Q: Astronomers study light breakdown to determine what the composition
is of a
star right?? okay . i understand that . so if a star is
made entirely of
one material that makes sense . . but stars are made of many things,
yet they
emit one kind of light . . how do astronomers determine the breakdown
of
light when the light is caused by many elements????
A: The spectral lines from many elements can appear in the same spectrum,
and
since elements all emit lines at completely different wavelengths,
you don't
tend to confuse one line with another. For example, you may get
a spectrum
that looks something like:
|| | ||| | | |
12 3 456 7 8
9
And you might be able to say, "Well, I can tell that those first two
lines are
from calcium, and the 3rd, 6th, and 8th are too. 4, 5, and 6
are caused by
helium, and 7 and 9 are caused by argon." You will be able to
make these
determinations by looking up what lines are caused by helium, calcium
and argon, and comparing these lines with what you actually observe.
Q: If we use the spectrum to tell us what something is made out of as
well as
to find out how it is moving, how do astronomers tell the difference
between
the two when they are looking at the same spectrum of light?
A: Good question! Usually the amount of redshift or blueshift
is pretty small,
so it doesn't confuse us too much. Astronomers can recognize
the *pattern* of
spectral lines and say, "Golly! That's the pattern caused by
hydrogen. It's
just shifted a little bit."
Example: Let's say there's an element called 'ploopium'.
If the spectrum
of ploopium in a laboratory looks like this
| || || ||| | |
then the redshifted spectrum of ploopium moving away from the observer
would look something like
| || || ||| | |
So the pattern is the same, just moved over a little bit.
In a few weeks we'll study exotic objects called quasars. When
they were
first studied around the late 50's and early 60's, it was thought that
they
may be made of some really bizarre stuff, since none of the spectral
lines
were familiar. Then somebody figured out that the spectral lines
were, in
fact, hydrogen lines, but *really* redshifted away from where they
usually
are on the spectrum. That meant that quasars are all travelling
away from
us *really* fast, but the *pattern* of the hydrogen lines (in other
words the
relative spacing between one line and the next) stayed the same, but
the
whole pattern was shifted towards the red end of the spectrum.
Q: I can see how a binary system orbits around each other, but how do
more
than 2 stars work? And the book says that such systems are very common
but
I don't see what that should be so.
A: Stars, like puppies, are tend to be born in litters. (Fortunately,
unlike
puppies, they don't poop all over the place or require neutering.)
But since
stars are born in clusters, they often form close enough to their neighbors
that they become bound by their mutual gravitational attraction.
This
is why binaries and trinaries are pretty common. When you have
more than one
star gravitationaly bound to another, they will all orbit around the
common center of mass.
By the way, there is a cluster of newborn stars (this kind of cluster
is
called an 'open cluster') visible to our naked eyes in the evening
sky!
It's called the Pleiades (pronounced PLEE-uh-deez). Here's how
to find it:
First find Orion. It's a pretty prominent constellation high
in the southern
part of the sky in the early evening. Look at the three stars
in Orion's
belt. Imagine that these three stars form a line. Follow
this line
to the right of Orion. The first star you'll find along that
line is the
red giant Aldebaran in the constellation Taurus. Keep looking
to the right.
The next thing you'll see is the Pleiades. Aldebaran is almost
right
between Orion and the Pleiades in the sky:
.
.
. . . . ....
..
^
Aldebaran
^
. The
Pleiades
.
^
Orion
Q: Also, a sort of general question: when astronomers are tracking stars
over
the course of months and years (for parallax measurements or binary
star
period of orbits) how do they ever manage to find the exact same one
again?
Won't the stars move and get confused with other stars?
I imagine with
computers it's easier to track them, but it seems pretty amazing people
could track the millions of stars...
A: Historically, this kind of thing was done by studying photographic
plates.
In otherwords, photographs of the stars were taken several months apart.
Most stars are too far away to detect a parallax shift. So what
astronomers
did was to look to see how much the closer stars seemed to move with
respect
to the more distant ones.
Here's an example of what they might have seen:
Photograph taken in December:
___________________
| . .
|
|
. |
| .
