These results are about the same as for the first midterm.

The mean score last time was about 117, but the lowest score was lower and the highest score was lower also.

At the right is the histogram of scores for the second midterm exam.Approximate letter grades are indicated on the histogram to give you an idea of the curve that will be applied.

Well, this test was perhaps a bit harder than the first one, but we did get a good spread in the scores, and a couple of students did very well and got all but 2 or 3 of the questions right.

Aregion

Bregion

Cregion

other

This is the score histogram for Midterm 2.

As I said in class, students whose scores improve significantly from the first to the second midterm exam, and whose scores remain at about the newly established level on the final, will have their first midterm exam scores weighted less heavily in determining their final grades.

As students have noted, the CANVAS system keeps your scores, with this test worth 180 points, but it does not revise the scores to reflect the correspondence of scores to letter grades that is indicated in the score histogram.Therefore, do not worry if the CANVAS system thinks you are not doing so well in the course so far. You can see how well you are doing by looking at where your score falls on the histogram.If you have questions, come aske me after the lecture.The histogram of total scores at this point in the course are given below:

Date: April 15, 2019TO: Astronomy 1001, Section 003 studentsFROM: Paul WoodwardSubject: score histogram at this point in the semester

After the scores on the second midterm exam came in, I produced a score histogram with indicated boundaries between the letter grades. As I have said in class, my intent with the exams is to have a good spread in the scores and also to have some of you get almost all the questions right. No one got a perfect score on the second midterm, but the results came out pretty well according to my intent.

The spread of scores on the exam, however, tends to produce difficulty for some students in assessing how they are doing, since it can result in relatively low point totals. To address this problem, I have created a second score histogram that you can use, together with looking up your scores on CANVAS, to see where you stand at this point in the semester and to see where you fall in the approximate mapping of scores to letter grades.

The number of points that an ideal student could have earned at this stage is 575, and the highest score in the class at this time is only a bit short of that ideal, at 546. The spread, however, is quite large. Your place in this tally of scores at this time can be strongly affected by even a single missed lab. I am not counting any points for the planetarium, since I did not include that in my syllabus explanation of which things count how much. Quite a few of you, according to CANVAS, missed lab J, and that can affect where you end up in the histogram. Perhaps your scores for that lab were not yet entered into the system. To address that possibility, I have revised the score histogram so that it does not include that particular lab, since it seems unlikely that so many of you actually did miss it. This brings the highest possible score down to 575 points, and I hope it gives a better reflection of the real score distribution at this point.I hope that this new score histogram is helpful in getting a sense of where you now stand. Go to CANVAS, find your present score (hover your mouse over the total expressed as a per-cent), then subtract from that your score for lab J and for the planetarium. The result is the score that shows your position on the histogram.Paul Woodward

Score Number of Students Approximate letter grade530-549 2 A510-529 2 A490-509 8 A470-489 3 A450-469 9 A430-449 13 B+/A-410-429 15 B390-409 10 B370-389 8 B350-369 10 C+/B-330-349 8 C310-329 4 C-/C290-309 1 C-270-289 1 ?000-269 7 -- this last category probably includes students who have dropped the course, or perhaps students taking the course pass/fail.

1

4

8

10

8

10

1513

9

3

8

2 2

7

000 -269

530 -549

510 -529

490 -509

470 -489

450 -469

430 -449

410 -429

390 -409

370 -389

350 -369

330 -349

310 -329

290 -309

270 -289

Cregion

other

Aregion

Bregion

1

As students have noted, the CANVAS system keeps your scores, with the 2 midterms worth 180 points each, but it does not revise the scores to reflect the correspondence of scores to letter grades that is indicated in the score histogram.Therefore, do not worry if the CANVAS system thinks you are not doing so well in the course so far. You can see how well you are doing by looking at where your score falls on the histogram.If you have questions, come ask me after the lecture.

Milky Way Kinematics, or how we discovered the geometry and dynamics of our own galactic environment.

Compiled by Paul R. Woodward and B. Kevin Edgar

The name “galaxy” comes from the Greek word for milk.

