Worksheet 6, Problem 4: The Stefan-Boltzmann Law

You observe two gravitationally bound stars (a binary pair). One is blue and one is yellow. They yellow star is six times brighter than the blue star. Qualitatively, compare their temperature and radii, i.e. which is hotter, which is smaller? Next, quantitatively compare their radii (to 1 significant figure).

To begin, qualitatively, we know that light from the yellow star has a larger wavelength than light from the blue star. Also, the energy corresponding to photons of larger wavelength is smaller than the energy for photons with smaller wavelength. Since temperature is directly proportional to energy and since energy is inversely proportional to wavelength, we can conclude that the temperature of the yellow star will be smaller than the temperature of the blue star.

We also know that $L\propto R^2T^4$. So if the temperature of the blue star is greater but the luminosity of the yellow star is greater, then it must be true that the radius of the yellow star is larger. 

So we have:

$T_B>T_Y$ and $R_Y>R_B$

Quantitatively, we can use the Stefan-Boltzmann law, which relates luminosity to radius and temperature. Let $L_Y$ be the luminosity of the yellow star and $L_B$ be the luminosity of the blue star. We know that $L_Y=6L_B$, so we can write the Stefan-Boltzmann law for each star:

$L_B=4\pi R_B^2\sigma T_B^4$      

      $L_Y=6L_B=4\pi R_Y^2\sigma T_Y^4$

Setting $L_Y=6L_B$ and canceling like terms, we get:

$R_Y^2 T_Y^4=6R_B^2 T_B^4$

Which can be written as a ratio of radii and temperatures:

$\dfrac{R_Y}{R_B} =\sqrt{6}\left(\dfrac{T_B}{T_Y}\right)^2$

From Wien's Law, we can solve for an expression for temperature in terms of wavelength:

$\lambda_{peak}=\dfrac{0.3\text{ cm K}}{T}\Longrightarrow T=\dfrac{0.3\text{ cm K}}{\lambda_{peak}}$

Plugging this into the expression we found for the ration of radii and temperatures, we get:

$\dfrac{R_Y}{R_B} =\sqrt{6}\dfrac{\left(\frac{0.3\text{ cm K}}{\lambda_Y}\right)^2}{\left(\frac{0.3\text{ cm K}}{\lambda_B}\right)^2}$

$\dfrac{R_Y}{R_B} =\sqrt{6}\left(\dfrac{\lambda_Y}{\lambda_B}\right)^2$

We can find the wavelengths of yellow light and blue light, which are approximately $\lambda_Y=570$ nm and $\lambda_B=450$ nm, as shown in the image below. 

Visible Light Spectrum
Image Credit: http://homepages.cwi.nl/~steven/Talks/2012/01-13-steven-colour/spectrum1.jpg

Putting these values into our equation for the ratio of radii:

$\dfrac{R_Y}{R_B} =\sqrt{6}\left(\dfrac{570}{450}\right)^2$

$\dfrac{R_Y}{R_B} \approx 4$               

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

Citations: 

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

http://en.wikipedia.org/wiki/Stefan%E2%80%93Boltzmann_law

Worksheet 5, Problem 2: Blackbodies- Wien's Law and the Raleigh-Jeans Tail

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 $F_{\nu}(T)$ over all frequencies to obtain the bolometric flux emitted from a blackbody, $F(T)$. You can do this by substituting the variable $u\equiv h\nu /kT$. 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 $\sigma$, which is also known as the Stefan-Boltzmann constant. If you're really into calculus, go ahead and show that $\sigma \approx 5.7\times 10^{-5} \text{ ergs s}^{-1}\text{ cm}^{-2}\text{ K}^{-4}$.

A blackbody is an object which absorbs all radiation incident upon it and then re-radiates this energy (isotropically) in a spectrum that is uniquely determined by the blackbody's temperature.

Planck Curve for a Blackbody
Image Credit:
http://quantumfreak.com/wp-content/uploads/2008/09/black-body-radiation-curves.png

We know that the equation for the intensity of radiation from a blackbody is given as:

$I_\nu (T)=\dfrac{2\nu ^2}{c^2}\dfrac{h\nu}{e^{\frac{h\nu}{kT}}-1}$

This equation gives us the energy per time, per area, per frequency, per solid angle. We want the equation for flux, $F_\nu (T)$, which gives us energy per time, per area, per frequency (without solid angles). So to 'cancel out' the solid angles, we must integrate $I_\nu (T)$ over all solid angles. Solid angles can be thought of as angles with a $\theta$ dimension and a $\phi$ dimension. Also, from Lambert's cosine law, we must scale the intensity by $\cos{\theta}$ since the blackbody is spherical, causing a surface area on the sphere to appear smaller to an observer who is not seeing the area from a head-on angle. Due to this fact, we multiply the intensity by $\cos{\theta}$, and since this is a spherical integral, we multiply by $\sin{\theta}$. So, integrating all solid angles viewed by an observer (a hemisphere), we get:

$F_\nu (T)=\displaystyle\int_{0}^{\frac{\pi}{2}}\int_{0}^{2\pi} I_\nu (T)\text{ }\cos{\theta}\sin{\theta}\text{ } d\phi d\theta$

          $=\displaystyle\int_{0}^{\frac{\pi}{2}}\int_{0}^{2\pi} \dfrac{1}{2}\text{ }I_\nu (T)\text{ }\sin{2\theta}\text{ } d\phi d\theta$

$=\pi\text{ } I_\nu (T)$                               

And now to get the bolometric flux, which is the energy per time, per area (independent of frequency), we must integrate over all frequencies.

