Worksheet 6, Problem 1: The Light of a Single Sun

1. Use a lamp fitted with a 100 Watt incandescent light bulb to estimate the luminosity of the Sun. Do this with the knowledge that:

• The Earth-Sun distance is 1 astronomical unit (AU), or a 1.5 x 10^13 cm
• An estimate of how far from the bulb you have to hold your hand in order for it to feel like a sunny Spring day;
• The fact that incandescent light bulbs convert electrical power into radiant energy with an efficiency of only about 3%.

As we dive further into the realm of black bodies, we begin to realize and appreciate their simplicity in characterizing the objects that we are dealing with in our exploration of space. In fact, in a simple experiment, we can roughly calculate the luminosity of the sun, with a few basic numbers and a 100 Watt incandescent light bulb. 

To approach this, we began covering our eyes and stepping up to our lit incandescent bulb. Carefully, we adjusted the distance of our hand away from the bulb, until the temperature resembled that of a warm spring day. As we've learned, our distance from the sun varies throughout the seasons, with the spring providing a rough average for the course of the year. Our group reached a consensus distance away from the bulb of 20cm. 

This value compares to the distance of 1 AU or \(1.5 \times 10^{13}\) cm that the earth is away from the sun on a warm spring day. So, you may ask, why is this information valuable? Essentially, what we have done here is matched the amount of radiance that reaches our body from the sun on a spring day to the amount of radiance that reached our hand from the bulb. Since this represents energy per area, we can consider this quantity to be the flux of the black bodies and as we've stated, we can set them equal to each other as follows: \[F_{sun} = F_{bulb}\] Since we know that flux is equal to luminosity per area, we can rewrite the above relationship as:\[\frac{L_{sun}}{SA_{sun-earth}}=\frac{L_{bulb}}{SA_{bulb-hand}}\] Since we know that black bodies radiate equally in all directions, the surface area that we are interested in is that of the sphere with a radius corresponding to the distance the object is away from the black body. The surface area of a sphere is: \(SA = 4\pi r^2\), where r is equal to the distance, d. Updating our relation above, we now have: \[\frac{L_{sun}}{4\pi (d_{sun-earth})^2}=\frac{L_{bulb}}{4\pi (d_{bulb-hand})^2}\] In trying to solve for the luminosity of the sun, the last remaining unknown is the luminosity of the bulb. Fortunately, we can easily calculate that based on the power output and efficiency of the incandescent bulb. Given that the bulb converts its 100 watts of electrical power to 3% radiant light, we can assume that the other 97\% is radiated as heat. \[L_{bulb} = 100 \, watts \times 97\% = 97 \, watts\] We can now solve for the luminosity of the sun: \[L_{sun} = \frac{L_{bulb}}{4\pi (d_{bulb-hand})^2}4\pi (d_{sun-earth})^2= \frac{L_{bulb}}{(d_{bulb-hand})^2} (d_{sun-earth})^2\] \[L_{bulb} = \frac{97\, watts}{ (20\,cm)^2} (1.5 \times 10^{13}cm)^2 \approx 5 \times 10^{25}\, watts\] Finally, we need to convert to ergs per second, and we have our answer! \[5 \times 10^{25}\, watts \, \times \frac{10^7 \,erg/s}{1\, watt} = 5 \times 10^{32}\, erg/s\]

Acknowledgements: Johnathan Budd & Willie Pirc

Mass Extinction and Dinosaurs!

Recently published research out of NYU may provide insight to the harrowing impact that dark matter may have on the fate of earth (http://astronomynow.com/2015/02/19/does-dark-matter-cause-mass-extinctions-and-geologic-upheavals/). The article studies how movement through dark matter effect the orbit of astroids and other objects in space. As stars, gases, dust, meteors, and dark matter intimately travel within a galaxy, their interactions have immense consequences. Below is an image of NGC 4565, a spiral galaxy similar to that of the Milky Way that shows the cluttered nature of space within a galaxy.

http://astronomynow.com/2015/02/19/does-dark-matter-cause-mass-extinctions-and-geologic-upheavals/

While dark matter still remains a mystery in many ways, the research from Professor Michael Rampino indicates that it's existence can throw comets off their orbit and even cause additional thermal heating to planets. As our solar system orbits the Milky Way every 250 million years, it follows a wave-like path that dips in and out of the galactic spiral approximately every 30 million years. During these intersections, our solar system passes through concentrations of dark matter, which leads to the increase in astroid activity and potential for biological destruction. Rampino believes that the earth can actually collect dark matter particles in its core, which can cause increased volcanic activity and increased sea levels. His research argues that over 30 million year periods, there is data in agreement with his claims. One of the most notable instances is the extinction of the dinosaurs that occurred approximately 66 million years ago (~two 30 million year periods).

http://abcnews.go.com/images/Technology/cb_dinosaurs_meteorite_nt_130215_wmain.jpg

 Whether Rampino is right about the role that dark matter plays in the periodic traumas that the earth endures, his research evokes a lot of questions about the role that dark matter plays in the universe, and the different proposed models behind it.

