BLUE – HOT!!! - Uplift Education

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Transcript BLUE – HOT!!! - Uplift Education

• The Electromagnetic Spectrum
In astronomy, we cannot perform experiments with
our objects (stars, galaxies, …).
The only way to investigate them, is by analyzing the
light (and other radiation) which we observe from them.
The study of the universe is so challenging, astronomers
cannot ignore any source of information; that is why they use
the entire spectrum, from gamma rays to radio waves.
Wavelength and Color
• The color of light depends upon its wavelength.
𝐸 = ℎν
BLUE – HOT!!!
• SHORT Wavelength
• HIGH Frequency
• HIGH ENERGY!
RED – COLD!!!
• LONG Wavelength
• LOW Frequency
• LOW ENERGY!
The Amazing Power of Starlight
Just by analyzing the light received from a
star, astronomers can retrieve information
about a star’s
1. Total energy output
2. Surface temperature
3. Radius
4. Chemical composition
5. Velocity relative to Earth
6. Rotation period
Color and Temperature
Stars appear in
different colors,
from blue (like Rigel)
Orion
Betelgeuse
via green / yellow (like
our sun)
to red (like Betelgeuse).
These colors tell us
about the star’s
temperature.
Rigel
Hot!
Through a telescope
the 'star' Albireo is
revealed to actually be
composed of two bright
stars. These stars have
very different surface
temperatures, as
indicated by their color.
Cool!
Cooler
stars!
Hotter
stars!
We only think of visible wavelengths as being
“light”, but a lot of familiar things involve light of
different wavelengths!
Size of
atomic
nucleus!
Size of Mt.
Everest!
Where Does Light Come From?
The following had been known during the
19th century:
 accelerated charges emit energy
 and hence produce light
If we picture an electron as in orbit around the
nucleus, it should radiate light
 changing direction is acceleration! (a force is
required from something to change direction)
This caused a major problem with classical physics
If the electron radiated due to its motion around the nucleus,
it would lose energy and soon spiral into the nucleus.
The world should collapse instantly!
1913, Niels Bohr formulated 3 rules regarding atoms:
1. Electrons can only
be in discrete orbits.
2. A photon can be emitted
or absorbed by an atom
only when an electron jumps
from one orbit to another.
3. The photon energy equals the energy difference
between the orbits.
The discrete (quantum) nature of the energy
"levels" of the electron gives Quantum
Mechanics its name.
Key Features of the Atoms/Ions Spectra
Spectral lines
Hydrogen spectrum
The greater the difference between the quantum numbers, the
larger the energy of the photon emitted or absorbed.
Most
prominent
lines in many
astronomical
objects:
Balmer lines
of hydrogen
Observations of the H-Alpha Line
Emission nebula, dominated
by the red Ha line.
Hydrogen and helium account for nearly all the nuclear matter in today's universe.
The abundance of hydrogen by mass, is 73% (Helium is 25%, All other elements 2%).
By atomic abundance, hydrogen is 90%, helium 9%, and all other elements 1%.
That is, hydrogen is the major constituent of the universe.
As we are already talkig about hydrogen let me mention
one very interesting fact.
I call it: a WOW story of time!!!!!!
spin-flip transition even though the atom is in its ground state
When a proton captures an electron to form hydrogen atom (after Big Bang) the spins
of the two particles either point in the same direction or opposite direction.
Due to the spin both electrons and protons behave like little magnets.
Hydrogen atom with antiparallel spins is more stable (the electron is more tightly
bound to the proton) than the atom with the two spins parallel because unlike poles
attract each other and like poles repel.
Since a anti-parallel-spin capture is three times as probable as an parallel-spin
capture, 75% hydrogen atoms in between the stars is with spins anti-parallel and 25%
with parallel-spin.
This means that the electron in the
hydrogen atom with its spin parallel to
that of the proton will ultimately flip over
so that its spin is anti-parallel to the
proton spin emitting 21-cm photon (radio
part of the spectrum).
We should not ordinarily expect to receive much 21-cm radiation because it takes
on the average about 11,000,000 years for an electron in an unmolested
hydrogen atom to flip over to an anti-parallel spin position. But there are so many
such atoms in interstellar space that the occasional emission of a 21-cm photon
from any one of them adds up to the observed intensity of the 21-cm line. The
collisions among the neutral hydrogen atoms that occur every few hundred years
keep the number of hydrogen atoms with parallel and antiparallel electrons spins
constant or in thermal equilibrium.
This hydrogen radio line with a wavelength of 21 cm was first predicted theoretically
in 1944 by the Dutch astronomer H. C. van de Hulst (highly unlikely to be seen in a
laboratory on Earth). It was observed in radio telescopes at Harvard and then
throughout the world in the early 1950s.
Most of what is known about the distribution of cold gas in the