. |
| .
|
| . .
|
|
. |
| .
|
|___________________|
Photograph of the same part of the sky taken in June:
___________________
| . .
|
|
. |
| .
. |
| .
|
| . .
|
|
. |
| .
|
|___________________|
Most of the stars haven't appeared to move, but two of them have.
Can
you spot which two? Of the two which one is closer to us?
Q: Why do variable stars exist and what causes them to change?
A: Some stars are variables because they're reacting to gunk getting
dumped on them by a binary companion.
Others are regular, pulsating variables (which have nothing to do with
'pulsars') When some stars run out of hydrogen fuel and leave
the main
sequence to become red giants, they go through a phase where they're
not
entirely stable.
I will quote the explanation of this phenomenon from "Astronomy Today"
by
Chaisson and McMillan:
"The structure of any star is determined in large part by how easily
radiation can travel from the core to the photosphere -- that is, by
the
opacity of the interior, the degree to which the gas hinders the passage
of light through it. If the opacity rises, radiation becomes
trapped, the
internal pressure increases, and the star 'puffs up.' If the
opacity falls,
radition can escape more easily, and the star shrinks. According
to theory,
under certain circumstances, a star can become unbalanced and enter
a state
in which the flow of radiation causes the opacity first to rise --
making
the star expand, cool, and diminish in luminosity -- and then to fall,
leading to the pulsations we observe."
Q: Can you please explain the stellar spectral classes and what each one is?
A: One of the first ways astronomers classified stars was by which
lines were the most prominent in the stars' spectra. Here is
a breakdown
of what each spectral class is: (Don't memorize this)
O: Strong HeII lines (HeII is ionized helium)
HeI lines moderately strong (HeI is non-ionized helium)
B: HeI lines strong
Hydrogen Balmer lines strong
A: Balmer lines strongest
CaII (ionized calcium) lines getting stronger
F: CaII lines continue to strengthen
Balmer lines continue to weaken
FeI (iron) and CrI (Chromium) lines appear
G: CaII lines even stronger
FeI and other metal lines stronger
K: CaII lines strongest
Spectrum dominated by lines caused by metals
M: Metal lines remain strong
Spectrum dominated by molecular lines, such as
TiO (titanium oxide).
Astronomers later figured out that there is a direct relationship between
which lines were the strongest (and weakest) and the temperature of
the
star. For example, A type stars are just the right temperature
for there
to be a high probability of finding hydrogen atoms such that the electrons
are in the 2nd energy level. Balmer lines (which were named after
the
famous physicist, Hugo M. Line) are caused by such hydrogen atoms.
For both
cooler and hotter stars, it is less likely to find such hydrogen atoms
in the
second energy level. Thus, even though there's hydrogen aplenty
in all stars,
the Balmer lines are weaker in stars that aren't A type.
Q: I'm still confused on how exactly absorption lines are formed on
the
spectrum. I understand how to "read" them and their significance, as
well
as what causes them in a general sense. However, the section entitled
"How
Temperature Affects a Star's Spectrum" starts getting into which elements
form which lines and how (i.e. light is absorbed when its energy matches
the energy difference between 2 electron orbits) the lines are formed.
Could you simplify this a bit? It would help a lot. Thanks!
A: The sun acts like a blackbody. That means it wants to emit
light in a
continuous rainbow, which means it wants to emit at every possible
wavelength
in some range, red, blue, green, you name it! But that light
has to pass
through the solar atmosphere. Consider little Charlie-Atom in
the solar
atmosphere. This atom might have an electron in a low energy
level. It
will jump up to a higher energy level if a photon with JUST THE RIGHT
ENERGY
comes along. If the photon has either too little or too much
energy, the
electron will ignore it. Let's say that it's green photons that
have just
the right energy to kick the electron up to higher energies.
As the
rainbow light from the sun passes through the atoms, the green light
will
get absorbed, but the other colors will pass through unmolested:
. . . .
purple ---> .. ..
. ---> purple
blue ---> . .. .. .
. ---> blue
green ---> . .
.. .
yellow ---> . . .
. ---> yellow
orange ---> .. . . .
. ---> orange
red ---> . .
