Download ASTR 314 : Survey of Astronomy Extragalactic Astronomy

ASTR 314 : Survey of Astronomy
Extragalactic Astronomy & Cosmology
Prof. D. DePoy
Office: MIST 240
Email: [email protected]
Webpage: http://faculty.physics.tamu.edu/depoy
How do you know the Earth Rotates ?
In 1851, Léon Foucault proved the
Earth’s rotation directly.
A pendulum swinging on the Earth
feels the rotation due to the Coriolis
Force.
FC = -2 m Ω x v
Versions of the Foucault Pendulum
now swings in the Panthéon of Paris
and the Houston Museum of Natural
Science
http://www.youtube.com/watch?
v=CpxDAdtJX2s
http://www.youtube.com/watch?
v=49JwbrXcPjc
How do you know the Earth Rotates ?
Coriolis Force affects rotation of weather patterns (cyclones rotate
counter clockwise in southern hemisphere; hurricanes rotate clockwise
in northern hemisphere).
How do you know the Earth Rotates ?
Hurricane Gustav
Aug 31, 2008
Coriolis Force affects rotation of weather patterns (cyclones rotate
clockwise in southern hemisphere; hurricanes rotate counter-clockwise in
northern hemisphere).
In Kepler’s day through the 19th century, we had only
relative distances to the Sun and Planets.
Estimated Mean Distances of the Planets from the
Sun (in Astronomical Units)
e
h
t
e
r
u
Mercury
0.389 eas 0.387
r
m
e
u
h
t
o
y
o
Venus
0.724
0.723
d
o
l
t
u
e
o
c
w
n
Earth
1.000
1.000
a
t
w
s
Ho Mars lute di 1.523ts ?! 1.524
o
e
s
n
b
a
aJupiter
pl 5.200
5.202
Saturn
Kepler
21st Century
9.510
9.539
Parallax: same idea as triangulation, derived from
Greek parallaxis, “the value of an angle”.
Example: How far is Rudder
Tower from the Albritton Bell
Tower ?
D = [ h / tan(θ) ]
for h=138 ft and θ=0.5o,
D= 1577.4 ft
θ=5o
D
h=138 ft
Parallax: same idea as triangulation, derived from
Greek parallaxis, “the value of an angle”.
Another example: how far away
is a car by its headlights ?
2θ
2w
D
D = [ w / tan(θ) ]
You subconsciously do this all the
time. Your brain judges how fast an
oncoming car is going by the change in
its angular size.
for w = 1 m and θ=0.5o, D=114.6 m
for w = 1 m and θ=5o, D=11.4 m
Parallax: same idea as triangulation, derived from
Greek parallaxis, “the value of an angle”.
d = B / tan(p)
First observation
1 AU
unmoving
background stars
Parallax
Position of star
180 days later
Position of star
on first
observation
180 days later
d = 1 AU / tan(p) ≈ 1 / p (radians) AU
≈ 57.3 / p (degrees) ≈ 206265 / p (arcsec) AU
Define: 1 parsec = 2.06265 x 105 AU
d = 1 / p(arcsec) pc.
Cygnus
Parallax
Example: The distance to 61 Cygni.
In 1838, after 18 months of
observations, Friedrich Bessel
announced a parallax angle to this
star of 0.316 arcseconds. This
corresponds to: d = 1 / 0.316 = 3.16 pc.
21st century value is 3.48 pc.
Estimated Mean Distances of the
Planets from the Sun
(in Astronomical Units)
Kepler
21st Century
Mercury
0.389
0.387
Venus
0.724
0.723
Earth
1.000
1.000
Mars
1.523
1.524
Jupiter
5.200
5.202
Saturn
9.510
9.539
The Answer: with Parallax
View from Pacific Ocean
N
Mars
View from United Kingdom
Tried by Jean Richer in 1672. Got an answer that 1 AU = 87 million miles.
(Present-day answer: 1 AU = 93 million miles.)
But, Richer’s data had lots of systematic errors, and no one took this seriously.
Venus comes much closer to the Earth than Mars.
But, when Venus is at its closest approach, it’s lost in the
Sun’s glare, so can’t see background stars.
But, can use the time that Venus begins transits.
View from Pacific Ocean
N
Venus
View from United Kingdom
Venus comes much closer to the Earth than Mars.
But, when Venus is at its closest approach, it’s lost in the
Sun’s glare, so can’t see background stars.
