Comparison to Wasp-10b

This post is going to compare the parameters we get from analyzing these graphs

both graphs provided on Worksheet 11.1

with actual values from exoplanet.org.  

First, let's start with the impact parameter, b.  

$b=1-\delta^{1/2}\frac{T}{\tau}$

In the light curve above, 

$\delta$=0.028  

T = 102 minutes

$\tau\approx$18 minutes

Using these values and the equation above, I found b=0.99, which is a lot higher than the value given by exoplanet.org, b=0.3.  I believe the reason for this difference is a difference in data.  For example, in the light curve above, the length of the transit looks to be a little longer than an hour and a half, but on exoplanets.org, the transit is at least half an hour longer than that.  

Next, let's look at the ratio of the radii of the planet and star.  

$\delta=\frac{R_P^2}{R_*^2}$

$\frac{R_P}{R_*}=\delta^{1/2}$

For this system, we find that the ratio is equal to 0.17, so the planet is 0.17 times as big as the star.  

On worksheet 11.1, it says that the star is 0.8$R_\odot$.  With this information and the ratio we just found, we can tell that the planet is about 1.36 Jupiter radii big.  This is closer to the website's value of about 1.08 Jupiter radii.  The reason for this difference could be that the website lists the sun as being 0.7 solar radii big.  

Now we'll look at the ratio $a/R_*$.

On the worksheet, not taking into account the impact parameter, I used the equation 

$\frac{a}{R_*}=\frac{P}{\pi T}$

This equation gives us a ratio value of about 13.3, which is pretty close to the website's value 11.9.  The difference can be attributed to the fact that the website found a semimajor axis value dependent on the impact parameter.  

Finally we'll look at the densities of the star and planet.

We all know that density equals mass over volume.  But the fun thing about the mass of a star is that it can be related to semimajor axis and period of the system using Kepler's Third Law.

$M_*=\frac{4\pi^2a_P^3}{GP^2}$

$\rho=\frac{3\pi a_P^3}{R_*^3GP^2}=\left ( \frac{a_P}{R_*} \right )^3\frac{3}{GP^2}$

When we substitute provided values into this equation, we find that the density of the star is about 1.45 grams per cubed centimeter, which is pretty darn close to the website's value of 1.51 g/cm^3.  

WS 11.2, Problem 3: Goldilocks Zones

A Goldilocks Zone, or a "habitable zone," is a distance from a star where it is neither too hot nor too cold for a planet to have liquid water.  This is the criterion for life that most people have a greed on, but we're not sure that it's the only or most important factor in determining whether or not life can exist on a planet, so it's not the end all be all of life on other planets.

Now let's fi gure out what stars make the best targets for both the radial velocity technique and the transit method, assuming we are primarily interested in fi nding planets in their stars' habitable zones. For a planet in the habitable zone of a star of mass $M_*$...

a)  How does the Doppler amplitude, $K_{HZ}$, scale with stellar mass?

To solve this problem, let's first recall the equation for the Doppler amplitude that we found on a previous worksheet, generously dropping unnecessary constants.

$K\sim\frac{1}{M_*^{4/6}P^{2/6}}$

This isn't as obvious as it might seem, though, because an object's period depends on its mass, so we have to do some more algebra.

$P^{1/3}\sim\left ( \frac{a_P^{3/2}}{M_*^{1/2}} \right )^{1/3}=\frac{a_P^{3/6}}{M_*^{1/6}}$

Of course, the semimajor axis of a planet, $a_P$, is also dependent upon its star's mass.

$a_P^{3/6}\sim M_*^{3/6}$

When you combine all of these scaling factors, you find that

$K\sim\frac{1}{M_*^{4/6}M_*^{2/6}}=\frac{1}{M_*}$

b)  How does the transit depth, $\delta_{HZ}$, scale with stellar mass?

For this, we have to remember that the depth of the transit depends on the ratio of the radii of the star and planet.

$\delta\sim\frac{R_P^2}{R_*^2}$

We can use mass-radius scaling relations to simplify this

$\delta\sim\frac{1}{R_*^2}$

$M\sim R$

$\delta\sim\frac{1}{M_*^2}$

c)  How does the transit probability, Prob$_{HZ}$, depend on stellar mass?

We learned on worksheet 11.1 that the probability that a planet will pass in front of its star is

$\frac{R_*}{a_P}

Now we have to remember both the mass-radius scaling relation and the relationship between a star's mass and $a_P$.

$M\sim R$

$a_P=\frac{M_*a_*}{M_P}$

Then we can rewrite the probability as

$\frac{M_*M_P}{M_*a_*}$

So the probability of a transit happening doesn't really depend on the star's mass at all.


d)  How does the number of transits (orbits) per year depend on stellar mass?

This wording of this question is a little unclear, but it reminds me of part (c) from problem 2 on this same worksheet.

If the Sun were half as massive and the Earth had the same equilibrium temperature, how many days would our year contain?

To solve this, we set up a proportion between the period of a habitable zone planet and the period of our Earth under the conditions specified in the problem.

$\frac{P_{HZ}}{P_\oplus}\sim\frac{a_{HZ}^{3/2}M_\odot^{1/2}}{M_*^{1/2}a_\oplus^{3/2}}$

If we drop the constants, we get

$P_{HZ}\sim\frac{a_{HZ}^{3/2}}{M_*^{1/2}}$

Now we can substitute in the equation for the semimajor axis of the habitable planet and simplify to get

$P_{HZ}\sim M_*$

But the problem asks for the relationship between a star's mass and the number of orbits of a planet, which is inversely proportional to its period, so

Transits$\sim\frac{1}{M_*}$

e)  Based on this analysis, what are the best kinds of target stars for the search for habitable zone planets? What factors did we ignore in this analysis?

