Blog 20: The Hubble Tuning Fork

The Hubble Tuning Fork

The Hubble Tuning Fork is a classification system developed by Edwin Hubble in 1926 as a theory of galaxy evolution. Today we view it as outdated, as there is evidence that spiral galaxies do not evolve from ellipticals, but it is a useful visual for galaxy types.

Ellipticals (E0-E7): 

Elliptical galaxies are spheroid or elongated spheres where stars have no uniform rotation around the center of the galaxy. Elliptical galaxies tend to have older, and therefore redder, stars. Elliptical galaxies are classified due to their shape. E0 galaxies appear almost perfectly spherical, at least from our perspective on Earth. E3 galaxies appear as slightly elongated ellipsoids, E5 slightly more, and E7 galaxies are extremely elongated.

Lenticular galaxies (S0):

Lenticular galaxies (charmingly named after the lentil bean) fall between elliptical and spiral galaxies. They have no spiral shape but do contain a bulge and thin disk, similar to a spiral galaxy.

Spiral Galaxies (Sa-Sc) and Barred Spiral Galaxies (SBa-SBc):

Spiral galaxies are galaxies like the Milky Way or Andromeda. They typically contain a disk around which stars revolve, a bulge, and spiral arms. Spiral galaxies contain younger, bluer, stars than ellipticals. Spirals are classified by how tightly their arms are wound with Sa spirals being the tightest wound and Sc being the loosest. Barred spirals have a bar of stars running across the nucleus out of which the arms extrude. Barred spirals are also classified by the distribution of their arms.

Irregular:

Irregular galaxies do not fit into any of these categories and are often the result of two galaxies colliding.

• E0: Image: M87
• E3: Image M86
• E5: Image: M59
• E7: Image
• S0: Image: Messier 102
• SA: Image: NGC 1302
• Sb: Image: M77 
• Sc: Image: NGC 628
• Sba: Image: M83
• Sbb: Image: Ngc 1300
• Sbc: Image: NGC 1365 

Blog 19: Exomoons

You've heard of exoplanets, now get ready for exomoons! If the goal of our search for other planets is to possibly find other life, or just to have more knowledge, the next logical step would be to look at moons. Even within our solar system, we have 182+ moons and of those moons, more have signs of liquid water (e.g. Europa and Enceladus) than planets (e.g. Mars). With all the exoplanets we have been finding lately, exomoons could fructuous line of research, if we have the technology to explore it.

Jupiter and some of its moons. Source

So how do we find exomoons? Some projects are already underway. Here at the Center for Astrophysics, research is already being conducted to look for anomalies in stars' light curves that could point to exomoons in addition to exoplanets. Today, I am going to talk about this astrobite, which discuses future, direct detection of exomoons.

How would one directly detect an exomoon? They key to exomoon detection is the compositional differences between a moon and the planet it orbits. This would lead to distinct spectrographs for a planet and its moon. By analyzing the combined spectrum of the planet-moon system, there could arise evidence for moons. For example, the figure below shows the spectra for an Earth-Moon analog system orbiting Alpha Centuri. The bottom half shows the percent of the flux density due to the moon in this system.

By analyzing a system's flux at different frequencies, different elements of the system can stand out. For example, in the infrared wavelength in the system above, the moon accounts for 99.8% of the total flux in the water band at ~2.7 microns. Analyzing these differences can also give us the spacial separation of a planet and its moon. The offset of the origin of peak flux in different wavelengths can point to the location of a moon (as shown in the figure below).

The one problem with this exciting new research is that we don't quite have the technology to resolve these systems. Therefore, these plans are for future technologies. For example, to find the Earth-Moon analog orbiting Alpha Centauri, one would need a spacial resolution of ~2 milliarcseconds in the infrared. The Hubble Space Telescope, by comparison,  can only resolve ~100 milliarcseconds in similar wavelengths. The image below describes the size of telescope needed to resolve nearby candidates.

