Archive-name: physics-faq/part4 Last-modified: 02-FEB-1995
These are the props. You own a barn, 40m long, with automatic doors at either end, that can be opened and closed simultaneously by a switch. You also have a pole, 80m long, which of course won't fit in the barn.
Now someone takes the pole and tries to run (at nearly the speed of light) through the barn with the pole horizontal. Special Relativity (SR) says that a moving object is contracted in the direction of motion: this is called the Lorentz Contraction. So, if the pole is set in motion lengthwise, then it will contract in the reference frame of a stationary observer.
You are that observer, sitting on the barn roof. You see the pole coming towards you, and it has contracted to a bit less than 40m. So, as the pole passes through the barn, there is an instant when it is completely within the barn. At that instant, you close both doors. Of course, you open them again pretty quickly, but at least momentarily you had the contracted pole shut up in your barn. The runner emerges from the far door unscathed.
But consider the problem from the point of view of the runner. She will regard the pole as stationary, and the barn as approaching at high speed. In this reference frame, the pole is still 80m long, and the barn is less than 20 meters long. Surely the runner is in trouble if the doors close while she is inside. The pole is sure to get caught.
Well does the pole get caught in the door or doesn't it? You can't have it both ways. This is the "Barn-pole paradox." The answer is buried in the misuse of the word "simultaneously" back in the first sentence of the story. In SR, that events separated in space that appear simultaneous in one frame of reference need not appear simultaneous in another frame of reference. The closing doors are two such separate events.
SR explains that the two doors are never closed at the same time in the runner's frame of reference. So there is always room for the pole. In fact, the Lorentz transformation for time is 
. It's the v*x term in the numerator that causes the mischief here. In the runner's frame the further event (larger x) happens earlier. The far door is closed first. It opens before she gets there, and the near door closes behind her. Safe again - either way you look at it, provided you remember that simultaneity is not a constant of physics.
References: Taylor and Wheeler's Spacetime Physics is the classic. Feynman's Lectures are interesting as well.
Item 24. Special Relativistic Paradoxes - part (b)
A Short Story about Space Travel:
Two twins, conveniently named A and B, both know the rules of Special Relativity. One of them, B, decides to travel out into space with a velocity near the speed of light for a time T, after which she returns to Earth. Meanwhile, her boring sister A sits at home posting to Usenet all day. When B finally comes home, what do the two sisters find? Special Relativity (SR) tells A that time was slowed down for the relativistic sister, B, so that upon her return to Earth, she knows that B will be younger than she is, which she suspects was the the ulterior motive of the trip from the start.
But B sees things differently. She took the trip just to get away from the conspiracy theorists on Usenet, knowing full well that from her point of view, sitting in the spaceship, it would be her sister, A, who was travelling ultrarelativistically for the whole time, so that she would arrive home to find that A was much younger than she was. Unfortunate, but worth it just to get away for a while.
What are we to conclude? Which twin is really younger? How can SR give two answers to the same question? How do we avoid this apparent paradox? Maybe twinning is not allowed in SR? Read on.
Paradox Resolved:
Much of the confusion surrounding the so-called Twin Paradox originates from the attempts to put the two twins into different frames --- without the useful concept of the proper time of a moving body.
SR offers a conceptually very clear treatment of this problem. First chose _one_ specific inertial frame of reference; let's call it S. Second define the paths that A and B take, their so-called world lines. As an example, take (ct,0,0,0) as representing the world line of A, and (ct,f(t),0,0) as representing the world line of B (assuming that the the rest frame of the Earth was inertial). The meaning of the above notation is that at time t, A is at the spatial location (x1,x2,x3)=(0,0,0) and B is at (x1,x2,x3)=(f(t),0,0) - always with respect to S.
Let us now assume that A and B are at the same place at the time t1 and again at a later time t2, and that they both carry high-quality clocks which indicate zero at time t1. High quality in this context means that the precision of the clock is independent of acceleration. [In principle, a bunch of muons provides such a device (unit of time: half-life of their decay).]
The correct expression for the time T such a clock will indicate at time t2 is the following [the second form is slightly less general than the first, but it's the good one for actual calculations]:
(1)
where 
is the so-called proper-time interval, defined by
.
Furthermore, 
is the velocity vector of the moving object. The physical interpretation of the proper-time interval, namely that it is the amount the clock time will advance if the clock moves by dx during dt, arises from considering the inertial frame in which the clock is at rest at time t - its so-called momentary rest frame (see the literature cited below). [Notice that this argument is only of heuristic value, since one has to assume that the absolute value of the acceleration has no effect. The ultimate justification of this interpretation must come from experiment.]
The integral in (1) can be difficult to evaluate, but certain important facts are immediately obvious. If the object is at rest with respect to S, one trivially obtains T = t2-t1. In all other cases, T must be strictly smaller than t2-t1, since the integrand is always less than or equal to unity. Conclusion: the traveling twin is younger. Furthermore, if she moves with constant velocity v most of the time (periods of acceleration short compared to the duration of the whole trip), T will approximately be given by
. (2)
The last expression is exact for a round trip (e.g. a circle) with constant velocity v. [At the times t1 and t2, twin B flies past twin A and they compare their clocks.]
Now the big deal with SR, in the present context, is that T (or , respectively) is a so-called Lorentz scalar. In other words, its value does not depend on the choice of S. If we Lorentz transform the coordinates of the world lines of the twins to another inertial frame S', we will get the same result for T in S' as in S. This is a mathematical fact. It shows that the situation of the traveling twins cannot possibly lead to a paradox within the framework of SR. It could at most be in conflict with experimental results, which is also not the case.
Of course the situation of the two twins is not symmetric, although one might be tempted by expression (2) to think the opposite. Twin A is at rest in one and the same inertial frame for all times, whereas twin B is not. [Formula (1) does not hold in an accelerated frame.] This breaks the apparent symmetry of the two situations, and provides the clearest nonmathematical hint that one twin will in fact be younger than the other at the end of the trip. To figure out which twin is the younger one, use the formulae above in a frame in which they are valid, and you will find that B is in fact younger, despite her expectations.
It is sometimes claimed that one has to resort to General Relativity in order to "resolve" the Twin "Paradox". This is not true. In flat, or nearly flat, space-time (no strong gravity), SR is completely sufficient, and it has also no problem with world lines corresponding to accelerated motion.
References: Taylor and Wheeler, Spacetime Physics (An *excellent* discussion) Goldstein, Classical Mechanics, 2nd edition, Chap.7 (for a good general discussion of Lorentz transformations and other SR basics.)