From parts (a) through (d), we can see that the Doppler amplitude and number of transits are both inversely proportional to stellar mass and the depth of the transit goes as the inverse square of the stellar mass.   We want all of these quantities to be as large as possible.  A high Doppler amplitude makes a transit more noticeable on an RV graph; a deep transit is more noticeable on a light curve; the more transits there are, the more opportunities we have to gather data.

Based on this, we can see that lower-mass stars make better targets for finding habitable-zone planets.


In doing this, we ignore the fact that lower-mass stars are harder to detect in general because their luminosities are so low.

WS 11.1, Problem 3: What's in a Light Curve?

This blog post is loosely modeled after problem 3 on worksheet 11.1.  I'm not going to answer the specific questions from the problem to generalize to all sets of data, but I am going to follow the path that the problem sets to discover what can be learned from a transit light curve.  This should be fun.

The problem gives us three pieces of information.

• The system's period is 3.1 days (not necessary since I'm not actually answering the parts of the problem, but now we know we can find a lot of the system's properties with its period).  
• This light curve

• And this radial velocity graph

The first part of the problem asks us to qualitatively describe the brightness distribution of the star.  We can tell this by looking at the bottom of the light curve.  The dots along the curve represent the amount of light in each image taken of the system.  The dots stay fairly close to the line, so the star's brightness is relatively evenly distributed.  But because the dip is more of a curve than a straight line, so we know that the center of the star is slightly more luminous than the edges.  This is to be expected due to the limb darkening effect, which basically just says that the center of the star will appear brighter.  (Yay for self-explanatory scientific terms!)

The second part asks us to find the ratio of the radii of the planet and star.  This is a little bit trickier than the first part because it involves actual math, but it's not that bad.  The theory behind the solution to this problem is that that the depth of the light curve depends on the ration of the areas of the two objects.  Actually solving the problem is a pretty logical process.  If the star is really big, the light curve will be pretty shallow.  In the same vein, the curve will be shallow if the planet is small (because it's not blocking a lot of the star's light).  So the depth of the curve goes as 

$\frac{R_P^2}{R_*^2}$

When the problem asks us to find the ratio of the objects' radii, we can find the average depth of the light curve and take its square root.  

The third part asks us to find the ratio of the planet's semimajor axis with the star's radius 

$\frac{a_P}{R_*}$

To solve this problem, we look at the duration of the transit.  If you believe that velocity is distance over time (which I do), then you can also believe that time is distance over velocity.  The distance that the planet has to travel to complete a transit is twice the star's radius, and it's velocity is $v_P$.  Lucky for us, the velocity of the planet is dependent upon its semimajor axis in the following way:

$v_P=\frac{2\pi a_P}{P}$

which means the ratio of the semimajor axis to the star's radius should be 

$\frac{a_P}{R_*}=\frac{P}{\pi t}$

where t is the duration of the transit.

It's important to realize that this equation is only true for equatorial transits (transits that happen along the equator of the star).  A more general equation would account for the transit time of a transit that travels along any chord on the star.

$T=T_{equatorial}(1-b^2)^{1/2}$

where b is called the impact parameter, which is basically the place on the star where the planet transits.  For equatorial transits, b is 0.  For transits where the planet just barely grazes the star--so it doesn't really cause any dip in the light curve at all--b is 1.  

$b=1-\delta^{1/2}\frac{T}{\tau}$

where $\tau$ is the time of egress--the time from the moment the planet first touches the star to the moment the last point of the planet crosses the first edge of the star--and delta is the depth of the transit.  

One of the parts of the question asks us to show that the ratio found above is related to density.  I started with the equation for density

$\rho_*=\frac{M_*}{4R_*^3}$

assuming that $\pi$ is 3.

But we can simplify this further by using the mass-radius scaling factor found in this post, which says that mass scales proportionally with radius.  

$\rho_*=\frac{1}{R_*^2}$

Because of this relationship, we can say that the planet's semimajor axis is related to the star's density in the following way:

$a_P=\frac{PR_*}{\pi t}=\frac{P}{\pi t sqrt{\rho_*}}$

If we know the radius of the planet (which we do, because we know the ratio of the radii of the two objects and the radius of the star), we can also find its density based on the 

$\rho_*=\frac{1}{R_*^2}$

equation.  

Here we have it.  Armed with three simple (I say simple, but in actuality, those graphs are not too easy to get) pieces of information, we can find the following properties

• ratio of the radii of the two objects in the system
• semimajor axis of the planet (which we can then use along with the mass ratios to find the semimajor axis of the star)
• the densities of both the star and the planet
• the impact parameter

Like I said at the beginning of this post, light curves are fun!

  

WS 10.2, Problem 2: Wobble, Baby, Wobble

a) Astronomers can detect planets orbiting other stars by detecting the motion of the star--its wobble|due to the planet's gravitational tug. Start with the relationship between $a_p$ and $a_*$  to find the relationship between speed of the planet and the star, $v_p$ and $v_*$ .

We started with the basic equation 

$a_*m_*=a_Pm_P$

We can also relate the semimajor axis and velocity in the following way

$v=\frac{2\pi a}{P}\Rightarrow a=\frac{vP}{2\pi}$

The first equation can now be rewritten and simplified as 

$\frac{v_*P}{2\pi}m_*=\frac{v_PP}{2\pi}m_P$

$v_*m_*=v_Pm_P$

b) Express the speed of the star, K, in terms of the orbital period P, the mass of the star $M_*$  and the mass of the planet.  Convert your units so that your expression is given as a speed in meters per second, with P measured in years, $M_*$ in solar masses and planet mass in Jupiter masses. 