A recent article in Ars Technica takes a look at the Milky Way's rotation using tracers across the galaxy. By marking these objects based on relative velocity, the author is able to get a distribution of the trajectory paths. By observing the behavior of the visual matter, the author illustrates the impact that undetected matter must have. Given the widely unknown nature of dark matter, and it's extensive impact on our universe as a whole, it's exciting to see a wide array of research going into the topic!

http://arstechnica.com/science/2015/02/updated-look-at-milky-ways-rotation-strengthens-the-case-for-dark-matter/

Worksheet 5, Problem 2:

Blackbodies are nice because they’re such simple objects. Their outward appearance is entirely determined by their temperature. If there were cows in space, astronomers would imagine them to be spherical blackbodies (and seriously, it wouldn’t be a bad approximation). In this exercise we’ll take advantage of the relative simplicity of blackbodies to derive some useful expressions that you’ll use during this term, and throughout your astronomy career.

A) In astronomy, it is often useful to deal with something called the “bolometric flux,” or the energy per area per time, independent of frequency. Integrate the blackbody flux \(F_v(T)\) over all frequencies to obtain the bolometric flux emitted from a blackbody, F(T). You can do this using by substituting the variable u = hν/kT. This will allow you to split things into a temperature-dependent term, and a term comprising an integral over all frequencies. However, rather than solving for the integral, just set the integral and all constants equal to a new, single constant called σ, which is also known as the Stefan-Boltzmann constant. If you’re really into calculus, go ahead and show that \(\sigma \approx 5.7 \times 10^{-5} erg*s^{-1}*cm^{-2}*K^{-4}\). Otherwise, commit this number to memory.

The beauty in treating objects in space as blackbodies is the ability to characterize their appearance entirely by temperature. For this problem, we get to examine several ways to quantize their radiation. Here, in part A, we are considering the "bolometric flux" which tells us the amount of energy passing through an area over time, independent of frequency. To arrive at this, we begin with the blackbody flux, \(F_v(T)\), and integrate over all frequencies.

First, let's consider the amount of radiation emitted from a given blackbody. This is given by: \[B_{\nu}(T) = \frac{2\nu^2}{c^2}\frac{h\nu}{e^{\frac{h\nu}{kT}}-1}\]
Now, to arrive at the amount of flux for a black body, we need to integrate over the face of one hemisphere. While a blackbody radiates evenly in all directions, we are only concerned with measuring the radiation toward us, since the radiation from the opposite hemisphere will be equal and opposite.

Similar to the latitude and longitude coordinates of the earth, a blackbody radiates out at solid angles, which are characterized by angles (\(\theta, \phi\)) with respect to our selected origin. As we consider how the radiation that we as an observer receive, we realize that for each point on the surface, only a portion of the radiation actually reaches us. By selecting the north pole of the black body as our origin, the solid angle conveniently behaves as a function of \(cos\phi\). To cover the full hemisphere, we will integrate from the north to south pole, or 0 to \(\frac{\pi}{2}\), and then around the circumference of the body, or 0 to \(2\pi\). Completing this we get:
\[F_{\nu}(T) = \int_{\phi=0}^{\frac{\pi}{2}} \int_{\theta=0}^{2\pi} B_{\nu}(T)cos(\theta)sin(\theta)d\theta d\phi\]
Evaluating this, we discover that: \[F_{\nu}(T) =\pi B_{\nu}(T)\]
Moving on to calculate the bolometric flux, we must now integrate over all frequencies, \(\nu\). \[F(T) = \int_{\nu=0}^{\infty}F_{\nu}(T) d\nu = $\int_{\nu=0}^{\infty}\pi B_{\nu}(T) d\nu\]\[F(T) =\int_{\nu=0}^{\infty} \frac{2\nu^2}{c^2}\frac{h\nu\pi}{e^{\frac{h\nu}{kT}}-1}d\nu\] To tackle this integral, we can utilize u-substitution, where \(x=\frac{h\nu}{kT}\). \[F(T)=\int_{x=0}^{\infty} \frac{2(\frac{xkT}{h})^2}{c^2}\frac{xkT\pi}{e^{x}-1}\frac{kT}{h}dx\] By setting this complicated integral with its constants to a new variable, \(\sigma\), known as the Stephan-Boltzmann constant, we get a bolometric flux of: \[F(t) =\sigma T^4\] In this, our constant \(\sigma \approx 5.7 \times 10^{-5} erg*s^{-1}*cm^{-2}*K^{-4}\).

