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Twin Paradox. The Theory of Relativity. About Relativity. As an object approaches the speed of light, time slows down. (Moving clocks are slow) (Moving rulers are short). A trip from Earth to Planet Hollywood. Homer stays on Earth.
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TwinParadox The Theory of Relativity
About Relativity • As an object approaches the speed of light, time slows down. • (Moving clocks are slow) • (Moving rulers are short)
A trip from Earth to PlanetHollywood • Homer stays on Earth. • Loner travels 10 light-years at 80% of the speed of light. (speed of light = c) • Beta (β) is the velocity of the object compared to the speed of light. (β = 4/5c = 0.8c) • Gamma (γ) is the effect of traveling at speeds close to light speed (c) has on time (t) or distance (x).
Homer Planet Hollywood Loner Earth
Effect of Time on Spaceship Velocity (v) is v = 4/5c = 0.8c therefore… β = 4/5 = 0.8 1 γ= (β² = 0.8² = 0.64) 1 - β² 1 - β² = 1 - 0.64 = 0.36 and the √ of 0.36 is 0.6 !! γ = 1 ÷ 0.6 = 1²/³ = 5/3
As Viewed From Earth • Without Relativity….. • x = vt or t = x/v • x is light years traveled • v is velocity. • t is time. • t = 10/0.8c = 12.5 years each way. • There and back makes the trip 12.5 x 2 or 25 years!!
As Viewed From Earth • With Relativity • Homer sees Loner’s clock is running slow by… • γ = 5/3 !! • Therefore Loner’s clock reads 25 years ÷ γ • 25 ÷ 5/3 = 15 years!!
Homer on Earth ages 25 years!! Loner, traveling at 80% the speed of light ages 15 years!! Physical Results of Trip
As Viewed From Spaceship Loner sees distance of planets contracted by γ = 5/3 In Loner’s frame distance is 10 light years ÷ 5/3 10 ÷ 5/3 = 6 light years. Therefore t = x/v = 6/0.8 = 7.5 years each way. There and back is 7.5 x 2 = 15 year trip for Loner!!
Cosmic Muons • Factor: 1 γ= 1 - v²/c² (β² = v²/c²)
Muon Lifetime • d = vt • t = 2 x 10(-6) seconds • v = 3 x 10(8) m/s • d = 2 x 10(-6) • 3 x 10(8) = 600m = 0.6km • A muon only travels 0.6 km into the atmosphere!
Question?? • Muons are created in the upper atmosphere. • Detectors tell us they reach the Earth’s surface. • Suppose the distance needed to travel is approximately 10 km. • If muons travel only 0.6 km, how do they reach the Earth’s surface??
With Relativity • suppose v = 0.999c 1 γ= = 22 1 - v²/c²
Earth’s Reference Frame • Muon’s clock runs slow by γ. • Therefore the muon lives 22x longer and travels 22x further! • So… d = (0.6 km) (22) = 13.2 km • 13.2 km > 10.0 km allowing for the muon to reach the Earth’s surface!!
Earth’s Reference Frame Muon heading towards Earth at 0.999c Homer notices that the muon’s “clock” is running slow
Muon’s Reference Frame • Earth’s atmosphere moves towards the muon at v = 0.999c, therefore the atmosphere is contracted by γ. • Thickness of atmosphere is 10km/22 or 0.45 km to the muon. • The muon travels 0.6 km and 0.6 > 0.45 so the muon reaches Earth!!
Muon’s Reference Frame Traveling at 0.999c The muon’s ruler is “shortened,” while The astronauts, barely Moving in comparison, See a thicker atmosphere. Astronaut’s atmosphere 10km Muon’s atmosphere 0.45km Earth