Vern J. OstdiekDonald J. Bord

Chapter 12Special Relativity

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V=1%c V=0.01c Y=1.00005V=10%c V=0.1c Y=1.005V=50% V=0.5c Y=1.155V=60% V=0.6c Y=1.250V=80% V=0.8c Y=1.667V=90% V=0.9c Y=2.294V=99% V=0.99c Y=7.088

Some values of Y for different speeds.

Dilated time Δt= YΔtoContracted length L=Lo/Y

Lo is the proper length= length measured between 2 points at rest with respect to the frame of Reference it is measured in. L=Lo/Y

Δto is the proper time =time elapsed between 2 events happening at the same place relative To the frame of reference it is measured in.The clock is at rest with respect to the frame. Δt= YΔto

The relative speed V is the same forThe 2 frames.v=L/Δto for the astronaut moving.The time runs slower. The timeBetween 2 event is shorter ΔtoSo the distance between 2 points isSmaller = L

In the frame not moving withRespect to 2 points, the distanceBetween them is= Lo larger than L. That means Δt Larger time (dilated). V=Lo/Δt The relative speed V for the 2 frames is the same: V=Lo/Δt = Lo/YΔto V=L/Δto=Lo/YΔto

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SO IS IT REALS ?

Cosmic rays (high energy particles) from the Sun interact with air molecules toProduce muons. These subatomic particles have a very short life.They decay into an electron and other particles in a time of 0.000002 seconds.If it was not for relativistic effect, these muons would not have time to reachEarth. The distance to cover is 10 kmBut as measured in our frame (Earth) these muons have a longer life Δt !!

This is because their speed is 90% the speed of life !Their time runs slower !!!

more:http://www.jlab.org/~cecire/muonexp.html

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ExampleWhat is the mean lifetime of a muon as

measured in the laboratory if it is traveling at 0.90c with respect to the laboratory? The mean lifetime of a muon at rest is 2.2×10-6 seconds=2.2microseconds=2.2μs.

(see previous table to find the gamma factor).

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ExampleExample 12.1

ANSWER:

Y=2.294Δt = 2.294x0.000002 s = 5 10-6s=5μs

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DISCUSSION:So the muon “thinks” it only exists for 2.2

µs (as measured by its moving clock).

But we on the ground think the muon exists for twice as long (as measured by our stationary clocks): 5.0 µs.

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Length contraction happens because v=space/time so if the clockruns slows in a moving frame with respect to us, the elapsed time is shorter than ours so the length is also shorter. V=space/time = SPACE/TIME

A spaceship traveling at the velocity of 1.8 108m/scovers a distance of 900kmAs measured by an observer on the Earth. (the distance is the proper distanceLo because the 2 points are at rest with respect to Earth )

A)Find V with respect of c (compute V/c with c = 3 108m/s). Use the previous table to find the Y factor.

B) What is the distance traveled in this time as measured by the pilot of theSpaceshipt ? (from the point of view of the astronaut, the 2 points (Earth and destination) are moving relative to him. So the distance isThe contracted length L .

C)How much time does it take to cover this distance as measured by the observerOn earth and as measured by the pilot ? CONVERT to meters(use V=distance/time solve for time)

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A)Y=2.294 if V=60%c

A) From Earth's frame of reference, the distance is the proper distance Lo.The 2 points are in rest with respect to eArth. So Lo=900kmL is contracted = Lo/Y = 720km

B) In Earth's frame of reference, therelative speed is V, the distance is LoAnd the elapsed time is Δt such as V = Lo/ Δt1.8 108m/s = 900,000 / Δt Δt=0.005 s = 5ms

In the spaceship Δto is the elapsed time and L is the distance.The relative speed is the same.1.8 108m/s = 720,000 / Δto Δto=0.004s=4ms

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Alpha centauri is a nearby star in our galaxy 4.3 light years away.That means that, as measured by person on earth, it would take light 4.3 years toReach this star. If a rocket leaves for Alpha Centauri and travels at a speed = 0.95cRelative to the earth, by how much will the passenger have aged , according toTheir own clock, when they reach their destination ?Assume that the Earth and the star are stationnary with respect to one another.

The astronaut “ sees” the 2 events (leaving Earth and reaching the star) at the samePlace. (just out his door) so he has the proper time Δto. The 2 points (Earth and star) are at rest with respect to earth so this is theProper distance Lo.

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Y=3.2

If the speed is c, the time is 4.3 years.How long it would take if the speed is 0.95 c ? the distance is the same.Time = Δt = 4.3 / 0.95= 4.5 years.

Earth's people age 4.5 years.

In the speacehip the time elapsed isΔt/Y = 1.4 years !!!

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Assignments:

1) A boat can travel with a velocity of 12m/s in still water is moving atMaximum speed against stream that flows with a velocity of 5m/s relativeTo the earth. What is the velcoity of the boat relative to the bank of the stream?

2) A plane that can travel a 460mph in still air is flying with a tail wind of 40mph.How is its speed with respect to the ground. How long does it take for thePlane to travel a distance of 750 miles (relative to earth ?)

3) A swimmer swims upstram with a velocity of 4m/s relative to the water. The Velocity of the current is 3.5 m/s (downstream). What is the velocity of theSwimmer relative to the bank ?

4) A ball is thrown with a velocity of 60mph down the aisle (toward the tail of thePlane) of a jetliner traveling wil a velocity of 260mph relative to the Earth.What is the velocity of the ball relative to the earth ?

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5) An astronaut aims a flashlight toward the tail of his spaceship, which isTraveling with a velocity of 0.5c relative to the earth. What is the velocityOf the light beam relative to the earth. (remember postulates of relativity).

5) The factor appears in many expressions derived

From the theory of relativity. Fins Y when v = 92% c (or v=0.92c)

6) An astronaut cooks a three-minute egg in his spaceship whizzing past earthAt a speed of 0.92c. How long has the egg cooked as measured by anObserver on earth ? (The observer on earth measured the dilated time ΔtBecause the 2 event : raw egg-uncooked egg) don't happen at the same placeWith respect to earth). The proper time is Δto is 3 minutes. Use 6).

7) An observer on earth notes that an astronaut on a spaceshipt puts in a 4hoursShift at the controls of the spaceship. How long is this shift as measured byThe astronaut himself. The spaceship is moving with a velocity of 0.8c relativeTo the earth ? (4 hours is Δt the dilated time because the events are movingFor Earth. You are solving for the proper time Δto as measured on ship)See table to get Y.

2

2

1

1

cv−

=Υ

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8) A spaceship that is 50m long as measured by its occupants is travelingAt a speed of 0.1c relative to the Earth. So 50m is the proper distance LoBecause the 2 points are not moving with respect to the astronaut. How long is the spaceship as measured by Mission control is Houston?The 2 points move relative to Earth so they measure L. By how much the length decreased as measured By Earth ? (in cm /mm)

NOTE From earth perspective even the meter sticks of the astronautsHave shrinked so it is consisten.

9)The crew of a spaceship traveling with a velocity of 0.5c relative to the earthMeasures the distance between 2 cities on earth (in a direction parallel to theirMotion) as 600km. (they are in motion relative to the 2 points so they measureL). What is the distance between these 2 cities as measured by people on Eath ?(on Earth the 2 cities don't move so they measure the proper distance Lo)

10) p. 353 1, 2,