Worksheet 11.2, Problem 1: Radiative Energy Transport

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 the 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. 

a) Inside the star, consider a mass shell of width \(\Delta r\), at a radius \(r\). This mass shell has an energy density \(u+\Delta u\), and the next mass shell out (at a radius \(r+\Delta r\)) 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 \(\Delta r\). Use this to derive an expression for \(L(r)\) in terms of \(\frac{du}{dr}\), the energy density profile. This is the diffusion equation describing the outward flow of energy. 

Mass Shell

We need to write an expression for the outward flow of energy, L(r), which we are told is equal to the total excess energy in the inner shell divided by the time to cross the shell's width. To find the excess energy, we need to multiply the energy density of the shell by the shell's volume:

\(\Delta E=V_{shell}[u-(u +\Delta u)]=-4\pi r^2\Delta r\Delta u\)

And to find the time to cross the shell's width, \(\Delta t\), we can use the \(d=rt\) equation:

\(\Delta t=\dfrac{\Delta r}{\overrightarrow{v}_{diff}}\)

And \(\overrightarrow{v}_{diff}\) is the diffusion velocity, which we solved for in the previous problem to get:

\(\overrightarrow{v}_{diff}=\dfrac{c}{\kappa \rho (r) \Delta r}\)

Where \(c\) is the speed of light, \(\kappa\) is the absorption coefficient, and \(\rho (r)\) is the mass density. So we get a time of:

\(\Delta t=\dfrac{\Delta r^2\kappa \rho (r)}{c}\)

Now writing out expression for the outward flow of energy, \(L(r)\), we get:

\(L(r)=-\dfrac{4\pi r^2 \Delta u c}{\Delta r \kappa\rho(r)}\)

Since the shell is infinitesimally small, we can write:

\(L(r)=-\dfrac{du}{dr}\dfrac{4\pi r^2 c}{\kappa\rho(r)}\)

We have now solved for the diffusion equation, which describes the outward flow of energy from a star.

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:

\(\dfrac{dT(r)}{dr}\propto -\dfrac{L(r)\kappa \rho (r)}{\pi r^2 acT^3}\)

where \(a\) is the radiation constant.

Plugging our equation for \(u(T(r))\) into the diffusion equation, we get:

\(L(r)=-\dfrac{du(T(r))}{dr}\dfrac{4\pi r^2 c}{\kappa\rho(r)}\)

         \( =-\dfrac{d}{dr}aT(r)^4\dfrac{4\pi r^2 c}{\kappa\rho(r)}\)

                  \( =-4aT(r)^3\dfrac{dT(r)}{dr}\dfrac{4\pi r^2 c}{\kappa\rho(r)}\)

Now, solving for \(T(r)\), we get:

  \(\dfrac{dT(r)}{dr}=-\dfrac{L(r)\kappa\rho(r)}{16\pi r^2acT^3}\)

So we have the relation: 

\(\dfrac{dT(r)}{dr}\propto -\dfrac{L(r)\kappa\rho(r)}{\pi r^2acT^3}\)

And now we have derived the equation for radiative energy transport.

I would like to thank Charles Law and Daniel Chen for their help in solving this problem. 

Worksheet 11.1, Problem 1: Photons Random Walking Out of a Star

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 Astronomy 16 students studying on a nice Spring day.

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 one photon take to travel a distance \(\Delta r\)?

Hint: If each step is a vector \(\overrightarrow{r_i}\), then \(\overrightarrow{D}=\sum\overrightarrow{r_i}\). However, this displacement is zero for a large number of random-walk journeys. Instead, calculate \(\overrightarrow{D}^2\) and take the square root of the result to be \(\Delta r=(\overrightarrow{D}^2)^{1/2}\). This is the root-mean-squared displacement, which is a scalar rather than vector quantity. Not that 'multiplying' two vectors isn't as simple as multiplying two scalars. You must take the dot product, which looks like \(\overrightarrow{A}\cdot \overrightarrow{B}=AB \cos{\theta}\), where \(\theta\) is the angle between the two vectors.

A random walk is a path consisting of a series of steps or movements in random directions. The motion of photons in stars as they travel from the star's center to its surface is modeled by a random walk. As they move through the star, they collide into other particles and electrons and bounce off, colliding into more particles. This process causes the random walk path that the photon travels. The image below shows an example of what the random walk of a photon in a star might look like.

Random walk of a photon through a star.
Image Credit:
http://www.atnf.csiro.au/outreach//education/senior/astrophysics/images/stellarevolution/rndmwlkphotonsm.jpg

For this problem, we want to find how many steps the photon has to take on average to travel a distance \(\Delta r\). The image below shows this set up, where each line is a vector \(\vec{r}_i\), with average magnitude l, and a net displacement \(\vec{D}=\sum \vec{r}_i\). 

