To a stationary observer, due to the contraction, the moving ladder is able to fit entirely inside the building as it passes through.
Because of its high velocity, the ladder undergoes the relativistic effect of length contraction, and becomes significantly shorter.
As a result, as the ladder passes through the garage, it is, for a time, completely contained inside it.
As an observer moving with the ladder is travelling at constant velocity in the inertial reference frame of the garage, this observer also occupies an inertial frame, where, by the principle of relativity, the same laws of physics apply.
A clear way of seeing this is to consider the doors, which, in the frame of the garage, close for the brief period that the ladder is fully inside.
We see that, as simultaneity is relative, the two doors did not need to be shut at the same time, and the ladder did not need to fit inside the garage.
In the frame of the garage, the ladder at any specific time is represented by a horizontal set of points, parallel to the x axis, in the red band.
In a more complicated version of the paradox, we can physically trap the ladder once it is fully inside the garage.
In the frame of the garage, we assume the exit door is immovable, and so when the ladder hits it, we say that it instantaneously stops.
As its relative velocity is now zero, it is not length contracted, and is now longer than the garage; it will have to bend, snap, or explode.
In both instances it is the acceleration-deceleration that differentiates the phenomena: it's the twin, not the earth (or the ladder, not the barn) that undergoes the force of deceleration in returning to the temporal (or physical, in the case of the ladder-barn) inertial frame.
This contradicts special relativity, which states that information can travel no faster than the speed of light (which is too fast for us to notice in real life, but is significant in the ladder scenario).
At this point the ladder is actually shorter than the original contracted length, so the back end is well inside the garage.
Depending on the physics, the ladder could break; or, if it were sufficiently elastic, it could bend and re-expand to its original length.
At sufficiently high speeds, any realistic material would violently explode into a plasma.
This early version of the paradox was originally proposed and solved by Wolfgang Rindler[1] and involved a fast walking man, represented by a rod, falling into a grate.
These aspects of the problem introduce complications involving the stiffness of the rod which tends to obscure the real nature of the "paradox".
The ring is now moving downward and to the left, and will be Lorentz-contracted along its horizontal length, while the bar will not be contracted at all.
The mathematical resolution of the bar and ring paradox is based on the fact that the product of two proper Lorentz transformations (horizontal and vertical) may produce a Lorentz transformation which is not proper (diagonal) but rather includes a spatial rotation component.
Figure 1: An overview of the garage and the ladder at rest
Figure 2: In the garage frame, the ladder undergoes length contraction and will therefore fit into the garage.
Figure 3: In the ladder frame, the garage undergoes length contraction and is too small to contain the ladder.
Figure 4: Scenario in the garage frame: a length contracted ladder passing through the garage
Figure 5: Scenario in the ladder frame: a length contracted garage passing over the ladder. Only one door is closed at any time
Figure 6: A Minkowski diagram of the ladder paradox. The garage is shown in light blue, the ladder in light red. The diagram is in the rest frame of the garage, with x and t being the garage space and time axes, respectively. The ladder frame is for a person sitting on the front of the ladder, with x
′
and t
′
being the ladder space and time axes respectively. The blue and red lines, AB and AC, depict the ladder at the time when its front end meets the garage's exit door, in the frame of reference of the garage and the ladder, respectively. Event D is the rear end of the ladder reaching the garage's entrance.
Figure 7: A ladder contracting under acceleration to fit into a length contracted garage
Figure 8: A Minkowski diagram of the case where the ladder is stopped all along its length, simultaneously in the garage frame. When this occurs, the garage frame sees the ladder as AB, but the ladder frame sees the ladder as AC. When the back of the ladder enters the garage at point D, it has not yet felt the effects of the acceleration of its front end. At this time, according to someone at rest with respect to the back of the ladder, the front of the ladder will be at point E and will see the ladder as DE. It is seen that this length in the ladder frame is not the same as CA, the rest length of the ladder before the deceleration.
Figure 1: A Minkowski diagram of the case where the ladder is stopped by impact with the back wall of the garage. The impact is event A. At impact, the garage frame sees the ladder as AB, but the ladder frame sees the ladder as AC. The ladder does not move out of the garage, so its front end now goes directly upward, through point E. The back of the ladder will not change its trajectory in spacetime until it feels the effects of the impact. The effect of the impact can propagate outward from A no faster than the speed of light, so the back of the ladder will never feel the effects of the impact until point F (note the 45° angle of the line A-F, corresponding to the speed of light transmission of information) or later, at which time the ladder is well within the garage in both frames. Note that when the diagram is drawn in the frame of the ladder, the speed of light is the same, but the ladder is longer, so it takes more time for the force to reach the back end; this gives enough time for the back of the ladder to move inside the garage.
A man (represented by a segmented rod) falling into a grate
The diagram on the left illustrates a bar and a ring in the rest frame of the ring at the instant that their centers coincide. The bar is Lorentz-contracted and moving upward and to the right while the ring is stationary and uncontracted. The diagram on the right illustrates the situation at the same instant, but in the rest frame of the bar. The ring is now Lorentz-contracted and rotated with respect to the bar, and the bar is uncontracted. Again, the ring passes over the bar without touching it.
