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Calculating length contraction

Suppose an astronaut, such as the twin in the twin paradox discussion, travels so fast that γ = 30.00 . (a) The astronaut travels from Earth to the nearest star system, Alpha Centauri, 4.300 light years (ly) away as measured by an earthbound observer. How far apart are Earth and Alpha Centauri as measured by the astronaut? (b) In terms of c , what is the astronaut’s velocity relative to Earth? You may neglect the motion of Earth relative to the sun ( [link] ).

In figure a, the earth and Alpha Centauri are shown as separated by a distance L naught and the earth’s clock shows a time delta t. A horizontally contracted space ship is moving to the right with velocity v. We are given the equation v = L naught / Delta t. In figure b, the earth and Alpha Centauri are shown as separated by a distance L. Both the earth and Alpha Centauri are moving to the left with velocity v and are contracted horizontally. The spaceship is stationary and not contracted. The ship’s clock shows a time delta tau. We are given the equation v = L / Delta tau.
(a) The earthbound observer measures the proper distance between Earth and Alpha Centauri. (b) The astronaut observes a length contraction because Earth and 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.

Strategy

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), the 4.300-ly distance between Alpha Centauri and Earth is the proper distance L 0 , because it is measured by an earthbound observer to whom both stars are (approximately) stationary. To the astronaut, Earth and Alpha Centauri are moving past at the same velocity, so the distance between them is the contracted length L . In part (b), we are given γ , so we can find v by rearranging the definition of γ to express v in terms of c .

Solution for (a)

For part (a):

  1. Identify the knowns: L 0 = 4.300 ly; γ = 30.00 .
  2. Identify the unknown: L .
  3. Express the answer as an equation: L = L 0 γ .
  4. Do the calculation:
    L = L 0 γ = 4.300 ly 30.00 = 0.1433 ly.

Solution for (b)

For part (b):

  1. Identify the known: γ = 30.00 .
  2. Identify the unknown: v in terms of c .
  3. Express the answer as an equation. Start with:
    γ = 1 1 v 2 c 2 .

    Then solve for the unknown v/c by first squaring both sides and then rearranging:
    γ 2 = 1 1 v 2 c 2 v 2 c 2 = 1 1 γ 2 v c = 1 1 γ 2 .
  4. Do the calculation:
    v c = 1 1 γ 2 = 1 1 ( 30.00 ) 2 = 0.99944

    or
    v = 0.9994 c .

Significance

Remember not to round off calculations until the final answer, 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 ) , and we see that v is approaching (not equaling) the speed of light. Because the distance as measured by the astronaut is so much smaller, the astronaut can travel it in much less time in her frame.

People traveling at extremely high velocities could cover very large distances (thousands or even millions of light years) and age only a few years on the way. However, like emigrants in past centuries who left their home, these people would leave the Earth they know forever. Even if they returned, thousands to millions of years would have passed on Earth, obliterating most of what now exists. There is also a more serious practical obstacle to traveling at such velocities; immensely greater energies would be needed to achieve such high velocities than classical physics predicts can be attained. This will be discussed later in the chapter.

Why don’t we notice length contraction in everyday life? The distance to the grocery store does not seem to depend on whether we are moving or not. Examining the equation L = L 0 1 v 2 c 2 , we see that at low velocities ( v < < c ) , the lengths are nearly equal, which is the classical expectation. But length contraction is real, if not commonly experienced. For example, a charged particle such as an electron traveling at relativistic velocity has electric field lines that are compressed along the direction of motion as seen by a stationary observer ( [link] ). As the electron passes a detector, such as a coil of wire, its field interacts much more briefly, an effect observed at particle accelerators such as the 3-km-long Stanford Linear Accelerator (SLAC). In fact, to an electron traveling down the beam pipe at SLAC, the accelerator and Earth are all moving by and are length contracted. The relativistic effect is so great that the accelerator is only 0.5 m long to the electron. It is actually easier to get the electron beam down the pipe, because the beam does not have to be as precisely aimed to get down a short pipe as it would to get down a pipe 3 km long. This, again, is an experimental verification of the special theory of relativity.

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Source:  OpenStax, University physics volume 3. OpenStax CNX. Nov 04, 2016 Download for free at http://cnx.org/content/col12067/1.4
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