B) The Wien Displacement Law: Convert the units of the blackbody intensity from \(B_{\nu}(T)\) to \(B_{\lambda}(T)\). IMPORTANT: Remember that the amount of energy in a frequency interval dν has to be exactly equal to the amount of energy in the corresponding wavelength interval dλ.

We are now considering the blackbody's radiation in terms of wavelength, rather than frequency. We can begin by knowing that: \[\int_{\lambda = 0}^{\infty} B_{\lambda}(T) d\lambda = $\int_{\nu = 0}^{\infty} B_{\nu}(T) d\nu\] We also know the relationship between \(\lambda\) and \(\nu\) to be \(\nu = \frac{c}{\lambda}\). Taking the derivative of this, we also obtain: \(\frac{d\nu}{d\lambda} = \frac{-c}{\lambda^2}\). By inserting this into our equality, we arrive at: \[\int_{0}^{\infty} B_{\lambda}(T) d\lambda = \int_{0}^{\infty} B_{\nu}(T) \frac{-c}{\lambda^2}d\lambda\]Considering our earlier assertion that the integrands are equal, this reduces to: \[B_{\lambda}(T)=B_{\nu}(T) \frac{c}{\lambda^2}\] Finally, expanding \(B_ {\nu}(T) \frac{c}{\lambda^2}\), we arrive at our final answer of: \[B_{\lambda}(T) = \frac{2hc^2}{\lambda^5(e^{\frac{hc}{\lambda k T}}-1)}\]
C) Derive an expression for the wavelength \(\lambda_{max}\) corresponding to the peak of the intensity distribution at a given temperature T. (HINT: How do you find the maximum of a function?) Once you do this, again substitute \(u = \frac{h\nu}{kT}\). The expression you end up with will be transentental, but you can solve it easily to first order, which is good enough for this exercise.

To determine the wavelength of maximum intensity, we can use our expression for \(B_{\lambda}(T)\) and find a maximum. To find this maximum, we will take the first derivative and find an inflection point (where the first derivative equals zero). Then, we will need to observe from the second derivative if the point is a maximum or a minimum.

First, we need to take the derivative of \(B_{\lambda}(T)\). \[B_{\lambda}(T)\frac{d}{d\lambda} =  \frac{2hc^2}{\lambda^5(e^{\frac{hc}{\lambda k T}}-1)}\frac{d}{d\lambda}\] \[B_{\lambda}(T)\frac{d}{d\lambda} =  \frac{-10hc^2}{\lambda^6(e^{\frac{hc}{\lambda k T}}-1)}+\frac{2h^2c^3e^{\frac{hc}{\lambda k T}}}{\lambda^7kT(e^{\frac{hc}{\lambda k T}}-1)^2}\] Before setting this to zero, we can simplify the expression by factoring out common terms: \[B_{\lambda}(T)\frac{d}{d\lambda} = (\frac{2hc^2}{\lambda^6(e^{\frac{hc}{\lambda k T}}-1)})(\frac{-5}{1}+\frac{hce^{\frac{hc}{\lambda k T}}}{\lambda k T(e^{\frac{hc}{\lambda k T}}-1)})\]
Again, we can substitute \(x = \frac{hc}{\lambda k T}\). We now have a much friendlier term of: \[B_{\lambda}(T)\frac{d}{d\lambda}= -5 +\frac{xe^x}{e^x-1}\]
Setting this to zero, we now get:\[B_{\lambda}(T)\frac{d}{d\lambda}= -5 +\frac{xe^x}{e^x-1}=0\]\[x=4\] Now substituting x back in for it's actual values, we get \[x= \frac{hc}{\lambda k T} = 4\]\[\lambda = \frac{hc}{4kt}\]
D) The Rayleigh-Jeans Tail: Next, let’s consider photon energies that are much smaller than the thermal energy. Use a first-order Taylor expansion on the term \(e^{\frac{h\nu}{kT}}\) to derive a simplified form of \(B_{\nu}(T)\) in this low-energy regime. (HINT: The Taylor expansion of \(e^x \approx 1+x\))