The Galaxy appears to us like a river of milk, and hence its name, the Milky Way.

Kinematics – describes the motion without consideration of the

causes leading to the motion.

Dynamics – describes the relationship between motion of objects

and its causes

Just What is a Galaxy?

Was everyting seen in telescopes part of the Milky Way?

Stars

Clusters

Supernova Remnants

Planetary Nebulae

Dark Nebulae

Reflection Nebulae

“Spiral Nebulae”

Was the Milky Way the entire Universe?

Debate raged well into the 20th Century

Mapping from the Inside

What is the Earth's place in the Universe

(As Usual, we first thought we were at the center)

What is the Sun's place in the Galaxy

(For a long time we thought the Sun was at the center of the Milky Way)

Starting With Star Counts

1. William Herschel

“Star Gauging”

683 regions around the sky

Assumed Galaxy uniformly filled with stars to edge

His data were almost completly uniform e.g. Sun at Center

William Herschel’s Universe, 1785

Starting With Star Counts continued

2. Jacobus Kapetyn 1906

First “Modern” attempt using photography

Assumed all stars have same intrinsic brightness

Determine distance based on apparent and intrinsic magnitudes

Bright stars uniformly distributed

Fainter stars concentrated in a band “Galactic Equator”

Conceived a lens shaped galaxy 25,000 light years thick

Solar System near the center

Map local stellar environment by means of star counts.

First plot positions on sky of stars brighter than some level 1.

Then plot positions of stars between brightness levels 1 and 2.

And so on.

If distribution of stars in space is uniform, each plot will be also.

Number of stars per square degree in each successive plot will increase in a fashion that we could predict.

Find stars concentrated in band of sky, the galactic equator.

Find that numbers of fainter stars do not increase away from the galactic equator so rapidly (as uniform distribution would imply).

This true also toward galactic equator, although shortfall is smaller.

Led to “Kapteyn Universe” with us in center.

1912 - Henrietta Levitt and Cepheid variables

Typical Cepheid Light Curve

Cepheid Period-Luminosity Relation - 1912

To determine the Absolute Magnitude (MV), essentially the luminosity, of a Cepheid: first find the Period of the Cepheid along the horizontal axis (x-axis), go up to the diagonal line which corresponds to the Period-Luminosity relation, then read across to the vertical axis (y-axis) to see what value of MV it corresponds to. Write the corresponding value of MV in the correct column on your Cepheid Data Sheet.

Cepheid period luminosity relation.

In 1917, after Henrietta Levitt’s discovery of the period-luminosity relation for the Cepheid variable stars in the Small Magellanic Cloud, Harlow Shapley proposed that these stars were pulsating.

The explanation that we can determine the intrinsic luminosity of one of these special, pulsating stars from measuring its pulsational period goes like this.

The pulsational period depends upon the size of the object. This is because the pulsation is a pressure-driven phenomenon. One period is the time it takes sound (a pressure wave) to propagate from the surface to the deep interior and back. The longer this distance is, the longer the period of the pulsation.

This should be familiar to you from the different sounds of large and small bells or of long and short vibrating strings (in a piano, violin, or guitar). Big bells sound low notes. Big stars have low frequency pulsations.

So we see the relationship between the pulsational period and the size of the pulsating star.

The size of the star is in turn related to its intrinsic luminosity, because the Cepheid variable stars are all giant stars that are located in a nearly vertical strip in the H-R diagram. This means that they all have about the same surface temperature. The properties of the black body spectrum then imply that their luminosities are given by a nearly constant factor determined by their nearly identical surface temperatures multiplied by the areas of their surfaces.

This is summed up by the period-luminosity relation for Cepheid variable stars that is shown on the previous slide.

Shapley applied Levitt’s relation to RR-Lyrae variable stars in globular clusters, to get their distances.

He did not realize that the period-luminosity relationship for this different class of variable stars is not the same as for the Cepheid variables studied by Henrietta Levitt. His globular cluster stars were very old, and contained relatively few heavy elements, so they had a different period-luminosity relation.

Because of this mistake, Shapley got systematic overestimates for the distances to these star clusters, but he got the right geometrical distribution.

