Eclipsing Binary Lab

Introduction/Purpose

In this lab, we observed a transiting binary star system using the Harvard Clay telescope.  Since the system is far too far away to dirtily resolve wither equipment, we measured the observed luminosity of the system as a function of time.  With this data, we can effectively find the masses, radii, and separation distance of the two stars.  

This is scientifically relevant because this system was recently discovered by Professor John Johnson (of AY16) and it is a great experience for undergraduates to do real research and data collection.

These binary systems are star systems in which two stars orbit one another.  This causes a very distinctive light curve when observed form Earth if one star passes in front of the other form our line of sight.

Here is a diagram of the system.  The two stars orbit one another around the dotted axis a distance a apart.  Imagine that the Earth is to the very far right side of your computer screen, then add a few lightyears.

This is what we would see from Earth (with an insanely powerful telescope) .  At this point in time, the smaller star is in front of the larger star.  This would cause a decrease in brightness rom the system.  Stars are great at emitting light, but they are also good at blocking it.  

A little bit later, we would see the system back at full brightness as the Stars no longer overlap.

Next, we would see a second dip in brightness as the small star passes behind the large one.  As it moves behind, the light from the system will decrease over time.

 The light curve will se a second low point when the mall star is fully behind the larger one.

As the small star begins to reemerge, the opposite of its disappearance will occur, there will be a steady increase in system brightness.

Next, the stars will not overlap, resulting in the system's return to full brightness.

Next we will see the start of the first decrease again as the small star transits the large star.  Then our cycle continues.

Methods/Theory

From a transit light-curve  and a Doppler shift light curve of each body, we can calculate the distance between the two stars, their masses, the time far a transit, their radii, and the period of the system.

The period is easy, since we just need to wait for a full cycle to elapse and check the time.  

Transit time is also, easy, by looking at our transit light curve, we can easily tell when the transit start and end.  

The relative radii of the two stars can be calculated using the depth of the transit.  When the smaller star passes in front of the larger, it will block a portion of the light from the star based on visible area.  If \(r_*\) is the smaller star's radius, and \(R_*\) is the larger's radius, then:
\[D = \frac{A_1}{A_2}=\frac{\pi r_*^2}{\pi R_*^2}=\frac{r_*^2}{R_*^2}\]

We can use Kepler's third law of motion and the known period to calculate the distance between the stars, \(a\).
\[P^2=\frac{4\pi^2 a^3}{G(M_*+m_*)}\]
\[a^3=\frac{P^2 G(M_*+m_*)}{4\pi^2}\]

Using \(p\) and \(a\) we can find orbital velocity.  With orbital velocity and the period of a transit, we can find the exact radii of both stars by relating this to our earlier relation.
\[t_{transit}=\frac{R_*+r_*}{v}\]
\[R_*+r_*=v t_{transit}\]
This makes sense, because when the stars are "tangent" from our point of view, the smaller needs to pass all the way across the larger, and then through it's own radius to fully expose the larger.  This is the total distance traveled, the total time for this is the transit duration, and the velocity is the orbital velocity of the transiting star.

To find the star's masses, we will use doppler data from the provided lab handout.  When a star is moving toward us, it's light will be blue-shifted, when it is moving away, its light will be redshifted.  This data lets us solve for the speed of each star, which we can use to find the true barycenter, and therefore their masses.

Observations

Observations were made using the Harvard Clay telescope on a series of (thankfully) clear nights.  Cambridge, MA is not the best place for precise observations due to the massive amount of Bostonian light and air pollution, as well as its frequent cloudy weather.  
We took data in the red spectrum of light for about an hour before, and an hour after the eclipse such that we could establish a baseline luminosity.  After our data was collected, we used sky flats to reduce disturbances from dust in the telescope.  The data was then analyzed using MaximDL as shown below.

http://www.fas.harvard.edu/~astrolab/object_field.png

The light from the object star(s) was compared to the brightness of the reference stars to remove noise from clouds or sky-wide fluctuations.  The resultant light curves were exported and combined in Excel for all of the groups giving the light curve shown below.

We used the period time to fold the normalized data into overlapping light curves to find the best fits of each transit.

This doppler data was provided in the lab manual.

Analysis

Radial Velocities:
Given from the doppler data: \(v_*=180 km/s \text{    }V_*=170 km/s\)

Transit Duration:
Using the light curve, each transit lasts approximately 1.5 hours.

Period:
From the light curve, the period appears to be 8.84 hours.

