মই প্যারাডক্স

Suppose you have a ladder and a garage, both of proper length $L.$ A man carrying the ladder moves toward the garage in an attempt to fit the ladder inside the garage, while his friend observes him from the side. Suppose that the man with the ladder is somehow moving with a very high speed $v,$ such that we must consider relativistic effects.

The friend, standing in the inertial frame of the garage, sees the man and ladder traveling with speed $v.$ Due to length contraction, the ladder appears to be shorter than the garage; thus, the friend concludes that the ladder should fit comfortably inside the garage.

The man, in the inertial frame of the ladder, instead sees his friend and the garage moving backward with speed $v.$ Therefore, due to length contraction, the garage appears to be shorter than the ladder; thus, the man concludes that the ladder will never be entirely inside the garage.

We believe that "The ladder fits inside the garage" should be an unambiguous true-or-false statement. In fact, suppose that the friend could activate doors at the entrance and exit--he can close the doors simultaneously while the ladder is fully inside, to demonstrate that the ladder fits! (The friend will then open the doors again once the ladder is about to exit the garage.) How could the two men possibly disagree on whether a door is open or closed?

So how can two observers disagree on the statement merely due to their speeds? This is the ladder paradox. You are encouraged to think about the paradox for a few minutes until you figure it out or are completely stumped before continuing.

The answer to the ladder paradox is actually quite simple--we can find it by examining special relativity more closely. Along with length contraction, another consequence of special relativity is the breaking of simultaneity! This is important because, when we measure the length of objects, we must measure the positions of both ends of the object at the same time.

The first step is to carefully define the notion of the ladder "fitting inside" the garage. The consistent way of doing this is as follows: there is some time at which both ends of the ladder lie between the ends of the garage.

In this problem, this means that the front end of the ladder passes through the garage exit after the back end of the ladder passes through the garage entrance. Call these two events $A$ and $B$ respectively. These events occur at times $t_A$ and $t_B,$ respectively, in the garage reference frame, and they occur at times $t'_A$ and $t'_B,$ respectively, in the ladder reference frame.

Suppose, for convenience, that the two people synchronize their clocks to $t=t'=0$ at the instant at which the front end of the ladder passes through the garage entrance. Now, let us calculate the times $t_A, t_B, t'_A, t'_B.$

The man's friend sees the ladder contracted to a length $L' = \frac{L}{\gamma}$ (where $\gamma = \frac{1}{\sqrt{1-v^2/c^2}} > 1$ for any $v\gt 0$). Thus, event $B$ occurs after the ladder has moved a distance $L'.$ Therefore, $t_B = \frac{L}{v\gamma}.$

Now, according to the friend, the man and the ladder must still travel a distance $L$ to cover the length of the garage. Thus, $t_A = \frac{L}{v} > t_B.$ It follows that, indeed, the front end of the ladder passes through the garage exit after the back end of the ladder passes through the garage entrance. So according to the friend, the ladder fits!

However, the man himself sees the garage contracted to a length $L' = \frac{L}{\gamma}.$ Thus, event $A$ occurs after the garage has moved backward a distance $L'$ at a speed $v.$ Therefore, $t'_A = \frac{L}{v\gamma}.$

The man still sees the ladder as having length $L.$ Thus, for the garage entrance to coincide with the back end of the ladder, the garage must travel a distance $L.$ Therefore, $t'_B = \frac{L}{v} > t'_A.$ It follows that the man sees the front end of the ladder passes through the garage exit before the back end of the ladder passes through the garage entrance! Thus, according to the man carrying the ladder, the ladder does not fit inside the garage.

The breaking of simultaneity is the key here--this is how the two men can disagree without contradiction. And, as one final question, you may ask: but what about the doors? We mentioned those earlier, and suppose that the friend closes the doors while the ladder is fully inside! Well, the breaking of simultaneity fixes this too. In the reference frame of the garage, the doors close simultaneously. But according to the man carrying the ladder, the door at the exit closes first, before the door at the entrance! Then, because the friend opens the doors afterward, the door at the exit opens (allowing the ladder to pass through the exit), and this occurs before the door at the entrance closes at all (at least in the ladder's reference frame)! This allows the ladder to pass through the garage without ever fitting but also without ever colliding with the doors.