$F(T)=\displaystyle\int_{0}^{\infty} F_\nu (T) \text{ }d\nu$      

                  $=\dfrac{2\pi h}{c^2} \displaystyle\int_{0}^{\infty} \dfrac{\nu^3}{e^{\frac{h\nu}{kT}}-1} d\nu$

Now let $u\equiv \frac{h\nu}{kT}$, where $\nu=\frac{ukT}{h}$ and $d\nu=\frac{kT}{h}du$. Substitute to get:

$F(T)=\dfrac{2\pi h}{c^2}\left( \dfrac{kT}{h}\right)^4 \displaystyle\int_{0}^{\infty} \dfrac{u^3}{e^{u}-1} du$

To complete this integral, which is called a Bose-Einstein integral, I used Mathematica. 

$F(T)=\dfrac{2\pi^5k^4T^4}{15c^2h^3}$

Let's define the coefficient of $T^4$ to be $\sigma$, which is known as the Stefan-Boltzmann constant:

$\sigma=\dfrac{2\pi^5k^4}{15c^2h^3}=\dfrac{2\pi^5(1.38\times 10^{-16}\text{ erg K}^{-1})^4}{15(3\times 10^{10}\text{ cm s}^-1)^2(6.626\times 10^{27}\text{ erg s})^3}$

$\sigma\approx 5.7\times 10^{-5}\text{ erg s}^{-1}\text{ cm}^{-2}\text{ K}^{-4}$

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

The relation between wavelength and frequency is $\lambda=\frac{c}{\nu}$, and we know that:

$F(T)=\displaystyle\int_{0}^{\infty} \dfrac{2\pi h}{c^2}\left( \dfrac{kT}{h}\right)^4 \dfrac{u^3}{e^{u}-1} du$

Changing $u$ to be in terms of $\lambda$, we get $u= \frac{hc}{\lambda kT}$ and $du=-\frac{hc}{\lambda^2kT}d\lambda$. Plugging these back into the expression above, we get:

$F(T)=\displaystyle\int_{\infty}^{0}\dfrac{2\pi h}{c^2}\left( \dfrac{kT}{h}\right)^4 \dfrac{\left(\frac{hc}{\lambda kT}\right)}{e^{\frac{hc}{\lambda kT}}-1} \left( -\dfrac{hc}{\lambda^2 kT}\right)d\lambda$

$F(T)=\displaystyle\int_{0}^{\infty}\dfrac{2hc^2}{\lambda^5 (e^{\frac{hc}{\lambda kT}}-1)}d\lambda$

So from this, we can conclude that $B_\lambda (T)$ is given by:

$B_\lambda (T)=\dfrac{2hc^2}{\lambda^5 (e^{\frac{hc}{\lambda kT}}-1)}$

c) Derive and expression for the wavelength $\lambda_{max}$ corresponding to the peak of the intensity distribution at a given temperature $T$. Once you do this, again substitute $u\equiv h\nu /kT$. The expression you end up with will be transcendental, but you can solve it easily to first order, which is good enough for this exercise. 

First, we must find the maximum of the function $B_\lambda (T)$ for a given temperature $T$. To do this, we can take the derivative of the function and set it equal to zero. 

$0=\dfrac{\partial B_\lambda (T)}{\partial \lambda}$                                        

$0=\dfrac{\partial}{\partial \lambda}\left(\dfrac{2hc^2}{\lambda^5 (e^{\frac{hc}{\lambda kT}}-1)}\right)$                  

                               $0=2hc^2\left(-5\lambda^{-6}(e^{\frac{hc}{\lambda kT}}-1)^{-1}+\frac{hc}{kT}\lambda^{-7}e^{\frac{hc}{\lambda kT}}(e^{\frac{hc}{\lambda kT}}-1)^{-2}\right)$

Cancel out like terms and let $u\equiv \frac{hc}{\lambda kT}$ so that we get:

 $0=-5+\dfrac{u e^{u}}{e^{u}-1}$

Letting Wolfram Alpha find an approximation for this value graphically, we find that $u\approx 4.965$. But $u= \frac{hc}{\lambda kT}$, so we can solve this for the peak wavelength.

