Day Lab (Part 4): The Astronomical Unit

Okay, so in the last 3 parts of this lab, we calculated the following values:

• Angular diameter of the Sun, $\theta_{\odot} = 0.5547°$
• Rotational speed of the Sun, $v_{\odot} = 1.86 \frac{km}{s}$
• Rotational period of the Sun, $P_{\odot} = 26.1$ days

But what we actually want to calculate is the astronomical unit (AU). How can we find it though?

Well, let's draw a diagram that relates the astronomical unit to the angular diameter:

Based on this diagram, we can relate the astronomical unit to the angular diameter of the Sun, $\theta_{\odot}$ and the radius of the Sun, $R_{\odot}$:

\begin{align}
\tan \left(\frac{\theta_{\odot}}{2}\right) &= \frac{R_{\odot}}{AU}\\
AU &= \frac{R_{\odot}}{\tan \left(\frac{\theta_{\odot}}{2}\right)}
\end{align}

Okay, so we have an equation for calculating AU, but it requires us to have the radius of the Sun, $R_{\odot}$. We don't have that value, but we can calculate it using the Sun's rotational speed, $v_{\odot}$ and the rotational period, $P_{\odot}$ using the equation for rotational velocity:

\begin{align}
v_{\odot} &= \frac{2\pi R_{\odot}}{P_{\odot}}\\
R_{\odot} &= \frac{v_{\odot} \times P_{\odot}}{2\pi}\\
R_{\odot} &= \frac{1.86 \frac{km}{s} \cdot \frac{10^5 \text{ cm}}{1 \text{ km}}\times 26.1 \text{ days} \cdot \frac{24 \text{ hours}}{1 \text{ day}} \cdot \frac{60 \text{ minutes}}{1 \text{ hour}} \cdot \frac{60 \text{ seconds}}{1 \text{ minute}}}{2\pi}\\
R_{\odot} &= 6.68 \times 10^{10} \text{ cm}
\end{align}

Now that we have the radius of the Sun, we can plug this value into the first equation to solve for the AU:

\begin{align}
AU &= \frac{R_{\odot}}{\tan \left(\frac{\theta_{\odot}}{2}\right)}\\
AU &= \frac{6.68 \times 10^{10} \text{ cm}}{\tan \left(\frac{0.5547°}{2}\right)}\\
AU &= 1.379 \times 10^{13} \text{ cm}
\end{align}

We have calculated the AU to be $1.379 \times 10^{13} \text{ cm}$, which is very close to the actual answer of  $1.5 \times 10^{13} \text{ cm}$, which is within 8% error. This can be due to the inaccuracies in measuring the sunspot latitudes, since the spherical grid was not very precise since it lacked small unit divisions. Also, measuring the angular diameter could be inaccurate due to human error of stopping the stopwatch at the accurate time.

Day Lab (Part 3): Rotational Period of the Sun

In the first two parts of the lab, we calculated the angular diameter of the Sun, $\theta_{\odot} = 0.5547°$ and the rotational velocity of the Sun, $v_{\odot} = 1.86 km/s$. Using these two values, we can calculate the time it takes for the Sun to make one full rotation, also known as it's period, $P_{\odot}$.

In order to calculate the full rotation of the Sun, it would make sense to use a marker on the Sun, and see how long it takes to see that marker again, which would indicate the period, $P_{\odot}$. Unfortunately, there are no permanent markers on the Sun, since the surface of the Sun is constantly changing. However, we do have temporary markers on the Sun, sunspots, which can be tracked as the Sun is rotating. Using sunspots, we can create a relationship between the the angular distance, $\theta$ a sunspot travels as it moves across the surface of the Sun over a given period of time, $t$, and the rotational period of the Sun, $P_{\odot}$ as it travels a full 360°.

This relationship can be modeled as follows:

\begin{align}
\frac{P_{\odot}}{360°} &= \frac{\Delta t}{\Delta \theta}\\
P_{\odot} &= \frac{\Delta t (360°)}{\Delta \theta}
\end{align}

Okay, so we have a model to calculate the period, but how do we measure the transit of the sunspots across the surface of the Sun?

Well, we could directly view sunspots by projecting an image of the Sun via mirrors on a piece of paper, and see how the sunspots that are visible over the course of a few days. However, when doing this lab, the day was rather cloudy, so we decided to use sunspot data from NASA in 2001.