Recalling that: \[B_{\nu}(T) = \frac{2\nu^2}{c^2}\frac{h\nu}{e^{\frac{h\nu}{kT}}-1}\] We can use this Taylor expansion approximation to rewrite this as: \[B_{photon}(T) = \frac{2\nu^2}{c^2}\frac{h\nu}{1+x-1}\] This then simplifies to: \[B_{photon}(T) = \frac{2\nu^2}{c^2}kT\]
E) Write an expression for the total power output of a blackbody with a radius R, starting with the expression for \(F_{\nu}\). This total energy output per unit time is also known as the bolometric luminosity, L.

The total power output of a blackbody can easily be found by using the bolometric flux that we found in part A. Since the bolometric flux is given as power per area, we just need to multiply this quantity by the surface area of the blackbody to arrive at the power output. Recall: \[F(t) =\sigma T^4\] Multiplying this by the surface area of a sphere, we get:\[L = F(T) \times surface\,area = \sigma T^4 4\pi R^2\] \[L = \sigma T^4 4\pi R^2\]

Acknowledgements: Johnathan Budd & Willie Pirc

Operation Starshade

The Jet Propulsion Lab is currently working to tackle one of the most challenging issues in space imaging, and their approach is pretty spectacular. Until now, astronomers have been limited to only indirect methods of detecting planets orbiting far off suns. The issue is that the light from the planets' sun is far greater than the light emitted by planets themselves. To overcome this issue, the JPL is working to suppress the star's light.

Their approach to this revolves around a revolutionary apparatus, called a starshade. With a design that looks straight out of science fiction, the starshade is in the early stages of becoming a reality. The starshade is capable of being incorporated in a two-part spacecraft that allows for the use of pre-existing space telescopes. Additionally, it's unique outer petal design causes less bending of light, which prevents the stars light from leaking around the edges. 

According to the project's lead engineer, Dr. Stuart Shaklan, "Less light bending means that the starshade shadow is very dark, so the telescope can take images of the planets without being overwhelmed by starlight" (http://planetquest.jpl.nasa.gov/video/15). 

Below is a series of images that depict how the starshade will operate.

Initially, the starshade and telescope portion of the spacecraft are attached as a two part apparatus.

http://planetquest.jpl.nasa.gov/video/15

 The starshade then begins to rotate open and detach from the telescope, using built in thrusters. 

http://planetquest.jpl.nasa.gov/video/15

 The starshade continues to spin open, while beginning to align itself to the proper position for starlight supression.

http://planetquest.jpl.nasa.gov/video/15

http://planetquest.jpl.nasa.gov/video/15

http://planetquest.jpl.nasa.gov/video/15

 With the starshade fully open, the individual outer petals rotate into position to compete the final form of the shade blanket. 

http://planetquest.jpl.nasa.gov/video/15

 Finally, the starshade and telescope move into position with their respective thrusters and align with their target for image capturing.

http://planetquest.jpl.nasa.gov/video/15

 With the starshade in place, the telescope can begin taking images with the light from the star suppressed, and the newly visible planets clearly in sight. 

http://planetquest.jpl.nasa.gov/video/15

The development of JPL's starshade hold potential for a major leap in space imaging and planet discovery. It would open the possibility of discovering planets that may be similar to earth and hold potential for hosting life forms.

Below is a video of simulation of the starshade in action!

https://www.youtube.com/watch?v=uxLfJVoGBYU

Worksheet 4, Problem 2: Telescopes

CCAT is a 25-meter telescope that will detect light with wavelengths up to 850 microns. How does the angular resolution of this huge telescope compare to the angular resolution of the much smaller MMT 6.5-meter telescope observing in the infrared J-band? Ask a TF in class or the internet about the meaning of “J-band.”

In Worksheet 4, Problem 1, we explored how wavelength and aperture width relate to the resolution of a telescope. Now, we get to put this to the test, and see just how significant their impact is. As a reminder, the relationship is as follows: \[\theta = \frac{\lambda}{D}\] \[\theta = angle\,from\,central\,axis\] \[\lambda = wavelength\] \[D = Distance\,between\,slits\]
We can now use this to calculate the angular resolution, \(\theta_{CCAT}\), of the CCAT:
\[\theta_{CCAT} = \frac{\lambda}{D} = \frac{850 \times 10^{-4}cm}{25\times10^2cm} = 3.4 \times 10^{-5}\]
For the MMT, we need to determine the wavelength of the J-Band. This band correspond to 1.5cm - 3cm. As an estimate of this, we can use a wavelength of 2cm.