A random walk, where each step has an average length, l, and a net displacement vector, \(\overrightarrow{D}\).
Image Credit: http://www2.ess.ucla.edu/~jewitt/images/random.gif

The distance \(\Delta r\) is the root-mean-squared displacement, given by: 

\(\Delta r=(\vec{D}^2)^{1/2}\)

First, calculating \(\vec{D}^2\), we must take the dot product, since \(\vec{D}\) is a vector:

\(\vec{D}\cdot\vec{D} = \left|\vec{D}\right|^2 = \left|\displaystyle\sum_{i=1}^N\vec{r}_i\right|^2 = \displaystyle\sum_{i=1}^N\left|\vec{r}_i\right|\left|\vec{r}_i\right|\cos{(0)} + \displaystyle\sum_{i,j}\left|\vec{r}_i\right|\left|\vec{r}_j\right|\cos{\theta}\)

Where N is the number of steps taken by the photon to travel the distance \(\Delta r\), and \(\theta\) is the angle between vectors \(\vec{r}_i\) and \(\vec{r}_j\). Since the vectors are pointing in random directions, we can take the average of \(\cos{\theta}\), which is 0. So the second term will be zero, leaving us with the term:

\(\left|\vec{D}\right|^2 = \displaystyle\sum_{i=1}^N\left|\vec{r}_i\right|\left|\vec{r}_i\right| = \left|\vec{r}_1\right|^2 + \left| \vec{r}_2\right|^2 + \ldots + \left|\vec{r}_N\right|^2\)

We already know that the average magnitude of each vector is l, so this gives us:

\(|\vec{D}|^2 =Nl^2\)

We also know that \(\Delta r=(\vec{D}^2)^{1/2}\), so we can rewrite our expression and solve for N:

\(\Delta r^2 =Nl^2\)

\(N=\left(\dfrac{\Delta r}{l}\right)^2\)

b) What is the photon's average velocity over the total displacement after many steps? Call this \(\overrightarrow{v}_{diff}\), the diffusion velocity. 

To solve for the diffusion velocity, we can use the basic equation, \(d=rt\). First, we must find the time it takes to travel all N steps:

\(t=\dfrac{Nl}{c}\)

Where c is the speed of light (the speed of each photon). Then, knowing that the photon has a displacement of \(\Delta r\), we can solve for the diffusion velocity:

\(\vec{v}_{diff}=\dfrac{\Delta r}{t}\)

From part (a), we found that \(\Delta r =\sqrt{N}l\). Plugging this in, we find:

\(\vec{v}_{diff}=\dfrac{\sqrt{N}lc}{Nl}=\dfrac{c}{\sqrt{N}}\)

Which can also be rewritten as:

\(\vec{v}_{diff}=\dfrac{lc}{\Delta r}\)

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 \(\sigma\). Give an analytic expression for "mean free path" relating these parameters. 

The image below shows a volume filled with targets, each with cross-sectional area \(\sigma\),  and with number density n, through which a photon travels.  

Image Credit: http://press.princeton.edu/chapters/s03_8457.pdf

The projected number of targets per unit area in the path of the photon will be ndx, and the fraction of the area that is covered by the targets is \(\sigma ndx\) (since each target has cross-sectional area \(sigma\)). So this means that the number of targets intersected by a path through the volume will be \(n\sigma dx\). Since the photons are limited to one interaction, we can set this equal to one. Here, dx will give us the typical distance a photon will travel between interactions, also known as the mean free path, l. Solving for this value, we get:

\(l=\dfrac{1}{n\sigma}\)

We can see that the units work for this: n is the number of targets per volume, so it has units of \(\text{cm}^{-3}\), and \(\sigma\) has units of area, so \(n\sigma\) has units of \(\text{cm}^{-1}\), giving l units of centemeters, which is what we wanted. 

d) The mean free path, l, can also be related to the mass density of absorbers \(\rho\), and the "absorption coefficient" \(\kappa\) (cross-sectional area of absorbers per unit mass). How is \(\kappa\) related to \(\sigma\)? Express \(\overrightarrow{v}_{diff}\) in terms of \(\kappa\) and \(\rho\) using dimensional analysis. 

The absorption coefficient is given to be the cross-sectional area of the absorbers (or targets) per unit mass, so it is given by:

\(\kappa =\dfrac{\sigma}{\overline{m}}\)

Which can be rewritten as:

\(\sigma = \kappa\overline{m}\)

Where \(\overline{m}\) is the typical mass of an absorber. \(\kappa\) has units of \(\frac{\text{cm}^2}{g}\). And thus we have our relation between \(\kappa\) and \(\sigma\). 

We also know that \(n=\frac{\rho}{\overline{m}}\). From parts (b) and (c) of this problem, we found that \(\vec{v}_{diff}=\dfrac{lc}{\Delta r}\) and \(l=\dfrac{1}{n\sigma}\). We can write these expressions in terms of \(\kappa\) and \(\rho\) to get:

\(l=\dfrac{1}{\rho\kappa}\)

Plugging this into our expression for the diffuse velocity, we get:

\(\vec{v}_{diff}=\dfrac{c}{\rho\kappa\Delta r}\)

 Plugging these relationships into the expression for the diffusion velocity, we get:
\(\vec{v}_{diff}=\dfrac{lc}{\Delta r}\)

Checking this for dimensional analysis:

\([l]=\)m
\([c]=\)m/s
\(\Delta r=\)m

So the dimensions simplify to be m/s, which is what we wanted.

We have now developed the tools to attack the problem of radiative diffusion. 

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 \(\sigma_T =7\times 10^{-25}\text{ cm}^2\) 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, \(\rho\), set equal to the mean Solar density. 

Note: 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 \(\lambda\) 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.

Here, the photons are undergoing Thomson scattering, which is when photons collide and are scattered by free, ionized elections. Since this is a low-energy process, it is only appropriate if the electrons aren't moving too fast, which is the case in the Sun. Also, since we assume that the Sun's interior is only made up of pure hydrogen, we can say that \(\overline{m}\approx m_p\) because hydrogen has only one electron and one proton.