Using this data, we can overlay a live recording of the surface of the Sun with graph of spherical coordinates, and measure the change in angular distance over a given period of time. In this lab, we recorded the transit of 3 sunspots, with the following data:

Sunspot 1: 

Trial 1: 03-04-2001 at 17:36 to 03-09-2001 at 11:12

$\Delta \theta = +70°$, $\Delta t = 408,960$ seconds 

Trial 2: 03-06-2001 at 00:00 to 03-08-2001 at 17:36

$\Delta \theta = +40°$, $\Delta t = 236,160$ seconds 

Trial 3: 03-05-2001 at 11:12 to 03-10-2001 at 00:00

$\Delta \theta = +67°$, $\Delta t = 291,680$ seconds 

Average: 

$\Delta \theta = 59°$, $\Delta t = 345,600$ seconds 

$P_{\odot} = \frac{\Delta t (360°)}{\Delta \theta} = \frac{345,600 (360°)}{59°} =$ 24.4 days

Sunspot 2: 

Trial 1: 03-22-2001 at 00:00 to 03-25-2001 at 11:12

$\Delta \theta = +50°$, $\Delta t = 299,520$ seconds 

Trial 2: 03-20-2001 at 17:36 to 03-27-2001 at 17:36

$\Delta \theta = +100°$, $\Delta t = 604,800$ seconds 

Trial 3: 03-22-2001 at 17:36 to 03-27-2001 at 00:14

$\Delta \theta = +60°$, $\Delta t = 369,480$ seconds 

Average: 

$\Delta \theta = 70°$, $\Delta t = 424,600$ seconds 

$P_{\odot} = \frac{\Delta t (360°)}{\Delta \theta} = \frac{424,600 (360°)}{70°} =$ 25.3 days

Sunspot 3: 

Trial 1: 04-06-2001 at 08:45 to 04-11-2001 at 00:00

$\Delta \theta = +60°$, $\Delta t = 400,500$ seconds 

Trial 2: 04-04-2001 at 19:12 to 04-10-2001 at 06:24

$\Delta \theta = +70°$, $\Delta t = 472,200$ seconds 

Trial 3: 04-07-2001 at 00:00 to 04-14-2001 at 11:12

$\Delta \theta = +90°$, $\Delta t = 645,120$ seconds 

Average: 

$\Delta \theta = 73.3°$, $\Delta t = 505,940$ seconds 

$P_{\odot} = \frac{\Delta t (360°)}{\Delta \theta} = \frac{505,940 (360°)}{73.3°} =$ 28.7 days

Having three measurements of rotational period, $P_{\odot}$, we can take the average of all three to get the rotational period of the Sun to be $P_{\odot} = 26.1$ days. The actual rotational period of the Sun at the equator is 24.7 days, but that is the sidereal rotation period. The method we used to get the period required following a fixed feature on the Sun, which is measured by synodic rotational period, which for the Sun is 26.24 days. The synodic rotational period of the Sun is very close to our measurements.

Day Lab (Part 2): Rotational Speed of the Sun

We said in the first part of the lab that in order to calculate the astronomical unit (AU), we will be using the Doppler shift technique. In order to use the Doppler shift technique, we need to know the angular diameter of the Sun, $\theta_{\odot}$, as well as its rotational speed. We got the angular diameter of the Sun in the previous part, which was $\theta_{\odot} &=  0.5547°$. Now, we need the rotational speed of the Sun.

Wait, what?!?! How can we tell how fast the Sun is rotating?

Well, we can use the spectral lines of the Sun, specifically the Sodium D line (NaD) spectral lines, to see if they are moving towards us or away from us using blueshift and redshift, respectively, of the light coming from the Sun as a result of the Doppler effect, and compare the movement of the NaD spectral light with the $H_2O$ Telluric line that is absorbed by the atmosphere, and therefore stationary.

In order to get the most accurate rotational speed of the Sun, we want to measure the movement of the spectral lines from the equator of the Sun. Unfortunately, we don't know what the orientation of the Sun is due to the tilt of the Earth, as well as the several mirrors that are used to project the solar disk into the spectrometer. Therefore, we will take 8 measurements of the NaD spectral lines, in the pairs that are opposite from one another:  (top, bottom), (left, right), (top-left, bottom-right), and (top-right, bottom-left). The orientation with the widest gap between the two pairs of NaD spectral lines would be at the solar equator.

After taking the picture of the NaD spectral lines from those 8 orientations, the resulting data showed the following:

Looking at the 8 measurements of the spectral lines, it is apparent that the greatest shift in the NaD spectra lines took place between the "top-right" and "bottom-left" orientation, indicated by the shift in the purple lines. Therefore, that is the orientation of the Sun that measures the spectral shift at the equator.

Knowing that the "top-right" and "bottom-left" orientation was the equatorial orientation, we can use the shift data to determine the exact redshift that occurs, and compare it to the Telluric spectral lines to calculate the rotational speed of the Sun.