Now we can calculate the angular resolution, \(\theta_{MMT}\), of the MMT:
\[\theta_{MMT} = \frac{\lambda}{D} = \frac{2cm}{6.5 \times 10^2cm} = 3 \times 10^{-3}\]

To then compare the angular resolution of the two telescopes, we can divide the two, and determine how many times greater the resolution of the CCAT is than that of the MMT.
\[\frac{\theta_{MMT}}{\theta_{CCAT}} = \frac{3 \times 10^{-3}}{3.4 \times 10^{-5}} \approx 100\]
This shows that the CCAT resolution is approximately 100 times that of the MMT.

Acknowledgements: Johnathan Budd & Willie Pirc

Worksheet 4, Problem 1: Double Slits

In one of the most elegant and fundamental experiments in physics, Thomas Young helped to demystify the behavior of light and photons through his double slit apparatus. In this problem, we explore some of these properties, as well as their relationship to the Fourier Transform, while also interpreting the implications on telescopes.

The configuration of Young's Experiment involves casting a light source through two slits in a screen and on to a phosphorescent backdrop. The pattern shown on the second screen reveals alternating dark and light bands. These occur as a result of constructive and destructive interference between waves of light traveling through the two slits. This phenomenon can is illustrated below.

http://www.askamathematician.com/2014/08/q-before-you-open-the-box-isnt-schrodingers-cat-alive-or-dead-not-alive-and-dead/

To help understand the distribution of this interference pattern, we can consider the path that each light source travels before reaching the final screen.

Simplified interference pattern distribution sketch 

a) Convince yourself that the brightness pattern of light on the screen is a cosine function. (HINT: Think about the conditions for constructive and destructive interference of the light waves emerging from each slit).

To approach this thought exercise, we began by drawing the image shown above of the simplified interference pattern distribution. The pattern created on the far screen is a result of the wavelength of the light source, \( \lambda\), the distance between slits, D, and the angle from the central axis, \(\theta\). At the central axis, equidistant between the two slits, the waves of light travel the same distance to the screen, and thus are in phase. This results in a constructive interference, which creates a light spot. However, at all other points on the screen, the two waves of light from the slits travel different distances to backdrop. When the distance traveled by the light is off by a full wavelength, there is again constructive interference from the waves. When the distance is off by a half wavelength, however, there is destructive interference, which results in a dark spot. This property of constructive and destructive interference of waves can be seen below.

http://method-behind-the-music.com/mechanics/physics/

To further understand this distribution pattern, we can apply some simple geometry to derive an equation that guides this behavior. From our simplified interference pattern sketch, we can see that a right triangle can be drawn whose shortest leg depicts the difference in distance traveled by the light waves. Using trigonometry, we can relate the sides of the triangle with the top angle, \(\theta\). \[Sin(\theta) = \frac{X}{D}\]
X represents the difference in distance traveled by the two waves, and for constructive interference to occur, we know that X must be equal to \( \lambda\). Since in the scale we are looking at, the distance between the two screens, L, is much larger than the wavelength of light, we know that the angle (\theta\) must be extremely small. As a result, we can make the following estimation. \[Sin(\theta) \approx \theta\]
This is known as the small angle approximation. This allows us to then simplify our relational equation to: \[\theta = \frac{\lambda}{D}\]
As we discussed above, when \(\theta = 0\), or \(\theta = \frac{\lambda}{D}\), we get constructive interference. Then, for each half phase of \(\lambda\), or \(\theta = \frac{\lambda +0.5}{D}\), we get destructive interference. Using the knowledge that this distribution will be continuous and periodic across the screen, as well as the fact that constructive interference occurs at \(\theta = 0\), we can represent this distribution as a cosine function.

b) Now imagine a second set of slits placed just inward of the first set. How does the second set of slits modify the brightness pattern on the screen?

Adding a second set of slits to the screen introduces two more waves to the interference area. Since the slits are positioned closer together, the value of D will be smaller. According to \(\theta = \frac{\lambda}{D}\), this will create a wider interference pattern than our first two slits. When we consider what effect this will have on the overall pattern, we can consider how this new interference pattern will interact with our original pattern. At \(\theta = 0\), there will still be constructive interference, which will further add to the peak from the original interference pattern. As we move away from the origin, however, we will start to destructive interference occurring from the wider waves, which will make the central peak narrower. Additionally, the constructive peaks that exist away from the origin will be smaller, since the waves aren't perfectly in sync. Ultimately, the greatest effect of these new slits will be making the central peak brighter and narrower.

c) Imagine a continuous set of slit pairs with ever decreasing separation. What is the resulting brightness pattern?