$u= \dfrac{hc}{\lambda kT}$

$\lambda =\dfrac{hc}{ukT}$

$\lambda=\dfrac{(6.626\times 10^{-27}\text{ erg s})(3\times 10^{10}\text{ cm/s})}{4.965(1.38\times 10^{-16}\text{ erg/K}) \text{ }T}$

$\lambda_{peak}=\dfrac{0.3 \text{ cm K}}{T}$

And we have now derived Wien's Displacement law.

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 $e^x \approx 1+x$.

We know that $B_{\nu}(T)$ is given by:

$B_\nu (T)=\dfrac{2\nu ^2}{c^2}\dfrac{h\nu}{e^{\frac{h\nu}{kT}}-1}$

Using the Taylor expansion for $e^x$, we can simplify this expression to:

$B_\nu (T)=\dfrac{2\nu ^2}{c^2}\dfrac{h\nu}{1+\frac{h\nu}{kT}-1}$

$=\dfrac{2\nu ^2}{c^2}\dfrac{h\nu}{\frac{h\nu}{kT}}$

Giving us a final answer of: 

$B_\nu (T)=\dfrac{2\nu ^2}{c^2}kT$

And now we have derived the Rayleigh-Jeans Tail. The Rayleigh-Jeans Tail is only accurate for larger wavelengths, or lower energy, and is quite inaccurate for smaller wavelengths and higher energies, as shown in the graph below. This discrepancy with observations in the smaller wavelengths is called the "ultraviolet catastrophe" since it predicts that blackbodies will emit radiation with infinite power in the shorter, ultraviolet wavelengths. This obviously is not the case, however.

Rayleigh-Jeans Tail and the Ultraviolet Catastrophe in comparison to the Planck Curve
Image Credit: http://hyperphysics.phy-astr.gsu.edu/hbase/imgmod/uvcatas.gif

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, $L$.

Power is given in units of energy per time. We know the bolometric flux, which has units of energy per area per time, so we can just integrate over the area to get energy per time. However, since we already know the surface area of a sphere is $4\pi R^2$, we can just multiply the bolometric flux by the area to find the bolometric luminosity. So we get:

$L=4\pi R^2 F(T)$

$L=4\pi R^2 \sigma T^4$

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

Citations: 

http://en.wikipedia.org/wiki/Lambert%27s_cosine_law
http://en.wikipedia.org/wiki/Rayleigh%E2%80%93Jeans_law
http://hyperphysics.phy-astr.gsu.edu/hbase/mod6.html

Free Form: Extreme Mass-Ratio Binary Stars

Recently, the Harvard-Smithsonian Center for Astrophysics discovered an extreme mass-ratio binary system that is especially interesting because their finding introduced a new class of binary stars: while one of the stars was fully formed, the other star was still in the process of formation.  

Before we can appreciate what this discovery means, however, we need to have a better understanding of binary stars. Binary stars are a gravitationally bound pair of stars, meaning that the two stars are orbiting around each other without escaping or falling into their gravitational pull on each other. These stars are useful because we can use their orbits to find the mass of each star, and then knowing their masses, we can indirectly find their radii and densities. Most stars in our galaxy are in binary systems, and particularly, more massive stars tend to have companion stars. But stars in binary systems usually have similar masses, but not always. This is why astronomers have been looking for binary systems of stars with very different masses, called extreme mass-ratio binaries, so that they can better understand how binary stars are formed and perhaps how they evolve. To give you an idea of the difference in masses in extreme mass-ratio binary systems, the more massive stars tend to be from 6-16 times the mass of the sun, and the and the less massive stars are around 1-2 times the mass of the sun.

The discovery of extreme mass-ratio binaries is somewhat nuanced since the more massive star would be much brighter, causing it to outshine and therefore hide the lighter star in its overpowering light. So in order to see the lighter star, astronomers look for eclipsing systems since when the lighter star is directly in front of the more massive star, the total brightness seen drops significantly. But for this to happen, there must be a perfect alignment of the two stars with respect to Earth. 

Image Credit: Robert Gendler and Josch Hambsch 2005
http://www.sciencedaily.com/releases/2015/02/150212102850.htm

This recent discovery of the extreme mass-ratio binaries in formation, depicted in the image above, is especially important because, not only have we found another class of binary stars, we also get to observe the beginning and evolution of a binary system in order to better understand massive stars and binaries in general.

Citations:

http://www.sciencedaily.com/releases/2015/02/150212102850.htm

https://www.cfa.harvard.edu/news/2015-06

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

Worksheet 4, Problem 2: Comparing 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 the infrared J-band? 

The angular resolution of a telescope is the telescope's ability to see objects or details that are an angular distance $\theta$ apart. The angular resolution of a telescope is given by the equation:

$\theta =\dfrac{1.2\lambda}{D}$

Where $\theta$ is the angular resolution in radians, $\lambda$ is the wavelength of the incident light, and $D$ is the diameter of the telescope's lens aperture. 