If we could achieve this technology, not only could we find these exomoons, but we could check for chemical compositions that could mean the moons are habitable or even containing biosignatures. The cliche saying to shoot for the moon could take on a new meaning in the near future. 

Tully-Fisher: Blog 18, Worksheet 6.1, Problem 4

Over time, from measurements of the photometric and kinematic properties of normal galaxies, it became apparent that there exist correlations between the amount of motion of objects in the galaxy and the galaxy’s luminosity. In this problem, we’ll explore one of these relationships. Spiral galaxies obey the Tully-Fisher Relation: \[L \sim v_{max}^4 \], where L is total luminosity, and vmax is the maximum observed rotational velocity. This relation was initially discovered observationally, but it is not hard to derive in a crude way:

Source

(a) Assume that \(v_{max} \sim σ \) (is this a good assumption?). Given what you know about the Virial Theorem, how should vmax relate to the mass and radius of the Galaxy?

The \(v_{max} \sim σ \) assumption is a pretty safe one to make because σ is basically the width of the possible velocities distribution. It has the same units as v_{max} and will always be off just by a constant.

In the previous problem, we derived \[M \approx \frac{σ^2R}{ G}\] a relationship that can be rearranged to show \[ v_{max}^2 \sim σ^2 \approx \frac{GM}{ R} \] So as v increases as mass increases and radius decreases.

(b) To proceed from here, you need some handy observational facts. First, all spiral galaxies have a similar disk surface brightnesses (<I> =  \( \frac{L}{R^2}\) ) (Freeman’s Law). Second, they also have similar total mass-to-light ratios (M/L).

This part is really just setting up definitions for part c: \[ <I> =   \frac{L}{R^2} \] can be rearranged to \[ R = \left(\frac{<I>}{L} \right)^{\frac{1}{2}} \] and \[ x = \frac{M}{L}\] which is \[ M = \frac{x}{L}\]

(c) Use some squiggle math (drop the constants and use ~ instead of =) to find the Tully-Fisher
relationship.

\[v_{max}^2 \sim \frac{M}{R} = \frac{ \frac{x}{L}}{\left(\frac{<I>}{L} \right)^{\frac{1}{2}}}= x <I>^{\frac{1}{2}} L^{\frac{1}{2}} \] Dropping the constants gives: \[L \sim v_{max}^4 \]

(d) It turns out the Tully-Fisher Relation is so well-obeyed that it can be used as a standard candle, just like the Cepheids and Supernova Ia you saw in the last worksheet. In the B-band (λcen ~ 445 nm, blue light), this relation is approximately: \[M_B = -10 log \left( \frac{v_{max}}{ km/s} \right) + 3\] Suppose you observe a spiral galaxy with apparent, extinction-corrected magnitude B = 13 mag. You perform longslit optical spectroscopy (ask a TF what that is), obtaining a maximum rotational velocity of 400 km/s for this galaxy. How distant do you infer this spiral galaxy to be?
\[M_B = -10 log \left( \frac{v_{max}}{ km/s} \right) + 3\] \[M_B = -10 log \left( \frac{400}{ km/s} \right) + 3\] \[M_B \approx -23 \] We know the distance formula from last week and can use it to find our answers: \[ d = 10^{\frac{m-M + 5}{5}} \] \[ d = 10^{\frac{13 - (-23) + 5}{5}} = 1.6 \times 10^8 \: pc \]

Good Ol' Virial Theorem: Blog 17, Worksheet 6.1, Problem 3

One of the most useful equations in astronomy is an extremely simple relationship known as the Virial Theorem. It can be used to derive Kepler’s Third Law, measure the mass of a cluster of stars, or the temperature and brightness of a newly-formed planet. The Virial Theorem applies to a system of particles in equilibrium that are bound by a force that is defined by an inverse central-force law ( \( F \propto 1/r^α\) ). It relates the kinetic (or thermal) energy of a system, K, to the potential energy, U, giving \[K = - \frac{1}{ 2} U\]

(a) Consider a spherical distribution of N particles, each with a mass m. The distribution has total mass M and total radius R. Convince yourself that the total potential energy, U, is approximately \[U \approx \frac{GM^2}{R} \] You can derive or look up the actual numerical constant out front. But in general in astronomy, you don’t need this prefactor, which is of order unity. 