The more slit pairs that we add with decreasing separation will amplify the effect described in part b, continuing to reduce the outer peaks, while making the central peak brighter and narrower. In a sense, this is increasing our resolution of discerning objects at a distance, since we can width of our central peak is generated by a smaller and smaller \(\theta\).

d) Notice that this continuous set of slits forms a “top hat” transmission function. What is the Fourier transform of a top hat, and how does this compare to your sum from the previous step?

The Fourier transform of a top hat is a Sinc function, which is depicted below.

http://www.diracdelta.co.uk/science/source/s/i/sinc%20function/source.html#.VONqQ1MbDmY

This agrees very well with our pattern from the sum of slits in part d. We can see the very narrow, intense central peak, and the greatly reduced outer peaks in this pattern.

e) For the top hat function’s FT, what is the relationship between the distance between the first nulls and the width of the top hat (HINT: it involves the wavelength of light and the width of the aperture)? Express your result as a proportionality in terms of only the wavelenght of light λ and the diameter of the top hat D.

In the Sinc function, the x-intercepts correspond to fully destructive interference, or dark spots on our interference pattern. These intercepts occur at \(\frac{\lambda}{D}\). This translates to a top hat width of  \(\frac{D}{\lambda}\).

This is an incredibly important equation, because it tells us how to optimize our aperture size and light source wavelength to get the best resolution, and narrowest peak. To do so, we would want to have a very large D value, and very small wavelength, λ.

f) Take a step back and think about what I’m trying to teach you with this activity, and how it relates to a telescope primary mirror.

This is vitally important to the design of telescopes, because it essentially tells us how to make telescopes with a high resolution. If we want to be able to distinguish between entities, we need to have the narrowest light intensity peaks possible. As we have shown, in order to do this, we would need to utilize a very wide telescope.

Acknowledgements: Johnathan Budd & Willie Pirc

Worksheet 2.1 Problem 3

This problem focuses on using dimensional analysis to derive an equation. I was particularly fond of this problem because it demonstrates how elegant and useful effective dimensional analysis can be. Additionally, it helps to break down a problem and illustrate the actual units dimensions that you are working with in a given equation, that may otherwise seem arbitrary.

This problem as to determine the relationship of pressure and density for a gas to determine the speed of sound traveling through the medium.

First, I approached this problem by determining the units for each of the constraints:
\[C_s = \frac{distance}{time} = \frac{cm}{s} \]
\[P = \frac{force}{area} = \frac{N}{cm^2} = \frac{kg * \frac{cm}{s^2}}{{cm}^2} = \frac{kg}{cm s^2}\]
\[\rho = \frac{mass}{volume} = \frac{kg}{{cm}^3} \]

With each of these terms defined by their units, we can start to tackle the relationship between pressure and density for determining the speed of sound in the gas.

First, lets consider our desired units: \[speed = \frac{distance}{time}\]

Next, lets consider the units of pressure from above: \[pressure = \frac{mass}{distance*{time}^2}\]

Finally, lets consider the units of density: \[density = \frac{mass}{volume}\]

Since we know that our desired units don't have mass, we should look to cancel out the mass units. Additionally, we know that time is on the bottom and distance is on the top. By a quick inspection, we can see that the dividing pressure by density will get us close to this.

\[ \frac{P}{\rho} = \frac{\frac{mass}{distance*{time}^2}}{\frac{mass}{{distance}^3}} = \frac{{distance}^3}{distance*{time}^2} = \frac{{distance}^2}{{time}^2}\]
From this, we see that the only difference between this and our final desired units is that both the top and bottom units are squared. To handle this, we can take the square root of both the top and bottom and achieve our desired results.
\[\sqrt{\frac{{distance}^2}{{time}^2}} = \frac{distance}{time}\]

Therefore...

\[C_s = \sqrt{\frac{P}{\rho}}\]

Collaborators on this problem included Willie Pirc and Johnathan Budd.

Worksheet 3 Problem 3

For this problem, we are imagining a star (AY Sixteenus) that is located at RA = 18 hours and Declination = +32 degrees. We are asked here to determine the LST of this star on the meridian as viewed from Cambridge.