Galaxy, including the mapping of the nearby spiral arms, has come
from detailed studies of the variation of 21-cm line of Hydrogen
emission across the sky. Stars radiate all freq. but visible light won't
penetrate the dust clouds and this 21 cm will, giving us informations.
From the data that have now been collected, we know that the dust in
interstellar space, which constitutes only about 1% of the interstellar
material, is almost entirely responsible for the dimming of the stars. The
size of a dust grain is about 0.000001 cm (about the size of the
wavelength of visible light) and only one such grain, on the average, is
present in each 10,000,000,000 cm3 of space.
Reading the intensity of 21 cm hydrogen line from different parts of
Universe we can get a fairly reliable picture of the distribution of gas
(hydrogen) and dust throughout the galaxy.
Analyzing Absorption an Emission Spectra
● Each element (atom/ion)
produces a specific set of
absorption (and emission) lines.
We call this the "spectral signature"
or “fingerprints” of an atom/ion.
● Allows the identification
of elements across the
galaxy and universe.
(If we mapped it and
can recognize it)
● Comparing the relative
strengths of these sets of
lines, we can study the
composition of gases.
Step from line spectrum to continuous spectrum
The energy levels get closer together as the
quantum numbers get larger.
Key Features of the Continuum Spectra
► If an electron is given enough energy (via a photon or by other
means) it can escape the atom. The electron is then "unbound" and the
quantization of energy levels disappears. The energy of an electron in
the continuum is not quantized.
A hot, dense object contains many "loose" electrons which can emit
photons of any energy.
► The light produced by a hot, dense object is called thermal
emission and it contains photons of all energies, i.e. light of all
colors, or wavelengths. The resulting "rainbow" is called a
continuous spectrum.
► As we heat up an object, we are giving the electrons more kinetic
energy, so they become able to emit more energy. The hotter the
object becomes,the brighter the continuous spectrum becomes.
THERMAL
RADIATION
Hot, so it emits light
Peak color (wavelength) shifts to shorter
wavelengths as an object is heated
increasing temperature
So, emitted spectrum tells us about temperature
Temperature Spectrum of Objects
• All objects emit a continuous spectrum
• You are giving off light right now!
If you could fill a teaspoon just with material as dense as the
matter in an atomic nucleus, it would weigh ~ 2 billion tons!!
just for fun – so cute
Kirchoff's Laws of Spectroscopy/Radiation
Kirchoff formulated these laws empirically in the mid-19th
century – didn’t explain why – in early 20th century: QM –
nature of atom – beginning of understanding origin of spectra
Kirchhoff did not know about the existence of energy levels in
atoms. The existence of discrete spectral lines was later
explained by the Bohr model of the atom, which helped lead to
quantum mechanics.
1. A hot solid, liquid or gas at high pressure produces a
continuous spectrum – all λ.
2. A hot, low-density / low pressure gas produces an
emission-line spectrum – energy only at specific λ.
3. A continuous spectrum source viewed through a cool,
low-density gas produces an absorption-line spectrum –
missing λ – dark lines.
Thus when we
see a spectrum
we can tell what
type of source we
are seeing.
Absorption Spectrum of Hydrogen Gas
Actual Examples of Emission Spectra
Element Spectrum
Argon
Helium
Mercury
Neon
Sodium
The Spectra of Stars
Inner, dense layers of a
star produce a continuous
(blackbody) spectrum.
Cooler surface layers absorb light at
specific frequencies.
=> Spectra of stars are absorption spectra.
To understand the nature, to interpret many beautiful
phenomena you have to have a tool. We are
introducing something that we know all about and then
we’ll compare the nature with that ideal case!!!!!!!
Blackbody
Radiation
A black body is a theoretical object that absorbs 100%
of the radiation that hits it. Therefore it reflects no
radiation and appears perfectly black.
In practice no material has been found to absorb all incoming
radiation, but carbon in its graphite form absorbs all but about
3%. It is also a perfect emitter of radiation. At a particular
temperature the black body would emit the maximum amount of
energy possible for that temperature. This value is known as the
black body radiation. It would emit at every wavelength of light
as it must be able to absorb every wavelength to be sure of
absorbing all incoming radiation.
• In 1900 Max Planck characterized the light coming from a
blackbody.
• The equation that predicts the radiation of a blackbody at different
temperatures is known as Planck's Law.
• The peak emission
from the blackbody
moves to shorter
wavelengths as the
temperature increases
(Wien’s law)
• The hotter the
blackbody the more
energy emitted per unite
area at all wavelengths.
Bigger object emit more
radiation.
Note that the peak shifts with temperature.
Wien's
Law
The wavelength of the maximum
emission of a blackbody is given by:
 peak
3
2.9