We can think about this sphere of particles with a uniform density in terms of thin shells, each with a width of dR.

We also know that \[ U = - \frac{GMm}{R} \] We can treat the entire sphere as our mass, M, and the shell as our mass, m. Then, we can rewrite each mass in terms of density: \[ \rho = \frac{M}{V} \] Where \( V = \frac{4}{3} \pi R^3\) So \[ \rho = \frac{M}{\frac{4}{3} \pi R^3} \] And we can rewrite the equation for potential energy of one shell as: \[ dU = - \frac{GMdm}{R} = - \frac{G(\frac{4}{3} \pi R^3 \rho) (4 \pi R^2 \rho) dR}{R}= - G \frac{16}{3} \pi^2 R^4 \rho^2 dR \] But now we have dU in terms of dR, so we can integrate to get U: \[ \int_{0}^{R} - \frac{16}{3} G \pi^2 R^4 \rho^2 dR =  - \frac{16}{3} G \pi^2 \rho^2  \int_{0}^{R} R^4 dR = - \frac{16}{3} \frac{R^5}{5} G \pi^2 \rho^2 - 0 \] But this still is not the answer we want. We need U in terms of M, not density so we can reuse  \[ \rho = \frac{M}{ \frac{4}{3} \pi R^3 } \] to get: \[ U = - \frac{16}{3} \frac{R^5}{5}\left( \frac{M}{\frac{4}{3} \pi R^3 } \right)^2 G \pi^2  \] Amazingly enough, this simplifies to the much nicer expression: \[ U = - \frac{3GM^2}{5R} \] This simplifies to our original expression: \[ U \approx - \frac{GM^2}{R} \]

(b) Now let’s figure out what K is equal to. Consider a bound spherical distribution of N particles (perhaps stars in a globular cluster), each of mass m, and each moving with a velocity of vi with respect to the center of mass. If these stars are far away in space, their individual velocity vectors are very difficult to measure directly. Generally, it is much easier to measure the scatter around the mean velocity if the system along our line of sight, the velocity scatter σ 2 . Show that the kinetic energy of the system is: \[K =  N \frac{3 }{2} m σ^2\]

Our traditional, classical mechanics, definition of kinetic energy is \[ K = \frac{1}{2} mv^2 \] where velocity is the total or average velocity of the system. That is very difficult to measure for a distant object here on Earth, because the entire system (e.g. a globular cluster) is moving away from us with the universe's expansion. To account for this we must change our frame of reference to the system.
 

So our definition for kinetic energy \[ K = \frac{1}{2} m \Sigma (v_o + \sigma_{vi} )^2 \] turns into  \[ K = \frac{1}{2} m \Sigma ( \sigma_{vi} )^2 =   \frac{1}{2} m \sigma_{vi}^2\] in the reference frame. However, this is a 1 dimensional representation of velocity scatter, as that is all we can observe - the cluster in one plane. To account for the 3 dimensions of space.

So our final expression would be \[ K =   \frac{3}{2} m \sigma_{vi}^2\]

   (c) Use the Virial Theorem to show that the total mass of, say, a globular cluster of radius R and stellar velocity dispersion σ is (to some prefactor of order unity): \[M = \frac{σ^2R}{ G}\]

Using the Virial Theorem, we can combine our two derived equations to find the answer. \[K = - \frac{1}{ 2} U\] \[\frac{3}{2} m \sigma_{v}^2 = -\frac{1}{2} \left( - \frac{GM^2}{R} \right) \] \[3M\sigma_{v}^2 = \frac{GM^2}{R} \]  \[M = \frac{σ^2R}{ 3G}\]  \[M \approx \frac{σ^2R}{ G}\]

The Great Debate: Blog 16

Like many discoveries in science, astronomy discoveries have not come into modern cannon uncontested. Many of our now-accepted theories began out of scientists trying to understand nebulous (get it?) data collected with limited technology. The question of the scale of the universe, which is still debated today, was a hot topic in 1920. 