.. . ---> red
. . .
\_______ _______/
\/
Atoms absorbing green photons
So when you pass the emerging light through the prism, it's missing
the
color green in its spectrum. There's just blackness there.
A figure much nicer than the one above shows the same thing in your
book. I'm
not sure on which page because my officemate has borrowed my copy of
the
book.
Q: Could you explain how to measure a star's motion using the Doppler
shift.
I think I basically got Doppler shift concept but it's still a little
fuzzy.
I'm not clear on the concepts behind the formula.
A: Lambda is the symbol usually used to denote wavelength. Since
I can't
type Greek letters, I'm going to use the letter 'L' instead.
Let Lo (read 'L-naught') denote the wavelength of some spectral line
as it's observed in a laboratory. This is called the 'rest wavelength'
since, in a laboratory, neither the gas emitting the light, nor the
equipment
measuring the light is moving.
Let L denote the wavelength of the same spectral line as it's actually
observed. If the emitter is not moving with respect to the observer,
L will
equal Lo. However, if the emitter is moving away from the observer,
L
will be bigger than Lo, i.e. L will be 'redshifted' since moving to
larger
wavelengths means moving closer to the color red on the spectrum).
If
the emitter is moving toward the observer, L will be shorter than Lo,
i.e.
the light will be 'blueshifted.' We can compare L and Lo, then,
to
determine how fast the emitter of light is moving with respect to the
observer using the formula
(L - Lo)/Lo = v/c
where v is the velocity of the emitter and c is the speed of light.
If you do a little algebra to this formula, you get
L = Lo + Lo(v/c)
You can interpret this formula as follows: The observed wavelength
L
is equal to the emitted wavelength Lo plus a small correction term
proportional to the velocity v of the emitter. If v = 0, then
L = Lo,
which makes sense.
Q: I just read the section in Chapter 12 on the Stefan-Boltzman
Law and I am
pretty confused. They say that the method is confusing but the
idea is
quite straight forward. I guess I am missing something.
Could you explain
its importance on what we are actually doing in the class. Also,
is this
for measuring the radius or the luminosity?
A: The Stefan-Boltzman law tells you what the luminosity of a blackbody
is
like. It says:
L = 4*pi*R^2*sigma*T^4
When you throw out the constants, you get
L is proportional to R^2 * T^4.
This means that the bigger and hotter your blackbody is, the more luminous
it will be.
This is most useful to astronomers to calculate the radius of a star:
Step 1: Use parallax or some other method to obtain distance d
to a star.
Step 2: Use distance d, brightness b, and the inverse-square
law
to calculate luminosity L.
Step 3: Use blackbody curve and Wien's law to calculate temperature
T.
Step 4: Use temperature T and luminosity L to calculate radius
R
using the Stefan-Boltzman law.
Q: What's the difference between in "d" in the
Brightness equation vs the R in the Luminosity
equation?
A: d stands for 'distance'; R stands for 'radius'.
Q: when a massive star explodes, what determines whether it forms a
neutron
star or it forms a black hole? is a neutron star just a black
hole that
isnt as dense?
A: While both neutron stars and black holes are very dense, exotic objects
(and pretty neat too, if you were to ask me!), they are not the same
thing.
Black holes have so much gravitational force around them that light
can't
escape from them; this is not true about neutron stars.
What determines whether neutron stars or black holes remain after a
supernova is how much mass is left behind. If it's less than
2 or 3 solar
masses, it'll be a neutron star. If it's more, there'll be enough
mass
to collapse down to a black hole.
Various factors can determine whether enough mass is left behind to
make
a black hole or not. One important factor is how much of the
exploding star
is made of heavy elements.
When a star explodes, it releases a buttload of energy (and I'm not
talking
about a measly I-just-had-way-too-many-chalupas-at-Taco-Bell buttload,
I
mean a real BUTTLOAD!) Some of this energy goes into blowing
the star to
smithereens. Some of this energy goes into breaking up heavy
atoms into
smaller ones. If too much energy goes into breaking up atoms,
and not enough
into blowing away the star, some of the star material can fall back
onto the
core, making it massive enough to become a black hole.