To find the diffusion timescale for a photon, we can use the basic \(d=rt\) equation again:

\(t_{diff}=\dfrac{\Delta r}{\vec{v}_{diff}}\)

Plugging the expressions we solved for above, we get:

\(t_{diff}=\dfrac{\kappa\rho\Delta r^2}{c}\)

In this context, we know that \(\kappa =\frac{\sigma _T}{m_p}\). Putting this into our equation:

\(t_{diff}=\dfrac{\sigma_T\rho\Delta r^2}{m_pc}\)

Now, we can plug in values. We know that \(\sigma_T =7\times 10^{-25}\text{ cm}^2\), \(\rho =1.14\text{ g cm}^{-3}\), \(\Delta r= R_\odot =6.96\times 10^{10}\text{ cm}\), \(m_p= 1.67\times 10^{-24}\text{ g}\), and \(c=3\times 10^{10}\text{ m s}^{-1}\). 

\(t_{diff}=\dfrac{(7\times 10^{-25}\text{ cm}^2)(1.14\text{ g cm}^{-3})(6.96\times 10^{10}\text{ cm})^2}{(1.67\times 10^{-24}\text{ g})(3\times 10^{10}\text{ m s}^{-1})}\approx 9.54\times 10^{10}\text{ s}\approx 3000\) years

So it would take a photon approximately 3000 years to travel from the center to the surface of the Sun.

I would like to thank Charles Law, Daniel Chen, and Andy Mayo for their help in solving this problem. 

Citations:
http://en.wikipedia.org/wiki/Random_walk
http://press.princeton.edu/chapters/s03_8457.pdf

Free Form: Music and the Voice

As a singer in the Radcliffe Choral Society, I thought it might be interesting to discuss the physics of music and the voice. We have already discussed light waves and light interference to some extent here. Sound waves work in a very similar manner, though they are different from light waves through their manner of propagation. While light waves are electromagnetic waves, sound waves are mechanical waves, meaning that they propagate via pressure and displacement through mediums like air or water. The volume of the sound is proportional to the amplitude of the sound wave, and the pitch is dependent on the frequency or the wavelength of the wave. Higher pitches correspond to higher frequencies and lower pitches to lower frequencies.

One thing that is interesting to consider is how harmonies work. The octave of a note corresponds to half or double the frequency of that note. In this case, the nodes of the sound waves lines up, as shown in the image below:

Image Credit: http://www.musicez.com/images/octave.gif

We say that octaves have a frequency ratio of 2:1. Other common music intervals include the major third and the perfect fifth, which correspond to frequency ratios of 5:4 and 3:2 respectively. Interestingly, your ear can feel differences in frequency if there is a lot of dissonance between two notes. Dissonance can occur when two notes are off by a small interval and their nodes don't match up. Dissonant sounds are often described as 'grating' or 'unstable.' You can hear the pitch waver due to the misalignment of the phases of the sound waves. If you click here, you will find an application for a virtual piano where you can create chords. You can play around with it and see if you can figure out what note combinations make dissonant chords.

Having explained a little bit about how sound and music works, we can take a look at how the voice creates sound. To make sound, your body begins with the lungs and the diaphragm, where it pushes air up to the vocal folds, located within the larynx, with enough pressure to cause the folds to vibrate. The vibrations of the vocal folds cause them to open and close, chopping up the air that is pushed through them into little pulses of compressed air. These pulses of air are what make up the sound that we hear. The image below shows how the vocal folds work.

Vocal folds releasing pulses of compressed air.
Image Credit: http://hyperphysics.phy-astr.gsu.edu/hbase/music/voice.html

Needless to say, the loudness of the sound is controlled by the amount of air pushed up by the lungs and diaphragm. A greater amount of air means more compressed air pulses, or in other words, larger amplitude of sound waves. We are able to adjust the pitch of the sound we make by using muscles in our larynx to stretch our vocal folds tighter or less tight, creating vibrations with higher or lower frequencies, and therefore, higher or lower pitches. The reason men have lower voices than females in general is because men have larger vocal folds. Men's vocal folds range from 17mm to 25mm in length, while women's vocal folds range from 12.5mm to 17.5mm. This causes the vibrations in the vocal folds of men to have lower frequencies than those of women.


Citations:
http://en.wikipedia.org/wiki/Sound
http://en.wikipedia.org/wiki/Music_and_mathematics
http://en.wikipedia.org/wiki/Interval_ratio
http://en.wikipedia.org/wiki/Consonance_and_dissonance
http://hyperphysics.phy-astr.gsu.edu/hbase/music/voice.html
http://en.wikipedia.org/wiki/Human_voice

AU Daytime Lab, Post 4: Measuring the Astronomical Unit

This is the final post regarding this lab. Now, we have found the angular size of the Sun, its rotational speed, and its rotational period. We have everything we need to calculate the distance to the Sun. 

Procedure:

First, we need to find the solar radius, which can be done using basic trigonometry.

Here, \(\alpha\) is half of the angular size of the Sun. We can solve for the radius, \(R_\odot\), by using the equation, distance equals rate times time, where the period is time, rotational speed is the rate, and the circumference is the distance.