10

T
peak
T (ºK)
Sun
500 nm
5800
People
Neutron Star
9x103 nm
310
2.9x10-2 nm
108
Radio
10 m
0.03 K
Microwave
1 cm
3K
Infrared
1 mm
300 K
Visible
500 nm
6000 K
Ultraviolet
100 nm
100,000 K
X-Ray
0.1 nm
10 M K
 peak in m
T in K
The spectrum
of a star
reveals it’s
temperature
Consequences of Wien's Law
Hot objects look blue.
Cold objects look red.
Except for their surfaces, stars
behave as blackbodies.
An interesting example of a black body radiation is the
thermal emission of the Earth (or any other body). This
thermal emission (also called infrared emission, due to its
characteristic wavelength) is due to the Earth's temperature.
In the picture, we can see the real emission compared to a
black body radiation of a body at a temperature of 280K. In
the picture it is also possible to see the absorption spectral
lines of oxygen and CO2.
What Color is Our 5800K Sun?
• The Sun emits all colors
• Blue-green is most intense
Peak Color (Wavelength) Depends
on Object’s Temperature
Thermal radiation can explain much of
this spectrum
Thermal radiation
at 6000K
Thermal radiation
at 225K
This is a spectrum of Mars! The 6000K radiation is
reflected light from the sun. The 225K radiation is
thermal emission from the planet.
But what is that other stuff?
Emission lines!
Extra light at very
specific wavelengths.
Absorption lines!
Light has been
removed at very
specific wavelengths.
Sometimes emission lines dominate the
light output.
The Cygnus supernova remnant 
emits almost all of its light as emission lines!
PS. A supernova remnant is the expanding
shell of hot gas left over after a star
explodes.
A caveat to the rules of thermal
radiation!
• Cooler objects can sometimes emit more light
overall.
• Decreasing temp means less light emitted per
unit area. An object can compensate by being
BIGGER.
• Lower “surface brightness”, but larger surface
area.
Hot.
Cool.
Same total light emitted
Cool, but big.
The Luminosity of a Star
How bright a star or galaxy really is is described by its luminosity.
The energy radiated by a star is emitted uniformly in all direction.
The total energy emitted by the star per second (i.e. power)
is called Luminosity of the star, L.
The luminosity of a 100 W lightbulb is (approximately) 100 W if you
measure over all wavelengths. (Most of it is in the infrared (HOT);
the part of the luminosity in visible wavelenghts is less.)
If we regard stars to be black body radiators, the luminosity L of a
star is given by Stefan-Boltzmann law for a black body:
L = 4π R2 σ T4
(W)
(all λ)
Doubling the temperature increases the luminosity by a factor of 16.
A hotter star is more luminous than a cooler one of the same radius.
A bigger star is more luminous than a smaller one of the same
temperature.
d
By the time the energy arrives at the
Earth it will be spread out over a sphere
of radius d (distance).
apparent brightness
L
b
4d 2
energy received per unit
time per unit area (W/m2)
It is easy to measure the apparent brightness of a star, a galaxy, a
supernova, ... by bolometer attached to a telescope.
radiation sensitive instrument
If d can be measured then the luminosity of the star can be
determened. This is very important property to know as it gives clues
to the internal structure of the star, its age and its future evolution.
Since we can't go to a star to measure its luminosity, we have to be
clever. If we know the distance to the star we can do it, because
there is a simple relation between the distance d to the star, the
apparent brightness b of the star, and the luminosity L of the star.
A person radiates ~ 100 W = 100 J/s
So that the energy output in a day is
Energy = power×time
= 100×24×3600 J
1cal
= 8640000J×
4186J
= 2064 cal
I don’t know exactly how to use this information to lose extra
weight that I gained recently.
How can we measure the Sun's luminosity?
► We can measure the intensity of sunlight at the Earth.
• This should include all wavelenghts.
• b ~ 1.4 x 103 W/m2.
► So we should multiply b by the area of this sphere:
L = 4  d2 b = 4 x 3.14 x (1.5 x 1011m)2 = 2.8 x 1023 m2
distance Sun-Earth can be measured
using different method
► So the luminosity is (1.4 x 103 W/m2) x (2.8 x 1023 m2)
≈ 3.90 x 1026 W .
Sometimes you’ll find luminosity of our Sun written as:
LΘ = 3.90 x 1026 W
There are a few stars that are
more luminous than the Sun.
For instance, Betelgeuse
has L ~ 14000 x Lsun.
There are lots
more low
luminosity stars
than high
luminosity stars.
Betelgeuse ("beetle juice"), a red supergiant star about 600 light years
distant - one of the brightest stars in the familiar constellation of Orion, the
Hunter - the first direct picture of the surface of a star other than the Sun.
While Betelgeuse is cooler than the Sun, it is more massive and over 1000
times larger. If placed at the center of our Solar System, it would extend past
the orbit of Jupiter (has an immense but highly variable, outer atmosphere ).
As a massive red supergiant, it is nearing the end of its life and will soon
become a supernova.
However:
knowing how bright a star looks doesn’t
really tell us anything about the star itself!
We cannot see their
size!
… if we anly knew
the distance
From Brightness to Luminosity
• If we know the
distance to a star,
we can calculate
the luminosity
(energy output)
It is very hard to
measure the
distance.
Parallax is the most
direct measure of
distance
We know how big the
Earth’s orbit is, we
measure the shift
(parallax), and then we
get the distance…
background stars
it does not make any significant
difference which distance you want to
talk about – from sun or from earth
sin p  tan p  p  rad 
or
D
d
p
Parallax
angle, p
½ of the total
angular shift.
For small angles:
sin p  D
d
click
me
p D
d
D=1 AU
July
January
D
d
p
D = 1 AU = 149597870691 m ≈ 1.5x1011 m
when talking about stars, parallax is very, very small number.
Parallaxes are expressed in seconds.
p  rad 


