# Length contraction

### Proper lengths and distances

When we speak of lengths in particular reference frames, we mean space-time separations between events that are simultaneous in that reference frame.

The proper length, $\lambda$, of an object is its length as measured in the reference frame in which it is at rest. This means that the time components of the two events that represent the endpoints of the rod are the same at any particular time.

Figure 4.8 - Length contraction

Suppose we have a rod of proper length $\lambda$, say, at rest in a stationary frame as shown by the dashed lines in figure 4.8, with one end at the origin and the other at $x = \lambda$ along the $x$-axis.

In order to measure the length of the rod in a frame moving at speed $v$ say, we need to measure it along a line of simultaneity in the frame, in this case on the $x'$-axis. Specifically, at the event $(t' = 0, x' = l')$.

Since $$t - vx = 0$$ on the $x'$-axis, this event has coordinates $(t = v \lambda, x = \lambda)$ in the rest frame.

Thus, the space-time separation between the two endpoints is $$-l'^2 = S^2 = \left(v \lambda\right)^2 - \lambda^2$$ and so the length, $l'$, as measured in the moving frame is $$l' = \sqrt{1 - v^2}\lambda = \frac{\lambda}{\gamma} \lt \lambda$$ hence the coordinate length (in the moving frame in this case) is less than the proper length (rest frame).

This is called length contraction. Note that this means that the proper length is greater than any coordinate length. As the speed of a coordinate observer increases to speeds close to the speed of light, the length of the rod in the rest frame decreases to zero.

Nb. We have drawn in the hyperbola $-l'^2 = t^2 - x^2$ to calibrate the two $x$-axes so that we can compare the lengths $\lambda$ and $l'$.

Figure 4.9 - Length contraction again

We could have considered a rod of length $\lambda$ at rest in the moving frame, as in figure 4.9.

The endpoint has coordinates $$(t' = 0, x' = \lambda)$$ in the moving frame and, using the Lorentz transformations, $$(t = \gamma v \lambda, x = \gamma \lambda)$$ in the rest frame.

The world-line of the endpoint, $x - vt = l$ cuts the line of simultaneity ($x$-axis) at $x = l$ so we get \begin{align*}l &= \left(\gamma \lambda\right) - v\left(\gamma v \lambda\right)\\&= \gamma \left(1 - v^2\right) \lambda\\&= \sqrt{1 - v^2}\lambda\\&= \frac{\lambda}{\gamma} \lt \lambda\end{align*} and we end up with length contraction again.

Is it possible for a 20ft car to fit into a 16ft garage?

From the garage's point of view, the answer is yes, if we make the car is travel at speed $v = \tfrac{3}{5}$ (by which we mean $\tfrac{3}{5}$th's of the speed of light). Then, $$\text{length of car } = \sqrt{1 - \left(\tfrac{3}{5}\right)^2}\left(20\text{ft}\right) = \frac{4}{5}\left(20\text{ft}\right) = 16\text{ft} = \text{ length of garage}$$

But, from the car's frame, where the garage is travelling at speed $v = -\tfrac{3}{5}$, the answer is no, since $$\text{length of garage} = \sqrt{1 - \left(\tfrac{3}{5}\right)^2}\left(16\text{ft}\right) = \frac{4}{5}\left(16\text{ft}\right) = 12.8\text{ft} \lt 20\text{ft} = \text{ length of car}$$

The solution to this ‘paradox’ is, as usual for these type of problems, the breakdown of simultaneity.

Figure 4.10 - Garage frame

In the garage frame, we are comparing the coordinate length of the car to the proper length of the garage.

In figure 4.10, the red strip represents the car moving at speed $v$ in the $x$-direction of the garage, represented by the blue strip. The bars represent the garage and the car as laid out in their own frames, at the time when the back of the car enters the garage.

The back of car enters the garage ($E_3$) and the front of car exits the garage ($E_2$) simultaneously in this frame, hence we can say that the car fits in the garage.

Nb. We have opened up the back of the garage in the diagram!!

Figure 4.11 - Car frame

In the car frame, we are comparing proper length of the car to the coordinate length of the garage.

In figure 4.11, the blue strip represents the garage moving at speed $-v$ in the $x$-direction of the car, represented by the red strip.

Now the key events $E_3, E_2$ are not simultaneous, in fact $E_2$ occurs before $E_3$ meaning the front of the car exits the garage before the back of the car enters.

There is no time in this frame where both the front and the back are in the garage simultaneously.