Primary Transit Depth:
\[D=\frac{\Delta f}{f}=\frac{..66}{1.33}=0.5\]

Secondary Transit Depth:
\[D=\frac{f_1}{f_2}=\frac{..54}{1.33}=0.4\]

Radius Ratio:

\[D=\frac{r_*^2}{R_*^2}\]

\[0.45=\frac{r_*^2}{R_*^2}\]

\[r_*=0.67R_*\]

Distance apart:
Using the period, we can find separation distance.
\[2\pi a_*= Pv_*\]
\[2\pi a_* = 32000sec \times 170km/s\]
\[a_*=8.6\times10^{10} cm\]
\[2\pi a_*= PV_*\]
\[2\pi A_* = 32000sec \times 180km/s\]
\[a_*=9.3\times10^{10} cm\]
\[a=a_*+A_*=8.6\times10^{10} +9.3\times 10^{10}=1.8\times 10^{11} cm\]

Mass Ratio:
This can be found using center of mass with velocities.
\[MV=mv\]
\[M170km/s=m180km/sec\]
\[M=1.06m\]

Masses:
Using the ratio and Kepler's Third Law, the individual masses can be solved for.
\[a^3=\frac{P^2 G(M_*+m_*)}{4\pi^2}\]

\[M+m=\frac{4 \pi^2 a^3}{P^2 G}\]

\[2.06m=\frac{4 \pi^2 (1.8/times 10^{11}cm)^3}{32000sec \times 6.7\times10^{-8}}\]

\[m=1.6\times 10^{33}g\]

\[M=1.7\times 10^{33}g\]

Radii:

\[R_*+r_*=v t_{transit}\]

\[R_*+r_*=350km/sec \times 1.5 hours\]

\[R_*+r_*=1.89\times10^{11}cm\]

\[r_*=0.67R_*\]

\[r_*=7.5\times 10^{10}cm\]

\[R_*=1.1\times 10^{11}cm\]

Results

\[m_*=1.6\times 10^{33}g=0.8M_{\odot}\]

Actual: \[m=0.50M_{\odot}\]

\[M_*=1.7\times 10^{33}g=0.85M_{\odot}\]

Actual: \[M=0.54M_{\odot}\]

\[r_*=7.5\times 10^{10}cm=1.07M_{\odot}\]

Actual: \[r=0.51R_{\odot}\]

\[R_*=1.1\times 10^{11}cm=1.57M_{\odot}\]

Actual: \[r=0.54R_{\odot}\]

Poetry and Astronomy

Walt Whitman's poem "When I heard the Learn'd Astronomer" reads:

"WHEN I heard the learn’d astronomer;
When the proofs, the figures, were ranged in columns before me;
When I was shown the charts and the diagrams, to add, divide, and measure them;
When I, sitting, heard the astronomer, where he lectured with much applause in the lecture-room,
How soon, unaccountable, I became tired and sick;
Till rising and gliding out, I wander’d off by myself,
In the mystical moist night-air, and from time to time,
Look’d up in perfect silence at the stars"

First published in his book Blades of Grass in 1885, Whitman's poem reminds the astronomy community to remember what they are studying.  In the poem Whitman's narrator recalls sitting in a lecture, and looking upon a presentation of math, formulas, and diagrams. The lecture hall applauds the astronomer's work, but the narrator becomes ill.  He quickly realizes that this is not the type of astronomy that he wanted to study.  Seeking haven, he leaves the lecture hall and runs out into the night to stare slightly at the stars.  

In this work, Whitman seeks not to criticize the glorious mathematics of astronomy, rather, he calls astronomers to remember the reality behind the equations.  It is easy to lose the beauty of the universe in the mathematics, but thankfully, we have labs to get us out of classroom, and back to our true subject matter: the stars.

Poem source: http://www.bartleby.com/142/180.html 

Worksheet 14.2, Problem 1: Tides

1. Draw a circle representing the Earth (mass M‘), with 8 equally-spaced point masses, m, placed around the circumference. Also draw the Moon with mass MK to the side of the Earth. In the following, do each item pictorially, with vectors showing the relative strengths of various forces at each point. Don’t worry about the exact geometry, trig and algebra. I just want you to think about and draw force vectors qualitatively, at least initially.