The telescopes, CCAT, has diameter $D=25$ m, and is observing light at $\lambda =850\times 10^{-6}$ m. Putting these values into the equation for angular resolution, we get:

$\theta _{CCAT} = \dfrac{1.2(850\times 10^{-6}\text{m})}{25\text{m}}$

$\theta_{CCAT} = 4.08\times 10^{-5}\text{ radians}$

           $= 2.3\times 10^{-3}\text{ degrees}$

This means that the CCAT can resolve objects or details within angular distance of $4.08\times 10^{-5}$ radians apart or $2.3\times 10^{-3}$ degrees apart.

The smaller telescope, the MMT, has a diameter of $D=6.5$ m, and is observing wavelengths in the J-band, which is the range of wavelengths centered on 1.25 micrometers. Plugging these values into the equation, we have:

$\theta _{MMT}= \dfrac{1.2(1.25\times 10^{-6}\text{m})}{6.5}$

$\theta_{MMT} = 2.3\times 10^{-7}\text{ radians}$

            $=1.2\times 10^{-5}\text{ degrees}$

The MMT can resolve much smaller objects than the CCAT telescope can at their respective wavelengths. In fact it is able to resolve objects that are 177 times smaller than those that the CCAT can.

I would like to thank Charles Law for his help in solving this problem and for proof reading this blog post.

Worksheet 4, Problem 1: Angular Resolution

To understand the basics of astronomical instrumentation, it is useful to go back to the classic Young's double slit experiment. Draw a double slit setup as a 1-D diagram. The two slits are separated by a distance $D$, and each slit is $w$ wide, where $w\ll D$ such that its transmission function is basically a delta function. There is a phosphorescent screen placed a distance $L$ away from the slits, where $L\gg D$. We'll be thinking of light as plane-parallel waves incident on the slit-plane, with a propagation direction perpendicular to the slit plane. Further, the light is monochromatic with a wavelength $\lambda$.

You may be familiar with Thomas Young's double slit experiment which demonstrated the wave nature of light. In this experiment, Young directed light through two small slits and placed a screen a little distance behind the slits that would detect the incident light from the slits. What he found was an interference pattern of bright and dark fringes similar to what you might see in intersecting water waves/ripples. This problem explores some of the physics behind the double slit experiment and applies it to observational astronomy and telescopes.

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).

Figure 1. Double Slit Experiment

First, using the fact that $L\gg D$, we can assume that $\theta$ is very small and that $r_1$ and $r_2$ are parallel. This makes the small triangle in the diagram above look like the following:

Figure 2. Approximations of Angles in Diagram for Double Slit Experiment

The light passing through the bottom slit has to travel further than the upper slit in order to reach the point $y$ on the screen shown in Figure 1. This difference in distance will cause an offset in the wavelengths of the light passing through the two slits, which, in turn, will cause interference in places where the peaks and troughs don't align. In order to understand the pattern of interference observed on the screen, we must find the extent of the offset. Since $\theta$ is small, we can approximate that the length of $r_2$ is greater than that of $r_1$ by $d$, and from Figure 2, we can see that $d=D\sin{\theta}$. 

Now, to find the effect of the offset, $d$, on the interference of the light, we can derive equations for the conditions of constructive (bright fringes) and destructive (dark fringes) interference. For constructive interference, the peaks must perfectly align, so $d$ has to be an integer multiple of the wavelength.

$d=n\lambda$

$D\sin{\theta}=n\lambda$          

Where $n=0,1,2,3\ldots$.

And for destructive interference, the peaks of one wave, should line up exactly with the troughs of the other, so they should be offset by half a wavelength.

$d=(n+\dfrac{1}{2})\lambda$

$D\sin{\theta}=(n+\dfrac{1}{2})\lambda$      

For $n=0,1,2,3\ldots$.

Since $\theta$ is small, we can make the approximation that $R\approx L$. Using this approximation, for constructive interference, we get:

$D\sin{\theta}=D\dfrac{y}{L}=n\lambda$

Which gives us the condition:

$y_{con}=n\dfrac{\lambda L}{D}$

And for destructive interference:

$y_{des}=(n+\dfrac{1}{2})\dfrac{\lambda L}{D}$

From this, we can already see a periodic nature for the constructive and destructive interference of the waves, which implies that the brightness pattern may be a trigonometric function. Since we have a bright fringe at $y=0$ for $n=0$, we would have a cosine function.

The pattern that would be seen on the screen is shown in the following figure, where the right pane depicts the bright and dark spots on the screen from the interference pattern.

Figure 3.
Image Credit: http://ipodphysics.com/prop-of-light-youngs-double-slit.php

Also, note that from the conditions we derived for constructive and destructive interference, we can see that the further away the screen is from the slits (for larger $L$) and the larger the wavelength of the light, the more spread apart the bright fringes are. However, the further the slits are apart (for smaller $D$), the closer the bright fringes are together.

Another way of understanding the cosine-form brightness pattern on the screen is through Fourier transforms. The two slits can be modeled as delta functions and taking their Fourier Transform, we get a cosine function, as shown in the image below.

