28.3 Length contraction  (Page 2/5)

Substituting for $\gamma$ gives an equation relating the distances measured by different observers.

If we measure the length of anything moving relative to our frame, we find its length $L$ to be smaller than the proper length $L_0$ 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 $\gamma =\text{30}\text{.}\text{00}$ . (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 $c$ , what is her velocity relative to the Earth? You may neglect the motion of the Earth relative to the Sun. (See [link] .)

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), note that the 4.300 ly distance between the Alpha Centauri and the Earth is the proper distance $L_0$ , 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 $L$ . In part (b), we are given $\gamma$ , and so we can find $v$ by rearranging the definition of $\gamma$ to express $v$ in terms of $c$ .

Solution for (a)

1. Identify the knowns. $L_0-\mathrm{4.300\; ly}$ ; $\gamma =\text{30}\text{.}\text{00}$
2. Identify the unknown. $L$
3. Choose the appropriate equation. $L=\frac{L_0}{\gamma }$
4. Rearrange the equation to solve for the unknown.
   $\begin{array}{lll}L& =& \frac{L_0}{\gamma }\\ & =& \frac{\mathrm{4.300\; ly}}{\text{30.00}}\\ & =& \text{0.1433 ly}\end{array}$

Solution for (b)

1. Identify the known. $\gamma =\text{30}\text{.}\text{00}$
2. Identify the unknown. $v$ in terms of $c$
3. Choose the appropriate equation. $\gamma =\frac{1}{\sqrt{1-\frac{v^2}{c^2}}}$
4. Rearrange the equation to solve for the unknown.
   $\begin{array}{lll}\gamma & =& \frac{1}{\sqrt{1-\frac{v^2}{c^2}}}\\ \text{30.00}& =& \frac{1}{\sqrt{1-\frac{v^2}{c^2}}}\end{array}$

   Squaring both sides of the equation and rearranging terms gives

   $\text{900}\text{.}0=\frac{1}{1-\frac{v^2}{c^2}}$

   so that

   $1-\frac{v^2}{c^2}=\frac{1}{\text{900}\text{.}0}$

   and

   $\frac{v^2}{c^2}=1-\frac{1}{\text{900}\text{.}0}=0\text{.}\text{99888}\text{.}\text{.}\text{.}.$

   Taking the square root, we find

   $\frac{v}{c}=0\text{.}\text{99944},$

   which is rearranged to produce a value for the velocity

   $\text{v=}0\text{.}\text{9994}c.$

Discussion

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 ( $\text{γ=}\text{30}\text{.}\text{00}$ ), and we see that $v$ 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.