(a)  What is the gravitational force due to the Moon, F⃗L,cen, on a point at the center of the Earth? Recall that vectors have both a magnitude (arrow length) and a direction (arrow head).
(b)  What is the force vector on each point mass, F⃗L, due to the Moon? Draw these vectors at each point.
(c)  What is the force difference, ∆F⃗, between each point and Earth’s center? This is the tidal force.
(d)  What will this do to the ocean located at each point?
(e)  How many tides are experienced each day at a given location located along the Moon’s orbital plane?
(f) Okay, now we will use some math. For the two points located at the nearest and farthest points from the Moon, which are separated by a distance ∆r compared to the Earth-Moon distance r, show that the force difference is given by
\[\Delta F = \frac{2GmM_L}{r^3}\Delta r\]
(HINT: recall that
\[lim_{\Delta x \rightarrow 0}\frac{f(x+\Delta x)-f(x)}{\Delta x} = \frac{d}{dx}f(x) \approx \frac{\Delta f(x)}{\Delta x}\]
(g) Compare the magnitude of the tidal force ∆FK caused by the Moon to ∆F@ caused by the Sun. Which is stronger and by how much? What happens when the Moon and the Sun are on the same side of the Earth?

(h)  How does the magnitude of ∆F caused by the Moon compare to the tidal force caused by Jupiter during its closest approach to the Earth (r = 4 AU)? 

(a)  What is the gravitational force due to the Moon, F⃗L,cen, on a point at the center of the Earth? Recall that vectors have both a magnitude (arrow length) and a direction (arrow head).

This is pretty simple.  We've dealt with gravity before.

\[F_g=\frac{GMm}{r^2}\]

\(G\) is Newton's gravity constant, \(M_E\) is the mass of Earth, \(M_L\) is the mass of our moon (Luna), and \(r\) is the distance between them.

\[F_g=\frac{GM_L M_E}{r^2}\]

(b)  What is the force vector on each point mass, F⃗L, due to the Moon? Draw these vectors at each point.

Let's let each point have mass \(m\).  Also, let's let \(d\) be the distance from each point to the center of the moon.

\[F_g=\frac{GM_L m}{d^2}\]

\(d\) will vary for each point, but its safe to say that the further away from the far east point on our diagram of Earth we measure from, the greater \(d\) becomes.  Due to this, our vectors are approximated and their differences are greatly exaggerated.

(c)  What is the force difference, ∆F⃗, between each point and Earth’s center? This is the tidal force.

We now subtract the force of each point upon the earth's center from the moon's force vector (the yellow vector) yielding the following diagram:

The math here is mainly boring 2D algebra, so I will omit it in favor of qualitative explanation.

At the 0 degree point, we subtract the yellow vector format the black, decreasing it's length, but resulting in the red vector towards the moon. 

At the 45 and -45 points vector subtraction calculus a bit of the horizontal component, but he vertical component is intact, yielding a net force towards the moon and towards the Earth-Moon orbital plane.

At the 90 and -90 points, the force on the point is almost equal to the force on the center, so the horizontal components cancel leaving only a vertical component towards the center of the Earth.

At the 135 and -135 points, the yellow vector is greater than the black, thus it reverses the direction of the resultant, giving us red vectors similar to the 45 points, but in the opposite direction.

At the 180 degree point, there are no vertical forces, and the yellow vector is greater then the back, resulting a a negative red resultant.

(d)  What will this do to the ocean located at each point?

This will pull the ocean in the direction of the force vector.  In the diagram below, the dashed line represents the highly exaggerated open level at around the Earth.

(e)  How many tides are experienced each day at a given location located along the Moon’s orbital plane?

From our diagram, we can see as the Earth rotates, a location on the orbital plane will pass through two high tides and two low tides per day.

(f) Okay, now we will use some math. For the two points located at the nearest and farthest points from the Moon, which are separated by a distance ∆r compared to the Earth-Moon distance r, show that the force difference is given by 
\[\Delta F = \frac{2GmM_L}{r^3}\Delta r\]
(HINT: recall that
\[lim_{\Delta x \rightarrow 0}\frac{f(x+\Delta x)-f(x)}{\Delta x} = \frac{d}{dx}f(x) \approx \frac{\Delta f(x)}{\Delta x}\]

Doing this with exact math will be a nightmare.  Bottom line is, I don't want to deal with algebra right now.  Let's use a slick trick instead.  We already have an expression for force on a point:

\[F_g=\frac{GM_L m}{r^2}\]

Lets' take the first derivative of this.

\[\frac{dF_g}{dr}=\frac{-2GM_L m}{r^3}\]

\[dF_g=\frac{-2GM_L m}{r^3}dr\]

We can ignore the negative, since it's just there to tell us the potential is negative.

\[\boxed{\Delta F_g=\frac{2GM_L m}{r^3}\Delta r}\]

That was a lot easier.

