT}$$ And once again we can find the answer in terms of density: $$M = \rho \frac{4}{3} \pi R^3$$   $$R_J = \frac{G  \left( \rho \frac{4}{3} \pi R^3 \right) \bar{m}}{3kT}$$ Which simplifies to   $$R = \sqrt{ \frac{9kT}{4 \pi G \rho \bar{m}}}$$

Acknowledgements: I worked with Barra and Dylan on this problem.

Worksheet 9, Problem 1: The Spatial Scale of Star Formation

The problem: The size of a modest star forming molecular cloud, like the Taurus region, is about 30 pc. The size of a typical star is, to an order of magnitude, the size of the Sun.

Part A: If you let the size of your body represent the size of the star forming complex, how big would the forming stars be? Can you come up with an analogy that would help a layperson understand this difference in scale? For example, if the cloud is the size of a human, then a star is the size of what?

Solve: To begin, we need to get everything on the same scale:

• Molecular Cloud Radius: $15 pc \frac{3 \times 10^8 cm}{1 pc} = 4.5 \times 10^{19} cm$
• Radius of the Sun: $7 \times 10^{10} cm$
• Radius of a person: $\approx 1 m = 100 cm$

Now we can set up a ratio: $$\frac{R_{\odot}}{R_{cloud}} = \frac{x}{R_{person}}$$ Solving for x gives: $$x = \frac{R_{\odot} R_{person}}{R_{cloud}} = \frac{7 \times 10^{10} \times 100}{4.5 \times 10^{19} } \approx 3 \times 10^{-8} cm$$

This is in the same order of magnitude of the radius of an atom! Imagine your whole body condensing into the size of an atom...

Part B: Within the Taurus complex there is roughly $3 \times 10^4 M_{\odot}$of gas. To order of magnitude, what is the average density of the region? What is the average density of a typical star (use the Sun as a model)? How many orders of magnitude difference is this? Consider the difference between lead (ρlead = 11.34 g cm^-3 ) and air (ρair =  0.0013 g cm^-3 ). This is four orders of magnitude, which is a huge difference!

What we know:

• $R_{\odot} = 7 \times 10^{10} cm$
• $R_T = 15 pc \frac{3 \times 10^8 cm}{1 pc} = 4.5 \times 10^{19} cm$
• Mass of the sun: $2 \times 10^{33} g$
• Mass of Taurus complex: $3 \times 10^4 M_{\odot} = 3 \times 10^4 2 \times 10^{33} = 6 \times 10^{37}$

Solve: To find density we know: $$\rho = \frac{M}{V} = \frac{M}{\frac{4}{3} \pi R^3}$$  

To find the density of the sun: $$\rho = \frac{M_{\odot}}{\frac{4}{3} \pi R_{\odot}^3}= \frac{2 \times 10^{33}}{\frac{4}{3} \pi (7 \times 10^{10})^3}  = 1.4 \frac{g}{cm^3}$$ And the density of the Taurus complex would be:   $$\rho = \frac{M_T}{\frac{4}{3} \pi R_T^3}= \frac{6 \times 10^{37}}{\frac{4}{3} \pi (4.5 \times 10^{19})^3}  = 1.6 \times 10^{-22}\frac{g}{cm^3}$$

This yields a difference of 22 orders of magnitude! I can't even imagine what those densities would look like.

Acknowledgements: I worked with Dylan and Barra on this worksheet.

Worksheet 8; Problem 5: Kelvin-Helmholtz timescale

The problem: The Virial Theorem states that half of a gravitationally-bound system’s potential energy goes into the kinetic energy of the system’s constituent particles. In the case of a cloud of gas, the cloud can shrink, but only if it loses energy by radiating. This occurs because as the radius shrinks, the potential energy becomes more negative, and therefore the particle motions must increase (higher kinetic energy). Particles moving faster leads to a higher gas temperature, and an increase in thermal emission.

We know that the Sun started from the gravitational collapse of a giant cloud of gas. Let’s hypothesize that the Sun is powered solely by this gravitational contraction, as was once posited by astronomers long ago. As it shrinks, its internal thermal energy increases, increasing its temperature and thereby causing it to radiate. How long would the Sun last if it was thermally radiating its current power output, $L_{\odot} = 4 \times 10^{33} erg \: s^{-1}$? This is known as the Kelvin-Helmholtz timescale. How does this timescale compare to the age of the oldest Moon rocks (about 4.5 billion years, also known as Gyr)?