Again let's start with the last equation from part (a)

$v_*m_*=v_Pm_P$

We can rearrange this to find the equation for the velocity of the star

$v_*= \frac{v_Pm_P}{m_*}$

We also know the following relationship between period and velocity

$v_P=\frac{2\pi a_P}{P}$

So now we can find K, the speed of the star

$K=v_*=\frac{m_P}{m_*}v_P=\frac{m_P}{m_*}\frac{2\pi a_P}{P}$

c) We can measure the velocity of a star along the line of sight using a technique similar to the way in which you measured the speed of the Sun's limb due to rotation. Specfi cally, we can measure the Doppler shift of stellar absorption lines to measure the velocity of the star in the radial direction, towards or away from the Earth, also known as the "radial velocity." What is the time variation of the line-of-sight velocity of the star as a planet orbits?

To someone who's grown up thinking that the solar system revolves around the sun, this problem might be a little bit conceptually difficult, because if the sun is truly the center of the solar system, it shouldn't move.  The point of this problem, however, is to show that the sun is not the exact center of the solar system (as I'm typing this, a bunch of dead white men are probably rolling in their graves because this problem debunks the work they devoted their lives to).  Instead, the sun wobbles--it moves away from the system's center of mass as the planet moves.  

If a planet is moving away from our viewpoint, the sun--assuming it only moves in our visual plane, and not up and down--is moving toward us.  If the planet is moving toward us, the sun has to compensate in its wobble and move away from the center of mass, or away from out viewpoint.  The sun's radial velocity would appear to be 0 (moving neither towards nor away from us) when the planet is at a point in its orbit when it's not moving towards or away from us, so when it's on the side of its orbital path.

d)  Sketch the velocity of a star orbited by a planet as a function of time. Denote the maximum velocity as K, and express the x-axis in terms of the number of periods, in intervals of P/4.

For this graph, we're starting at the point where the radial velocity is zero. The K is at it's peak hi and low at every other 1/4 of a period.  

e) What is the velocity amplitude, K, of an Earth-mass planet in a 1-year orbit around a Sun-like star?

This is another simple plug-and-chug problem where we take the equation from part (b) and substitute the sun's values where necessary.  

Thanks to WolframAlpha, I was able to find a K value of about 9.4 cm/s.  Earth's pretty freaking slow. 

WS 10.2, Problem 1: Center of Mass

In the first part of this problem, we were asked to draw a picture illustrating the relationship between the masses of a star and planet and their center of mass.

The relationship between the center of mass and and masses is

$x_{COM}=\frac{(-a_*m_*)+(a_Pm_P)}{m_*+m_P}$

If we set the center of mass to 0, like it is in the picture, the equation becomes 

$a_*m_*=a_Pm_P$

b) In a two-body orbital system the variable a is the mean semimajor axis, or the sum of the planet's and star's distances away from their mutual center of mass.  Label this on your diagram. Now derive the relationship between the total mass, , orbital period P and the mean semimajor axis a, starting with the Virial Theorem for a two-body orbit (assume circular orbits from here on).

So we start with the Virial Theorem like the problem told us to.  

$K=-\frac{1}{2}U$

$\frac{1}{2}(m_*v_*^2+m_Pv_P^2)=\frac{Gm_*m_P}{a}$

In order to conserve momentum, the periods of the star and planet have to be the same.  

$v_*=\frac{2\pi a_*}{P}$

$v_P=\frac{2\pi a_P}{P}$

The star is so much more massive than the planet, so it moves a lot more slowly.  When the two are compared, the velocity of the star is practically 0.  

The distance from the center of mass to the planet, $a_P$ is also a lot larger than the distance to the star, $a_*$.  Since $a=a_*+a_P$, $a=a_P$.  

We can now rewrite the Virial Theorem as 

$m_P\left ( \frac{4\pi^2a_P^2}{P^2} \right )=\frac{Gm_*m_P}{a_P}$

and we can rearrange it to say

$m_*=\frac{4\pi^2a_P^3}{GP^2}$

which is basically just Kepler's Third Law of Motion, so yay!

c) By how much is the Sun displaced from the Solar System's center of mass (a.k.a. the Solar System \barycenter") as a result of Jupiter's orbit? Express this displacement in a useful unit such as Solar radii.

This should be a pretty simple plug and chug problem.  

$a_\odot M_\odot=a_{Jup}M_{Jup}$

$a_\odot=\frac{a_{Jup}M_{Jup}}{M_\odot}$

Plugging in all of those values, we find that the sun is displaced $5\times 10^4$cm, or $7\times 10^{-7}$ solar radii.  

WS 10.1, Problem 2: White (I still don't know why the plural is spelled this way) Dwarfs

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

a) 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?

This part is pretty much just a bunch of algebraic manipulation of the Virial Theorem.

$K=-\frac{1}{2}U$

The above equation, though, is only useful for relating the energies of one electron.  In order to find the relationship for the whole white dwarf, you have to multiply by the total number of electrons.  

$N_e\frac{1}{2}m_e v^2 = \frac{3}{10}\frac{GM^2}{R}$

but because of the relationship between momentum and velocity, 

$N_e\frac{1}{2}m_e \frac{p_e}{m_e}^2 = \frac{3}{10}\frac{GM^2}{R}$

$N_e\frac{p_e^2}{m_e} = \frac{3}{5}\frac{GM^2}{R}$

b)  Express the relationship between the kinetic energy of electrons and their number density n (Hint: what is the relationship between an object's kinetic energy and its momentum?)

So, we start with the last equation from part (a).  