Figure 4. The Fourier transform of two delta functions is representative of the interference pattern by two slits
Image Credit: Gray's Stellar Photospheres

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? 

Interference patterns are created by 'adding' the light waves together (a wave can be represented as a sinusoidal function and the overlap of waves which causes interference can be though of as the sum of each light wave's sine function). Due to this fact, we can treat the second set of slits as a separate occurrence, and then add the brightness patterns of both sets of slits together to find what the brightness pattern would be for all four slits. As we found in part (a), the smaller the distance between the two slits, the more spread apart the brightness pattern is, meaning that the second set of slits just inside of the first set will have a smaller separation between slits and therefore a wider resulting cosine function as its brightness pattern. When the two interference patterns are added together, we get something that resembles this:

Figure 5. Superposition of Cosines with Different Wavelengths

So the superposition of the two interference patterns caused the central maximum brightness (at $n=0$) to increase in intensity and the other bright fringes decrease in intensity.

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

If we continue the process discussed in part (b) by adding more cosine functions that are even more spread each time, eventually, the resulting brightness pattern will be focused more towards the center, with a very bright central maximum $(n=0)$ and almost imperceptible secondary maxima. The resulting brightness pattern will look like a sinc function as depicted below:

Figure 6. The sinc function is representative of the brightness pattern that would be seen on the screen.

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 set?

We can use Fourier transforms to find the resulting brightness pattern on the screen. As the number of slits goes towards infinity with their separation nearing zero, the resulting system of slits approaches a "top hat," or a larger opening with width $D$. 

Figure 7. Limit as number of slits approaches infinity and separation approaches zero

The Fourier transform of this top hat is the sinc function, as shown in the image below:

Figure 8. Fourier Transform of Top Hat
(Note: for consistency, we will refer to $W$ in the image as $D$)
Image Credit: Gray's Stellar Photospheres

Notice that the distance from the central peak to the first dark fringe, or null, is given as $\dfrac{1}{D}$.

e) For the top hat function's Fourier transform, 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 you result as a proportionality in terms of only the wavelength of light $\lambda$ and the diameter of the top hat $D$.

As explained in part (d), the slit approaches a top hat shape, which is essentially a slit with width $d$. The following image shows what the brightness pattern would look like and includes the equation of the condition for minima, or nulls.

Figure 9. Single Slit Diffraction
(Note: for consistency, we call the width of the slit $d$ instead of $a$, and the distance to the screen $L$ instead of $D$)
Image Credit: http://www.gitam.edu/eresource/Engg_Phys/semester_1/optics/diffrac_files/sinslit.gif

Incidentally, the equation for nulls in a single slit setup is the same as the equation for the double slit setup. The distance from the central maximum to the nulls is given by:

$y=n\dfrac{\lambda L}{D}$

Additionally, we know from part (a) that larger wavelengths result in a more spread apart brightness pattern, and from the Fourier transform in part (d) we saw that the distance to the first null is inversely proportional to the distance $D$ between the two outer slits.

Using the conclusions from the single slit setup and from our observations in parts (a) and (d), we can see that the distance to the first null is directly proportional to wavelength and inversely proportional to the width of the slit.

$y\propto \dfrac{\lambda}{D}$

f) Take a step back and think about what this activity is trying to teach you and how it relates to a telescope primary mirror. 

Applying the this problem to telescopes, a telescope's primary mirror is essentially the screen in the double slit experiment, and the telescope's aperture is like the slit with width $D$. This means that the resulting image will have a brightness pattern which can be determined using a similar method as we did for the top hat slit with a width. The difference in this case is that the 'slit' is in two dimensions, so the Fourier transform will be a jinc function, shown below:

Figure 10. The jinc function is representative of the brightness pattern that would be seen from the telescope.
Image Credit: (Left) http://upload.wikimedia.org/wikipedia/en/e/e6/Airy-3d.svg (Right) http://upload.wikimedia.org/wikipedia/commons/1/14/Airy-pattern.svg

The telescope's ability to resolve objects further away (its angular resolution), with smaller angular size, is determined by the central maximum peak (which is the central bright dot in the right image in Figure 10. The brighter (and therefore narrower) the peak, the more focused the image in the telescope and the greater the telescopes ability to resolve. This implies that the angular resolution $\theta$ is directly proportional to the distance to the null $y$ in the brightness pattern. As we found in part (e), $y$ is directly proportional to wavelength and indirectly proportional to width, or diameter $D$ in this case. So we have:

$\theta \propto \dfrac{\lambda}{D}$

This gives us the minimum possible angular size that the telescope can resolve at a given wavelength. 

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

Free Form: Exoplanets and Kepler-432b

Recently, a Jupiter-like exoplanet, or extrasolar planet, called Kepler-432 b was discovered with unusually high density and a mass six times larger than Jupiter's but a size that is the same as Jupiter's. In fact, Kepler-432 b is one of the most massive and densest planets we have observed. Moreover, the orbit of Kepler-432 b around a red giant (a star in a later stage of its evolution) is very small and strangely shaped for a planet so massive. The image below shows what Kepler-432 b's orbit around the red giant looks like in comparison to the orbit of Mercury around our Sun.

