Einstein's theory of special relativity describes the relationship between space and time; in particular, they are not independent quantities, but rather two sides of the same coin, namely spacetime. There are two postulates in special relativity:

These postulates are in direct contradiction with the traditional notions of absolute space and time that provided the background for pre-20th century physics, but experiments have proven them to be correct.

The coordinates defining the position of an event in spacetime depend on the choice of inertial reference frame. The two postulates mentioned earlier imply that these coordinates must transform in a certain way known as the Lorentz transformation when switching reference frames. Suppose that $S$ and $S\'$ are two reference frames with coordinates $\left(x\,y\,z\,t\right)$ and $\left(x\'\, y\'\, z\'\, t\'\right)$, such that the origin of both frames corresponds to the same event, and $S\'$ is moving at speed $v$ in the $x$-direction relative to $S$. The coordinates are related via:

In particular, if an object accelerates to a speed $v$ close to the speed of light, relative to an observer, then that observer will see that the object has shrunk by a factor of $\mathrm{\γ}$  relative to its rest length. Similarly, it will appear to be aging more slowly.

As mentioned earlier, suppose there is a pole of length 20 m and a barn of width 10 m.

Exactly how fast would the pole have to be going for it to fit inside the barn from the barn's perspective?

Let the common origin $\left(x\,t\right) \= \left(x\'\,t\'\right) \= \left(0\,0\right)$ in both frames be the event when the leading end of the pole gets to the back door of the barn (assuming the pole is moving through the barn from front to back). We hypothesize that at the same time in the barn's frame of reference, the trailing edge of the pole coincides with (has the same $x$-coordinate as) the front door of the barn. The coordinates for the second event in the barn's frame are $\left(x\,t\right) \=\left(-10\, 0\right)$, and in the pole's frame of reference they are $\left(x\'\,t\'\right)\= \left(-20\, t\'\right)$. Plugging these values into the second equation of the Lorentz transformation yields:

$-20 \mathrm{m} \= \mathrm{\γ} \left(-10 \mathrm{m}-v \cdot 0\right)$,

which requires $\mathrm{\γ}\=2$, and thus $v \= \frac{\sqrt{3}}{2}c\=0.866 c$.

From the pole's point of view, what time is it when the end of the pole finally makes it into the barn?

The answer is just the value of $t\'$ stated earlier. To find it, use the first equation in the Lorentz transformation:

$t\' \= 2 \left(0-\frac{\frac{\sqrt{3}}{2}c \cdot \left(-10 \mathrm{m}\right)}{c^2}\right)\=\frac{5 \sqrt{3}\mathrm{m}}{c}\= 1.67 \times {10}^{-8}\mathrm{s}$ .

Admittedly, this is a really short time after $t\'\=0$, when the front of the pole was already hitting the end of the barn, but given the speed involved it is still an important difference, and it proves that the pole did not quite fit into the barn (in the pole's reference frame).

From the pole's perspective, how much of the pole had still not fit into the barn when the front of the pole hit the back of the barn?

We need to find the $x\'$ coordinate at time $t\'\=0$, so we use the second equation of the inverse Lorentz transformation, noting that the $x$ coordinate for this event is -10 m:

$x \= \mathrm{\γ} \left(x\'\+v t\'\right)$.

Plugging in the numbers gives:

$-10 \mathrm{m}\= 2 \left(x\'\+0\right)$,
$x\' \= -5 \mathrm{m}$.

In other words, from the pole's perspective, only one quarter of the pole can fit into the barn!