\(2\pi R_\odot=v_\odot P_\odot\)

Solving for the radius:

\(R_\odot = \dfrac{v_\odot P_\odot}{2\pi}\)

We also know that:

\(\tan{\alpha}=\dfrac{R_\odot}{1\text{ AU}}\)

Plugging in our equation for \(R_\odot\) and setting \(\alpha\) equal to half of the angular size, \(\theta\), we get:

\(\tan{\frac{\theta}{2}}=\dfrac{\frac{v_\odot P_\odot}{2\pi}}{1\text{ AU}}\)

Now solving for the AU:

\(1\text{ AU}=\dfrac{v_\odot P_\odot}{2\pi\tan{\frac{\theta}{2}}}\)

Using the small angle approximation, \(\tan{\theta}\approx \theta\), we get:

\(1\text{ AU}=\dfrac{v_\odot P_\odot}{\pi\theta}\)

We can now plug in the values we solved for in the three previous posts:

\(\theta = 9.59\times 10^{-3} \pm 1.53\times 10^{-4}\) radians

\(v_\odot= 1.31\pm 0.04\) km/s

\(P_\odot = 27.6\pm 0.98\text{ days}= 2.38\times 10^6 \pm 8.47\times 10^4\) seconds

So we get a final answer of:

\(1\text{ AU}=\dfrac{(1.31\text{ km/s})( 2.38\times 10^6\text{ s})}{\pi \text{ }9.59\times 10^{-3}}\)

\(1\text{ AU}= 1.03\times 10^8\text{ km}\)

Error Analysis and Discussion:

Now we have to account for the error propagation due to the error in our variables that we plugged in.

When multiplying quantities X, Y, Z, with errors \(\delta X\), \(\delta Y\), \(\delta Z\):

\(R=\dfrac{X\cdot Y}{Z}\)

The error propagates in the following form:

\(\delta R= |R| \sqrt{\left(\dfrac{\delta X}{X}\right)^2+\left(\dfrac{\delta Y}{Y}\right)^2+\left(\dfrac{\delta Z}{Z}\right)^2}\)

Plugging in the variables, we get:

\(\delta R= (1.03\times 10^8\text{ km}) \sqrt{\left(\dfrac{0.04\text{ km/s}}{1.31\text{ km/s}}\right)^2+\left(\dfrac{0.98\text{ days}}{27.6\text{ days}}\right)^2+\left(\dfrac{1.53\times 10^{-4}\text{ rad}}{9.59\times 10^{-3}\text{ rad}}\right)^2}\)

\(\delta R= 5.10\times 10^6\) km

So we have a final answer of:

\(1\text{ AU}=1.03\times 10^8\pm 5.10\times 10^6\text{ km}\)

             \(=1.03\times 10^{13}\pm 5.10\times 10^{11}\text{ cm}\)

The Astronomical Unit is a value that has been very accurately measured. According to Google, \(1\text{ AU}=1.496\times 10^8\) km. So our answer has a percent error of about 30%. This is relatively high, but it is to be expected due to our large percent error in finding the rotational speed. Recall that there are several possible factors that might have contributed to this error. There was potential for human error in both the calculation of the angular size of the sun and the rotational period of the Sun. Additionally, the sensitivity of the spectrograph might have created error if it was bumped or shifted in any way. Moreover, the large shift in the Telluric line, where there should have been little or none, might be indication of a large uncertainty in our data.  It is likely that most of the error in our measurement is a result of the uncertainties in our calculations of rotational speed. Future experiments should work to improve this second step in the lab.

Conclusion: 

Now we have successfully measure the Astronomical Unit! The Astronomical Unit is generally used to describe stellar system scale distances. It is also very important in regards to measuring other, larger distances, since the AU is a basis in the cosmic distance ladder, as shown in the image below.

Cosmic Distance Ladder

Also, the parsec, which is often used to describe bigger distance scales, is defined by the AU. Furthermore, the AU was used to calibrate other methods for finding distances to objects in the universe. This means that if there were an error in our measurement of the AU, our estimations for many other distance scales would be off too, and we would have a false understanding of some of our universe. Fortunately, we know the distance to the Sun to a very small uncertainty of 0.000,000,003. Since the mid 1600s, Huygens already had a very good estimate for the Astronomical Unit, and by 1895, we had a fairly accurate measurement by Simon Newcomb. From this lab, you can see that solving for the AU is something that can be done using only simple geometry and a few tools from an astronomy laboratory.


I would like to thank Charles Law and the TFs Allyson and Andrew for their help in this lab.

Citations:
http://en.wikipedia.org/wiki/Astronomical_unit
http://en.wikipedia.org/wiki/Propagation_of_uncertainty

AU Daytime Lab, Post 3: Determining the Rotational Period of the Sun

This is the third post in our series of four on the measurement of the AU. In this post, we will calculate the rotational period of the Sun based on observations of the motion of sunspots across the Sun as the Sun is rotating.

Sunspots are temporary, visibly darker spots on the photosphere of the Sun. They occur in places with concentrated magnetic fields, causing convection, and thus resulting in reduced surface temperature. Sunspots have temperatures of about 3,000 to 4,500 K while surrounding areas have temperatures of 5,780 K. This large difference in temperature allows the sunspots to be distinctly visible as darker spots. In the image below, clearly see the sunspots. 

Image Credit:
http://science.nasa.gov/media/medialibrary/2008/09/30/30sep_blankyear_resources/midi512_blank_2001.gif

Since the sunspots are near the surface of the Sun, they move with the Sun's rotation, and can be used to find the rotational period of the Sun.

Equipment:

Unfortunately, due to a series of cloudy days, we were unable to directly observe the sunspots from the heliostat, so instead we used previous data images from the Solar and Heliospheric Observatory (SOHO). On the Observatory's website, there is an application with a slider bar that can slide through the images of the Sun over time. With the application open on the computer screen, we placed a grid with the lines of latitude and longitude that was printed on transparency paper over the image of the Sun on the screen. The image of the Sun was scaled to fit the grid. The screen with the image of the Sun and the grid placed over it would look something like the following: 

Image Credit:
http://sohowww.nascom.nasa.gov/classroom/docs/Spotexerweb.pdf

Procedure:

On the slider bar application, I chose a sunspot towards the left of the screen and recorded its latitude and longitude using the grid. Then, following this same sunspot, I used the slider bar to track its motion, recording its position at two times as it moved across the Sun. On the grid that we used, each line represented 10 degrees of angular displacement. I repeated this process for six different sunspots. The image below gives an example of what this progression of sunspots looks like.

