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. ► 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).

Earth Home r Loner Planet Hollywood

Effect of Time on Spaceship Velocity (v) is v = 4/5c = 0.8c therefore… β = 4/5 = 0.8 γ =γ = β² 1 - β² = = 0.36 and the √ of 0.36 is 0.6 !! (β² = 0.8² = 0.64) γ = 1 ÷ 0.6 = 1²/³ = 5/3

As Viewed From Earth Without Relativity….. Without Relativity….. x = vt or t = x/v x = vt or t = x/v x is light years traveled x is light years traveled v is velocity. v is velocity. t is time. t is time. t = 10/0.8c = 12.5 years each way. t = 10/0.8c = 12.5 years each way. There and back makes the trip There and back makes the trip 12.5 x 2 or 25 years!! 12.5 x 2 or 25 years!!

As Viewed From Earth With Relativity With Relativity Homer sees Loner’s clock is running slow by… Homer sees Loner’s clock is running slow by… γ = 5/3 !! γ = 5/3 !! Therefore Loner’s clock reads 25 years ÷ γ Therefore Loner’s clock reads 25 years ÷ γ 25 ÷ 5/3 = 15 years!! 25 ÷ 5/3 = 15 years!!

Physical Results of Trip Homer on Earth ages 25 years!! Loner, traveling at 80% the speed of light ages 15 years!!

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: γ =γ = v²/c² (β² = v²/c²)

Muon Lifetime d = vt d = vt t = 2 x 10(-6) seconds t = 2 x 10(-6) seconds v = 3 x 10(8) m/s v = 3 x 10(8) m/s d = 2 x 10(-6) 3 x 10(8) = 600m = 0.6km d = 2 x 10(-6) 3 x 10(8) = 600m = 0.6km A muon only travels 0.6 km into the atmosphere! 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 γ =γ = v²/c² = 22

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 Earth Muon’s atmosphere 0.45km Astronaut’s atmosphere 10km Traveling at 0.999c The muon’s ruler is “shortened,” while The astronauts, barely Moving in comparison, See a thicker atmosphere.

The End!!