Ladder paradox

Resolution

The solution to the apparent paradox lies in the relativity of simultaneity: what one observer (e.g. with the garage) considers to be two simultaneous events may not in fact be simultaneous to another observer (e.g. with the ladder). When we say the ladder "fits" inside the garage, what we mean precisely is that, at some specific time, the position of the back of the ladder and the position of the front of the ladder were both inside the garage; in other words, the front and back of the ladder were inside the garage simultaneously. As simultaneity is relative, then, two observers disagree on whether the ladder fits. To the observer with the garage, the back end of the ladder was in the garage at the same time that the front end of the ladder was, and so the ladder fit; but to the observer with the ladder, these two events were not simultaneous, and the ladder did not fit.

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 now look at these events in the frame of the ladder. The first event is the front of the ladder approaching the exit door of the garage. The door closes, and then opens again to let the front of the ladder pass through. At a later time, the back of the ladder passes through the entrance door, which closes and then opens. 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.

The situation can be further illustrated by the Minkowski diagram below. The diagram is in the rest frame of the garage. The vertical light-blue band shows the garage in spacetime, and the light-red band shows the ladder in spacetime. 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.

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. One example is the bold blue line segment, which lies inside the blue band representing the garage, and which represents the ladder at a time when it is fully inside the garage. In the frame of the ladder, however, sets of simultaneous events lie on lines parallel to the x′ axis; the ladder at any specific time is therefore represented by a cross section of such a line with the red band. One such example is the bold red line segment. We see that such line segments never lie fully inside the blue band; that is, the ladder never lies fully inside the garage.

Shutting the ladder in the garage

In a more complicated version of the paradox, we can physically trap the ladder once it is fully inside the garage. This could be done, for instance, by not opening the exit door again after we close it. 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. By this time, the entrance door has also closed, and so the ladder is stuck inside the garage. 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.

Again, the puzzle comes from considering the situation from the frame of the ladder. In the above analysis, in its own frame, the ladder was always longer than the garage. So how did we ever close the doors and trap it inside?

It is worth noting here a general feature of relativity: we have deduced, by considering the frame of the garage, that we do indeed trap the ladder inside the garage. This must therefore be true in any frame – it cannot be the case that the ladder snaps in one frame but not in another. From the ladder's frame, then, we know that there must be some explanation for how the ladder came to be trapped; we must simply find the explanation.

The explanation is that, although all parts of the ladder simultaneously decelerate to zero in the garage's frame, because simultaneity is relative, the corresponding decelerations in the frame of the ladder are not simultaneous. Instead, each part of the ladder decelerates sequentially, from front to back, until finally the back of the ladder decelerates, by which time it is already within the garage.

As length contraction and time dilation are both controlled by the Lorentz transformations, the ladder paradox can be seen as a physical correlate of the twin paradox, in which instance one of a set of twins leaves earth, travels at speed for a period, and returns to earth a bit younger than the earthbound twin. As in the case of the ladder trapped inside the barn, if neither frame of reference is privileged – each is moving only relative to the other – how can it be that it's the traveling twin and not the stationary one who is younger (just as it is the ladder rather than the barn which is shorter)? In both instances it is the acceleration-deceleration that differentiates the phenomena: it is 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.

Ladder paradox and transmission of force

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 rigid in everyday life. But being completely 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 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). 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; 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.

Man falling into grate variation

This early version of the paradox was originally proposed and solved by Wolfgang Rindler and involved a fast walking man, represented by a rod, falling into a grate. 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.

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 front of the rod is first accelerated downward (shown in cell 3 of the 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. 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.

For this non-rigid behaviour of the rod to become apparent, both the rod itself and the grate must be of such a scale that the traversal time is measurable.

Bar and ring paradox

A problem very similar but simpler than the rod and grate paradox, involving only inertial frames, is the "bar and ring" paradox. The rod and grate 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; and 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 rod which tends to obscure the real nature of the "paradox". The "bar and ring" paradox is free of these complications: 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. 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 (horizontal and vertical) may produce a Lorentz transformation which is not proper (diagonal) but rather includes a spatial rotation component.