Left sodium absorption line, with a pixel shift of 3.98 pixels.

Right sodium absorption line, with a pixel shift of 4.21 pixels.

In order to compare the pixel shift of absorption between the sodium absorption lines, we need to look at the Telluric spectral line of the Earth's atmosphere:

The Telluric Line has a pixel shift of 2.09 pixels. 

There should be no shift in the Telluric Line, because they are part of the Earth's atmosphere, and therefore are not redshifting. This is most likely due to the scattering of light in the Earth's atmosphere.

Based on the data from these graphs, the difference between the left and right sodium lines is 335.01 pixels separating the two, which is a separation of 5.97 angstroms, since the conversion factor is 0.017821 angstroms per pixel.

We can use the spectral separation of 5.97 angstrom to calculate the rotational velocity using the Doppler equation, which gives us an average rotational velocity between the two sodium spectral lines of the sun, $v_{\odot} = 1.86 km/s$.

Since the actual rotational speed of the Sun is 2 km/s, the calculated rotational velocity has a 7% error, which is rather good.

Day Lab (Part 1): Angular Size of the Sun

The Astronomical Unit (AU) is a fundamental unit of measurement that measures the distance between the Earth and the Sun. The AU is the measure-stick by which many astronomical distances are based-off, therefore, it is imperative that we have an accurate measurement of the AU.

In this lab, we will be using the Doppler shift technique, using the rotational speed of the Sun and its angular diameter in the sky, to measure the AU.

The first thing we need is the angular diameter of the Sun. The angular diameter is angular measurement describing how large an object appears from a given point of view. In this case, our point of view is the Earth, and we're measuring the size of the Sun as it appears in the sky.

In order to measure the size of the Sun as it appears in the sky, we can use a proportion with the rotation of the Earth. We know that the Earth rotates a full 360° in 24 hours, or an Earth day, in which the Sun makes a full 360° shift and is back to it's original position in the sky. Therefore, one way to measure how large the Sun appears in the sky is to see how long it takes the Sun to shift across the sky by the length of its own diameter. Then, we can relate the time it takes for the Sun to move across its own diameter to the time it takes the Earth to rotate a full 360° to figure out what portion of the 360° the Sun takes up in the sky.

The proportion to figure out the angular diameter of the Sun, $\theta_{\odot}$ can be set up as follows:

\begin{align}
\frac{360°}{\text{Time for full rotation of Sun in the sky}} &= \frac{\theta_{\odot}}{\text{Time for Sun to shift across the sky by its diameter}}\\
\theta_{\odot} &= \frac{360° \times \text{Time for Sun to shift across the sky by its diameter}}{\text{Time for full rotation of Sun in the sky}} \\
\theta_{\odot} &= \frac{360° \times t}{24 \text{ hours}} \\
\theta_{\odot} &= \frac{360° \times t}{24 \text{ hours} \times \frac{60 \text{ minutes}}{\text{1 hour}} \times \frac{\text{60 seconds}}{\text{1 minute}}}\\
\theta_{\odot} &= \frac{360° \times t}{86400 \text{ seconds}}
\end{align}

Now that we have an equation for finding the angular diameter, we just need to measure how long it takes the Sun to move across its diameter. In this lab, we had the following data:

Trial 1 : 130.28 seconds

Trial 2 : 133.22 seconds

Trial 3 : 134.62 seconds

Trial 4 : 131.50 seconds

Trial 5 : 136.06 seconds

Average: 133.136 seconds

The average time for the Sun to move along its diameter is 133.136 seconds. We can plug in this value into the equation for the angular diameter of the Sun for the $t$ value.

\begin{align}
\theta_{\odot} &= \frac{360° \times t}{86400 \text{ seconds}}\\
\theta_{\odot} &= \frac{360° \times 133.136 \text{ seconds}}{86400 \text{ seconds}}\\
\theta_{\odot} &=  0.5547°
\end{align}

Therefore, the angular diameter of the Sun is $\theta_{\odot} =  0.5547°$.

Fritz Zwicky: Dark Matter in the Coma Cluster

Source: http://upload.wikimedia.org/wikipedia/commons/1/1d/Hubble_close-up_on_the_Coma_Cluster.jpg

In 1933, Fritz Zwicky made an observation in the Coma Cluster that defied anything known in astronomy at the time. The Coma Cluster is a cluster of over 1000 observed galaxies. Looking through Zwicky's paper, we see that Zwicky observed the individual galaxies moving away from each other at very fast speeds. However, the gravitational effects of the total mass of the cluster could not account for these large velocities. Therefore, Zwicky concluded that there must be other matter besides the "luminous matter" that was present in the Coma Cluster, which we now refer to as dark matter.