Orbit of Kepler-432 b Compared to Mercury
Image Credit: http://www.sciencedaily.com/releases/2015/02/150212114243.htm

Due to the shape of the orbit, which causes the planet to be closer to the star at some points of its orbit and further at others, Kepler-432 b has very extreme seasons. One year on Kepler-432 b is equivalent to 52 Earth days, and the temperatures between seasons vary from 500 degrees Celsius to 1000 degrees Celsius. Since the planet's orbit is not centered around its star, the summer season (closest to the star) is significantly shorter than the winter season (furthest from the star).

The reason this finding is interesting for astronomers is because lately, they have been finding more Jupiter-like planets orbiting older stars like red giants. Their findings have shown that the orbits of these Jupiter-like planets around red giants and other evolved stars have different characteristics than their orbits around regular, main-sequence stars. Particularly, astronomers have noted that they don't find many such planets in close orbit around giant stars. Their predictions for why this might be the case include two possibilities: either the giant stars pull the planets inward and swallow them or the Jupiter-like planets form around regular, intermediate-mass stars but don't move inward on their orbits. The finding of Kepler-432 b obviously shows that it is possible for such planets to be in close orbit around giant stars, but astronomers postulate that within the next 200 million years, the planet will be completely sucked in by the red giant.

By studying this topic, astronomers might better understand the evolution of orbiting planets as their stars evolve from main-sequence stars. This new understanding can, in turn, be used to predict what our Solar System might look like as our Sun reaches the end of its main sequence. More generally, this is one of the goals of extrasolar planetary astronomy. Through observing the characteristics of other planets and their orbital systems, we can begin to make sense our own Solar System and how it may change over time. We can also work the other way and use our Solar System to make predictions on the characteristics of other solar systems and planets. One example of this is the search for extraterrestrial life and other inhabitable planets.


Citations:
http://www.sciencedaily.com/releases/2015/02/150212114243.htm
http://www.aanda.org/articles/aa/pdf/2015/01/aa25146-14.pdf
http://www.aanda.org/articles/aa/pdf/2015/01/aa25145-14.pdf

Worksheet 3, Problem 2: Local Sidereal Time

The Local Sidereal Time (LST) is the right ascension that is at the meridian right now. 

LST = 0:00 is at noon on the Vernal Equinox (the time when the Sun is on the meridian March 20th, for 2015).

a) What is the LST at midnight on the Vernal Equinox?
b) What is the LST 24 hours later (after midnight in part 'a')?
c) What is the LST right now (to the nearest hour)?
d) What will the LST be tonight at midnight (to the nearest hour)?
e) What LST will it be at Sunset on your birthday?

Local Sidereal Time (LST) is a measure of time on Earth with respect to the fixed stars. As opposed to solar time, which is the time it takes for the Sun to return to the meridian, sidereal time is the time it takes for a fixed star to return to the meridian. A sidereal day is approximately 23 hours and 56 minutes long, so it is about 4 minutes shorter than a solar day. Also, we are told in the question that on March 20th, at noon, LST = 0:00. Using these two facts, we can create a general equation to solve for the LST at a specific date and time.

Let $d$ be the number of solar days after the Vernal Equinox (March 20th), $h$ be the number of hours into the solar day in consideration, and $m$ be the number of minutes into the hour. Since a sidereal day is 4 minutes shorter than a solar day, for every solar day, we have completed a sidereal day and 4 minutes. So we can create an equation that solves for the number of minutes the Local Sidereal Time is ahead of the solar time $(x)$.

$x=4(d+\dfrac{h}{24}+\dfrac{m}{1440}+\dfrac{1}{2})$

In this equation, we are multiplying the number of days by 4 minutes since this is time offset between the solar and sidereal day. We divide the hours by 24 and minutes by 1440 (number of minutes in a day) to find the fractional day. The added $\frac{1}{2}$ is to account for the fact that LST is 0:00 twelve hours into the solar day.

a) I assumed that this meant the midnight following noon. This is 12 hours after noon, which is half a day. Since 4 minutes are added for a full day, we can deduce that there will be 2 minutes added for half a day. Here, the problem is simple enough that we do not need to use our equation. So at 12:00am solar time, it will be 12:00 + 0:02 = 12:02am in LST.


b) 24 hours following the midnight after Vernal Equinox, the LST will further increase by 4 minutes since a day has passed. So we have, 12:02 + 0:04 = 12:06am in LST.


c) Right now, it is February 10th and it is 1:00pm. It is 326 days and 13 hours after the Vernal Equinox. This means that $d=326$, $h=13$, and $m=0$. Plugging into our equation, we get:

$x=4(326+\dfrac{13}{24}+\dfrac{0}{1440}+\dfrac{1}{2})=1308$ mins

Converting this into hours, we divide by 60 to get:

$\dfrac{1308\text{ mins}}{60\text{ mins}}=21.8\approx 22$ hours

This means that we are 22 hours ahead of the time right now, which is 13:00 + 22:00 = 11:00am.

d) At midnight tonight, it will be 327 days after the Vernal Equinox. So we have $d=327$, $h=0$, and $m=0$. Plugging these values in, we get:

$x=4(327+\dfrac{0}{24}+\dfrac{0}{1440}+\dfrac{1}{2})=1310$ mins

Converting this into hours, we divide by 60 to get:

$\dfrac{1310\text{ mins}}{60\text{ mins}}=21.833\approx 22$ hours

This means that it is 22 hours past midnight (0:00), which is 22:00, or 10:00pm in LST.

e) My birthday is on May 5th, which is 45 days following the Vernal Equniox, and sunset occurs at 7:56pm (19:56). Putting these values into our equation, we get:

$x=4(45+\dfrac{19}{24}+\dfrac{56}{1440}+\dfrac{1}{2})\approx 186$ mins

Converting this into hours, we divide by 60 to get approximately 3 hours. This means that the LST is 3 hours ahead of the solar time, so we have:

7:56pm + 3:00 = 10:56pm $\approx$ 11:00pm (23:00)

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

Worksheet 3, Problem 1: The Sidereal Day vs. Solar Day on Mars

What is the difference between sidereal and solar day on Mars if Mars has the same rotation period and orbits at 1.5 AU?

We would like to solve for $\phi$ since this how much more Mars must rotate after a sidereal day to complete a solar day. Since a sidereal day is relative to the fixed stars which have an arbitrary and large distance from Mars, we can assume that they are infinitely far away, which makes the lines extending from Mars to the stars horizontal. Therefore, we can approximate that $\phi\approx\theta$. To solve for $\theta$, which is the angle subtended by one rotation of Mars about the Sun (given to be 24 hours), we must first find the period of Mars' orbit around the Sun. We can do this using Kepler's Third Law.

$P^2=\dfrac{4\pi ^2d^3}{GM_{tot}}$

We were not given the total mass of the system, however, so we must solve for the period of Mars using a ratio to the known period of Earth and the assumption that the total mass of the system is approximately equal for both the Sun and the Earth and the Sun and Mars. Letting $P_E$ and $P_M$ be the period of Earth and Mars respectively, $d_E$ and $d_M$ be the orbital radii, and $M_E$ and $M_M$ be the total mass of their respective systems, we have:

$\big(\dfrac{P_M}{P_E}\big)^2=\dfrac{\frac{4\pi ^2d_M^3}{GM_M}}{\frac{4\pi ^2d_E^3}{GM_E}}$

$\big(\dfrac{P_M}{P_E}\big)^2=\dfrac{d_M^3}{d_E^3}$

$P_M=P_E\sqrt{\dfrac{d_M^3}{d_E^3}}$

Plugging in the values $P_E=365$ days, $d_E=1$ A.U., and $d_M=1.5$ A.U., we get:

$P_M=(365\text{ days})\sqrt{\dfrac{(1.5\text{ A.U.})^3}{(1\text{ A.U.})^3}}=670$ days

Now, we must find $\theta$, the number of degrees Mars rotates about the Sun in one day. Assuming circular motion:

$\theta = \dfrac{360^{\circ}}{670\text{ days}}\approx 0.54^{\circ}$

Since we determined that $\phi\approx\theta$, we know that it takes 0.54 degrees longer to complete a solar day. So to find the difference between the sidereal and solar day, we just need to find how long it takes to rotate $0.54$ degrees.

$\dfrac{24\text{ hours}}{360^{\circ}}=\dfrac{t}{0.54^{\circ}}$

$t=0.036\text{ days}\approx 2\text{ mins}$

Therefore, the solar day is approximately 2 minutes longer than the sidereal day on Mars.

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

Worksheet 2.1, Problem 6: Calculating the Work Done by a Fly in CGS Units

How much energy does a fly exert if it does a single push-up in cgs (centimeters, grams, seconds) units. This should be about equal to the standard cgs unit of energy. Assume a fly is approximated by a cube 0.25 cm on a side, and is made of water (density of 1 g cm$^2$).

$s = 0.25$ cm
$\rho = 1$ g cm$^{-3}$

First, we must find the mass of the fly. To do this, we can estimate the size of the fly to find its volume, and then multiply by its density. We are already given that a fly can be considered as a cube with a side-length of 0.25 cm and with the density of water. Solving for the volume, we get:

$V=s^3=0.25^3\approx 0.015cm^3$

And multiply by density to find the mass:

$m=V\rho=(0.015cm^3)(1g\cdot cm^{-3})=0.015g$

Then to find the energy exerted by the fly, we must solve for the work done by the fly. Let's assume that a push-up is the upward lift of the fly's body and that the fly must lift its body 0.25cm.

