how do you know the earth rotates ? how do you measure the...
TRANSCRIPT
How do you know the Earth Rotates ?
How do you measure the distances to objects in the Solar System and nearby stars ?
What is the Nature of Light ?
Thursday, January 26, 2012
How do you know the Earth Rotates ?
Thursday, January 26, 2012
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
Thursday, January 26, 2012
How do you know the Earth Rotates ?
Hurricane GustavAug 31, 2008
Thursday, January 26, 2012
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 GustavAug 31, 2008
Thursday, January 26, 2012
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)
Thursday, January 26, 2012
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 ?!
Thursday, January 26, 2012
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
Thursday, January 26, 2012
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
Thursday, January 26, 2012
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
Thursday, January 26, 2012
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
2wYou subconsciously do this all the
time. Your brain judges how fast an oncoming car is going by the change
in its angular size.
Thursday, January 26, 2012
Parallax: same idea as triangulation, derived from Greek parallaxis, “the value of an angle”.
d = B / tan(p)
pd2B
Thursday, January 26, 2012
d = 1 AU / tan(p) ≈ 1 / p (radians) AU≈ 57.3 / p (degrees) ≈ 206265 / p (arcsec) AU
Define: 1 parsec = 2.06265 x 105 AUd = 1 / p(arcsec) pc.
Parallax
1 AU
unmoving background
starsFirst observation
Position of star on first
observation
Thursday, January 26, 2012
d = 1 AU / tan(p) ≈ 1 / p (radians) AU≈ 57.3 / p (degrees) ≈ 206265 / p (arcsec) AU
Define: 1 parsec = 2.06265 x 105 AUd = 1 / p(arcsec) pc.
Parallax
1 AU
unmoving background
stars
180 days later
Position of star 180 days later
Thursday, January 26, 2012
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
Thursday, January 26, 2012
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
Thursday, January 26, 2012
The Answer: with Parallax
NView from Pacific Ocean
View from United Kingdom
Mars
Thursday, January 26, 2012
The Answer: with Parallax
NView 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.
Thursday, January 26, 2012
NView 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.
Thursday, January 26, 2012
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.
Thursday, January 26, 2012
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.
Thursday, January 26, 2012
NView 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”
Thursday, January 26, 2012
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.
NView from Pacific Ocean
View from United Kingdom
Venus
“I see Venus begin Transit at Time T1”
“I see Venus begin Transit at Time T2”
Thursday, January 26, 2012
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.
NView 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.
Thursday, January 26, 2012
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).
Thursday, January 26, 2012
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 !
Thursday, January 26, 2012
The Wave Nature of Light
Double-Slit Experiment of Thomas Young (1773-1829)
Thursday, January 26, 2012
Constructive Interference Destructive Interference
Thursday, January 26, 2012
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-
Thursday, January 26, 2012
Constructive Interference
Destructive Interference
Barrier withDouble Slits
Coherent Light
From Single Slit
Light Propagation Direction
Screen
Intensity Distribution of Fringes
Thursday, January 26, 2012
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)
Thursday, January 26, 2012
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
Thursday, January 26, 2012
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
Thursday, January 26, 2012
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
Thursday, January 26, 2012
Thermal Radiation (Blackbody Radiation)
Thursday, January 26, 2012
Thursday, January 26, 2012
The temperature of lava can be estimated from its color, typically 1000-1200 K.
Thursday, January 26, 2012
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).
Thursday, January 26, 2012
Thermal Radiation:•Hotter objects emit more light at all
frequencies.
•Hotter objects emit photons with higher average energy (higher frequencies).
Thursday, January 26, 2012
Blackbody Radiation
T=10,000 K T=8000 K T=5800 KT=3000 K
Thursday, January 26, 2012
Blackbody Radiation
T=10,000 K T=8000 K T=5800 KT=3000 K
Observed Spectra of Vega-type StarSolar-type Star
Thursday, January 26, 2012
Blackbody Radiation
T=10,000 K T=8000 K T=5800 KT=3000 K
Observed Spectra of Vega-type StarSolar-type Star
Here Stars Look almost exactly like
blackbodies
Thursday, January 26, 2012
Blackbody Radiation
T=10,000 K T=8000 K T=5800 KT=3000 K
Observed Spectra of Vega-type StarSolar-type Star
Here Stars Look almost exactly like
blackbodies
Lots of absorption from atoms in the stars’ atmospheres(more next week)
Thursday, January 26, 2012
Thursday, January 26, 2012
Hottest stars (blue): T=50,000 K
Coldest stars (red):T=3,000 K
Sun: T=5,800 K
Thursday, January 26, 2012
During Influenza Season, many airports around the world screen for people with temperatures using
Infrared cameras:
Thursday, January 26, 2012
λ = (0.0029/T)
The peak wavelength (in meters) is related to the temperature (in Kelvin) as:
Thermal Radiation: Wien’s Law
Thursday, January 26, 2012
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 !
Thursday, January 26, 2012
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
Thursday, January 26, 2012
Blackbody RadiationNote: 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
Thursday, January 26, 2012
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⊙2 σ
( )1/4
= 5777 K
Thursday, January 26, 2012
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
Thursday, January 26, 2012
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 10,000100
Log
Inte
nsity
-1
0
1
2
3
4
5
Thursday, January 26, 2012
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 10,000100
Log
Inte
nsity
-1
0
1
2
3
4
5
Experiments showed this !
Thursday, January 26, 2012
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 10,000100
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
Thursday, January 26, 2012
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.
Thursday, January 26, 2012
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
Thursday, January 26, 2012
Blackbody Radiation
T=10,000 K T=8000 K T=5800 KT=3000 K
Thursday, January 26, 2012
Blackbody Radiation
T=10,000 K T=8000 K T=5800 KT=3000 K
Observed Spectra of Vega-type StarSolar-type Star
Thursday, January 26, 2012
Blackbody Radiation
T=10,000 K T=8000 K T=5800 KT=3000 K
Observed Spectra of Vega-type StarSolar-type Star
Here Stars Look almost exactly like
blackbodies
Thursday, January 26, 2012
Blackbody Radiation
T=10,000 K T=8000 K T=5800 KT=3000 K
Observed Spectra of Vega-type StarSolar-type Star
Here Stars Look almost exactly like
blackbodies
Lots of absorption from atoms in the stars’ atmospheres(more next week)
Thursday, January 26, 2012
• The effect of the coriolis force is one way we can prove the Earth is Rotating.
• We measure the distances to planets and nearby stars using Parallax.
• Light acts like a wave which carries energy and can exert radiation pressure.
• At sources radiate light as blackbodies following a Planck Function , which specifies the amount of light emitted per frequency and is dependent only on the object’s temperature.
What have we learned ?
Thursday, January 26, 2012