astr 314 : survey of astronomy extragalactic astronomy &...
TRANSCRIPT
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ASTR 314 : Survey of Astronomy�Extragalactic Astronomy & Cosmology
Prof. D. DePoy Office: MIST 240
Email: [email protected]
Webpage: http://faculty.physics.tamu.edu/depoy
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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
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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).
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How do you know the Earth Rotates ?
Coriolis Force affects rotation of weather patterns (cyclones rotate clockwise in southern hemisphere; hurricanes rotate counter-clockwise in northern hemisphere).
Hurricane Gustav Aug 31, 2008
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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
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)
How would you measure the
absolute distance to other
planets ?!
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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 ?
θ=5o
D
h=138 ft
D = [ h / tan(θ) ]
for h=138 ft and θ=0.5o, D= 1577.4 ft
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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
D
D = [ w / tan(θ) ]
for w = 1 m and θ=0.5o, D=114.6 m
for w = 1 m and θ=5o, D=11.4 m
2w You subconsciously do this all the
time. Your brain judges how fast an oncoming car is going by the change in
its angular size.
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Parallax: same idea as triangulation, derived from Greek parallaxis, “the value of an angle”.
d = B / tan(p)
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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.
Parallax
1 AU
unmoving background stars
First observation
Position of star on first
observation
180 days later
Position of star 180 days later
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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.
Cygnus
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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
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The Answer: with Parallax
N View from Pacific Ocean
View from United Kingdom
Mars
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.
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N View from Pacific Ocean
View from United Kingdom
Venus
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.
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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.
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N View from Pacific Ocean
View from United Kingdom
Venus
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 T1”
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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.
N View from Pacific Ocean
View from United Kingdom
Venus
“I see Venus begin Transit at Time T1”
“I see Venus begin Transit at Time T2”
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.
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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.
Cygnus
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 !
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The Wave Nature of Light
Double-Slit Experiment of Thomas Young (1773-1829)
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Constructive Interference Destructive Interference
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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-
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Constructive Interference
Destructive Interference
Barrier with Double Slits
Coherent Light From Single Slit
Light Propagation Direction
Screen
Intensity Distribution of Fringes
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The Wave Nature of Light
nλ (n=0,1,2,3... ), constructive interference
(n-1/2) λ (n=0,1,2,3... ), destructive interference
d sinθ = {
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)
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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
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.
S
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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
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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
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Thermal Radiation (Blackbody Radiation)
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The temperature of lava can be estimated from its color, typically 1000-1200 K.
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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).
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Blackbody Radiation
T=10,000 K T=8000 K T=5800 K T=3000 K
Observed Spectra of Vega-type Star Solar-type Star
Here Stars Look almost exactly like blackbodies
Lots of absorption from atoms in the stars’ atmospheres (more next week)
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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 !
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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
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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
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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
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
T⊙ =
L⊙
4π R⊙ σ ( )
1/4 = 5777 K
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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).
0 L
Standing waves of λ = 2L, L, 2L/3, 2L/4, 2L/5, ...
E/M waves must satisfy E=0 at wave edges.
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
Bλ : Intensity (units of Energy per second per area per wavelength)
c : speed of light
BUT, when you integrate this over all wavelengths, it diverges to infinity, called the “ultraviolet catastrophe” !
E
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The Problem with Blackbody Radiation: Classical Physics (before 20th century) could not explain it !
A more complete derivation provided by Rayleigh and James Jeans in 1905.
Rayleigh-Jeans: Bλ(T)≈2c kT / λ4
Wavelength [nm]
1000 10,000 100
Log
Inte
nsity
-1
0
1
2
3
4
5
Experiments showed this ! Wien developed this empirical
relation : Bλ(T)≈ (a / λ5) e-b/λT
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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:
2hc2 / λ5
(ehc/λkT - 1) Bλ(T) =
2hν3 / c2
(ehν/kT - 1) Bν(T) = or
This result greatly influenced the development of Quantum Mechanics.
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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.
8πhc / λ5
(ehc/λkT - 1) uλ dλ = (4π / c ) Bλ(T) dλ =
8πhν3 / c3
(ehν/kT - 1)
dλ
Similarly, the specific energy density, uν, is the E/M energy per unit volume between ν and ν+dν.
uν dν = (4π / c ) Bν(T) dν = dν
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Blackbody Radiation
T=10,000 K T=8000 K T=5800 K T=3000 K
Observed Spectra of Vega-type Star Solar-type Star
Here Stars Look almost exactly like blackbodies
Lots of absorption from atoms in the stars’ atmospheres (more next week)