36.7 Length Contraction

We’ve seen that relativity requires us to rethink our idea of time. Now let’s turn our attention to the concepts of space and distance. Consider the rocket that traveled from the sun to Saturn in Example 36.4. FIGURE 36.25a shows the rocket moving with velocity v through the solar system reference frame S. We define $L = Δ x = x_{Saturn} - x_{sun}$ as the distance between the sun and Saturn in frame S or, more generally, the length of the spatial interval between two points. The rocket’s speed is $v = L / Δ t$ , where $Δ t$ is the time measured in frame S for the journey from the sun to Saturn.

The figure labeled (“a”) illustrates the rocket trip between the Sun and Saturn in frame S.

The figure labeled (b) illustrates the rocket trip between the Sun and Saturn in frame S prime.

Full Alternative Text

FIGURE 36.25b shows the situation in reference frame $S'$ , where the rocket is at rest. The sun and Saturn move to the left at speed $v = L' / Δ t'$ , where $Δ t'$ is the time measured in frame $S'$ for Saturn to travel distance $L'$ .

Speed v is the relative speed between S and $S'$ and is the same for experimenters in both reference frames. That is,

v = \frac{L}{Δ t} = \frac{L'}{Δ t'}

(36.12)

The time interval $Δ t'$ measured in frame $S'$ is the proper time $Δ τ$ because both events occur at the same position in frame $S'$ and can be measured by one clock. We can use the time-dilation result, Equation 36.9, to relate $Δ τ$ measured by the astronauts to $Δ t$ measured by the earthbound scientists. Then Equation 36.12 becomes

\frac{L}{Δ t} = \frac{L'}{Δ τ} = \frac{L'}{\sqrt{1 - β^{2}} Δ t}

(36.13)

The $Δ t$ cancels, and the distance $L'$ in frame $S'$ is

L' = \sqrt{1 - β^{2}} L

(36.14)

Surprisingly, we find that the distance between two objects in reference frame $S'$ is not the same as the distance between the same two objects in reference frame S.

Frame S, in which the distance is L, has one important distinction. It is the one and only inertial reference frame in which the objects are at rest. Experimenters in frame S can take all the time they need to measure L because the two objects aren’t going anywhere. The distance L between two objects, or two points on one object, measured in the reference frame in which the objects are at rest is called the proper length $ℓ$ . Only one inertial reference frame can measure the proper length.

We can use the proper length $ℓ$ to write Equation 36.14 as

L' = \sqrt{1 - β^{2}} ℓ \leq ℓ

(36.15)

This “shrinking” of the distance between two objects, as measured by an experimenter moving with respect to the objects, is called length contraction. Although we derived length contraction for the distance between two distinct objects, it applies equally well to the length of any physical object that stretches between two points along the x- and $x' -$ axes. The length of an object is greatest in the reference frame in which the object is at rest. The object’s length is less (i.e., the length is contracted) when it is measured in any reference frame in which the object is moving.

An ariel photograph of the Stanford linear accelerator. — The Stanford Linear Accelerator (SLAC) is a 2-mi-long electron accelerator. The accelerator’s length is less than 1 m in the reference frame of the electrons.

EXAMPLE 36.5 The distance from the sun to Saturn

In Example 36.4 a rocket traveled along a line from the sun to Saturn at a constant speed of 0.9c relative to the solar system. The Saturn-to-sun distance was given as $1.43 \times 10^{12} m$ . What is the distance between the sun and Saturn in the rocket’s reference frame?

MODEL Saturn and the sun are, at least approximately, at rest in the solar system reference frame S. Thus the given distance is the proper length $ℓ$ .

SOLVE We can use Equation 36.15 to find the distance in the rocket’s frame $S' :$

\begin{matrix} L' & = & \sqrt{1 - β^{2}} ℓ = \sqrt{1 - 0 . 9^{2}} (1.43 \times 10^{12} m) \\ = & 0.62 \times 10^{12} m \end{matrix}

REVIEW The sun-to-Saturn distance measured by the astronauts is less than half the distance measured by experimenters on earth. $L'$ and $ℓ$ are different because space is different in two reference frames moving relative to each other.

The conclusion that space is different in reference frames moving relative to each other is a direct consequence of the fact that time is different. Experimenters in both reference frames agree on the relative velocity v, leading to Equation 36.12: $v = L / Δ t = L' / Δ t'$ . We had already learned that $Δ t' < Δ t$ because of time dilation. Thus $L'$ has to be less than L. That is the only way experimenters in the two reference frames can reconcile their measurements.

To be specific, the earthly experimenters in Examples 36.4 and 36.5 find that the rocket takes 5300 s to travel the $1.43 \times 10^{12} m$ between the sun and Saturn. The rocket’s speed is $v = L / Δ t = 2.7 \times 10^{8} m / s = 0.9 c$ . The astronauts in the rocket find that it takes only 2310 s for Saturn to reach them after the sun has passed by. But there’s no conflict, because they also find that the distance is only $0.62 \times 10^{12} m$ . Thus Saturn’s speed toward them is $v = L' / Δ t' = (0.62 \times 10^{12} m) / (2310 s) = 2.7 \times 10^{8} m / s = 0.9 c$ .