$KE=N_e\frac{p_e^2}{m_e}$

In the problem, we're given the following information

• $\Delta p\Delta x>\frac{h}{4\pi}\Rightarrow \Delta p\sim \frac{1}{\Delta x}$

• $\Delta p\approx p$

• Volume\sim \Delta x^3$

If number density is number over volume, 

$n_e=\frac{N_e}{V}=\frac{N_e}{\Delta x^3}\sim \Delta x^{-3}$

We went through all of that so that we could find a relationship between the $p_e^2$ in the equation from part (a) and the number density.  

$p_e^2\sim n_e^{2/3}$

The total number of electrons, $N_e$, is equal to the number of protons, which is equal to 

$\frac{M_*}{m_p}$

After all of that, we can write 

$KE=\frac{M_*n_e^{2/3}}{m_p m_e}$

c) What is the relationship between $n_e$ and the mass M and radius R of a WD?

We kind of started solving this part in part (b).

$n_e=\frac{N_e}{V}$

We can get rid of a lot of constants in this equation to end up with 

$n_e\sim \frac{M}{R^3}$

d) Substitute back into your Virial energy statement, aggressively yet carefully drop constants, and relate the mass and radius of a WD.

First, let's recall the the equation from part (a).

$N_e\frac{p_e^2}{m_e} = \frac{3}{5}\frac{GM_*^2}{R_*}$

Based on all of the relationships we've found in parts (b) and (c), we can rewrite this equation as 

$\frac{M_*n_e^{2/3}}{m_p m_e}=\frac{3}{5}\frac{GM_*^2}{R_*}$

First, let's get rid of the constants we don't really need to have a basic understanding of the relationships between these quantities 

$M_*=\frac{M_*^2}{R_*}$

Finally, after simplifying, we find that the mass and radius of a White Dwarf are inversely proportional 

$M\sim R^{-1}$

WS 10.1, Problem 1: 'Til the Fusion Ends

We've talked about the birth of stars in molecular clouds. We also briefly discussed the main
sequence, on which stars are in hydrostatic equilibrium owing to energy generated by nuclear fusion in their cores. Now let's investigate what happens when a star like the Sun can no longer support itself via nuclear fusion.

a) At what rate is the Sun generating energy?

At first, we were confused by this question.  We knew it couldn't possibly be as simple as we were making it.  Then we realized we were being typical Harvard students and overthinking things, and that luminosity is given in ergs per second.  So, the sun is generating energy at a rate of

$L_\odot=4\times 10^{33}$ ergs/s

b) If fusion converts matter into energy with a 0.7% e fficiency, and if the Sun has 10% of its mass available for fusion (in the core only), how long does it take to use up its fuel supply? What is the general relationship between the mass of a star and its main-sequence lifetime?

The first step was to find the mass of the sun available for fusion.  

$(.1)(M_\odot)=(1\times 10^-1)(2\times 10^{33}g)= 2\times 10^{32}g$

The second step was to find the energy produces by fusion in the core using the equation for energy that I always knew, but never really knew when to use.  

$E=M_{fus} c^2=(2\times 10^{32}g)(3\times 10^{10}cm/s)$

But, the question states that the sun's fusion is only 0.7% efficient, so we have to multiply the energy by the efficiency.  

$E=M_{fus} c^2=(7\times 10^{-3})(2\times 10^{32}g)(3\times 10^{10}cm/s)=1.4\times 10^{51} ergs$

Finally, using dimensional analysis, we were able to find the final equation for the time.  

$t=\frac{E}{L_\odot}=4\times 10^{17}s\approx 1\times 10^{10} yrs$

d) The core will collapse until there is a force available to hold it up. One such force is supplied by degeneracy pressure. The pressure inside of a white dwarf star is provided by the motion of electrons. The electrons are in a tough situation: they can't occupy the exact same state, but there's not much room for them to coexist easily inside of a dense white dwarf. As a result, they must always be in motion to avoid other electrons (roughly speaking). This e ffect becomes important when the inter-particle spacing is of order the de Broglie wavelength $\lambda$, which is related to the momentum via 

$\lambda=frac{h}{v}$  

For a stellar core of a given temperature, which particles reach this critical density fi rst: electrons or protons?

The foundation of our solution to this problem was the equation for momentum:

$p=mv$

We had to compare the momentums of the electrons and photons in the star.  

$p_e=m_e v_e$

$p_p=m_p v_p=1800m_e v_p$ 

Because protons are just so much more massive than electrons, the momemtum of a proton is a lot bigger than the momentum of an electron.  Therefore, protons have a shorter de Broglie wavelength, so the protons reach their critical density first.   

e) If a typical white dwarf has roughly half the mass of the Sun and the radius of the Earth, what is the typical density of a white dwarf in grams per cubic centimeter? What is the volume of white dwarf material that weighs as much as a car?

This was pretty simple. 

$\rho_{WD}=\frac{m}{v}=1.5\times 10^6 g/cm^3$

We then figured that the average car is about 1 metric ton, so about $1\times 10^6$g.  

$v=\frac{m_{car}}{\rho_{WD}}=0.67$ cm^3

Free Form: Spaghetti shouldn't be a part of physics, but...

A while back, I wrote a blog post on G2, a gas cloud about to be eaten by our galaxy's black hole, Sagittarius A*.  In the post, I wrote about spaghettification (because a blog post on black holes without spaghettification is hardly a blog post at all) and lamented the lack of good spaghettification GIFs on the internet.  Professor Johnson saw my cry and told me that I should send an email to James Guillochon, an Einstein Fellow at the Harvard CfA who focuses on collisions in astrological systems.  So I did, and we set up a meeting to talk about the more intricate details about spaghettification.