But, can use the time that Venus begins transits.
http://www.astro.umd.edu/openhouse/gallery/2004-06-08VT/other/VenusTransitGCS.gif
Venus comes much closer to the Earth than Mars.
But, when Venus is at its closest approach, it’s lost in the
Sun’s glare, so can’t see background stars.
But, can use the time that Venus begins transits.
View from Pacific Ocean
N
Venus
“I see Venus begin Transit at Time T1”
View from United Kingdom
Venus comes much closer to the Earth than Mars.
But, when Venus is at its closest approach, it’s lost in the
Sun’s glare, so can’t see background stars.
But, can use the time that Venus begins transits.
“I see Venus begin Transit at Time T2”
View from Pacific Ocean
N
Venus
“I see Venus begin Transit at Time T1”
View from United Kingdom
Royal Society sponsored an exhibition in 1768 to Tahiti to measure Venus’ transit of the Sun.
This led to a measurement of the AU within 10% of the present-day value. Subsequent
observations of Mars, Venus, and asteroids confirmed and refined this measurement. Humanity
now had a yardstick for the AU. Parallax
Example: The distance to 61 Cygni.
In 1838, after 18 months of
observations, Friedrich Bessel
announced a parallax angle to this
star of 0.316 arcseconds. This
corresponds to: d = 1 / 0.316 = 3.16 pc.
21st century value is 3.48 pc.
1 parsec = 2.06265 x 105 AU
= 3.09 x 1016 m
= 1.92 x 1013 miles
=3.26 lightyears (lyr).
Therefore, d(61 Cygni) = 10.3 lyr !
Cygnus
The Wave Nature of Light
Double-Slit Experiment of Thomas Young (1773-1829)
Constructive Interference
Destructive Interference
http://www.youtube.com/watch?
v=P_rK66GFeI4&eurl=http://video.google.com/videosearch?
hl=en&client=safari&rls=en-us&q=wave
%20interference&um=1&ie=UTF-
Light Propagation Direction
Destructive
Interference
Screen
Coherent Light
From Single Slit
Barrier with
Double Slits
Constructive Interference
Intensity Distribution of Fringes
The Wave Nature of Light
d sinθ =
{
nλ
(n=0,1,2,3... ), constructive
interference
(n-1/2) λ
(n=0,1,2,3... ), destructive
interference
Young found that blue light has λ = 400 nm = 4000 Å
red light has λ = 700 nm = 7000 Å
where 1 Å = 10-10 m (1 Ångstrom)
Radiation Pressure
Like all waves, light carries both energy and
momentum in the direction of propagation. The
amount of energy carried is described by the
Poynting vector:
S = (1/μ0) E x B
S
where S has units of W m-2 (energy per unit
area). The average Poynting vector is given by
the time-average E and B fields.
<S> = (1 / 2μ0) E0B0
where E0 and B0 are the amplitude of the waves.
Radiation Pressure
The Radiation Pressure depends on if the
light is absorbed or reflected. Absorption, force is in direction of light’s
propagation: (absorption)
Reflection, force is always perpendicular to
surface (reflection)
Radiation Pressure, as a means of space travel ?!!
http://www.youtube.com/watch?
v=eq2DATxcft0&feature=fvw
http://www.youtube.com/watch?
v=Wfa1ggUlKnk&NR=1
Thermal Radiation (Blackbody Radiation)
The temperature of lava can be estimated from
its color, typically 1000-1200 K. Blackbody Radiation
Any object with a temperature above T=0 K emits light of all
wavelengths with varying degrees of efficiency.
An IDEAL emitter is an object that:
1. Absorbs all light energy incident upon it and 2. Emits this energy with a characteristic spectrum of a “Black Body”.
Stars and Planets are approximately blackbodies (as are gas clouds,
and other celestial objects).
Blackbody Radiation
Observed Spectra of Vega-type Star
Solar-type Star
Lots of absorption
from atoms in the
stars’ atmospheres
(more next week)
T=10,000 K
T=8000 K Here Stars Look almost T=5800 K
exactly like blackbodies T=3000 K
Blackbody Radiation
Wilhelm Wien (1864-1928) received
the Nobel Prize for his contribution
to our understanding of Blackbody
Radiation.
Through experimentation, Wien
discovered that the peak emission of
a wavelength, corresponds to a
wavelength λmax which relates to the
temperature as:
λmax T = 0.002897755 m K
Wien’s Displacement Law !