The sunspots move across the sun in a relatively short period of time.
Image Credit: http://sohowww.nascom.nasa.gov/classroom/docs/Spotexerweb.pdf

Results and Analysis:

The following table shows the data collected for each sunspot.

To solve for the rotational period, we can create a ratio of the angular distance traveled to the time taken to travel this distance:

\(\dfrac{360^\circ}{P_\odot}=\dfrac{\Delta \theta}{\Delta t}\)

Where \(P_\odot\) is the rotational period of the Sun, \(\Delta\theta\) is the change in longitude, and \(\Delta t\) is the time to travel this longitude. Solving for the rotational period, we get:

\(P_\odot =\dfrac{360^\circ t}{\Delta\theta}\)

Now, using each data point, we can calculate the rotational period of the Sun:

So we have an average rotational period of 27.6 days.

Error Analysis and Discussion:

From Wikipedia, the rotational period is about 24.47 days, so we are off by 3.13 days, which is a percent error of 11%. Possible factors that contributed to this error include inexact measurements of the location of sunspots on the grid. Also, the latitude affects the rotational period. The rotational period at the poles is larger than the rotational period at the equator. 

Calculating the standard error of the rotational periods using the methods described in the first blog post of this series, we get a sample standard deviation of \(s=2.4\), and a standard error of:

\(SE_{\overline{x}}=\dfrac{2.4}{\sqrt{6}}\)

\(SE_{\overline{x}}=0.98\) days

I would like to thank Charles Law and the TFs, Allyson and Andrew for their help in the lab.

Citations:

http://sohowww.nascom.nasa.gov/

http://sohowww.nascom.nasa.gov/classroom/docs/Spotexerweb.pdf

http://en.wikipedia.org/wiki/Sunspot

http://en.wikipedia.org/wiki/Solar_rotation

http://en.wikipedia.org/wiki/Sun

AU Daytime Lab, Post 2: Determining the Rotational Speed of the Sun

This is the second post of the series of four posts on this lab. In this post, we discuss how to find the rotational speed of the Sun using Doppler shifts due to the Sun's rotation. A Doppler shift is a change in the frequency of a wave due to motion of the emitting source relative to the observer. If the emitting source is moving towards the observer the frequency appears to be larger, and if the source is moving away from the observer, it appears to be smaller. This occurs because as the source moves with respect to the observer, the crests of each wave are emitted at a closer or further position from the observer than the previous one depending on the direction of the source's motion. This causes the second crest to reach the observer either before or after it would have had the source been stationary, thus giving the appearance of a different frequency than the one emitted. In astronomical terms with regards to light, an increase in frequency is called a blueshift and a decrease is called a redshift.

In the case of the Sun, since it is rotating, one side of the Sun is moving towards us and the other side is moving away from us, causing a redshift in the Sun's spectrum on one side, and a blueshift on the other, as shown in the image below.

Image Credit:
http://a-levelphysicstutor.com/images/waves/dopp-redshift03.jpg

If we find the Doppler shift in the spectrum of the Sun's light on either side of the Sun, we can find the relative speed of either side of Sun, which is the rotational speed of the Sun.

The Sun emits light at many wavelengths, however, so we must choose a specific region to observe. The NaD absorption lines in the Sun's spectrum are well known and are caused by sodium. They occur at 5889 and 5896 Angstroms. Using a sodium lamp, we can see where the sodium lines are without a Doppler shift, and use this knowledge to calibrate the spectrograph. Then using the Telluric absorption lines which won't be Doppler shifted since they are caused by the \(\text{H}_2\text{O}\) in the Earth's atmosphere, we can align the spectrum from the Sun to the non-Doppler shifted sodium spectrum and compare their positions to find the rotational speed of the Sun.

Equipment:

A spectrograph was used to measure the spectrum of the Doppler shifted sodium absorption lines and the Telluric lines. The spectrograph separates the light into a frequency spectrum which is then recorded by a CCD camera.

Spectrograph
Image Credit: http://www.fas.harvard.edu/~astrolab/astrolab.html#heliostat

We used the Heliostat to reflect the sunlight into laboratory and into the spectrograph. The mirrors had to be aligned correctly so that the reflected light was positioned into the spectrograph and then the Heliostat was set to follow the Sun's motion through the sky. 

Heliostat
Image Credit: http://www.fas.harvard.edu/~astrolab/astrolab.html#heliostat

As already mentioned, a sodium lamp was used to calibrate the spectrograph and to have a comparison for the Doppler shifted spectrum of the Sun. Finally, the softwares, MaximDL and Excel were used to perform the analysis of the data. 

Procedure:

First, for calibration, we placed the sodium lamp in from of the slit of the spectrograph and adjusted the slit width until we had a non-saturated, sharp image from the CCD camera. The camera itself was also moved so that the spectral lines were centered in the image. The quality of the image could be tested through short exposures on MaximDL until the desired image was found. It is important to note that no sunlight should get through the spectrograph. 

After calibration, the sodium lamp was removed and the sunlight from the Heliostat was directed into the slit. A test image was taken to make sure everything was still aligned correctly and adjustments were made if necessary. 

Since we don't know the direction of the Sun's rotation, we had to take measurements at four pairs of points: top, bottom; left, right; top left, bottom right; bottom left, top right. The pair of points with the largest difference in spectral lines would be the ones closest to the line of the equator, and therefore the direction of rotation. 

