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?

How do you know the Earth Rotates?

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=49jwbrxcpjc

How do you know the Earth Rotates?

What is the effect? The coriolis force is a =2v ω magnitude is a =2vω sin θ maximum particles motion is perpendicular to Earth s rotation (e.g., straight up if you re on the equator), and zero when parallel (e.g., straight up when you re at the pole) everywhere else we can take an average that sin θ 1 2 so that a cor vω

What is the effect? A particle feels this effect for the time it is in flight. If the particle is in flight for a time Δt, the velocity will be altered by a fractional amount v v a cor t v ω t where ω -1 = 1 day / 2π ~ 4 hrs ~ 14,000 s So, if travel time, Δt << 14,000 s, then no visible effect. Example: During WW1, Germans used a huge artillery gun to bombard Paris from a distance of 120 km. Gun had a muzzle velocity of 1.6 km/s, and shells reached an altitude of 40 km with a flight time of Δt = 170 s. The deflection is then d = 1 2 a cor( t) 2 vω( t) 2 /2 2km

Coriolis Force affects wind patterns Blue: flow from high to low pressure Red: Coriolis force in Northern Hemisphere

Coriolis Force affects wind patterns Blue: flow from high to low pressure Red: Coriolis force in Northern Hemisphere Coriolis Force affects rotation of weather patterns - you need to include the effects of atmospheric pressure. As air flows into low-pressure regions it is deflected to one side or the other (depends on the hemisphere). Cyclones rotate clockwise in southern hemisphere; hurricanes rotate counterclockwise in northern hemisphere (shown here)

How do you know the Earth Rotates? Hurricane Gustav Aug 31, 2008

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).

How do you know the Earth Rotates? Hurricane Ike Sept 4, 2008

How do you know the Earth Rotates? Hurricane Ike before making landfall

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) Kepler 21st Century Mercury 0.389 0.387 Venus 0.724 0.723 Earth 1.000 1.000 How would you measure the Mars 1.523 1.524 absolute distance to other planets?! Jupiter 5.200 5.202 Saturn 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? h=138 ft θ=5 o D

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=0.5 o, D= 1577.4 ft h=138 ft θ=5 o D

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? D 2θ 2w D = [ w / tan(θ) ] for w = 1 m and θ=0.5 o, D=114.6 m for w = 1 m and θ=5 o, D=11.4 m

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=0.5 o, D=114.6 m for w = 1 m and θ=5 o, D=11.4 m

Parallax: same idea as triangulation, derived from Greek parallaxis, the value of an angle. 2B d p d = B / tan(p)

First observation Parallax unmoving background stars 1 AU 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 10 5 AU d = 1 / p(arcsec) pc.

Parallax unmoving background stars 1 AU Position of star 180 days later 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 10 5 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

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. 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 Cygnus 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 10 5 AU = 3.09 x 10 16 m = 1.92 x 10 13 miles =3.26 lightyears (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 http://www.youtube.com/watch?v=ovzkfmuxznc

Light Propagation Direction Coherent Light From Single Slit Destructive Interference Barrier with Double Slits Screen Constructive Interference Intensity Distribution of Fringes

The Wave Nature of Light d sinθ = { nλ (n-1/2) λ (n=0,1,2,3... ), constructive interference (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 S 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.

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).

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 K T=3000 K

Blackbody Radiation Observed Spectra of Vega-type Star Solar-type Star T=10,000 K T=8000 K T=5800 K T=3000 K

Blackbody Radiation Observed Spectra of Vega-type Star Solar-type Star Here Stars Look almost exactly like blackbodies T=10,000 K T=8000 K T=5800 K T=3000 K

Blackbody Radiation Observed Spectra of Vega-type Star Solar-type Star Lots of absorption from atoms in the stars atmospheres (more next week) Here Stars Look almost exactly like blackbodies T=10,000 K T=8000 K T=5800 K T=3000 K

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:

Thermal Radiation: Wien s Law The peak wavelength (in meters) is related to the temperature (in Kelvin) as: λ = (0.0029/T)

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!

Note: as T increases, the blackbody emits more radiation at all wavelengths. Blackbody Radiation Josef Stefan (1835-1893) Ludwig Boltzmann (1844-1906) Related the Luminosity of a Blackbody to the Surface Area of the object: L = A σ T 4 For a sphere: L = (4π R 2 ) σ T 4 Stefan-Boltzmann constant: σ = 5.670400 x 10-8 W m -2 K -4

Example: Luminosity of the Sun, L =3.839 x 10 26 W Radius of the Sun, R =6.95508 x 10 8 m Using: L = (4π R 2 ) σ T 4 T = L ( ) 1/4 = 5777 K 4π R 2 σ Radiant Flux (or surface flux) = L / Area = L / (4π R 2 ) for a sphere Fsurf = σ T 4 = 6.316 x 10 7 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: E Bλ(T) 2c kt / λ 4 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 4 Log Intensity 3 2 1 0 A more complete derivation provided by Rayleigh and James Jeans in 1905. Rayleigh-Jeans: Bλ(T) 2c kt / λ 4-1 100 1000 10,000 Wavelength [nm]

The Problem with Blackbody Radiation: Classical Physics (before 20th century) could not explain it! 5 4 Experiments showed this! Log Intensity 3 2 1 0 A more complete derivation provided by Rayleigh and James Jeans in 1905. Rayleigh-Jeans: Bλ(T) 2c kt / λ 4-1 100 1000 10,000 Wavelength [nm]

The Problem with Blackbody Radiation: Classical Physics (before 20th century) could not explain it! 5 4 Experiments showed this! Wien developed this empirical relation : Bλ(T) (a / λ 5 ) e -b/λt Log Intensity 3 2 1 0 A more complete derivation provided by Rayleigh and James Jeans in 1905. Rayleigh-Jeans: Bλ(T) 2c kt / λ 4-1 100 1000 10,000 Wavelength [nm]

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) = 2hc 2 / λ 5 (e hc/λkt - 1) or Bν(T) = 2hν 3 / c 2 (e hν/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 / λ 5 (e hc/λkt - 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 = 8πhν 3 / c 3 (e hν/kt - 1) dν

Blackbody Radiation T=10,000 K T=8000 K T=5800 K T=3000 K

Blackbody Radiation Observed Spectra of Vega-type Star Solar-type Star T=10,000 K T=8000 K T=5800 K T=3000 K

Blackbody Radiation Observed Spectra of Vega-type Star Solar-type Star Here Stars Look almost exactly like blackbodies T=10,000 K T=8000 K T=5800 K T=3000 K

What have we learned? 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.