I don't have the best background in computers, so the chances of me actually creating my own spaghettification GIF--the original purpose of the meeting with Dr. Guillochon-- are pretty slim.  But from the meeting, I did get the knowledge necessary to understand how such a GIF would work.

Spaghettification happens because the tidal forces from the massive object (in this case, Sagittarius A*) are stronger than the gravitational binding forces of the spaghettified object.  That object gets ripped apart and the individual pieces then move on their own Keplerian orbits.  The first step to making this GIF is to think of the object being spaghettified as millions and millions of point masses.  Then, one could use a computer program to assess the gravitational potential energy of each individual point.

The picture above is of the situation right before spaghettification.  Already, because of tidal forces, the particles closest to Sag A* are slower than the ones on the opposite side because Sag A* is pulling more on the particles closest to it.  

When I get more computer skills (which should probably happen eventually), I'm going to give the world a real spaghettification GIF.  Until then, I'll have to be satisfied with the fact that I know, in theory, how that program will work.

WS 9.2: ALL of the Relationships

The point of this worksheet, as seen in the title of this blog post, was to find all of the relationships between a star's mass, temperature, and luminosity and its radius.

In question 1, we were asked to recall two equations from previous classes.

$\frac{dP}{dr}=-\frac{GM_r\rho_r}{r^2}$

$\frac{dT}{dr}=-\frac{L_rk_B\rho_r}{16\pi r^2acT^3}$

Pretty easy, right? Well, it gets harder.

Question 2 asked us to derive the equation for the relationship between mass and radius, $\frac{dM}{dr}$.  

We wrote down the equation for the mass of an object, which is actually the rearranged equation for the density of an object.  

$M=\frac{4\pi r^3\rho_r}{3}$

Then we found its derivative 

$\frac{dM}{dr}=4\pi r^3\rho_r$

Still pretty easy to understand

Question 3 asked us to find the relationship between density and gas pressure.

Obviously, I saw the words "gas" and "pressure" and immediately thought of the ideal gas law

$P=nk_BT$

But, since the question wanted to include density, we had to find a way to put a $\rho$ in the equation.  Since $\rho=n\bar{m}$, the ideal gas law can be written as 

$P=\frac{\rho}{\bar{m}}k_BT$

Question 5 asked us to rewrite the all of the relationships.  

We were given that $P_C\sim \frac{M^2}{R^4}$.

$\frac{dP}{dr}\sim \frac{P}{r}=-\frac{GM_r\rho_r}{r^2}$*

If you substitute in the full equation for $\rho$ and rearrange for P, you do in fact find that 

$P_C\sim \frac{M^2}{R^4}$

Next we'll do the temperature-radius relationship.

$\frac{dT}{dr}\sim \frac{T}{r}=-\frac{L_rk_B\rho_r}{16\pi r^2acT^3}$

Again, if you substitute in for $\rho$ and rearrange, you find 

$T^4\sim \frac{LM}{R^4}$

Now, it's mass' turn.

$\frac{dM}{dr}\sim \frac{M}{r}=4\pi r^3\rho_r$

Substituting and rearranging gives you 

$\frac{M}{R}\sim\frac{M}{R}$

which is so obvious that it's not actually helpful.  So, the trick is not to substitute for $\rho$ and you get 

$\rho\sim\frac{M}{R^3}$

Question 6 asks us to summarize all the relationships found above by using power laws

a)  $R\sim M$ which you can find by simplifying the relationship found in Question 5

b) $L\sim \frac{R^4}{M}$ but because of the relationship in part (a), we can rewrite this as $L\sim M^3$

c) $L\sim \frac{M}{\rho}$ which makes sense because if a star isn't particularly massive, a dense star will be really small, and therefore not very luminous.  If we do the substitution for $\rho$ one last time, and we use the relationship from part (a), we find that $L\sim M^5$

e) For low-mass stars, we start with 

$L\sim M^2T^4$

which is the from equation for the surface luminosity of a star.  

But we can substitute using the mass-luminosity relation found in part (c) and get

$L\sim L^{2/5}T^4$

When you rearrange, you get 

$L\sim T^{20/3}$

For massive stars, we once again start with 

$L\sim M^2T^4$

but now we substitute the mass-luminosity relation in part (b)

$L\sim M^{2/3}T^4$

$L\sim T^12$

See? I told you it would get harder.  

* You can say that $\frac{dP}{dr}\sim\frac{P}{r}$ because we assume that the initial pressure is 0 and the final pressure (the pressure at the surface) is not.  So since we assume that dP=P(final)-P(initial), we can posit make that leap.  

Gravitational Time Dilation

Free form posts are always the hardest because I never know what to write.  It's one of those "water, water everywhere, but not a drop to drink" problems.  So, instead of writing my blog posts the other day, I decided to procrastinate by watching Stargate (SG1, otherwise known as the best Stargate) and there was a great episode on gravitational time dilation.

In the episode, the team opens the Stargate to a planet whose galaxy is actually being eaten by a black hole.  The strong gravity from the black hole comes through the Stargate and affects the compound here on earth.  The cool part is that because the compound is now experiencing such strong gravity, time inside the compound starts to move more slowly than time outside it.

I wanted to know if this was in any way realistic, so I did some googling!

The simple answer is: yes.  Time fluctuation die to gravity is actually a thing that happens in real life.  Albert Einstein predicted in his theory of General Relativity and it has since been tested that if two observers are standing in two regions with different gravitational potential, time will actually move faster for one than for the other.

They tested this by placing atomic clocks at two different elevations.  It took a while, but the clock at the higher altitude (which had more gravitational potential) eventually got ahead of the one at a lower altitude.  Basically, the closer something is to a massive body, the slower time will move (or the slower it will appear to move when viewed by someone further away from that massive body.