Solar regions that were observed

Next, the spectral images taken by the CCD camera were opened on MaximDL and a narrower box (about 10 pixels wide) was drawn across each of the eight images spanning the entire width of the image (from left to right). The data acquired from MaximDL through this process was then exported and saved for each image. 

Using an Excel Template that was provided, the exported data for each of the eight images was transferred into their appropriate columns in the "Raw Data" sheet. On the "Normalized Plots" sheet, the data was plotted, and we could see the shifts between the pairs of points. The pair that had the largest separation was selected, and in this case, we found that the left and right points had the largest difference. The corresponding columns for the pair with the largest shift was then copied and pasted into the columns on the "Shift Data" sheet and larger plots of these two columns were made on the "Analysis" sheet. A second order polynomial was then fit to the plots and the range of the x-values was adjusted so that the polynomial and data points matched up as well as possible. The same thing was done for the plot of the Telluric lines and any shifts in the Telluric lines were accounted for in the shifts of the NaD lines. Additionally, a conversion factor for pixels to Angstroms was found using Excel. Finally, using the Doppler equation, we were able to solve for the difference in velocities on either side, and this difference was divided by two in order to account for the different frames of reference. And thus the solar rotational speed was found.

Results and Analysis:

The following graph is what was shown in the "Normalized Plots" sheet. It is the plot of the spectra of each image, where the x-axis is the pixel count, and the y-axis is intensity.

From this plot, we can see that the red lines have the largest difference, corresponding to the left and right images of the Sun. The two largest dips are spectral absorption lines associated with NaD. We chose to perform our analysis on the left-most absorption line. The following image is a close up of the left absorption line.

Left Absorption Line

Here, we can see more clearly that the black lines, corresponding to the bottom and top readings, are the most separated and that the red lines corresponding to the left and right readings are the least separated. So we know that the Sun is rotating in the top-bottom direction.

Now we can do an closer analysis of both absorption lines in the NaD spectrum for the top and bottom images. The individual plots of the NaD absorption lines are shown below.

Left absorption line with corresponding 2nd order polynomials with a shift of 4.91 pixels

Right absorption line with corresponding 2nd order polynomials with a shift of 4.78 pixels

For calibration, we have to look at the shift in the Telluric lines, which come from the water in Earth's atmosphere, so that we can account for it in our calculation of the Doppler shifts. There should be no shift in the Telluric lines, however we do observe one. This is not caused by a Doppler shift since they Earth's atmosphere is not moving with respect to us. A possible explanation is that some of the light is being diffracted as it passes through the atmosphere, before reaching us.

Telluric line with corresponding 2nd order polynomials with a 1.99 pixel shift

Finally, we need to find the conversion factor from pixels to Angstroms before we can continue our calculations. We know that the two sodium absorption lines should be separated by 5.97 Angstroms, and Excel shows that there are 332 pixels separating the absorption lines. So, on Excel, we calculated that there is a conversion factor of 0.018 Angstroms per pixel.

Now, to do the calculations, we first need to subtract the shift in the Telluric line from the shifts in the left and right NaD lines to account for the effects of the atmosphere. So we get shift of 2.91 on the left NaD line and 2.79 on the right NaD line. Then we convert these from pixels into angstroms to get shifts of 0.052 Angstroms on the left and 0.05 Angstroms on the right.

We are ready to apply the Doppler equation to solve for velocities:

\(\dfrac{\Delta v}{c}=\dfrac{\Delta \lambda}{\lambda}\)

Where c is the speed of light, \(\Delta v\) is the difference in velocity of the two sides of the Sun. \(\Delta\lambda\) is the difference in wavelengths between the two lines, and \(\lambda\) is the emitted wavelength. We want to find half of \(\Delta v\), which will give us the rotational speed:

\(v_\odot =\dfrac{\Delta v}{2}\)

So we get:

So we have an average solar rotational speed of 1.31 km/s. 

Error Analysis and Discussion:

According to Wikipedia, the rotational velocity is 2 km/s. So our answer is off by about 0.7 km/s, which is a percent error of about 35%. There are several possible explanations for this large error. First of all, any bumping of the equipment or small mistakes in analysis could cause an error. Additionally, the sample size was small, and perhaps if we had done more trials, we would have had better results. We might have also been incorrect in our assumption that the top and bottom points of observation were on the equator. Also, the shifts in the Telluric lines, which were larger than one might expect, would have added more error to our data as well. Future experiments should be more careful in performing observations and should have more trials.

To calculate the standard error, we use the same method from our first post for this lab, which gives us a standard deviation of 0.057 km/s and a standard error of 0.04 km/s.

I would like to thank Charles Law and the TFs Allyson and Andrew for their help in completing this lab. 

Citations:

http://en.wikipedia.org/wiki/Sun

http://en.wikipedia.org/wiki/Doppler_effect

AU Daytime Lab, Post 1: Determining the Angular Size of the Sun

This is the first post in a series of four posts on a lab in which we are measuring the distance to the Sun, known as the Astronomical Unit (AU). We will do this using only basic geometry and a few tools in the laboratory. The accuracy of this measurement is important since the AU is the basis of the cosmic distance ladder and has been used to calibrate other methods of distance measurement. An error in the measurement of the AU would throw off many of our estimations for other objects in the Universe.

In our calculation of the AU, we will use rotational speed found by the Doppler shift due to the Sun's rotation, the rotational period found by the motion of sunspots across the Sun, and the angular size of the Sun found using a "sundial" lens. In this post, we will discuss the measurement of the angular size of the Sun, which is the angle that the diameter of the Sun subtends in the sky from Earth. The image below shows the angular size of the sun, represented as \(\theta\).

