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.