This works with black holes because when people watch something fall into a black hole, it takes a very long time.  Once something crosses the event horizon--even light--it can't escape, so observers see something very similar to one of those penny funnel things.

This makes it look like time is slowing down.  

So, was Stargate right?  In a loose sense, yes.  For distant observers, time does appear to slow down in regions close to very massive objects, like black holes.  But if you take into account their solution to the problem--Replicators, devices used to speed up time--they are completely wrong.  

But that will never stop me from watching the show.  

H-R Diagrams

From the Wikipedia page on H-R Diagrams

Put simply, the Herzsprung-Russel Diagram is a graph that shows the relation between a star's luminosity and its effective (surface) temperature.

The diagram increases in luminosity going up the y-axis.  It increases in temperature moving left along the x-axis.  "Why does the x-axis increase to the left?" you might ask (because I definitely did).  It increases to the left because the x-axis used to be B-V color instead of luminosity, and the more negative B-V values corresponded to higher temperatures.

The diagram breaks down into three basic categories.

Main Sequence

Generally speaking, stars spend most of their lives on the main sequence.  They leave or evolve from the main sequence by cooling down (but not getting smaller/losing mass) and becoming a giant.  They can also lose enough energy that they condense and become a white dwarf.  The ultimate fate of a star depends on its mass.

White Dwarfs

These are in bottom, left-hand side of the graph because they're really hot, but not very bright.  They evolve from low-mass stars who move off the main sequence because they run out of hydrogen to burn and don't have enough energy to resist collapse.

Giants

These are in the top, right-hand side of the diagram.  When high mass stars run out of hydrogen to burn, they puff up and become giants.  In the process of puffing up, they lose some of their temperature.

WS 9.1, Problem 1: Random Walks

Consider a photon that has just been created via a nuclear reaction between two protons in the center of the Sun. The photon now starts a long and arduous journey to the Earth to be enjoyed by Ay16 students studying on a calm Spring day.

a)  The photon does not freely travel to the surface of the Sun. Instead it random walks, one collision at a time. Each step of the random walk traverses an average distance w, also known as the mean free path. On average, how many steps does the photon take to travel a distance $\Delta r$?

As always, our first step was to draw a picture illustrating a photon's random walk journey over a distance, $D=\Delta r$.

We were given the hint that if each step with length w is a vector, $\vec{r_i}$, then the vector 

$\vec{D}=\sum \vec{r_i}$

We're asked to find the number of steps it takes a photon to travel $\Delta r$, which is a scalar quantity.  So the hint also told us that we should find 

$\Delta r=\left ( \vec{D}^2 \right )^{1/2}=\left ( \sum \vec{r_i}^2 \right )^{1/2}$

To solve this, we have to find the dot products of each vector, $\vec{r_i}$ and then take the square root of the scalar sum of those dot products.  This could be a really tedious process, but the concept of randomness actually helps us out here.  The sum of these dot products is 

$\vec{D}^2=r_1^2+r_2^2+r_3^2...r_N^2$

because all of the terms that aren't a vector squared are the dot puct between two vectors.  Since each vector is completely random, there's a chance that any two vectors are perpendicular to each other, and therefore have a dot product equal to 0.  Yes, it's not at all realistic, but neither is assuming that everything in space is a spherical blackbody.  

So, now we can rewrite the last equation as 

$\vec{D}^2=\sum_{i}^{N}r_i^2$

$= \sum_{i}^{N}w^2$

because each random walk has a length, w.

$=Nw^2$

Plugging this into the equation for $\Delta r$, we find that

$\Delta r=D=\left ( Nw^2 \right )^{1/2}=w\sqrt{N}$

We can rearrange this equation to find N, the number of steps it takes the photon to travel $\Delta r$.

$N=\left ( \frac{D}{w} \right )^2$

b)  What is the photon's average velocity over this distance? Call this $v_{diff}$ , the di ffusion velocity.

We started with the general idea that velocity equals distance over time.  The distance, $\Delta r$, is stated in the problem.  The tricky part was finding the time.

We're dealing with photons--light particles, so the speed of light has to be involved.  The problem with a random walk, though, is that the photon keeps hitting things, so it's only moving at the speed of light for little bursts at a time.  More specifically, it's only moving at the speed of light for the time it takes the photon to move a distance, w.

With this knowledge, we found the time it takes a photon to move one step,

$t_i=\frac{w}{c}$

To find the total time, multiply by the total number of steps, N, so that 

$t=N\frac{w}{c}$

$=\frac{D^2w}{w^2c}=\frac{D^2}{wc}$

because we have to put N in terms of known values, like $D=\Delta r$ and w.  

This leaves us with this equation for velocity:

$v_{diff}=\frac{wc}{\Delta r}$

c)  The `mean free path' w is the characteristic (i.e. average) distance between collisions. Consider a photon moving through a cloud of electrons with a number density n. Each electron presents an effective cross-section $\sigma$ . How are these parameters related?

We started solving this problem by making a list of each quantity and their units.

$\sigma$=cm$^2$

w=cm

n=cm$^{-3}$

We originally tried to solve the problem only using dimensional analysis, but that's not the most effective method.  So, instead we stopped to think about how the parameters should effect each other.

A high number density, n, should decrease the mean free path, w, because there are more particles for the photon to hit, so it can't travel as far without hitting something.

A high effective cross section, $\sigma$, should also decrease the mean free path, w, because the particles are bigger, so they're harder to avoid and the photon can't go as far without hitting something.