Angular Size of the Sun
Image Credit: http://www.ifa.hawaii.edu/~barnes/ast110_06/homework/hw08_fig1.gif

Equipment:

For this portion of the lab, we only need a lens attached to a window facing the Sun, which focuses the Sun's light to a sharper image on an easel. We call this lens a "sundial." The image below shows the setup, and you can see the image of the sun projected onto the easel. It goes without saying that this lab is best done on a sunny day. 

Lab Setup with Image of Sun through Sundial
Image Credit: http://www.fas.harvard.edu/~astrolab/angulardiameter.html

To solve for the angular size of the Sun, we need to measure the time that it takes for the image of the Sun to traverse its own diameter. Knowing that the Earth completes one rotation in 24 hours (86,400 seconds), we can therefore say that the Sun makes a complete rotation (360 degrees) in the sky with respect to the Earth in 86,400 seconds. We can solve for the diameter of the Sun by creating a ratio of the angular distance traveled by the Sun to the time it takes the Sun to move this angular distance.

\(\dfrac{360^\circ}{86400\text{ secs}}=\dfrac{\theta}{t}\)

Where \(\theta\) is the angular size of the Sun, and \(t\) is the time it took the image of the Sun to traverse it's diameter. So solving for the angular size, we get:

\(\theta = \dfrac{360^\circ t}{86400\text{ secs}}\)

Procedure:

The sundial lens and easel were positioned so that there is a circular and focused image of the Sun on the easel. We could see that the image was moving slowly to the right as its position in the sky was changing with the Earth's rotation. We made a mark on the easel to the right of the Sun's image and began a timer when the right side of the image reached the mark. Then, we stopped the timer when the left side of the Sun's image crossed the mark (after the image traversed its diameter). This was done three times and their average was taken in order account for errors in measurement. The picture below shows the motion of the Sun's image.

The circle to the left of the line is where the timer is started,
and the circle to the right is where the timer is stopped.
The arrow indicates direction of motion.

Results and Analysis:

The traversal times from our three trials are shown in the table below:

From this data, we can find that there is an average traversal time of 131.84 seconds. Now, putting this into our equation for the angular size, we get:

\(\theta = \dfrac{360^\circ t}{86400\text{ secs}}\) 

                 \( = \dfrac{360^\circ \times 131.84\text{ secs}}{86400\text{ secs}}\)

\(\theta = 0.55^\circ\)           

So we have found that the Sun has an angular diameter of 0.55 degrees, or \(9.59\times 10^{-3}\) radians.

Error Analysis and Discussion:

According to Wikipedia, the angular size of the Sun is about 32 arcminutes, which is equal to about 0.53 degrees or \(9.31\times 10^{-3}\) radians. So our answer is only off by about 0.02 degrees or \(2.8\times 10^{-4}\) radians, which is a percent error of approximately 3%. 

There are several factors that may have caused this error, including human error in timing, or perhaps imperfect alignment of the sundial and the easel. We can calculate the standard error, which is an estimate of the standard deviation of some sample that describes the variability of the individual measurements. The standard error is given by:

\(SE_{\overline{x}}=\dfrac{s}{\sqrt{n}}\)

Where \(s\) is the sample standard deviation, \(n\) is the sample size, and \(\overline{x}\) is the sample mean. We already know that \(\overline{x}=131.84\) seconds, and \(n=3\).

The sample standard deviation is given by:

\(s=\sqrt{\dfrac{1}{n-1}\displaystyle\sum _{i=1}^n (x_i -\overline{x})}\)

\(s=\sqrt{\dfrac{(135.90-131.84)^2+(130.78-131.84)^2+(128.83-131.84)^2}{3-1}}\)

\(s=3.65\)

Plugging values in, we get:

\(SE_{\overline{x}}=\dfrac{3.65}{\sqrt{3}}=2.11\) seconds

So we have a standard error of \(1.53\times 10^{-4}\) radians.

I would like to thank Charles Law, and the TFs Allyson and Andrew for their help in this lab.

Citations:

http://en.wikipedia.org/wiki/Sun

http://en.wikipedia.org/wiki/Standard_deviation

http://en.wikipedia.org/wiki/Standard_error

Virial Theorem in the Coma Cluster

Fritz Zwicky was born in Bulgaria in 1898 to Swiss parents. He worked at the California Institute of Technology for most of his life. Zwicky is responsible for several important discoveries in the field of astrophysics, but his most notable discovery was that of dark matter.

Fritz Zwicky
Image Credit:
http://www.cosmotography.com/images/starburst/images/fritz_zwicky.jpg

In 1933, Zwicky, made an important observation that gave the first implications for the existence of dark matter. While observing the Coma Cluster (a cluster of gravitationally bound galaxies) and applying Virial Theorem to the galaxies in the cluster, Zwicky found that the velocities of the galaxies were much larger than expected from the expected mass in the cluster, implying that there must be more mass (or gravitational potential) in the cluster that was unseen. It is due to the unobservable nature of this mass that we call it dark matter. 

Fortunately, we have enough of an understanding of the Virial Theorem to follow the calculations Zwicky made to reach his conclusions. 

Coma Cluster
Image Credit: http://upload.wikimedia.org/wikipedia/commons/7/7d/Ssc2007-10a1.jpg

First, Zwicky made the reasonable assumption the the cluster was in a stationary state (neither expanding nor contracting), meaning that it was virialized. As we have gone over before, this means that the total kinetic energy is equal to half of the total potential energy in the system.