Q: Would it be easier to measure a star's parallax from somewhere else
in our
solar system? For example, from jupiter....would it make a difference?
A: Very perceptive! It would indeed be easier from Jupiter to
measure
a star's parallax. Remember, the star appears to change position,
not because
it itself is moving, but because WE are, here on Earth. If we
were at
Jupiter, the change in a star's apparent position would change even
more
since Jupiter's orbit is a lot wider than Earth's orbit.
Q: Why is the absorption lines in a hot star's spectrum spaced out into
a
pattern?
A: The pattern is caused by the energy levels in the atoms in the star's
atmosphere. If all of the atoms had equally spaced energy levels,
all
the spectral lines we see would be evenly spaced. Since energy
levels
aren't arranged that neatly, neither are the spectral lines.
Q: Quick question. How, exactly, are the temperature of a star
and the
luminosity related? I think the relation had something to do
with T^4,
but I can't see how the units of temperature and luminosity are
reconciled...
A: The Stefan-Boltzmann law says
L = 4pi * R^2 * sigma * T^4
So the hotter the star, the more luminous it is. The units work
out because
of the units of sigma.
Q: --This is more of a clarification question, but when we analyze emission
and absorption spectra, we are using the ONLY the wave model
(electromagnetic) of light, correct, and NOT the particle model (photons)?--
A: We actually have to accept both! In order to account for the
absorption
or emission of one photon, we have to accept that we can count photons,
which means they're particle-like. But the fact that they have
wavelengths
which can be spread apart by a prism means they're wave-like too.
Q: Tom, can you clarify something for me. I think I
heard the professor say that radiation has to do with
the neutron leaving the nucleus and releasing energy.
Is that true? And also, currently are there any cures
for radiation, of a person happens to be in a
radiative zone?
A: The word 'radiation' usually has one of two meanings. The first
use,
and the one used most often by astronomers, means any kind of electromagnetic
wave. EM radiation includes radiowaves, microwaves, infrared,
visible light,
ultra-violet, x-rays, and gamma-rays.
The other meaning of 'radiation' means 'that which is given off by something
which is radioactive.' In other words, it applies to things with
a
____
\ /
____\/____
\ / \ /
\/ \/
symbol associated with it.
The three types of radiation given off by radioactive substances are
alpha-rays (high energy helium nuclei), beta-rays (high energy electrons
or positrons), and gamma-rays (high energy photons).
Sometimes, in radioactive elements, the nucleus will decay to a lighter
element and in the process give off an alpha paricle. I
believe this is what Prof. Basri was talking about.
As for cures to radiation poisoning... I'm not an expert, but I think
there are drugs that help you will the short-term effects of radiation,
such as nausea. However, long-term effects like cancer still,
unfortunately,
have no cure.
Q: Oh, and the sunny weather has made me think about this
idea that if you wear white clothing, the clothes will
reflect all the wavelengths and leave you a bit
cooler. It just struck me that visible spectrum is
only a minuscule part of sunlight, and that heat comes
in the form of infrared wavelengths. So even if a
person wears white clothing, the clothing won't fight
off very much heat, would it, because the infrared is
still there?
A: Fortunately for us, our atmosphere blocks out most wavelengths of
light
that aren't in the visible range, including a lot of infrared (but
not all).
It's possible, though, that your white t-shirt bounces away infrared
photons as efficiently as it bounces off visible photons.
Don't throw away your air conditioner and bleach all your clothes to
stay
perfectly cool, though! You can't do anything about the fact
that sunlight
heats up the air, and the air can then transfer heat to your nice,
clean
white t-shirt just by touching it!
Q: I don't quite understand how to measure parallax of a star
by using a
binary star. What is the proper procedure for this?
A: You don't need a binary star to measure its parallax, just
a close star.
(Close, in this case, means within about 200 lightyears). By
looking at
where the star appears to be when Earth is on one side of its orbit
around
the sun, and then the other side, we can triangulate the position of
the star.
Q: If you can't get the actual mass of a star from a spectroscopic
binary
system why is it useful?