These two statements led us to the idea that n and $\sigma$ should both be inversely proportional to w, so we tried the following relation:

$w=\frac{1}{n\sigma}$

We tested the units, and after finding that this relation gave us centimeters, we figured we were correct!

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

To solve this problem, we followed the same process as in part (c).  

w = cm

$\rho$ = $\frac{g}{cm^3}$

$\kappa = $\frac{cm^2}{g}$

$sigma$ = cm$^2$

Using dimensional analysis, we found that 

$w = \frac{1}{\rho\kappa}=\frac{1}{n\bar{m}\kappa}$

This makes sense because the photon shouldn't be able to go very far in denser regions.  

The relation between $\kappa$ and $\sigma$ can be given as 

$\sigma = \kappa\bar{m}$

if we substitute these equations into the equation for $v_{diff}$ from part (b), we find that 

$v_{diff}=\frac{c}{\rho\kappa\Delta r}$

$\frac{c}{\frac{\rho}{\bar{m}}\sigma\Delta r}$

e)  What is the diff usion timescale for a photon moving from the center of the sun to the surface?

To solve this part of the problem, I'll use the equation we found in part (b) for the time it takes a photon to move a distance, $\Delta r=D$

$t_{diff}=\frac{D^2}{wc}$

$=\frac{D^2\frac{\rho}{\bar{m}}\sigma}{c}$

By plugging in the values for the radius and density of the sun and the mass of a hydrogen atom, you get 

$t_{diff}=8\times 10^{10}$ seconds

Free Form: IRAS08339+6517

One month ago, I found out that I had been accepted to do research at the Charlottesville site of the National Radio Astronomy Observatory.  As if that weren't exciting enough, the topic of research was something I actually found really exciting.

My mentor for the summer spent his grad school years researching the evolution of high-redshift galaxies.  This summer, we'll be researching a galaxy, IRAS08339+6517, which is at a redshift of about z=.02, about 80 megaparsecs away.  The galaxy is currently undergoing high amounts of star formation, and the project aims to answer the following questions:

• What causes star formation?  Is it mergers of different galaxies?  Accretion from the inter-galactic medium?
• Is the formation spread out across the entire galaxy or restricted to certain locations?
• Is there an Active Galactic Nucleus, and, if so, how does it influence the rate of star formation?

Astronomers have already found a lot of data on this galaxy.  For example, it's at an RA of 8 hours and 38 minutes, with a declination of $65^o$ and 7 minutes.  Its radial velocity (it's velocity away from our line of sight) is 5730 km/s.  They've taken spectra of the galaxy.

Both from the NED database

We'll be observing in the far infrared region of the electromagnetic spectrum.  We'll be taking more spectra of the galaxy and using them to determine the chemical composition of the stars being formed.  From that, we'll hopefully be able to figure out where most of the stars are coming from.

We're studying this particular galaxy because it's the perfect compromise between being really far away (so the information we're getting from studying it is actually hundreds of thousands of years old) but close enough that we can get good data.  When we answer the questions above and others like them, we'll have a better idea of what star formation was like closer to the big bang.  We're not exactly sure what we'll find, but we know that it's not going to be the same as the processes of star formation in nearby galaxies.

WS 8, Problem 1: Spacial Scales!

The size of a modest star forming molecular cloud, like the Taurus region, is about 30 pc. The size of a typical star is, to an order of magnitude, the size of the Sun.

a) If you let the size of your body represent the size of the star forming complex, how big would the
forming stars be? Can you come up with an analogy that would help a layperson understand
this di fference in scale? For example, if the cloud is the size of a human, then a star is the size
of what?

Our first step was to draw a big formless shape on the board, representing a molecular cloud with a radius of 30 parsecs.  We then converted that into centimeters so that we could more easily compare it to the size of the sun.

We found that the Taurus region is about $10^9$ times as large as our Sun.

If we take this down to a human size scale, assuming the average sphere-shaped (because everything's a sphere) has a radius of about 1 m, the typical star would have a radius of $1\times 10^{-9}$ m, or 1 nanometer.

To give a point of reference, this is a close-up image of a human hair

Science Picture Company

which has a width of of about 10 micrometers.  This means that, if I were a star forming molecular cloud, the average star formed from my gases would be smaller than $\frac{1}{1000}$ of a single hair. 

b)  Within the Taurus complex there is roughly $3\times 10^4 M_\odot$ of gas. To order of magnitude, what is the average density of the region? What is the average density of a typical star (use the Sun as a model)? How many orders of magnitude diff erence is this? Consider the di fference between lead and air.  

The average density of the region can be found using 

$\rho=\frac{M}{V}=\frac{M}{\frac{4}{3}\pi r^3}$

Just by plug-and-chugging provided values into the above equation, we find that the average density of the Taurus region is $1.5\times 10^{-23}$ grams per cm cubed.  

We plugged in the sun's values to find the average density of the sun, which we found to be $5\times 10^{-1}$ grams per cm cubed.  

This means that the sun is $10^{22}$ times as dense as the Taurus region.  This ratio is a lot bigger than the ratio of the densities of lead and air, which are of the same order of magnitude.  

WS 8, Problem 2: Forming Stars

Giant molecular clouds occasionally collapse under their own gravity (their own "weight") to form stars. This collapse is temporarily held at bay by the internal gas pressure of the cloud, which can be approximated as an ideal gas such that P=nkT, where n is the number density of gas particles within a cloud of mass M comprising particles of mass $\bar{m}$ (mostly hydrogen molecules) and k is the Boltzmann constant $k=1.4\times 10^{-16}$ erg/K.


a) What is the total thermal energy, K, of all of the gas particles in a molecular cloud of total
mass M? (HINT: a particle moving in the i$^{th}$ direction has $E_{thermal}=\frac{1}{2}mv^2=\frac{1}{2}kT$. This fact is a consequence of a useful result called the Equipartition Theorem.)