Globular clusters have spherical distribution centered on a position about 30,000 light years from us (the modern, corrected distance) in the direction of the galactic equator.

Thus we live in a disk, far from its center.

Globular clusters orbit center in all directions, disk stars all go around in one direction and in a single plane.

1918

Shapley Curtis Debate (1920)the Scale of the Universe

Main Questions:

* What is the nature of the `”nebulae'”?

* What is the size of our Galaxy?

* Is the Sun in the center of the Galaxy?

Shapley believed:

* diameter of our Galaxy was 300,000 light-years

* Sun was not at Galaxy's center, but 60,000 light-years away

* Milky Way was so large, it was the entire universe

* spiral nebulae were gaseous clouds repelled by Milky Way's light pressure

Curtis believed:

* diameter of our Galaxy was 30,000 light-years (10x smaller!)

* Sun was very close or at center of Galaxy

* spiral nebulae were galaxies -- island universes in their own right* Did NOT Believe that Cepheids had a strong period-luminosity

relation

The Andromeda Galaxy

Edwin Hubble 1924

* located Cepheids in nearest major spiral nebula

*derived distances larger than even Shapley's estimate of Milky Way's size

* Shapley admits he was wrong about spiral nebulae (an issue that he did not consider at the heart of the 1920 debate)

at the 100in on Mount Wilson

Trumpler's Discovery

1930 Robert Trumpler was studying star clusters

Noticed some appear fainter than expected based on their assumed size

Concluded (correctly) that space is filled with dust and gas which scatters light

This scattering is called Interstellar Extinction

Therefore Herschel and Kapteyn were only seeing the local stars

1930 Trumpler’s Discovery

He took advantage of wartime blackout conditions during World War II, which reduced light pollution at Mount Wilson Observatory, to resolve stars in the center of the Andromeda galaxy for the first time, which led him to define distinct “populations” for stars (Population I and Population II).

Walter Baade Mt Wilson 100 inch Hooker Telescope

Milky Way in Sagittarius

Milky Way in Sagittarius; Spitzer Infrared image

How by looking at stars in the sun’s vicinity, we are able to deduce the structure and dynamics of the Galactic disk:

See small number of “high velocity stars” near us moving rapidly relative to rest of material in disk.

All appear to be going in about same direction.

Bertil Lindblad said they are a population with little rotation about the Galaxy, and hence with large radial velocities.

We are rotating past this small population of stars.

We can measure this rate of shear in disk star motions(how much faster or slower stars orbit with distance toward or away from galactic center).

Also need to measure orbital velocity at radius of sun.

Bertil Lindblad figured out how to do this last thing, later confirmed by Jan Oort.

Measure dispersion of stellar velocities in two directions, along orbit and in direction of galactic center.

2 dispersions would be equal if galaxy rotated like a solid body.

Measure amount that they are unequal, and then can calculate orbital period of stars near us. This turns out to be related to the Oort constants mentioned earlier.

Answer is about 230 million years.

Use sun’s orbital velocity to “weigh” the galaxy inside sun’s orbit.

Result is 130 billion solar masses.

The Visible Sky

NGC 891, a spiral galaxy in the constellation Andromeda that is seen edge on, probably looks pretty much the way our galaxy would look when viewed from outside and edge on.

This galaxy looks pretty symmetrical, and pretty

thin.

Mapping our Galaxy by observing neutral hydrogen emission:

Henk van de Hulst realized that neutral hydrogen gas emits radio waves of 21 cm wavelength.

This radiation goes clear across the Galaxy without absorption.

Small radio telescope left in Holland by German army was then used to map the entire galactic disk.

This gave full “rotation curve” and revealed spiral arms as well.

At 21 cm, you can see forever, but result is confusion.

Consider single line of sight on diagram.

Only aid in handling confusion is different Doppler shifts of radiation from different regions along line of sight.

Gives a “line profile,” and is an art to interpret one.

The 21-centimeter Sky

To interpret line profile, use model of gas motions in galactic disk.

Simplest is circular orbits everywhere.