A: You can still get a couple of useful things:
1. The RATIO of the masses of the two stars
2. A LOWER LIMIT on the total mass of the two stars.
For example, after studying some binary, and astronomer could proclaim,
"Well, I don't know what the individual masses are. But I do
know that one
of the stars is three times as massive as the other one, and that the
two of them together are at least four times as massive as the sun!
I've
earned a Snickers bar!"
Q: If there were a small star brighter than a large star, then
if the large star were in front of the small one, would this system
be
dimmer than if it were the other way around.
A: Correct! When its the other way around (like it was in
your homework)
and the big star is brighter, it gets a little more complicated.
Two things come into play here:
1. Which star is intrinsically brighter?
2. How much of the bigger star is eclipsed when the little
one passes in front?
Even if the bigger star is the brighter, if only a small fraction of
it
is blocked when the little star is in front, then the brightness drops
the
most when the bigger star is in front.
Q: what's the easiest way to determine a star's relative mass and size
from a
H-R diagram?
A: To determine mass, if it's a main sequence star, you just use the
mass-luminosity relationship, L = M^3 (where L must be in solar luminosities
and M must be in solar masses). If it's not a main sequence star
it's much
more tricky.
To get size, remember that as you go up and to the right on the HR diagram
size gets bigger. There's a picture in your book (I don't know
the page)
which shows line of constant radius across the HR diagram. These
lines
run diagonally, from up and to the left to down and to the right.
Q: Why do similar atoms, (hydrogen and) helium for instance, have different
intesities in their spectral lines, shouldn't they all be the same?
A: Each individual atom, molecule, and ion has a completely different
set
of possible energy levels available to its electrons. Even though
two
atoms may have a similar number of electrons, neutrons, and protons,
their
energy levels will still be completeley different.
Q: If red giants can eventually become black holes, where do they appear
on the
H-R Diagram?
A: Black holes have no luminosity, so they wouldn't appear on the HR
diagram.
Q: here's my question. I dont clearly understand how temperature affects
a
star's spectrum and how abundant elements can create weak spectral
lines.
A: Here's a quote from the solutions I wrote for problem set 5:
"You can also tell what temperature a star is by looking at what absorption
lines are in its spectrum and how strong they are. For example,
consider the
Balmer lines. These spectral lines are caused by hydrogen at
a star's surface.
Specifically, Balmer absorption lines occur because hydrogen atoms
in which the
electrons are in the 2nd energy level absorb photons, exciting the
electrons
to a higher energy level. Thus for a star to have Balmer lines,
there must be
a lot of hydrogen with electrons in the 2nd energy level present at
the star's
surface. But if the star is too cold, the atoms at its surface
won't be very
excited, and you'll tend to find electrons only in the 1st energy level
most of
the time. On the other hand, if the star is too hot, then the
electrons will
tend to already by excited, or even stripped away from the hydrogen
nuclei, and
you will never see a Balmer line. Thus to see a Balmer line,
the temperature
has to be, as Goldilocks would say, just right. This is true
for *any* given
spectral line. So all we need to do is look at which spectral
lines are
present and which aren't in order to determine what temperature the
star must
be for that to be so."
Q: Also, I was wondering if you could explain eclipsing binary stars.
A: This is kind of a big topic. What would you like to know?
Q: What happens to the partner star when one of the binary stars go
supernova?
I mean does it blow up too?
A: It wouldn't blow up too. Usually, it would continue to evolve
as it would
have otherwise. Perhaps, if it intercepted and absorbed some
of the mass
blown off of the exploding star it would evolve faster, since it would
have
increased in mass, and the more massive a star is, the faster it evolves.
Q: Let's suppose Earth is turning into a Red Giant and eventually engulfs
Mercury as astronomers predict. I'm wondering, as the sun
expands into a Red
Giant and grows, does this mean its mass increase? If so, would
our Sun have
a new defined Roche Limit, thereby "stripping" the planets instead
of
engulfing them first?
A: The sun won't be gaining any mass when it expands; it'll simple become
less dense. So there won't be an increase in the sun's roche
limit. So
instead of getting ripped apart by tidal forces, Earth will simply
get
fried to a crisp. Oh boy!