First, we thought about what the equation would be for the energy of one molecule moving around the cloud.  We're given the equation for the energy of a molecule moving in one direction, and since it could move in three different directions (up/down, forwards/backwards, and left/right since we live in 3-dimensional space), we multiplied that equation by 3 to give the energy of a single molecule moving in any and all directions.

$K=\frac{3}{2}k_BT$

To find the energy of every molecule moving in any and all directions, multiply the above equation by the number of molecules, N, in the cloud.  We are given the average mass of each molecule, $\bar{m}$, and N and M  are related in the following way

$N=\frac{M}{\bar{m}}$

so, substituting this into the first equation, we find that 

$K=\frac{M}{\bar{m}}\frac{3}{2}k_BT$

b) What is the total gravitational binding energy of the cloud of mass M?

To solve this, we recalled the equation for binding (or potential) energy that we found on the last worksheet. 

$U=-\frac{3}{5}\frac{GM^2}{R}$

But there's a problem with this equation--we don't know R, the radius of the cloud.  We do, however, know the cloud's average density, $\rho$, and from that, we can find R.

$\bar{\rho} =n\bar{m}=\frac{M}{V}=\frac{M}{\frac{4}{3}\pi R^3}$

When we rearrange this equation, we can find an equation for R that depends on known values.  

$R=\left ( \frac{M}{4n\bar{m}} \right )^{1/3}$

The $\pi$ goes away in the simplified version because, as we've heard many times in this class, $\pi =3$, so it and the 3 cancel each other out.

Now that we have a way to write R in terms of known values, we can rewrite the equation for the binding energy.  

$U=-\frac{3}{5}\frac{GM^2}{\left ( \frac{M}{4n\bar{m}} \right )^{1/3}}$

$=-\frac{3}{5}GM^{5/3}(4n\bar{m})^{1/3}$

c) Relate the total thermal energy to the binding energy using the Virial Theorem, recalling that

you used something similar to kinetic energy to get the thermal energy earlier.

This part asked us to relate the thermal and binding energies, so we were basically relating our answers from parts (a) and (b).  It's a lot of algebra that starts with 

$K=-\frac{1}{2}U$

$\frac{M}{\bar{m}}\frac{3}{2}k_BT=\frac{3}{10}GM^{5/3}(4n\bar{m})^{1/3}$

Through a lot of algebra that would be pretty boring to show here, we got to this final, simplified equation

$5k_BT=GM^{2/3}(4n)^{1/3}\bar{m}^{4/3}$

d)  If the cloud is stable, then the Virial Theorem will hold. What happens when the gravitational

binding energy is greater than the thermal (kinetic) energy of the cloud? Assume a cloud of

constant density $\rho$.  

In our discussion, our group decided that the thermal energy was the energy pushing out from the center, resisting the binding energy, which is pushing in.  

If one goes by the Virial Theorem (and I will, because I'm not at the stage in my life where I can really go against it), the thermal and binding energies will cancel each other out (have a net force of 0) when U=2K.  So when U>2K, the binding energy is stronger than the thermal energy and there's a negative net force, so the cloud will start to collapse.  

e) What is the maximum mass, $M_J$, that the cloud can have before it collapses? This is known

as the "Jeans Mass."

To solve this, we found the mass at which the binding energy is twice the thermal energy, so we set up the following equation

$10k_BT=GM^{2/3}(4n)^{1/3}\bar{m}^{4/3}$

and rearranged to find an equation for $M_J$.  

$M_J=\frac{10^{3/2}k_B^{3/2}T^{3/2}}{2G^{3/2}n^{1/2}\bar{m}^2}$

f)  What is the minimum radius, $\lambda_J$, that the cloud can have before it collapses? This is known as the "Jeans Length."

Once again, we were faced with the challenge of finding the radius of the cloud in terms of known values.  So, just like before, I did this by relating density and volume.  I followed the same steps I followed in part (b), but instead of solving for R, I solved for M

$M=4r^3n\bar{m}$

and substituted it into the first equation from part (e).  

$10k_BT=G(4r^3n\bar{m})^{2/3}(4n)^{1/3}\bar{m}^{4/3}$

$=4Gr^2n\bar{m}^2$

and solved for r, or $\lambda_J$.

$r=\lambda_J=\left ( \frac{10k_BT}{4Gn\bar{m}^2} \right )^{1/2}$

Originally, our group used the volume-mass relationship to solve for mass in terms of radius.  We assumed that the Jeans Length would be the length at which the mass was equal to the Jeans Mass.  But I realized that probably isn't the case because I don't know for a fact that mass changes proportionally with radius in a giant molecular cloud.  So, instead, I solved for the Jeans Length independently of the Jeans Mass.



It's time for the next installment of the Selfies with Astro series.  This photo was taken at the request of Professor Johnson, who then pointed out that it wasn't actually a selfie because I didn't take the picture myself.  I promise to fix this problem in the future.

Star Formation: MAPOD

Moiya's Astronomy Picture of the Day

Trifid Nebula

Photo Credit: Spitzer Space Telescope

Explanation:  Also known as Messier 20 or NGC 6514, the Trifod Nebula is around $1.6\times 10^{3}$pc away by the constellation Sagittarius.  Since its discovery in 1764, the nebula has become a favorite for amateur astronomers because it's relatively bright even through strong binoculars.  Astronomers have found over 100 newborn stars and 30 still-forming stars.  The gas used in many of the still-forming stars is believed to be left over from the formation of Type O star, shown as the white spot in the center of the picture.