In order to see how Zwicky came to this conclusion, let's walk through the steps that he did. However, we will replace some of his older terms with modern terminology to get a better understanding of how the existence of dark matter was first detected.

First, Zwicky claims that the Coma Cluster "has mechanically reached a stationary state", which implies that the Coma Cluster is in equilibrium, and can therefore be modeled with the Virial Theorem. Zwicky models the Virial Theorem as follows:
\begin{align}
\overline{\varepsilon_k} = -\frac{1}{2} \overline{\varepsilon_p}
\end{align}
Which can be rewritten using modern terminology as:
\begin{align}
K = -\frac{1}{2} U
\end{align}
where $K$ is the total kinetic energy of the system, and $U$ is the total potential energy of the system.

Zwicky then tells us some of the details about the assumptions he made about the Coma Cluster which we can also use to help with the calculations. For example, the Coma Cluster can be assume to have a uniform density, has a radius $R$ of 1 million light years ($1 \times 10^{24}$ cm), and is comprised of "800 individual nebulae each of a mass corresponding to $10^9$ solar masses". Using this information, Zwicky calculated the total mass $M$ of the system (note that a solar mass is $2 \times 10^{33}$ g):

\begin{align}
M = \text{number of nebulae} \times 10^9 \text{solar masses} \times \text{solar mass} = 800 \times 10^9 \times (2 \times 10^{33}) = 1.6 \times 10^{45} \text{ g}
\end{align}

So now that we know the mass of the Coma Cluster, we can figure out how fast the cluster is going. Zwicky did this by finding solving for the equation $\overline{\varepsilon_p} = \frac{\Omega}{M}$ for the average potential energy of the system, and $\overline{\varepsilon_k} = \frac{v^2}{R}$ for the average kinetic energy of the system, where $\Omega$ is the total potential energy of the system, $M$ is the total mass of the system, $v$ is the velocity of the system, and $R$ is the radius of the Coma Cluster. Zwicky used these variables to solve for the velocity $v$ of the system. We can solve for the velocity using the Virial Theorem as follows. It is important to note that in the Virial Theorem, the $U = \Omega = -\frac{3}{5}\frac{GM^2}{R}$, which are both variables representing the total gravitational potential energy of the Coma Cluster:

\begin{align}
K &= -\frac{1}{2} U\\
\frac{1}{2} M v^2 &= -\frac{1}{2} \left( -\frac{3}{5} \frac{GM^2}{R} \right)\\
v^2 &= \frac{3}{5}\frac{GM}{R}\\
v &= \sqrt{\frac{3}{5}\frac{GM}{R}}\\
v &= \sqrt{\frac{3}{5}\frac{(6.67 \times 10^{-8} \frac{\text{cm}^3}{g \cdot s})(1.6 \times 10^{45} \text{ g})}{1 \times 10^{24} \text{ cm}}}\\
v &= 8 \times 10^6 \frac{m}{s}
\end{align}

Interestingly enough, Zwicky shows that based on the observations of the doppler shift, the galaxies were moving away at an average speed of 1000 $\frac{km}{s}$. In order for this to happen, the Coma Cluster must be at least 400 times denser than the calculated mass! This suggests that perhaps the mass calculated for the Coma Cluster was not all of the observed mass. Zwicky argued that perhaps there was other matter present in the Coma Cluster, dark matter, which was present alongside the "luminous matter", and that the mass of the dark matter was not taken into account.

However, the other possibility was that Zwicky assumed to the contrary that the Coma Cluster was at stationary equilibrium, and if that was not the case, the Virial Theorem would not hold. Therefore, if the Coma Cluster was not at stationary equilibrium, Zwicky argued that "the entire available potential energy appears as kinetic energy", as represented:

\begin{align}
K = -U
\end{align}

Even under this constraint, the calculations are only affected by a factor of 2, and the results don't change much from the initial calculation. This suggests that even if the Coma Cluster was not in stationary equilibrium, and that the total kinetic energy of the system was equal to the total potential energy of the system, the observed speeds of the galaxies was still too fast for the cluster to stay together. At the observed speeds, the Coma Cluster would fly apart, unless there was gravitational influence from matter that we did not account for. Again, this suggested that dark matter might exist in the Coma Cluster.

Another proposition that Zwicky made was that, assuming that the mass of the Coma Cluster was actually the mass calculated before, $M$, without any other "dark matter" in the system, and that the observed velocities of an average 1000 km/hr were real, then the Coma Cluster would break down into 800 individual galaxies flying apart from each other.