"The Great Debate" as it has been named, took place between two astronomers on April 26 1920. The "contestants" were Harlow Shapley, a young upcoming astronomer at Mt. Wilson Solar Observatory, and Heber D. Curtis, an older, more established professor at the Lick Observatory. Shapely argued in favor of the Milky Way being the entire scale of the universe while Curtis argues in favor of the universe being composed of multiple, separate galaxies. The debate centered around "nebulae" in the Milky Way, such as Andromeda, and whether on not they were a part of our galaxy or separate entities.

Shapley and Curtis - Source

Both scientists had previously published their findings on the scale and composition of the universe in different journals. Curtis's research focussed on star count analysis and spectral distance estimates. He concluded that the Milky Way is about 10 kpc in diameter and lens shaped. He found the nebulae observed had a similar number of novae to the rest of the galaxy, and therefore must be their own galaxies. On the other hand, Shapley argued for a model where one massive Milky Way (100 kpc across) composed the entire universe. He looked at Cepheid variables in globular clusters to make distance determinations. Shapely used the already known distance to the M31 cluster to estimate the relative distances to other clusters and determined them to be at the edge of the Milky Way. Additionally, the data of Adriaan van Maanen, a friend of Shapley's at Mt. Wilson Observatory, found that the Pinwheel galaxy was rotating so quickly that you could observe one rotation in a matter of years. If this galaxy lay beyond the Milky Way, this would mean the galaxy was rotating so fast that it violated the speed of light! This provided further evidence for Shapley's case.

Source

On April 26, 1920 both scientists gave separate talks proposing their ideas during the day and came together for a discussion at night, where they could provide counter arguments to each other's points. Who "won" is unclear. There is a common misconception that Shapely won, as he became more famous, though many scientists present agreed with Curtis's argument.

Today we know that both scientists were right in some ways and wrong in others. Shapely got the order of magnitude of the universe correct and our relative placement in it. Curtis correctly concluded that the "nebulae" were other galaxies. This was confirmed when Edwin Hubble studied Cepheid variables in Andromeda and concluded it lies outside of the radius of the Milky Way. Alas, Maanen's measurements were proven to be incorrect, the rotation of the Pinwheel galaxy cannot be measured in years and the high recessional velocities observed are actually evidence for the expanding universe, a debate that occurred  a few years later.


Sources:
http://cosmos.phy.tufts.edu/~zirbel/ast21/handouts/Curtis-Shapley.pdf
http://apod.nasa.gov/diamond_jubilee/papers/trimble.htmlhttp://astronomy.nmsu.edu/geas/lectures/lecture27/slide01.html
http://astronomy.nmsu.edu/geas/lectures/lecture27/slide01.html

Graphing Cepheid Magnitudes: Blog 15, Worksheet 5.2

As we previously learned, Cepheid variables are a special class of stars that radially pulsate in a predictable way. In 1908, Henrietta Swan Leavitt discovered that there is a distinct relationship between a Cepheid’s luminosity and pulsation period by examining many stars in the Magellanic Clouds. Henrietta was a member of “Harvard’s computers,” a group of women hired by Edward Pickering to analyze stellar spectra and light curves. In this worksheet, we will use Henrietta’s original data set to find our own Period-Luminosity relation for Cepheid variables. The data files for this activity will be located on Canvas under the name “Cepheid variables.csv.”

