Length Contraction by Heather Jacobson

Length Contraction: the length of an object is measured to be shorter when it is moving relative to the observer than when it is at rest i.e. moving objects seem shorter

A rocket leaves Earth, headed for Neptune at some fraction of the speed of light EARTH NEPTUNE ROCKET V/C This is what it would look like from Earth’s frame of reference

That situation is not the “proper” situation In the proper situation, the two planets are moving in the same direction to and from the rocket. To the people on the rocket, they aren’t moving EARTH V/C NEPTUNE V/C

Because of time dilation, the time interval for the people on the spaceship is less than for the observers from Earth Since the people on the spaceship measure the same speed as the Earth observers but less time, they must also measure the distance as less So, let’s derive a formula to represent this...

We know that Δt =Lo/v And Δt = Δto/ (1- v²/c²)^½ If L = vΔto then L = vΔto = vΔt(1- v²/c²)^½ = Lo(1- v²/c²)^½ L = Lo(1- v²/c²)^½ L = the contracted length Lo = the proper length (at rest)

In your data packet, they will give you these formulas: ã= 1/ (1- v²/c²)^½L = Lo/ã

A painting is hung on the wall of a spaceship going 0.9 the speed of light The painting is 1.00m tall and 1.50m wide What are dimension of the picture according to the people on the spaceship? What are the dimensions as seen by an observer on Earth? (a)1.00m tall, 1.50m wide (b) L = Lo(1- v²/c²)^½ = (1.50 m)(1- (0.90)²)^½ So, the picture has dimensions 1.00 m by 0.65 m

Its height isn’t affected because length contraction is only observed relative to the direction of velocity

Another concept to consider... You travel past a building at 2.5x10 8 m/s This is the same as saying the building is traveling past you at the same speed but in the opposite direction…WOW 2.5x10 8 m/s

As you run past, you can see the side of the building even though you are directly in front of it What the building looks like at rest What the building looks like when moving at a high speed

A spaceship moves past you at a speed of the speed of light. You measure its length to be 65.5 m. What would its length be at rest? L = 65.5L = Lo(1- v²/c²)^½ Lo = (65.5)/(1- (0.700)²)^½ = 91.7 m 91.7

A jet flies past you at a speed of 0.750c. You measure it to be 24.5 m long and 5.80 m high. What would its length and height be at rest? To the person in the jet, how fast do you appear to be moving? If the jet travels 12 km, what would the person on the jet measure that distance to be? a) L = Lo(1- v²/c²)^½ L = 24.5 m Lo = (24.5)/( ²)^½ = 37.0 m c) L = Lo(1- v²/c²)^½ Lo = m L = (12 E 3)( ²)^½ = 7900 m 37.0m, 0.750c, 7900m

Four-Dimensional Space-Time space takes up three dimensions and time is the fourth dimension here’s the weird part: time and space can intermix Ex: A person on a train eats a meal served on a 20-cm- diameter plate. The meal lasts 15 minutes according to a clock on the train. To an observer on Earth, the meal takes 20 minutes but the plate is only 15 cm in diameter. You can think of it as time dilation and length contraction being balanced. What the meal appears to lose in size it makes up for in the length of time it lasts. Space (or length) is exchanged for time.

So time is just like a spatial dimension?