(g) Compare the magnitude of the tidal force ∆F caused by the Moon to ∆F caused by the Sun. Which is stronger and by how much? What happens when the Moon and the Sun are on the same side of the Earth?

More math! 

Let's set this up as a ratio:

\[\frac{\Delta F_L}{\Delta F_S}=\frac{2GM_Lmr_S^3}{2GM_Smr_L^3}\]

\[\frac{\Delta F_L}{\Delta F_S}=\frac{M_Lr_S^3}{M_Sr_L^3}\]

Let's fill in some numbers from the internet.

\[\frac{\Delta F_L}{\Delta F_S}=\frac{(7.3 \times 10^{25}g)(1.5\times 10^{13}cm)^3}{(2\times 10^{33}g)(3.8\times 10^{10}cm)^3}\]

\[\boxed{\frac{\Delta F_L}{\Delta F_S}=2.2}\]

Therefore the tidal force from the moon is 2.2 times the tidal force from the sun, which is why we pretty much ignore the Sun's tidal force.

If the three spheres align, the forces will all add, resulting in lower low tides and higher high tides.

(h)  How does the magnitude of ∆F caused by the Moon compare to the tidal force caused by Jupiter during its closest approach to the Earth (r = 4 AU)? 

This is the same as above with different numbers.

\[\frac{\Delta F_L}{\Delta F_J}=\frac{2GM_Lmr_J^3}{2GM_Jmr_L^3}\]

\[\frac{\Delta F_L}{\Delta F_J}=\frac{M_Lr_J^3}{M_Jr_L^3}\]

Let's fill in some numbers from the internet.

\[\frac{\Delta F_L}{\Delta F_J}=\frac{(7.3 \times 10^{25}g)(6\times 10^{13}cm)^3}{(1.9\times 10^{30}g)(3.8\times 10^{10}cm)^3}\]

\[\boxed{\frac{\Delta F_L}{\Delta F_J}=1.5\times 10^5}\]

Therefore the tidal force from the moon is 150000 times the tidal force from Jupiter, which is why we never even bother to consider tJupiter's tidal force.

I worked on his problem with G. Grell and S. Morrison.

Worksheet 14.1, Problem 2: The Mass-Radius Relation for White Dwarfs

A white dwarf can be considered a gravitationally bound system of massive particles.

(a)  Express the kinetic energy of a particle of mass m in terms of its momentum p instead of the usual notation using its speed v.
(b)  What is the relationship between the total kinetic energy of the electrons that are supplying the pressure in a white dwarf, and the total gravitational energy of the WD?
(c)  According to the Heisenberg uncertainty Principle, one cannot know both the momentum and position of an election such that \(\Delta x \Delta p >\frac{h}{4\pi}\). Use this to express the relationship between the  kinetic energy of electrons and their number density ne (Hint: what is the relationship between an object’s kinetic energy and its momentum? From here, assume p = ∆p and then use the Uncertainty Principle to relate momentum to the volume occupied by an electron assuming Volume ~ (∆x)3.)
(d)  Substitute back into your Virial energy statement. What is the relationship between ne and the mass M and radius R of a WD?
(e)  Now, aggressively yet carefully drop constants, and relate the mass and radius of a WD.
(f)  What would happen to the radius of a white dwarf if you add mass to it? 

(a)  Express the kinetic energy of a particle of mass m in terms of its momentum p instead of the usual notation using its speed v.

Let's start with kinetic energy, where \(v\) is velocity, \(K\) is kinetic energy, and \(m\) is mass.

\[K=\frac{1}{2}mv^2\]

Momentum (\(p\)) is \(p=mv\). So:

\[\boxed{K=\frac{p^2}{2m}}\]

(b)  What is the relationship between the total kinetic energy of the electrons that are supplying the pressure in a white dwarf, and the total gravitational energy of the WD?

We are back to once again, you guessed it, the one and only: Virial Theorem.

\[K=-\frac{1}{2}U\]

\[\frac{1}{2}Mv^2=\frac{GM^2}{2R}\]

\[N_e m_e v^2=\frac{GM^2}{R}\]

Or:

\[\frac{N_e p^2}{2m_e}=\frac{GM^2}{2R}\]

(c)  According to the Heisenberg uncertainty Principle, one cannot know both the momentum and position of an election such that \(\Delta x \Delta p >\frac{h}{4\pi}\). Use this to express the relationship between the  kinetic energy of electrons and their number density ne (Hint: what is the relationship between an object’s kinetic energy and its momentum? From here, assume p = ∆p and then use the Uncertainty Principle to relate momentum to the volume occupied by an electron assuming Volume ~ (∆x)3.)