180  60  60 p  sec 
so
1
 648000 1

p  rad 
p  sec 
16
1
149597870691

648000
3
.
08

10
m
d  m 


p  sec 
p  sec 
1 pc = 3.09 X 1016 m
1
d  pc  
p  sec 
Parsec: distance where a star
shifts by 1” over a 1/2 year
“Parsec” is short for
parallax arcsecond
• Even the nearest star has a tiny parallax!
► First measured in 1838
► The closest bright star Alpha Centauri
4.3 light-year
0.75 pc
Parallax has its limits…
The farther away an object gets, the
smaller its shift. Eventually, the shift
is too small to see.
In 1989, the European Space Agency (ESA) launched a
satellite called Hipparcos to accurately measure the
positions and motions of nearly 120,000 stars - plus about
another million or so stars with good, but lower precision.
Parallaxes give us distances to stars up
to perhaps a few thousand light years.
Beyond that distance, parallaxes are so
small than they cannot be measured
with contemporary instruments.
Astronomers use more indirect methods
beyond a few thousand light years.
Suppose I observe with my telescope two red stars A and B that are
part of a binary star system.
Star A is 9 times brighter than star B.
What can we say about their relative sizes and temperatures?
Since both are red (the same color), the spectra peak at
the same wavelength. By Wien's law
 peak
3
2.9

10

T
 peak in m
T in K
then they both have the
same temperature.
By our law governing Luminosity, radius,
and temperature of an object (star!)
L = 4π R2 σ T4
(W)
It must be that star A is bigger in size (since it is the same
temperature but 9 times more luminous). How much?
Star A is 9 times brighter and as they are the same distance away
from Earth star A is 9 times more luminous:
LA 4RA2TA4

LB 4RB2TB4
RA2
 9 2
RB
 RA  3 RB
So, Star A is three times
bigger than star B.
Suppose I observe with my telescope two stars, C and D, that
form a binary star pair.
▪ Star C has a spectral peak at 350 nm - deep violet
▪ Star D has a spectral peak at 700 nm - deep red
What are the temperatures of the stars?
By Wien's law
 peak
3
2.9

10

T
 peak in m
T in K
Thus we have for star C,
3
3
2.9

10
2.9

10
TC 

 8300 K
9
 peak
350 10
and for star D
3
3
2.9

10
2.9

10
TD 

 4150 K
 peak
350 10 9
If both stars are equally bright (which means in this case they
have equal luminosities since the stars are part of a pair the
same distance away), what are the relative sizes of stars C and D?
LC 4RC2 TC4

LD 4RD2 TD4
RC2 83004
RC2
4
 1  2
 2  2
4
RD 4150
RD
RD2  16 RC2  RD  4 RC
So that stars C is 4 times smaller than star D.
The Doppler Effect
observed change in wavelength of an EM wave due to
relative velocity betwee the source and the detector
The light of a
moving source is
blue/red shifted by
0 = actual wavelength
emitted by the source
D/0 = vr/c
D  Wavelength change
vr = radial velocity
λ of EM waves coming from stars are very often
greater than those obtained in the laboratory
emitted from same elements (He, H).
Redshift – shift toward greater wavelength
Edwin Hubble in the 1920's.
Based on calculations using the Doppler effect, it appears
that nearby galaxies are moving away from us at speed of
about 250,000 m/s. The distant galaxies are moving away at
speeds up to 90 percent the speed of light. The universe is
moving apart and expanding in all directions.
The more distant a galaxy, the faster it is moving away.
Some galaxies, however, are
moving toward us, and their light
shows a blue shift.
Our
Sun
supercluster of galaxies,
BAS11
v=0.07c d = 1 billion LY