For this problem, it is important to consider what LST is really telling us. LST is the local sidereal time, which is defined by a far away star. Fortunately for us, AY Sixteenus is exactly that, a far off star. By definition, AY Sixteenus will always be present for an observer at the zenith at 18:00 LST, as long as they are in the northern hemisphere. While the solar time may very throughout the year, the LST will not, and the star will always be present at LST 18:00.

The second part of this problem deals with solar time, rather than strictly Local Sidereal Time, which makes things a bit more interesting. This asks on what date will the star be on the meridian at midnight. To solve this, we want to consider on what date will LST 18:00 be equal to midnight.

Let's begin with the knowledge that LST is equal to 0:00 at noon on March 20th. Furthermore, LST will be equal to ~12:00 at midnight on March 20th. Since we are interested in when AY Sixteenus will be present overhead at midnight, we need to gain ~6 hours in LST for this to occur.

Since LST gains 4 minutes each solar day, we can determine how many days it will take to gain 6 hours:

\[ 6\,hours * \frac{60\,min}{1\,hour}* \frac{1\,day}{4\,min} = 90\,days\]

Knowing that AY Sixteenus will be present overhead 90 days following the Vernal Equinox, we can simply count 90 days past March 20th and discover that this will occur on June 18th.



Collaborators on this problem included Willie Pirc and Johnathan Budd.

Worksheet 3 Problem 2

After I learned a bit about the initially complex notion of sidereal time, I began to understand it's advantages in measuring time from a far off object. Rather than measuring a day based on the amount of time the earth takes to complete one rotation with respect to the sun, sidereal time considers the amount of time it takes the earth to complete a full rotation with respect to a far off star. In a previous problem that I worked through (which is not posted here) I worked out the difference between a sidereal and solar day on earth. Since it is the basis of how I will approach this problem, I'll give a quick overview of how this was determined.

The earth revolves around the sun over the course of 1 year (365 days). As an approximation, this means that the earth travels ~1 degree in its orbit each day. From a far off star, this corresponds to an additional degree of rotation each day. Knowing this, we can determine how much longer a solar day is than a sidereal day:\[ \frac{1\,degree}{1\,day} \times \frac{1\,day}{360\,degrees} \times \frac{24\,hours}{1\,day} \times \frac{60\,min}{1\,hour} = 4\,\frac{min}{day} \]
This problem has several parts that examine the Local Sidereal Time (LST) at different points throughout the year. Local Sidereal Time is defined as the right ascension that is at the meridian right now. At noon on the Vernal Equinox, LST = 0:00.

a) What is the LST at midnight on the vernal Equinox?

Since the a sidereal day is 4 minutes shorter than a solar day, at midnight on the Vernal Equinox, 12 hours will have passed, plus an additional 2 minutes, due to the difference in time. Therefore, the Local Sidereal Time will be 12:02 at midnight on the Vernal Equinox.

b) What is the LST 24 hours later (after midnight in part 'a')?

With another 24 hours passing, LST will gain an additional 4 minutes on top of the answer to part a. Again, since there are 4 additional minutes to a solar day, Sidereal Time must gain an additional 4 minutes per day. \[ 12:02 \,+ \,24 \,+\, 0:04\, = \,36:06\, = \,12:06 \]

12:06

c) What is the LST right now (to the nearest hour)?

Current time: 10pm, February 8th

Since the Vernal Equinox is on March 20th, 2015 it is still 40 days away. Again, using our 4 minute offset, we can determine the number of additional minutes that LST will gain to this point.
\[ 40\,days \times \frac{4\,min}{1\,day} \times \frac{1\,hour}{60\,min} = 2:40\,hours \]

Knowing that noon on March 20th, 2015 corresponds to LST 0:00, we know that noon today corresponds to that time minus 160 minutes. \[ 0:00 - 2:40 = 21:20\] Since the current time, however, is not noon, but 10pm, we need to add in an additional 10 hours, plus 2 more minutes for the difference in today's LST. \[21:20 + 10:02 = 31:22 = 7:22 = 7:00\, rounded\]

7:00

d) What will the LST be tonight at midnight (to the nearest hour)?

Tonight at midnight occurs 2 hours later than the previous time that we calculated in part c. The additional amount of time that occurs in the 2 hours will be negligible for this problem, so the answer is simply \[ 7:22 + 2:00 = 9:22 = 9:00\, rounded \]

9:00

e) What LST will it be at sunset on your birthday?

