The ladder paradox (or barn-pole paradox) is a thought experiment in special relativity. It involves a ladder travelling horizontally and undergoing a length contraction, the result of which being that it can fit into a much smaller garage. On the other hand, from the point of view of an observer moving with the ladder, it is the garage that is moving and the garage will be contracted to an even smaller size, therefore being unable to contain the ladder at all. This apparent paradox results from the assumption of absolute simultaneity. In relativity, simultaneity is relative to each observer and thus the ladder can fit into the garage in both instances.

Figure 1: An overview of the garage and the ladder at rest

The problem starts with a ladder and an accompanying garage that is too small to contain the ladder. Through the relativistic effect of length contraction, the ladder can be made to fit into the garage by running it into the garage at a high enough speed.

Figure 2: In the garage frame, the ladder undergoes length contraction and will therefore fit into the garage.

However, both the ladder and garage occupy their own inertial reference frames, and thus both frames are equally valid frames to view the problem. From the reference frame of the garage, it is the ladder that moves with a relative velocity and so it is the ladder that undergoes length contraction.

Conversely, through symmetry, from the reference frame of the ladder it is the garage that is moving with a relative velocity and so it is the garage that undergoes a length contraction. From this perspective, the garage is made even smaller and it is impossible to fit the ladder into the garage.

Figure 3: In the ladder frame, the garage undergoes length contraction and seems too small to contain the ladder.

## Relative simultaneity

Figure 4: Scenario in the garage frame: a length contracted ladder entering and exiting the garage
Figure 5: Scenario in the ladder frame: a length contracted garage passing over the ladder

The solution to the apparent paradox lies in the fact that what one observer (e.g., the garage) considers as simultaneous does not correspond to what the other observer (e.g., the ladder) considers as simultaneous (relative simultaneity). A clear way of seeing this is to consider a garage with two doors that swing shut to contain the ladder and then open again to let the ladder out the other side.

From the perspective of the garage, the length-contracted ladder is short enough to fit entirely inside. The instant the ladder is fully inside the garage, the front and back doors close simultaneously. Then, since the ladder is still moving at considerable speed, the front and back doors simultaneously open again to allow the ladder to exit.

From the perspective of the ladder, the back (right) door closes and opens, then after the garage passes over the ladder, the front (left) door closes and opens.

The situation is illustrated in the Minkowski diagram below. The diagram is in the rest frame of the garage. The vertical light-blue-shaded band shows the garage in space-time, the light-red band shows the ladder in space-time. The x and t axes are the garage space and time axes, respectively, and x′ and t′ are the ladder space and time axes, respectively. The ladder is moving at a velocity of $v=c\sqrt{1/2}$ in the positive x direction, therefore $\gamma=\sqrt{2}$. (From the ladder's point of view, its speed in the x′ direction is the same.)

Since light travels at very close to one foot per nanosecond, let’s work in these units, so that $c \approx 1 \mbox{ft/ns}$. The garage is a small one, G=10 feet long, while in the ladder frame, the ladder is L=12 feet long. In the garage frame, the front of the ladder will hit the back of the garage at time $t_A=G/v\approx14.14\mbox{ ns}$ (if $t_D = t_O = 0$ is chosen as the reference point). This is shown as event A on the diagram. All lines parallel to the garage x axis will be simultaneous according to the garage observer, so the dark blue line AB will be what the garage observer sees as the ladder at the time of event A. The ladder is contained inside the garage. However, from the point of view of the observer on the ladder, the dark red line AC is what the ladder observer sees as the ladder. The back of the ladder is outside the garage.

Figure 6: A Minkowski diagram of 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 x and x′ axes are each 5 feet (1.5 m) long in their respective frames, and the t and t′ axes are each 5 ns in duration.

## Resolution

Figure 7: A ladder contracting under acceleration to fit into a length contracted garage

In the context of the paradox, when the ladder enters the garage and is contained within it, it must either continue out the back or come to a complete stop. When the ladder comes to a complete stop, it accelerates into the reference frame of the garage. From the reference frame of the garage, all parts of the ladder come to a complete stop. We shall assume that all parts of the ladder accelerate simultaneously.[1][2]

From the reference frame of the ladder, it is the garage that is moving, and so in order to be stopped with respect to the garage, the ladder must accelerate into the reference frame of the garage. All parts of the ladder cannot accelerate simultaneously because of relative simultaneity. What happens is that each part of the ladder accelerates sequentially,[1][3] front to back, until finally the back end of the ladder accelerates when it is within the garage, the result of which is that, from the reference frame of the ladder, the front parts undergo length contraction sequentially until the entire ladder fits into the 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.