We will use the cutoff case for this problem, so:

\[\Delta x \Delta p =\frac{h}{4\pi}\]

We are told that \(\Delta p \approx p\), so:

\[p\Delta x =\frac{h}{4\pi}\]

We are also told that an election takes up the space \(\Delta x^3\), so the number density of electrons can be shown as:

\[n_e=\frac{1}{\Delta x^3}\]
We can solve this for \(\Delta x\) and insert it into our Heisenberg equation:

\[\frac{p}{n_e^{\frac{1}{3}}} =\frac{h}{4\pi}\]

Earlier we established that \(K=\frac{N_e p^2}{2m_e}\)

We can solve this for \(p\).

\[p=\sqrt{\frac{2m_eK}{N_e}}\]

Now we can throw this into the Heisenberg:

\[\frac{\sqrt{\frac{2m_eK}{N_e}}}{n_e^{\frac{1}{3}}} =\frac{h}{4\pi}\]

Let's clean up some exponents:

\[\frac{2m_eK}{N_e n_e^{\frac{2}{3}}} =\frac{h^2}{16\pi^2}\]
Finally let's solve for K.

\[K=\frac{n_e^{\frac{2}{3}}h^2 N_e}{32m_e\pi^2}\]

(d)  Substitute back into your Virial energy statement. What is the relationship between ne and the mass M and radius R of a WD?

Our statements is that:

\[K=-\frac{1}{2}U\]

\[\frac{n_e^{2/3}h^2 N_e}{32m_e\pi^2}=\frac{GM^2}{2R}\]

Let's get rid of the fractions and simplify:

\[n_e^{2/3}h^2 N_e R=16GM^2 m_e\pi^2\]

This is truly messy, but it's the best we can get.

(e)  Now, aggressively yet carefully drop constants, and relate the mass and radius of a WD.

Now here is the fun part.

\[n_e^{2/3}h^2 N_e R=16GM^2 m_e\pi^2\]

\[n_e^{2/3} N_e R=M^2\]

We need to get rid of \(n_e\), so we should remember that for every electron in the star, there is a proton with mass.  So:

\[n_e \sim \frac{M}{R^3}\]

This makes sense because a number density whose item has mass is pretty much regular density with the mass ignored.  Luckily the mass is just a constant, so we can drop that.

\[n_e^{2/3} \sim \frac{M^{2/3}}{R^2}\]

Let's substitute this in:

\[\frac{M^{2/3}N_e}{R^2} R\sim M^2\]

\[\frac{N_e}{R} \sim M^{4/3}\]

We already established that proton number is tied to election count, so \(N_e \sim M\).

\[\frac{M}{R} \sim M^{4/3}\]

\[R \sim M^{-1/3}\]

Or:

\[\boxed{M \sim \frac{1}{R^3}}\]

(f)  What would happen to the radius of a white dwarf if you add mass to it? 
Since the Mass and Radius are inversely proportionate (with cube thrown in),  if you added mass, the radius would actually decrease.

I worked with G. Grell and N. James on this problem.

Daily Paper Summary: "Was Our Moon’s Formation Likely or Lucky?"

In a blog post on Astrobites, Michael Zevin, grad student at Northwestern University studying Physics and Astronomy, discusses the finding of two competing papers on the likelihood of our moon's formation.

Traditionally, scientists have accepted a theory in which a Mars-sized object (named Theia) collided with a young Earth, ripping off material from both objects into a disk of matter that clumped into the moon and Earth.  This theory was supported by the similarity of the composition of both the Earth and moon specifically in O-17 content.

Unfortunately, computer models of this collision propose that the majority of lunar material should come from Theia, not Earth.  This is clearly at odds with collision theory.  So, which is correct?

The blog post is inconclusive as paper 1 states that 5% of Earth-like planets have been struck by objects of similar composition.  Paper 2 claims that the figure is 50%.  

Each has validity, so we aren't really sure what to believe.  It's somewhat concerning that we still don't even know where our own Moon came from, but where there are questions, there is research funding!

Picture from http://www.novacelestia.com/images/earth_impact_moon.jpg
Blog post: http://astrobites.org/2015/04/08/was-our-moons-formation-likely-or-lucky/