\(K=-\frac{1}{2}U\)

He also assumed that the density, \(\rho\), was constant, which while not true, is okay since he was only trying to make an estimation. The radius, R, of the Coma Cluster is approximately one million light years, which is equal to \(10^{24}\) cm. The mass, M, was also estimated by finding the number of galaxies and multiplying by the mass of a galaxy:

 \(M=800\times 10^9M_\odot\times 2\times 10^{33}\text{ g M}_\odot^{-1}\approx 1.6\times 10^{45}\) g

So the potential energy, assuming spherical shape, is given by:

\(U=-\dfrac{3GM^2}{5R}\)

\(U=-\dfrac{3\times (6.674\times 10^{-8}\text{ cm}^3\text{ g}^{-1}\text{ s}^{-2})\times (1.6\times 10^{45}\text{ g})^2}{5\times 10^{24}\text{ cm}}\)

\(U\approx -10^{59}\text{ g cm}^2\text{ s}^{-2}\)

Where G is the gravitational constant.

Since we know that the Coma Cluster is virialized, we can use Virial Theorem to calculate the average velocity:

\(K=-\frac{1}{2}U\)

\(\frac{1}{2}M\langle v\rangle ^2=-\frac{1}{2} (-10^{59})\)

\(\langle v\rangle =\sqrt{\dfrac{10^{59}\text{g cm}^2\text{ s}^{-2}}{1.6\times 10^{45}\text{ g}}}\)

\(\langle v\rangle \approx 8\times 10^6\text{ cm s}^{-1}=80\text{ km s}^{-1}\)

However, through observations on the Doppler shifts in the spectrum of the  galaxies of the Coma Cluster, Zwicky found that there was a large spread of velocities in the cluster, with average velocities of at least 1000 km/s, and some velocities even as large as 1500 to 2000 km/s. In order for Virial Theorem to come up with velocities this large, we would need to have a total mass at least 400 times larger than our previous estimate. This is a very large difference, however, so it seems unlikely that our estimate based on observations of galaxies could be this far off. 

There are four possible explanations discussed by Zwicky as to why we might be observing velocities that are much larger than expected. His first, and most favorable, explanation is the existence of dark matter. There could be some form of matter in the universe that is unobservable but has gravitational effects. While today we have a large amount of evidence that speaks to the validity of this claim, at the time, Zwicky's idea of dark matter was not accepted, and would not be accepted for another fifty years. 

Coma Cluster
Image Credit:
http://upload.wikimedia.org/wikipedia/commons/0/02/Coma_Cluster_of_Galaxies_%28visible%2C_wide_field%29.jpg

Zwicky's second explanation for this discrepancy in theory and observation was that it was possible that the Coma Cluster was not virialized, and that instead, the total kinetic energy was equal to the total potential energy.

\(K=-U\)

But this only accounts for a factor of two from our previous findings. So there still remains the problem of a large proportion of mass that is unaccounted for. 

Another possibility is that the observations were correct, and that Zwicky's estimate for mass based on the luminous (visible) matter was valid. If this were true, however, the large velocities of the constituent galaxies in the cluster would cause them to break away from each other's gravitational pull, and the cluster would fall apart, leaving the 800 galaxies free, flying through space at their velocities of 1000 to 2000 km/s. If this were the case, however, we would probably be able to observe other free galaxies that are not associated with clusters (called field galaxies) which have similar velocities; but thus far, none have been seen to even have velocities over 200 km/s. So it seems unlikely that this possibility would hold true. 

His last explanation was that perhaps the observations of the velocities were incorrect. It is possible that the change in observed wavelength of light was not caused by motion (it was not a Doppler Shift), but instead by a gravitational force that was distorting the light observed. This kind of shift in wavelength is called gravitational redshift, or Einstein redshift. The equation to find redshift, z, is given as:

\(z=\dfrac{\Delta \lambda}{\lambda}\approx -\dfrac{GM}{c^2R}\)

Where c is the speed of light, \(\lambda\) is the emitted wavelength of the light, and \(\Delta\lambda\) is the shift in wavelength (the difference of the observed wavelength and emitted wavelength). From this equation, using our previous estimate for mass, we can solve for the redshift, and then using the relation, \(v\approx cz\), we can solve for the velocities of the galaxies in the Coma Cluster.

\(z\approx \dfrac{(6.674\times 10^{-8}\text{ cm}^3\text{ g}^{-1}\text{ s}^{-2})\times (1.6\times 10^{45}\text{ g})}{(3\times 10^{10}\text{ cm s}^{-1})^2\times 10^{24}\text{ cm}}\)

\(z=3.5\times 10^{-8}\)

Note that Zwicky's calculation may have been more careful, so I reported his calculation for z.

Now solving for velocity, v:

\(v=cz\approx (3\times 10^{10}\text{ cm s}^{-1})(3.5 \times 10^{-8})\approx 1000\text{ cm s}^{-1}=10\text{ m s}^{-1}\)

This certainly doesn't explain the large velocities observed. We could, instead, work the other way, starting with the large velocities and solving for the mass necessary to create this gravitational redshift. However, this method would require a mass that is much greater than before, which again implies that there must be a large proportion of dark matter.

From Zwicky's discussion of the possible explanations for the large velocities of galaxies in the Coma Cluster, it is apparent that there must be some form of matter that is unobservable but has gravitational effects which could cause these larger velocities. The nature of this dark matter is still unknown, but we have since found evidence that suggests that dark matter is the predominant form of matter in our universe. 

Citations:

http://en.wikipedia.org/wiki/Fritz_Zwicky

http://en.wikipedia.org/wiki/Coma_Cluster

http://en.wikipedia.org/wiki/Dark_matter