What if the back door (the door the ladder exits out of) is closed permanently and does not open? Suppose that the door is so solid that the ladder will not penetrate it when it collides, so it must stop. Then, as in the scenario described above, in the frame of reference of the garage, there is a moment when the ladder is completely within the garage (i.e., the back of the ladder is inside the front door), before it collides with the back door and stops. However, from the frame of reference of the ladder, the ladder is too big to fit in the garage, so by the time it collides with the back door and stops, the back of the ladder still has not reached the front door. This seems to be a paradox. The question is, does the back of the ladder cross the front door or not?

The difficulty arises mostly from the assumption that the ladder is rigid (i.e., maintains the same shape). Ladders seem pretty rigid in everyday life. But being rigid requires that it can transfer force at infinite speed (i.e., when you push one end the other end must react immediately, otherwise the ladder will deform). This contradicts special relativity, which states that information can only travel at most the speed of light (which is too fast for us to notice in real life, but is significant in the ladder scenario). So objects cannot be perfectly rigid under special relativity.

In this case, by the time the front of the ladder collides with the back door, the back of the ladder does not know it yet, so it keeps moving forwards (and the ladder "compresses"). In both the frame of the garage and the inertial frame of the ladder, the back end keeps moving at the time of the collision, until at least the point where the back of the ladder comes into the light cone of the collision (i.e., a point where force moving backwards at the speed of light from the point of the collision will reach it). At this point the ladder is actually shorter than the original contracted length, so the back end is well inside the garage. Calculations in both frames of reference will show this to be the case.

What happens after the force reaches the back of the ladder (the "green" zone in the diagram) is not specified. Depending on the physics, the ladder could break into a million pieces; or, if it were sufficiently elastic, it could re-expand to its original length and push the back end out of the garage.

## Man falling into grate variation

A man (represented by a segmented rod) falling into a grate

This paradox was originally proposed and solved by Wolfgang Rindler[1] and involved a fast walking man, represented by a rod, falling into a grate.[4] It is assumed that the rod is entirely over the grate in the grate frame of reference before the downward acceleration begins simultaneously and equally applied to each point in the rod.

From the perspective of the grate, the rod undergoes a length contraction and fits into the grate. However, from the perspective of the rod, it is the grate undergoing a length contraction, through which it seems the rod is then too long to fall.

In fact, the downward acceleration of the rod, which is simultaneous in the grate's frame of reference, is not simultaneous in the rod's frame of reference. In the rod's frame of reference, the bottom of the front of the rod is first accelerated downward (not shown in drawing), and as time goes by, more and more of the rod is subjected to the downward acceleration, until finally the back of the rod is accelerated downward. This results in a bending of the rod in the rod's frame of reference. It should be stressed that, since this bending occurs in the rod's rest frame, it is a true physical distortion of the rod which will cause stresses to occur in the rod.

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.

The above paradox is complicated: It involves non-inertial frames of reference since at one moment the man is walking horizontally, and a moment later he is falling downward. It involves a physical deformation of the man (or segmented rod), since the rod is bent in one frame of reference and straight in another. These aspects of the problem introduce complications involving the stiffness of the bar which tends to obscure the real nature of the "paradox". A very similar but simpler problem involving only inertial frames is the "bar and ring" paradox (Ferraro 2007) in which a bar which is slightly larger in length than the diameter of a ring is moving upward and to the right with its long axis horizontal, while the ring is stationary and the plane of the ring is also horizontal. If the motion of the bar is such that the center of the bar coincides with the center of the ring at some point in time, then the bar will be Lorentz-contracted due to the forward component of its motion, and it will pass through the ring. The paradox occurs when the problem is considered in the rest frame of the bar. 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. How can the bar pass through the ring?

The resolution of the paradox again lies in the relativity of simultaneity (Ferraro 2007). The length of a physical object is defined as the distance between two simultaneous events occurring at each end of the body, and since simultaneity is relative, so is this length. This variability in length is just the Lorentz contraction. Similarly, a physical angle is defined as the angle formed by three simultaneous events, and this angle will also be a relative quantity. In the above paradox, although the rod and the plane of the ring are parallel in the rest frame of the ring, they are not parallel in the rest frame of the rod. The uncontracted rod passes through the Lorentz-contracted ring because the plane of the ring is rotated relative to the rod by an amount sufficient to let the rod pass through.

In mathematical terms, a Lorentz transformation can be separated into the product of a spatial rotation and a "proper" Lorentz transformation which involves no spatial rotation. The mathematical resolution of the bar and ring paradox is based on the fact that the product of two proper Lorentz transformations may produce a Lorentz transformation which is not proper, but rather includes a spatial rotation component.