Then is single point along line of sight, the so-called tangent point, where gas has maximum velocity relative to us.

This maximum velocity measures speed of galactic rotation at one radius, radius of tangent point.

Whole series of such measurements gives the “rotation curve.”

Rest of the line profile:

2 locations on line of sight can produce each other velocity.

Radiation from near location would extend farther above and below galactic plane than radiation from far location.

Radiation from tangent point gives thickness of disk at that radius.

From all tangent points, get idea of disk thickness trend in radius.

This helps untangle near and far point problem.

This done in 1968 by Frank Kerr.

Some guess work, but result fairly clear.

We live in a spiral galaxy.

No doubt about it.

This sometimes allows us to choose one of the locations unambiguously. Sometimes not.

Can do same thing with CO emission from much denser, colder gas.

Get much flatter distribution.

Also see a “molecular ring” in inner disk, where spiral arms strongest.

Ring of Molecules around the core of the Milky Way

Molecular Clouds

Molecular Clouds

On CO diagram, see rotation curve traced by maximum velocities.

As near center, max velocity does not go up, but goes down.

Not like rotation curve for solar system.

So all mass in central region not concentrated at center.

In CO diagram, see ridges of bright emission along diagonal lines.

These are the spiral arms.

It takes a complex model to understand the shapes of these ridges.

This is a big, big deal.

The spiral arms are the crests of a WAVE.

Kind of like white caps on a lake.

Explained by Lin and Shu in 1968.

Believed by everyone 10 years later.

Based upon a large number of modern observations, this is the present consensus idea of how our galaxy would look if we could go outside of it and view it face on. The sun is located on the Orion Spur, between the 2 principal spiral arms.

NGC628 .

Except for the inner bar, our galaxy’s

inner region probably

would look pretty much

like this galaxy if we could see it face

on.

Size and Parts of the Milky Way

Diskbetween 100,000 and 160,000 light years

in diameterup to 2000 light years thick

Central Bulge around the nucleus6500 light years in diameter

HaloThe globular clusters trace out the halo

population

Typical orbit of disk star. Vertical velocity is small compared to orbital velocity. Oscillates up and down about center plane of disk.

Stars in the central bulge swarm around the center in all directions.

Weighing the Galaxy

Hard to see rotation of galaxy, because we are revolving with it.

But stars closer to center revolve faster; stars farther out go slower.

Like planets going around sun (remember Kepler’s laws?)

So our galactic disk is rotating differentially.

Rotation curve: plot orbital velocity trend with radius.

For solid body, like merry-go-round, get straight line.

For solar system, with Kepler’s laws and nearly all mass at center, get rapidly falling curve.

For galactic disk, encompass ever more mass as go out, so curve does not drop off.

If look toward and away from galactic center, we see that stars generally orbit faster and slower, respectively.

Jan Oort observed bright red giant stars near sun.

Got positions and line-of-sight velocities (Doppler shifts).

Concluded these stars are in disk.

Positions gave disk thickness and velocities gave mass in disk.

This mass was twice what you get from adding up visible matter.

Thus “mass-to-light ratio” of disk stars near us 5 times solar value.

Mostly have faint dwarfs, so no surprise.

Visible matter in galactic halo has only few % of disk mass.

Dynamical arguments say halo 10 times disk mass.

So mass-to-light ratio in halo is likely to be huge.

Nearby disk stars go around galactic center in almost circular orbits.

Typical orbit does not close on itself.

Determining the mass of an object

• Use small orbiting object.

• Kepler’s third law:P2 = a3

• Newton’s law of gravitation:P2 = a3 [4π2 / G (m1 + m2)]

• If one mass dominates, plot P2 against a3 and slope gives M:P2 = (4 π2 / GM) a3

• Solve for M in terms of the other quantities:M = (4 π2 / G) (a3/P2)

• Measure G in the laboratory, then put in a and P for a planet to get M for the sun.

• If measure a as an angle on the sky, then need to use the distance to the object M to convert a into distance units.

Use stars in disk to measure mass of Galaxy inside their orbits.

Galaxy is very flat (with small bulge in center).Assume circular orbits and axisymmetric distribution of mass.

