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  • Describe proper length.
  • Calculate length contraction.
  • Explain why we don’t notice these effects at everyday scales.
A long isolated double-lane road banked by barren land on both sides.
People might describe distances differently, but at relativistic speeds, the distances really are different. (credit: Corey Leopold, Flickr)

Have you ever driven on a road that seems like it goes on forever? If you look ahead, you might say you have about 10 km left to go. Another traveler might say the road ahead looks like it’s about 15 km long. If you both measured the road, however, you would agree. Traveling at everyday speeds, the distance you both measure would be the same. You will read in this section, however, that this is not true at relativistic speeds. Close to the speed of light, distances measured are not the same when measured by different observers.

Proper length

One thing all observers agree upon is relative speed. Even though clocks measure different elapsed times for the same process, they still agree that relative speed, which is distance divided by elapsed time, is the same. This implies that distance, too, depends on the observer’s relative motion. If two observers see different times, then they must also see different distances for relative speed to be the same to each of them.

The muon discussed in [link] illustrates this concept. To an observer on the Earth, the muon travels at 0.950 c size 12{c} {} for 7.05 μ s size 12{c} {} from the time it is produced until it decays. Thus it travels a distance

L 0 = v Δ t = ( 0.950 ) ( 3.00 × 10 8 m/s ) ( 7.05 × 10 6 s ) = 2.01 km

relative to the Earth. In the muon’s frame of reference, its lifetime is only 2.20 μ s . It has enough time to travel only

L = v Δ t 0 = ( 0 . 950 ) ( 3 . 00 × 10 8 m/s ) ( 2 . 20 × 10 6 s ) = 0 .627 km .

The distance between the same two events (production and decay of a muon) depends on who measures it and how they are moving relative to it.

Proper length

Proper length L 0 size 12{L rSub { size 8{0} } } {} is the distance between two points measured by an observer who is at rest relative to both of the points.

The Earth-bound observer measures the proper length L 0 size 12{L rSub { size 8{0} } } {} , because the points at which the muon is produced and decays are stationary relative to the Earth. To the muon, the Earth, air, and clouds are moving, and so the distance L size 12{L} {} it sees is not the proper length.

In part a observer observes from ground frame of reference a muon above earth with speed v in the rightward direction. The distance between the muon and the place where it disintegrates is two point zero one. In part b the system is shown in motion having velocity v in the leftward direction. So, the cloud and ground are displaced zero point six two seven kilo meter in the opposite direction.
(a) The Earth-bound observer sees the muon travel 2.01 km between clouds. (b) The muon sees itself travel the same path, but only a distance of 0.627 km. The Earth, air, and clouds are moving relative to the muon in its frame, and all appear to have smaller lengths along the direction of travel.

Length contraction

To develop an equation relating distances measured by different observers, we note that the velocity relative to the Earth-bound observer in our muon example is given by

v = L 0 Δ t . size 12{v= { {L rSub { size 8{0} } } over {Δt} } } {}

The time relative to the Earth-bound observer is Δ t size 12{Δt} {} , since the object being timed is moving relative to this observer. The velocity relative to the moving observer is given by

v = L Δ t 0 . size 12{v= { {L rSub { size 8{0} } } over {Δt} } } {}

The moving observer travels with the muon and therefore observes the proper time Δ t 0 size 12{Δt rSub { size 8{0} } } {} . The two velocities are identical; thus,

L 0 Δ t = L Δ t 0 . size 12{ { {L rSub { size 8{0} } } over {Δt} } = { {L} over {Δt rSub { size 8{0} } } } } {}

We know that Δ t = γ Δ t 0 size 12{Δt=γΔt rSub { size 8{0} } } {} . Substituting this equation into the relationship above gives

Questions & Answers

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Divide with 3.6
Mateo
multiply by (km/1000m) x (3600 s/h) -> 3.6
Muhammad
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No it has something to do with measurements bro... What we did today in class
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Friday bro... But the topics we did are in this app... Just try to master them quickly before the test dates... Are you done with the Maths sheet
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I eat ass
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I'll work on the maths sheet tomorrow bra @Sacky Malyenge but I'll try mastering them
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I'll eat your mom's ass with a side of tendies
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@Fillemon Nanwaapo
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electromagnetic field is a special type of field been produced by electric charges..!!! like the word electro from Electricity and the word magnetic from Magnetism.. so it is more of a join field..!!!
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Anderson
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Sacky
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Clifford Reply
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Clifford
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solve it please
Festus
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Really
Lawal
Practice Key Terms 2

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Source:  OpenStax, College physics. OpenStax CNX. Jul 27, 2015 Download for free at http://legacy.cnx.org/content/col11406/1.9
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