Source

1. The data file, “Cepheid variables.csv,” contains data for 25 Cepheid variables located in the Small Magellanic Cloud (SMC). Each line contains a specific Cepheids: (1) ID number, (2) Maximum apparent magnitude, (3) Minimum apparent magnitude and (4) Period. Calculate the mean apparent magnitude for each Cepheid.

Done! It is pretty easy to do in excel.

2. The distance to the SMC is about 60 kpc, where kpc = 1000 pc. Convert your mean apparent magnitudes into mean absolute magnitudes. Plot the Cepheid mean absolute magnitudes as a function of period. This plot should look exponential.

3. It is often handy to plot exponential (or power-law) functions with one or more logarithmic axes, which “straightens out” the data. Magnitudes are already exponential, so we don’t need to adjust that axis. Plot the Cepheid mean absolute magnitudes as a function of log(Period). Verify that the plot now looks linear.

4. Now that the data look linear, we can estimate the parameters of a linear relation, \(M_V (P) = A log_{10}(Period) + B \). A and B are “free parameters” that allow the function to match the data.

This would make A = -2.033 and B  = -.2782 for a final expression of approximately \[M_V (P) = -2.0 log_{10}(Period) - 2.8 \]

Stellar Magitude: Blog 14, Worksheet 5.1, Problem 2

Okay, let’s use your new-found knowledge of magnitudes. In the first problem of the wroksheet, we derived a relationship that looks something like this:

Source

Now lets put that knowledge to use: 

(a) Suppose you are observing two stars, Star A and Star B. Star A is 3 magnitudes fainter than Star B. How much longer do you need to observe Star A to collect the same amount of energy in your detector as you do for Star B? 

From question 1, we know: \[ \frac{F_1}{F_2} \approx 10^{0.4( m_2 - m_1)} \] For this problem we have  \[ \frac{F_1}{F_2} \approx 10^{0.4 \cdot 3} \approx 15.6 \] So we would have to observe for about 16 times longer.

(b) Stars have both an apparent magnitude, m, which is how bright they appear from the Earth. They also have an absolute magnitude, M, which is the apparent magnitude a star would have at d = 10 pc. How does the apparent magnitude, m, of a star with absolute magnitude M, depend on its distance, d away from you? 

For this problem, we are going to look back to a few weeks ago and use our definition of flux:
\[F_M  = \frac{L}{4 \pi d^2} \] \[F_m = \frac{L}{4 \pi r^2} \]  and  \[ \frac{F_1}{F_2} \approx 10^{0.4( M - m)} \] \[ \frac{d^2}{r^2} = 10^{0.4(M- m)} \] \[ \frac{2}{5}(M - m) = log_{10} \left( \frac{d^2}{r^2} \right) \] We know d=10pc, so  \[ \frac{2}{5}(M - m) = log_{10} \left( \frac{100}{r^2} \right) \]  \[ \frac{2}{5}(M - m) = log_{10}(100) - log_{10}(r^2) \] \[ \frac{2}{5}(M - m) = 2 - 2log_{10}(r) \] \[ m= M + 5log_{10}(r) -5\]

(c) What is the star’s parallax in terms of its apparent and absolute magnitudes?

We know from blog 2 that, \[ \theta = \frac{1 \: AU}{r}\] And we can find r from our work above \[ r = \left( \frac{100}{10^{0.4(M-m)}} \right)^{\frac{1}{2}}\] So \[ \theta = \frac{1 \: AU}{\left( \frac{100}{10^{0.4(M-m)}} \right)^{\frac{1}{2}}}\] but we need to multiply by 1000 to get our answer in AU/kpc so it is really  \[ \theta = \frac{1 \: AU}{\left( \frac{100}{10^{0.4(M-m)}} \right)^{\frac{1}{2}}}\cdot (1000) \] This simplifies to \[\theta = \frac{1000 \left( 10^{0.4(M-m)} \right)^{\frac{1}{2}}}{10} = 100 \left( 10^{0.4(M-m)} \right)^{\frac{1}{2}} \]