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Ladder paradox - Wikipedia, the free encyclopedia

Ladder paradox

From Wikipedia, the free encyclopedia

The ladder paradox (or barn-pole paradox) is a thought experiment in special relativity. If a ladder travels horizontally it will undergo a length contraction and will therefore fit into a garage that is shorter than the ladder's length at rest. 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. The garage will therefore need to be larger than the length at rest of the ladder in order to contain it.

Suppose we have a ladder that is longer than the length of the garage, but it is traveling at a fast enough speed that, from the frame of reference of the garage, its length is contracted to less than the length of the garage. Then from the frame of reference of the garage, there is a moment in time when the ladder can fit completely inside the garage, and during that moment one can close and open both front and back doors of the garage, without affecting the ladder.

However, from the frame of reference of the ladder, the garage is much shorter than the length of the ladder (it is already shorter at rest, plus it is contracted because it is moving with respect to the ladder). Therefore there is no moment in time when the ladder is completely inside the garage; and there is no moment when one can close both doors without it hitting the ladder. This is an apparent paradox.

Figure 1a: The problem, part 1 - A length contracted ladder being contained within a garage
Figure 1a: The problem, part 1 - A length contracted ladder being contained within a garage
Figure 1b: The problem, part 2 - A length contracted garage being too small to contain the same long ladder
Figure 1b: The problem, part 2 - A length contracted garage being too small to contain the same long ladder


Contents

[edit] Relative simultaneity

Figure 1c: The solution from the garage's point of view: A length contracted ladder entering and exiting a two-door garage
Figure 1c: The solution from the garage's point of view: A length contracted ladder entering and exiting a two-door garage

The solution to this dilemma 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.

Figure 1d: The solution from the ladder's point of view: A length contracted two-door garage passing over a ladder
Figure 1d: The solution from the ladder's point of view: A length contracted two-door garage passing over a ladder

From the perspective of the ladder what happens is that first the back door closes and opens and then, after the garage passes over the ladder, the front 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 tD = tO = 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 1e: A Minkowski diagram of ladder paradox No. 1. 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 long in their respective frames, and the t and t′ axes are each 5 ns in duration.
Figure 1e: A Minkowski diagram of ladder paradox No. 1. 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 long in their respective frames, and the t and t′ axes are each 5 ns in duration.


Using the one-dimensional Lorentz transformation (see special relativity) we can get:

Event Garage Frame Ladder Frame
A\, t_A=\frac{G}{v}\approx 14.14\mbox{ ns}\,\!

x_A=L=10\mbox{ ft}\,\!

t'_A=\frac{G}{\gamma v}=10\mbox{ ns}\,\!

x'_A=0\mbox{ ft}\,\!

B\, t_B=t_A\approx14.14\mbox{ ns}\,\!

x_B=G-\frac{L}{\gamma}\approx 1.514\mbox{ ft}\,\!

t'_B=18.48\mbox{ ns}\,\!

x'_B=-L=-12\mbox{ ft}\,\!

C\, t_C\approx2.142\mbox{ ns}\,\!

x_C\approx-6.971\mbox{ ft}\,\!

t'_C=10\mbox{ ns}\,\!

x'_C=-12\mbox{ ft}\,\!

D\, t_D=12\mbox{ ns}\,\!

x_D=0\mbox{ ft}\,\!

t'_D=16.97\mbox{ ns}\,\!

x'_D=-12\mbox{ ft}\,\!

[edit] Relativistic trains passing

Another dramatisation of the same effect is to consider two relativistic trains approaching each other along a single track, with only a short length of double track for the trains to pass, at a station platform.

The platform is shorter than the rest length of the trains (just as the garage is shorter than the ladder); but the trains can just pass (from the perspective of an observer on the platform), thanks to the relativistic length contraction: the head of each train reaches the end of the platform just as the tail of the other train clears it, going in the other direction.

From the train perspective, it is now the platform which is contracted. As before, the nose of the train reaches the end of the platform just as the tail of the other train clears it. However, in this perspective, the train still has its full rest-length. So the tail of the first train is still in the section of single-track - the front of the platform has not yet reached it. But there is no collision, because the oncoming second train is even more length contracted than the platform, and so its nose has not yet reached the front of the platform. There is just time for tail of the first train to clear the front of the platform, as the nose of the second train reaches it.

So a crash is again averted, in this frame because of the relativity of simultaneity: what is simultaneous to an observer standing on the platform is not simultaneous for an observer on one of the trains.

[edit] Resolution of the paradox if the ladder collides with the back door

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 space-time 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 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.
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 space-time 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 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.

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 the 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 hasn't 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 real 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 pretty 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 doesn't know it yet, so it keeps moving forwards (and the ladder kind of "compresses"). In both the frame of the garage and the initial 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.


[edit] Resolution of the paradox if the entire ladder stops simultaneously

Figure 1 - 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 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.

The explanation in the previous section depends on the ladder compressing because of the finite speed of propagation of force. One (failed) attempt to get around this is to say, what if somehow the ladder along its entire length simultaneously stops (not due to the force of the collision, but just spontaneously) in the frame of the garage (line AB in the diagram). Then in the frame of the garage the ladder does not need to compress.

However, from the perspective of the ladder this cannot be true (again, because simultaneity is relative). In the frame of the ladder, when its front hits the back door, the ladder is AC. In this frame, the back end hasn't stopped yet. In the frame of the ladder, the pieces of the ladder stops sequentially from front to back. The ladder does not compress as much as in the previous section, but it still compresses enough that the back of the ladder is within the garage by the time that the back end stops.

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


[edit] Man falling into grate variation

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

This paradox was originally proposed and solved by Wolfgang Rindler ("Length Contraction Paradox": Am. J. Phys., 29(6) June 1961) and involved a fast walking man, represented by a rod, falling into a grate.

From the perspective of the grate the man undergoes a length contraction and falls into the grate. However from the perspective of the man it is the grate undergoing a length contraction and seems too small for the man to fall through.

Using the solution of relative simultaneity, we can see that from the perspective of the grate every part of the man falls into the grate at the same time while from the perspective of the man his front falls into the grate before his rear, as can be easily seen if the man is represented by a segmented rod. This causes him to bend into the grate, which is similar to the ladder contracting to fit into the garage. In fact, if the man hits a bar of the grate as he falls in, since he is travelling forwards as well as falling and hitting the bar brings him to a halt, he undergoes the same contraction that the ladder experiences as well as bending into the grate.

[edit] Recent criticism

However, the above "man falling into grate" solution has recently been criticised in an article by Van Lintel and Gruber in the Eur.J.Phys.26 (Jan. 2005), p.19: The earlier solution is interpreted to imply that proper stiffness could be affected by relative speed. The more recent solution is that the rod's (unaffected) proper stiffness is related to the thickness of the rod, which thickness implies a time delay before any part of the upper surface of the rod can start falling. In this particular case, the rod will even arrive at the other side of the hole before any part of the upper surface "feels" the effect of the hole.



[edit] See also

[edit] References

  • Rindler, Wolfgang (2001). Relativity: Special, General and Cosmological. Oxford University Press. ISBN 0-19-850836-0. 

External Link:
Special Relativity Animations The animated train-and-tunnel paradox is an analog of the pole (train) and barn (tunnel) paradox.

[edit] Further reading

  • Edwin F. Taylor and John Archibald Wheeler, Spacetime Physics (2nd ed) (Freeman, NY, 1992)
- discusses various apparent SR paradoxes and their solutions


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