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 Chapter 28.2 Example 1 illustrates this concept. To an observer on the Earth, the muon travels at [latex]{0.950 \textbf{c}}[/latex] for [latex]{7.05 \;\mu \text{s}}[/latex] from the time it is produced until it decays. Thus it travels a distance

relative to the Earth. In the muon’s frame of reference, its lifetime is only 2.20μs2.20μs. It has enough time to travel only

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.

The Earth-bound observer measures the proper length [latex]{L_0}[/latex], 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 [latex]{L}[/latex] it sees is not the proper length.

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

The time relative to the Earth-bound observer is [latex]{\Delta t}[/latex], since the object being timed is moving relative to this observer. The velocity relative to the moving observer is given by

The moving observer travels with the muon and therefore observes the proper time [latex]{\Delta t_0}[/latex]. The two velocities are identical; thus,

We know that [latex]{\Delta t = \gamma \Delta t_0}[/latex]. Substituting this equation into the relationship above gives

Substituting for [latex]{\gamma}[/latex] 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 [latex]{L}[/latex] to be smaller than the proper length [latex]{L_0}[/latex] 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.

Example 1: Calculating Length Contraction: The Distance between Stars Contracts when You Travel at High Velocity

Suppose an astronaut, such as the twin discussed in Chapter 28.2 Simultaneity and Time Dilation, travels so fast that [latex]{\gamma = 30.00}[/latex]. (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 [latex]{c}[/latex], what is her velocity relative to the Earth? You may neglect the motion of the Earth relative to the Sun. (See Figure 3.)

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 [latex]{L_0}[/latex], 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 [latex]{L}[/latex]. In part (b), we are given [latex]{\gamma}[/latex], and so we can find [latex]{v}[/latex] by rearranging the definition of [latex]{\gamma}[/latex] to express [latex]{v}[/latex] in terms of [latex]{c}[/latex].

Solution for (a)

1. Identify the knowns. [latex]{L_0 - 4.300 \;\text{ly}}[/latex]; [latex]{\gamma = 30.00}[/latex]
2. Identify the unknown. [latex]{L}[/latex]
3. Choose the appropriate equation. [latex]{L = \frac{L_0}{\gamma}}[/latex]
4. Rearrange the equation to solve for the unknown.
   [latex]\begin{array}{r @{{}={}}l} {L} & {\frac{L_0}{\gamma}} \\[1em] & {\frac{4.300 \;\text{ly}}{30.00}} \\[1em] & {0.1433 \;\text{ly}} \end{array}[/latex]

Solution for (b)

1. Identify the known. [latex]{\gamma = 30.00}[/latex]
2. Identify the unknown. [latex]{v}[/latex] in terms of [latex]{c}[/latex]
3. Choose the appropriate equation. [latex]{\gamma =}[/latex] [latex]{\frac{1}{\sqrt{1 - \frac{v^2}{c^2}}}}[/latex]
4. Rearrange the equation to solve for the unknown.
   [latex]\begin{array}{r @{{}={}}l}{\gamma} & {\frac{1}{\sqrt{1 - \frac{v^2}{c^2}}}} \\[1em] {30.00} & {\frac{1}{\sqrt{1 - \frac{v^2}{c^2}}}} \end{array}[/latex]

   Squaring both sides of the equation and rearranging terms gives

   [latex]{900.0 =}[/latex] [latex]{\frac{1}{1 - \frac{v^2}{c^2}}}[/latex]

   so that

   [latex]{1 -}[/latex] [latex]{\frac{v^2}{c^2}}[/latex] [latex]{=}[/latex] [latex]{\frac{1}{900.0}}[/latex]

   and

   [latex]{\frac{v2}{c2}}[/latex] [latex]{= 1 -}[/latex] [latex]{\frac{1}{900.0}}[/latex] [latex]{= 0.99888 \dots}[/latex]

   Taking the square root, we find

   [latex]{\frac{v}{c}}[/latex] [latex]{= 0.99944},[/latex]

   which is rearranged to produce a value for the velocity

   [latex]{v = 0.9994c}.[/latex]

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 ([latex]{\gamma = 30.00}[/latex]), and we see that [latex]{v}[/latex] 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.

People could be sent very large distances (thousands or even millions of light years) and age only a few years on the way if they traveled at extremely high velocities. But, like emigrants of centuries past, they would leave the Earth they know forever. Even if they returned, thousands to millions of years would have passed on the Earth, obliterating most of what now exists. There is also a more serious practical obstacle to traveling at such velocities; immensely greater energies than classical physics predicts would be needed to achieve such high velocities. This will be discussed in Chapter 28.6 Relativistic Energy.

Why don’t we notice length contraction in everyday life? The distance to the grocery shop does not seem to depend on whether we are moving or not. Examining the equation [latex]{L = L_0 \sqrt{1 - \frac{v^2}{c^2}}}[/latex], we see that at low velocities ([latex]{ vFigure 4.) 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 the Earth are all moving by and are length contracted. The relativistic effect is so great than the accelerator is only 0.5 m long to the electron. It is actually easier to get the electron beam down the pipe, since the beam does not have to be as precisely aimed to get down a short pipe as it would down one 3 km long. This, again, is an experimental verification of the Special Theory of Relativity.