My birthday occurs on August 1st. In Boston, the sunset on August first will be at 8:05pm. Since August 1 occurs 134 days after the Vernal Equinox, we can calculate how far LST has shifted during this time: \[ 134 \,days \times \frac{4 \,min}{1 \,day} \times \frac{1\, hour}{60\, min} = 8:56 \,hours \]

Noon on August 1st will correspond to 0:00 + 8:56 = 8:56.

Then, adjusting to the time of sunset, we will have an additional 8:05. \[8:56 + 8:05 = 17:01\]

17:01

Collaborators on this problem included Willie Pirc and Johnathan Budd.

Worksheet 2.1 Problem 2

In order for a human being to register light with their eyes, a certain number of photons must enter. This problem examines just how powerful a light bulb must be in order for a person to see the light from a distance of 1 kilometer away in a dark cave.

Before diving into the mathematics of this problem, it is important to first consider what we know about the layout of this problem. First, we know that the number of photons that needs to reach the eye is ~10 photons. Additionally, we know that the eye has a refresh rate of approximately 10 Hz. Since the power output of the bulb will be dependent on the number of photons produced as a function of time, we will need to use both of these pieces of information to determine the output wattage.

Next we will want to consider the relevant equations to this problem.

The first equation is \( E = h\nu \). This tells us that the energy of a photon, E, is equal to the frequency, \(\nu\), times Planck's constant \(h=6.6 x 10^{-27}\frac{erg}{s} \).

The next equation is the surface area of a sphere, \(SA = 4 \pi r^{2} \). This is important because the light being produced by the bulb is isotropically radiated (going out evenly in all directions).

Now we can begin diving into the problem.

Since the light travels 1 km from the bulb and we are interested in the amount of light hitting the eye, we can set up the following proportion: \[ \frac{P_{eye}}{P_{bulb}} = \frac{SA_{eye}}{SA_{bulb}} \] Putting this in terms of our desired quantity of \(P_{bulb} \), our equation becomes: \[ P_{bulb} = \frac{P_{eye} SA_{bulb}}{SA_{eye}} \] Now, we will need to fully define each of the terms in the equation to solve for the Power of the bulb.

We can further define the power of the eye ( \(P_{eye} \) ) as the amount of energy reaching it per unit time, which we have declared to be 10 photons ( \(E_{eye} \) ) within the refresh rate of the eye ( \(T_{eye} \) ). This equation would be:\[ P_{eye} = \frac{E_{eye}}{T_{eye}} \] Additionally, we can define the energy of the eye using our initial energy equation. This gives:\[ E_{eye} = h\nu \] This, however, only provides the energy of a single photon. Since we want to know the value of 10 photons, we will multiply this by 10, and use the equation: \[ E_{eye} = 10 h\nu \]
To find the frequency of the light, we can use the following relationship:\[ \nu = \frac{c}{\lambda} \]
The surface area of the eye we can approximate to be \( 1 cm^2 \).

Using the surface area of a sphere, we can also determine the surface area of the bulb's light output with a 1 km radius. This is given by the equation: \[SA_{bulb} = 4 \pi r^2 \]
Inserting these into our equation for the power of the bulb, we now have a more complete equation of:
\[ P_{bulb} = \frac{10 h c 4 \pi r^2}{SA_{eye} \lambda T_{eye}} \]
The last step is to plug in numbers for for each term, as we have defined them.
\[h = 6.6 x 10^{-27}\frac{erg}{s} \]
\[c = 3.0 x 10^{10}\frac{cm}{s} \]
\[r = 10^{5} cm^2 \]
\[SA_{eye} = 1 cm^2 \]
\[\lambda = 500 x 10^{-7} cm \,(visible\, light) \]
\[T_{eye} = \frac{1}{10} {s} \]
This ultimately produces a result of: \[ P_{bulb} = 48 \frac{erg}{s} \]
I found this problem interesting, because it turned out to be quite straight forward determining the power output of the bulb at such a distance using basic physics concepts!



Collaborators on this problem included Willie Pirc and Johnathan Budd.

Hello, World!

Hi everyone!

My name is Tyler Kugler, and I am a senior in Eliot House studying Electrical Engineering with a secondary in Computer Science. I'm really excited to start exploring the world of astrophysics and begin understanding more about the construction of the universe. As a senior approaching graduation, I wanted to take a course vastly different than all others that I've previously taken, so here I am! Beyond the classroom, I captain the club tennis team, and spend my summers surfing back home on Cape Cod. Can't wait for an exciting semester!

- Tyler