Blackbody Radiation
Wilhelm Wien (1864-1928) received
the Nobel Prize for his contribution
to our understanding of Blackbody
Radiation.
Through experimentation, Wien
discovered that the peak emission of
a wavelength, corresponds to a
wavelength λmax which relates to the
temperature as:
λmax T = 0.002897755 m K
Wien’s Displacement Law !
λmax
Blackbody Radiation
Note: as T increases, the
blackbody emits more radiation
at all wavelengths.
Josef Stefan
(1835-1893)
Ludwig Boltzmann
(1844-1906)
Related the Luminosity of a Blackbody
to the Surface Area of the object:
L = A σ T4 For a sphere:
L = (4π R2) σ T4
Stefan-Boltzmann constant: σ = 5.670400 x 10-8 W m-2 K-4
Example: Luminosity of the Sun, L⊙=3.839 x 1026 W
Radius of the Sun, R⊙=6.95508 x 108 m
Using: L = (4π R2) σ T4
L⊙ ( 4π R
T⊙ = ⊙
1/4
)
σ
= 5777 K
Radiant Flux (or surface flux) = L / Area = L / (4π R2) for a sphere
Fsurf = σ T⊙4 = 6.316 x 107 W m-2
Wien’s Displacement Law: λmax = (0.002897755 m K) / T⊙ = 5.016 x 10-7 m = 501.6 nm
The Problem with Blackbody Radiation:
Classical Physics (before 20th century) could not explain it !
Lord Rayleigh (1842-1919), born John William Strutt, 3rd Baron Rayleigh),
did initial research into blackbodies. Awarded Nobel Prize in 1904.
Considered a hot oven (blackbody) of of size, L, at temperature, T, which
would then be filled with E/M radiation (light). E/M waves must satisfy E=0 at wave edges.
Standing waves of λ = 2L, L, 2L/3, 2L/4, 2L/5, ... In Classical Physics, each mode (different
standing wave) should receive equal energy
amount, kT, where k, is Boltzmann’s
constant. Rayleigh’s derivation gave:
Bλ(T)≈2c kT / λ4
E
Bλ : Intensity (units of Energy per second
per area per wavelength)
c : speed of light
0
L
BUT, when you integrate this over all
wavelengths, it diverges to infinity, called
the “ultraviolet catastrophe” !
The Problem with Blackbody Radiation:
Classical Physics (before 20th century) could not explain it !
5
Experiments showed this !
Log Intensity
4
Wien developed this empirical
relation : Bλ(T)≈ (a / λ5) e-b/λT
3
2
A more complete derivation provided by
Rayleigh and James Jeans in 1905. 1
Rayleigh-Jeans: Bλ(T)≈2c kT / λ4
0
-1
100
1000
Wavelength [nm]
10,000
The Problem with Blackbody Radiation:
Quantum Physics solves it !
Max Planck (1858-1947), German Physicist, solved the mystery
of blackbody radiation with the following radical suggestion.
A standing E/M wave could not acquire just any arbitrary
amount of energy. Instead, the E/M wave can only have allowed
energy levels that were integer multiples of a minimum wave
energy.
This minimum energy, a quantum, is given by E=hν=hc / λ,
where h is the “Planck” constant, h=6.62606876 x 10-34 J s.
This gives the formula for the intensity of blackbody radiation:
Bλ(T) =
2hc2 / λ5
(ehc/λkT
- 1)
or
Bν(T) =
2hν3 / c2
(ehν/kT - 1)
This result greatly influenced the development of Quantum Mechanics.
The Problem with Blackbody Radiation:
Quantum Physics solves it !
Similarly, the specific energy density, uλ, of E/M radiation is the
energy per unit volume between λ and λ+dλ.
uλ dλ = (4π / c ) Bλ(T) dλ =
8πhc / λ5
(ehc/λkT
- 1)
dλ
Similarly, the specific energy density, uν, is the E/M energy per
unit volume between ν and ν+dν.
8πhν3 / c3
uν dν = (4π / c ) Bν(T) dν =
dν
hν/kT
(e
- 1)
Blackbody Radiation
Observed Spectra of Vega-type Star
Solar-type Star
Lots of absorption
from atoms in the
stars’ atmospheres
(more next week)
T=10,000 K
T=8000 K Here Stars Look almost T=5800 K
exactly like blackbodies T=3000 K