How do you know the Earth Rotates ? How do you measure the...

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

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

Hurricane GustavAug 31, 2008

Coriolis Force affects rotation of weather patterns (cyclones rotate clockwise in southern hemisphere; hurricanes rotate counter-clockwise in northern hemisphere).

Hurricane GustavAug 31, 2008

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)

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 ?!

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 ?

h=138 ft

Example: How far is Rudder Tower from the Albritton Bell

Tower ?

h=138 ft

D = [ h / tan(θ) ]

for h=138 ft and θ=0.5o, D= 1577.4 ft

Another example: how far away is a car by its headlights ?

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

Another example: how far away is a car by its headlights ?

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.

d = B / tan(p)

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

unmoving background

starsFirst observation

Position of star on first

observation

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

unmoving background

180 days later

Position of star 180 days later

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

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

NView from Pacific Ocean

View from United Kingdom

The Answer: with Parallax

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.

http://www.astro.umd.edu/openhouse/gallery/2004-06-08VT/other/VenusTransitGCS.gif

“I see Venus begin Transit at Time T1”

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

d = 1 / 0.316 = 3.16 pc.

Cygnus

1 parsec = 2.06265 x 105 AU= 3.09 x 1016 m

= 1.92 x 1013 miles=3.26 lightyears (lyr).

Parallax

d = 1 / 0.316 = 3.16 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 !

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-

Constructive Interference

Destructive Interference

Barrier withDouble Slits

Coherent Light

From Single Slit

Light Propagation Direction

Screen

Intensity Distribution of Fringes

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)

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.

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

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).

Thermal Radiation:•Hotter objects emit more light at all

frequencies.

•Hotter objects emit photons with higher average energy (higher frequencies).

Blackbody Radiation

T=10,000 K T=8000 K T=5800 KT=3000 K

Blackbody Radiation

T=10,000 K T=8000 K T=5800 KT=3000 K

Observed Spectra of Vega-type StarSolar-type Star

Blackbody Radiation

T=10,000 K T=8000 K T=5800 KT=3000 K

Here Stars Look almost exactly like

blackbodies

Blackbody Radiation

T=10,000 K T=8000 K T=5800 KT=3000 K

blackbodies

Lots of absorption from atoms in the stars’ atmospheres(more next week)

Hottest stars (blue): T=50,000 K

Coldest stars (red):T=3,000 K

Sun: T=5,800 K

During Influenza Season, many airports around the world screen for people with temperatures using

Infrared cameras:

λ = (0.0029/T)

The peak wavelength (in meters) is related to the temperature (in Kelvin) as:

Thermal Radiation: Wien’s 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 !

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 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

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⊙ =

4π R⊙2 σ

( )1/4

= 5777 K

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).

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” !

A more complete derivation provided by Rayleigh and James Jeans in 1905.

Rayleigh-Jeans: Bλ(T)≈2c kT / λ4

Wavelength [nm]

1000 10,000100

Wavelength [nm]

1000 10,000100

Experiments showed this !

Wavelength [nm]

1000 10,000100

Experiments showed this !Wien developed this empirical relation :

Bλ(T)≈ (a / λ5) e-b/λT

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.

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)

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

Blackbody Radiation

T=10,000 K T=8000 K T=5800 KT=3000 K

Blackbody Radiation

T=10,000 K T=8000 K T=5800 KT=3000 K

Blackbody Radiation

T=10,000 K T=8000 K T=5800 KT=3000 K

blackbodies

Blackbody Radiation

T=10,000 K T=8000 K T=5800 KT=3000 K

blackbodies

Lots of absorption from atoms in the stars’ atmospheres(more next week)

• 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 ?

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