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L = L 0 γ . size 12{L= { {L rSub { size 8{0} } } over {γ} } } {}

Substituting for γ size 12{γ} {} gives an equation relating the distances measured by different observers.

Length Contraction

Length contraction L size 12{L} {} is the shortening of the measured length of an object moving relative to the observer’s frame.

L = L 0 1 v 2 c 2 . size 12{L - L rSub { size 8{0} } sqrt {1 - { {v rSup { size 8{2} } } over {c rSup { size 8{2} } } } } } {}

If we measure the length of anything moving relative to our frame, we find its length L size 12{L} {} to be smaller than the proper length L 0 size 12{L rSub { size 8{0} } } {} that would be measured if the object were stationary. For example, in the muon’s reference frame, the distance between the points where it was produced and where it decayed is shorter. Those points are fixed relative to the Earth but moving relative to the muon. Clouds and other objects are also contracted along the direction of motion in the muon’s reference frame.

Calculating length contraction: the distance between stars contracts when you travel at high velocity

Suppose an astronaut, such as the twin discussed in Simultaneity and Time Dilation , travels so fast that γ = 30 . 00 size 12{γ="30" "." "00"} {} . (a) She travels from the Earth to the nearest star system, Alpha Centauri, 4.300 light years (ly) away as measured by an Earth-bound observer. How far apart are the Earth and Alpha Centauri as measured by the astronaut? (b) In terms of c size 12{c} {} , what is her velocity relative to the Earth? You may neglect the motion of the Earth relative to the Sun. (See [link] .)

In part a the distance between the earth and the alpha centauri is measured as L-zero. A clock given in this figure is showing a time delta-t. A spaceship flying with velocity of v equals L-zero over delta-t from the earth to the star is shown.  Part b shows the spaceship frame of reference from which the distance L between the earth and star is contracted as they seem to move with same velocity in opposite direction. In part b the clock shows less time elapsed than the clock in part a.
(a) The Earth-bound observer measures the proper distance between the Earth and the Alpha Centauri. (b) The astronaut observes a length contraction, since the Earth and the Alpha Centauri move relative to her ship. She can travel this shorter distance in a smaller time (her proper time) without exceeding the speed of light.


First note that a light year (ly) is a convenient unit of distance on an astronomical scale—it is the distance light travels in a year. For part (a), note that the 4.300 ly distance between the Alpha Centauri and the Earth is the proper distance L 0 size 12{L rSub { size 8{0} } } {} , because it is measured by an Earth-bound observer to whom both stars are (approximately) stationary. To the astronaut, the Earth and the Alpha Centauri are moving by at the same velocity, and so the distance between them is the contracted length L size 12{L} {} . In part (b), we are given γ size 12{γ} {} , and so we can find v size 12{v} {} by rearranging the definition of γ size 12{γ} {} to express v size 12{v} {} in terms of c size 12{c} {} .

Solution for (a)

  1. Identify the knowns. L 0 4.300 ly ; γ = 30 . 00
  2. Identify the unknown. L size 12{L} {}
  3. Choose the appropriate equation. L = L 0 γ size 12{L= { {L rSub { size 8{0} } } over {γ} } } {}
  4. Rearrange the equation to solve for the unknown.
    L = L 0 γ = 4.300 ly 30.00 = 0.1433 ly alignl { stack { size 12{L= { {L rSub { size 8{0} } } over {γ} } } {} #= { {4 "." "300"" ly"} over {"30" "." "00"} } {} # =0 "." "1433"" ly" {}} } {}

Solution for (b)

  1. Identify the known. γ = 30 . 00 size 12{γ="30" "." "00"} {}
  2. Identify the unknown. v size 12{v} {} in terms of c size 12{c} {}
  3. Choose the appropriate equation. γ = 1 1 v 2 c 2 size 12{γ= { {1} over { sqrt {1 - { {v rSup { size 8{2} } } over {c rSup { size 8{2} } } } } } } } {}
  4. Rearrange the equation to solve for the unknown.
    γ = 1 1 v 2 c 2 30.00 = 1 1 v 2 c 2 alignl { stack { size 12{γ= { {1} over { sqrt {1 - { {v rSup { size 8{2} } } over {c rSup { size 8{2} } } } } } } } {} #"30" "." "00"= { {1} over { sqrt {1 - { {v rSup { size 8{2} } } over {c rSup { size 8{2} } } } } } } {} } } {}

    Squaring both sides of the equation and rearranging terms gives

    900 . 0 = 1 1 v 2 c 2 size 12{"900" "." 0= { {1} over {1 - { {v rSup { size 8{2} } } over {c rSup { size 8{2} } } } } } } {}

    so that

    1 v 2 c 2 = 1 900 . 0 size 12{1 - { {v rSup { size 8{2} } } over {c rSup { size 8{2} } } } = { {1} over {"900" "." 0} } } {}


    v 2 c 2 = 1 1 900 . 0 = 0 . 99888 . . . . size 12{ { {v rSup { size 8{2} } } over {c rSup { size 8{2} } } } =1 - { {1} over {"900" "." 0} } =0 "." "99888" "." "." "." } {}

    Taking the square root, we find

    v c = 0 . 99944 , size 12{ { {v} over {c} } =0 "." "99944"} {}

    which is rearranged to produce a value for the velocity

    v= 0 . 9994 c . size 12{ ital "v="0 "." "9994"c} {}


First, remember that you should not round off calculations until the final result is obtained, or you could get erroneous results. This is especially true for special relativity calculations, where the differences might only be revealed after several decimal places. The relativistic effect is large here ( γ= 30 . 00 size 12{ ital "γ=""30" "." "00"} {} ), and we see that v size 12{v} {} is approaching (not equaling) the speed of light. Since the distance as measured by the astronaut is so much smaller, the astronaut can travel it in much less time in her frame.

Questions & Answers

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