Q: i was wondering about the spectral classes of the stars.
How did the
first astromers or whoever who came up with the idea go about to classify
the classes of stars. I know the spectral class is determined
by the
spectrum, but where did they get the letters, O B A F G K M from.
did they
just make it up?
A: Annie Jump Cannon, who worked at Harvard early this century
(and it's
still the 20th century until 1/1/2001!) originally categorized spectra
into A-type, B-type, C-type, and so on. As understanding of stellar
spectra
grew, types were lumped together (which is why we have 8 spectra types
instead of 26) and the types were put into a more logical order, having
to do with the temperature of the stars. In the end, we're left
with
O, B, A, F, G, K, M, and L.
Q: How is it that some stars are cooler than others but more luminous,
and
others are hotter than others but dimmer?
A: Luminosity depends on radius, also. Even though red giants
are cool at the
surface, they're frickin' huge! So they're very luminous.
And white dwarfs
start out really hot (after all, they're the cores left behind by stars
who
shed their outer layers, and the core is the hottest part of the star),
but
they're pretty dinky (about the size of Earth) so even though they're
hot,
they're dim.
Q: I was wondering, how is the Stefan-Boltzmann constant derived?
A: The value of the S-B constant was probably first determined by figuring
out that for blackbodies, L is proportional to R^2 * T^4, and then
doing
lab experiments to figure out what the constant of proportionality
was.
Q: Can we tell how hot the star is by looking at the
colors on its spectrum? So if the spectrum has more
purple and blue on it, that means it's hotter than the
spectrum of a star with more red on it?
A: Yes! For a detailed answer to this question see the solution
to #4 of
homework #5.
Q: With which formulas does it matter whether a body is moving towards/away
from you?
A: If the version of the doppler formula you use is
(L-Lo)/Lo = v/c
and you get v < 0, that means the velocity is towards you.
If you get
v > 0, the velocity is away from you.
If you use the version
delta-L / Lo = v/c
Then you'll get a positive v in either case. In order to figure
out
if its toward or away from you, you need to consider whether the doppler
effect is redshifting or blueshifting the light (i.e. are the wavelengths
appearing larger or shorter?)
Q: Chapter 12: Could you go over exactly how you can determine mass
from
spectroscopic binaries. I know you did it in discussion but I've forgotten.
A: You cannot determine the individual masses from spectrascopic binaries.
You can determine the ratio of the masses. You can measure the
velocities of each star using the doppler shift. The ratio of
the velocities
will then tell you the ratio of the masses. (This is because
the more
massive the star, the slower it'll orbit around the other one.)
You can also get a lower limit to the masses. You know the component
of the
velocity heading toward you (from looking at the doppler shifts of
the stars)
and you can measure the period of the orbits. By knowing the
periods and
velocities, you can get an idea of the size of the orbit. Using
the size
of the orbit and the period, you can get a lower limit to the total
mass.
You can't, however, exactly calculate the total mass, since you don't
know
the tilt of the binary's orbit with respect to us.
Q: Chapter 13: How do white dwarfs generate energy? Is it from just
the act of
contraction? What happens to white dwarfs when they burn out? Black
dwarfs?
A: Once a star is in the white dwarf stage, it doesn't really generate
any
more energy. It glows simply because its hot! Remember
that the core is by
far the hottest part of a star, and the cores of red giants are extremely
hot!
When the red giant stops fusing in the core and sheds its planetary
nebula,
what is left is the hot core. This is the birth of a white dwarf.
One way to think about white dwarfs is to compare them with the burners
on an
electric stove. Imagine turning the burner on to 'high.'
After a while the
burner will get really hot and will glow. Now turn the burner
off. The
burner will still glow for a while, and you sure don't want to touch
it
because you know it's still hot, despite the fact that you've turned
it off.
Similarly, the white dwarf glows because it's hot, even though it's
turned
off (in the sense that there's no longer any nuclear fusion generating
new
energy).
After a while (billions of years) the white dwarf will cool off by radiating
all of its energy away into space, and you'll be left with a cool,
dark,
ball of carbon and oxygen. This is called a black dwarf.