(Axisymmetric means that the mass density is constant along every circle centered on the Galaxy’s rotation axis. Thus, if you were to spin the Galaxy a bit, an observer would see no difference, since an axisymmetric Galaxy looks the same as you go out along any radial line.)

Also assume orbits of all stars lie in same plane (flat Galaxy).

In infrared light that penetrates dust clouds, our galaxy looks really flat.

To get a rough measure of the mass from velocities of orbiting stars, we need only assume circular orbits and an axisymmetric distribution of

mass. We will also assume all orbits lie within a single plane.

The Milky Way, photographed in diffuse infrared radiation by the COBE satellite.

In infrared light that penetrates dust clouds, our galaxy looks really flat.

To get a rough measure of the mass from velocities of orbiting stars, we need only assume circular orbits and an axisymmetric distribution of

mass. We will also assume all orbits lie within a single plane.

The Milky Way, photographed in infrared radiation by the UMass 2 micron all-sky survey.

We assume that the mass of the galaxy is distributed evenly around each ring near a given radius (we assume that the galaxy is axisymmetric). Then chunks A and B of the

typical ring shown

have equal masses. They pull on the star equally in the directions indicated by the arrows. If we add up these two forces, the components pulling up and down on the

page cancel, leaving only the component directed toward the galactic center.

A

B

Each chunk, A, of a given ring has an equal-mass mirror chunk, B, so that the total force exerted on the star from each whole ring of matter is directed toward the galactic center. We get this total force by adding up the contributions from all the chunks, A.

Therefore the total force on the star from all the matter closer than the star to the galactic center is directed toward the galactic center. With a computer, or using the calculus, we could add up all

these force contributions, from all the chunks in all the rings, to get the total.

A

B

Similarly, we can show that the total force on our star from each ring located outside its radius is directed away from the galactic center. However, it takes only a much smaller

mass located at the center of the disk to cancel this outward pull.

To find the mass contained in each annulus of our galaxy, given a series of measured stellar revolution velocities at increasing radii from the center, we must solve what is

called an “inverse problem.”

A

B

Given a set of choices for the masses of the various annuli (rings), Newton’s laws and some calculus (or some computer-enabled algebra) allow us to calculate the accelerations, and hence the rotation

velocities of all the stars whose orbits define the annuli.

But what we really want to do is to compute things the other way around. We know the stellar velocities but not the masses of the annuli. Again, we can use a computer to make thousands or

even millions of trials in a matter of minutes, giving us the answer.

The rotation velocity of this star determines the mass in the

shaded ring.

Are our assumptions true for other galaxies?Nearby spiral galaxy M83 seen nearly face-on.Prominent spiral arms. Not very axisymmetric.But we see light distribution, not mass distribution.Spiral arms traced out by bright, massive stars.They are incredibly bright for their mass.If look in infrared, get better idea of mass distribution.Then see it is nearly axisymmetric.

See that edge-on galaxies are extremely flat.

Can also use companion galaxies to “weigh” whole galaxies.

Could use two small companions of Andromeda Galaxy.

Could use Magellanic Clouds to weigh our Galaxy.Weighing an irregular galaxy like the Large Magellanic Cloud

would require that we generalize our methods, since it is neither axisymmetric nor flat.

We have built our model of our own galaxy partly by looking at other ones in order to get a better perspective. This one is M83, 14 million

light years away from us.

Clearly, if our galaxy is like M83, shown here, our

assumption of mass spread out evenly around concentric rings

is in trouble.

Happily, what we see in this picture is the distribution of

light in M83, not the distribution of mass.

The Andromeda Galaxy, which is probably quite like our own.Its nuclear bulge is about 12,000 light years across.

This galaxy has two companions, that we could

use to help determine its mass

The Magellanic

Cloudsas viewed

from Australia

These two companions of our own galaxy should have motions

that we could relate to our galaxy’s mass.

The Large Magellanic Cloud,

a satellite of our own Milky Way

We could use the velocities of stars in this irregular

galaxy to help to determine its mass, but we would have to generalize

our method.