Solve: 

What we know:

• $M_{\odot} = 2 \times 10^{33} g$
  
• $R_{\odot} = 7 \times 10^{10} cm$
• $L_{\odot} = 4 \times 10^{33} erg \: s^{-1}$
• $G = 6.67 \times 10^{-8} \frac{cm^3}{gs^2}$

From the information provided in the problem, we can see that the units for luminosity are ergs per second, from this we can conclude that $$Luminosity\: \times time \: = \: energy$$ We also know from the virial theorem that $$K = \ \frac{1}{2}U = - \frac{GM^2}{2R}$$ Combining these two equations gives $$L \times t = \frac{GM^2}{R}$$ Solving for t gives: $$t = \frac{GM_{\odot}^2}{R_{\odot}L_{\odot}}$$ Now we can plug in the values we know.   $$t = \frac{6.67 \times 10^{-8} \times (2 \times 10^{33} )^2}{ 7 \times 10^{10} \times4 \times 10^{33}} \approx 9.5 \times 10^{14} s \: or \: 3 \times 10^7 years$$ In comparison, the age of the oldest moon rocks is $4.5 \times 10^9 years$ which is about 300 times older than our estimate, which makes me grateful the sun is not solely powered by gravitational contraction.

Acknowledgements: I worked with Barra in this problem.

Worksheet 8, Problem 2: Revisiting Derivations

Introduction: This week we are learning about star formation, which leads to some of the most beautiful images in astronomy.

Source

From the worksheet:
One of the most useful equations in astronomy is an extremely simple relationship known as the Virial Theorem. It can be used to derive Kepler’s Third Law, measure the mass of a cluster of stars, or the temperature and brightness of a newly-formed planet. The Virial Theorem applies to a system of particles held together by a force that varies according to the inverse central-force law $F \alpha \frac{1}{r^{\alpha}}$

The problem: Consider a spherical distribution of particles, each with a mass mi and a total (collective) mass $\sum_{i}^{N} m_i = M$, and a total (collective) radius R. Convince yourself that the total potential energy, U, is approximately $$U \approx - \frac{GM^2}{R}$$ You can derive or look up the actual numerical constant out front. But in general in astronomy, you don’t need this prefactor, which is of order unity.

Solve: We can think about this sphere of particles with a uniform density in terms of thin shells, each with a width of dR.

We also know that $$U = - \frac{GMm}{R}$$ We can treat the entire sphere as our mass, M, and the shell as our mass, m. Then, we can rewrite each mass in terms of density: $$\rho = \frac{M}{V}$$ Where $V = \frac{4}{3} \pi R^3$ So $$\rho = \frac{M}{\frac{4}{3} \pi R^3}$$ And we can rewrite the equation for potential energy of one shell as: $$dU = - \frac{GMdm}{R} = - \frac{G(\frac{4}{3} \pi R^3 \rho) (4 \pi R^2 \rho) dR}{R}= - G \frac{16}{3} \pi^2 R^4 \rho^2 dR$$ But now we have dU in terms of dR, so we can integrate to get U: $$\int_{0}^{R} - \frac{16}{3} G \pi^2 R^4 \rho^2 dR =  - \frac{16}{3} G \pi^2 \rho^2  \int_{0}^{R} R^4 dR = - \frac{16}{3} \frac{R^5}{5} G \pi^2 \rho^2 - 0$$ But this still is not the answer we want. We need U in terms of M, not density so we can reuse   $$\rho = \frac{M}{ \frac{4}{3} \pi R^3 }$$ to get: $$U = - \frac{16}{3} \frac{R^5}{5}\left( \frac{M}{\frac{4}{3} \pi R^3 } \right)^2 G \pi^2$$ Amazingly enough, this simplifies to the much nicer expression: $$U = - \frac{4GM^2}{5R}$$ The actual answer is $$U = - \frac{3GM^2}{5R}$$ So something must have gone wrong with our constants, but either way both simplify to our original expression: $$U \approx - \frac{GM^2}{R}$$

Acknowledgements: I worked with Barra and April on this worksheet.