$W=mgh=(0.015g)(1000cm\cdot s^{-2})(0.25cm)\approx 3.8ergs$

We found that the energy exerted by the fly is around 3.8 ergs. This is likely an overestimate since we were told in the question that 1 erg is a reasonable answer. The error in our answer could have resulted from a large approximation of the density of the fly and/or its volume. Perhaps treating the fly as a sphere with a diameter of 0.25cm might have been a better estimate since this would decrease the volume by a factor of two.

I would like to thank Charles Law for his assistance in solving this problem and in proofreading this blog post.

Image of fly: http://images.wisegeek.com/common-fly.jpg

Worksheet 2.1, Problem 4: Dimensional Analysis to Derive Kepler's Third Law Using

Imagine you are on a desert island (without wireless) and for some reason you need Kepler's Third Law of motion. Using dimensional analysis, what is the form of this equation, which relates the period, total mass, and separation of a two-body gravitational orbit? Use only the units and a tad of physics, e.g. what constants are likely involved? (Note: there are many paths to this solution)

We are given that Kepler's Third Law relates period $(P)$, total mass $(M_{tot})$, and the distance between the two objects $(d)$. The units of each are seconds $(s)$, grams $(g)$, and centimeters $(cm)$ respectively. We can already see that in order to relate the period to mass and distance, another constant is needed. This constant is the gravitational constant, $G$, whose units can be found using the equation for the gravitational force between to objects.

$F=\dfrac{GM_1M_2}{d^2}$

Solving for $G$ we get:

$G=\dfrac{Fd^2}{M_1M_2}$

This means that the units of $G$ are:

$[G]=\dfrac{(g\cdot cm/s^2)(cm^2)}{g^2}$

$[G]=\dfrac{cm^3}{g\cdot s^2}$

Now, in finding the form of Kepler's Third Law, we should consider what is actually happening between these two masses. The orbiting mass is gravitationally bound to the object it is revolving around. 

Let's imagine two scenarios of an object orbiting another object. In each scenario, the distance between the orbiting mass and the central mass is the same, but in one scenario, the total mass of the system is greater than that of the other scenario. We know by laws of gravity that objects with larger mass will have stronger gravitational force and therefore stronger gravitational attraction. This means that in the scenario with greater total mass, the orbiting object would have to revolve at a faster pace in order to keep from falling into the central mass, meaning that the period would be smaller than that of the other scenario. This implies that mass is inversely proportional to the period (so it should go in the denominator of Kepler's equation).

Similarly, imagine two scenarios of an object orbiting another object, but this time the total mass of the system is the same in both, and the distances differ. Again, by our understanding of gravity, we know that the force of gravity is weaker at larger distances, so in the scenario where the separation of the masses is larger, the orbiting mass will revolve at a slower rate, meaning that the period will be larger. This implies that the separation is directly proportional to the period (so it should go in the numerator of the equation).

So we have:

$[P]= [G]^a\dfrac{[d]^b}{[M_{tot}]^c}$

$s=\bigg(\dfrac{cm^3}{g\cdot s^2}\bigg)^a\dfrac{(cm)^b}{(g)^c}$

In order to make units cancel, we need $G$ to be in the denominator to take care of the $cm$ in the numerator and $g$ in the denominator. But $G$ has $cm^3$, so we must cube the distance $d$ so that $cm$ no longer remain. The grams in $G$ and $M_{tot}$ cancel each other out, and we are left with $s^2$, so we need to take a square root and we will be left with seconds, which is what is desired.

So $a=-1$, $b=3$, and $c=1$. This leaves us with the form:

$P\propto\sqrt{\dfrac{d^3}{GM_{tot}}}$

Which can be rewritten as:

$P^2\propto\dfrac{d^3}{GM_{tot}}$

And to check our answer:

$[P]^2=\dfrac{[d]^3}{[G][M_{tot}]}$

$s^2=\dfrac{cm^3}{\dfrac{cm^3}{g\cdot s^2}\cdot g}=s^2$

This answer agrees with the full form of Kepler's Third Law, which is given as:

$P^2=\dfrac{4\pi ^2d^3}{GM_{tot}}$

I would like to thank Charles Law for his help in solving this problem and in proofreading this blog post.

Hello World!

My name is Deanna Emery and I am a sophomore at Harvard University majoring in physics and astrophysics. I am currently taking the class Astro 16: Stellar and Planetary Astronomy taught by Prof. John Johnson. This blog will be dedicated to discussing and solving some of the interesting problems worked on in class and any other related topics. I took this class out of interest in the subject and as an introduction to a field of astronomy.

A little bit about myself: I was born and raised in a small town in Vermont, living in the middle-of-nowhere, where stars in the night sky are clear and bright, unaffected by light pollution. When I was little, like many other kids my age, I wanted to be an astronaut when I grew up. While I may not end up becoming an astronaut, I hope to be a researcher in either physics or astrophysics. In addition to my interests in science, I also enjoy singing, and am part of the Radcliffe Choral Society, an all-women's choral group at Harvard.