Special relativity predicts some surprising
things, one of which is that you can have a pair of twins, right
there, in front of you, at the same time, who have dramatically
different ages! 

Now that already sounds a bit paradoxical but the twins situation we
are about to investigate was called a paradox by critics of
Einstein’s work because it appeared to give results which disagree
with the basic postulates of his theory, namely that (putting it
simply) things should look the same from the points of view of all
(inertial) observers. 

The explanation I give here is inspired by
the one given in “Relativity and Common Sense” by Hermann Bondi. 

You should be aware of the method of
measuring relative velocity described here. 

The situation is
actually a little easier to illustrate if we have our usual three observers,
A, B and C. 

A is one of the twins
and the other twin is a sort of combination of B and C (all will become
clear). 

A stays at home on
earth and B and C are in separate spaceships. 

B and C are moving towards A with the same
speed, v, as shown here. (All diagrams from A's point of view) 



at the instant when B is very close to A,
(next diagram) they both set their clocks to zero. 



As B moves away from A he/she sends back
flashes of light towards A at regular intervals of 10 seconds,
on B's clock, of course. The first flash is sent at t = 10s. 

We will assume that v is such that A sees
the flashes at 20 second intervals (that is, k = 2 which means that
v is about 0.6c). 



The next point of interest is when B and C
meet. 

Let this meeting occur
at the instant when B sends the tenth flash. 

At this point, C has a quick look at B's
clock and sets his/her clock to the same time. 



C now takes over from B and does the
flashing (if you'll pardon the expression), sending pulses of light
towards A at 10 second intervals, on C's clock, of course. 



We must now note that, as C is moving
towards A, the pulses will be received by A at 5 second
intervals. 



Finally (in our analysis) C meets A at
the instant that C's tenth flash occurs. 



For an animated illustration of his thought
experiment, click here. 



We will now consider two events in
this situation. 

The two events are: 

 B meets A and 

 C meets A 



What are the times between these two events
as measured by 

1. A 

2. B and C combined 



1. A has observed 10
flashes of light at 20s intervals,
followed by 10 flashes at 5s
intervals, therefore 

Total time according to A =
250s 



2. B counted 10 intervals of 10s
and C did the same, 10 by 10s,
therefore 

Total time according to B and C together
= 200s 



We chose to investigate this situation using
A, B and C so as to be able to have only inertial observers
(which means that the ideas of special relativity theory can be
used). 



Now let us consider a nearly identical
situation in which we really do use just two observers (the twins of
the title). 



Again, A stays at home. 

B goes on a long, fast trip in some kind of
spaceship and then returns home. 

We will imagine that the rocket motor is
very powerful so that it can accelerate B to very high speed in a
fairly short time. 

B then travels for a long time at constant
speed. 

The very powerful rocket then cause another
acceleration in order to change the sense of the motion and take B
back home (again, at constant speed for most of the journey). 

So, if we assume that both the periods of
acceleration are of very short duration compared with the total time
of the round trip, we have something similar to the situation
involving A, B and C. 

We see that, when B
gets back home, he/she will find that the stayathome twin A has
aged more! 

If this sounds strange, note that in the
twins experiment only A is an inertial observer. 

B goes through a period of acceleration
(that is, a time during which B is not an inertial observer). 

This is what makes the difference. 



Putting the
calculations above in more general terms: 

Let T be the time period of the transmitted
flashes. 

Let t_{A} be the total time
according to A 

Let t_{B} the total time according
to B (or B and C combined) 

Let n be the number of flashes sent in each
half of the journey (or, more precisely, the number of time periods
between flashes) 



Then we can write 



and 



where k is given by 





Notice that, if the relative speed, v, is
very high such that k >> 1, this approximates to 



So, for an amusing illustration, let's make
v = 0.99c which corresponds to k = about 200. 

We will then have 



which means that, if B returns home 6 months
older, A will be 50 years older! 



PS 

If you think this sounds strange, it has
been verified by actual (not thought) experiment using twin
atomic clocks. 

One clock "stayed at home" and the other went on a trip
around the world. 

When they got back together, they were found to differ in their
readings by just what was predicted by relativity theory. 

See here for more detail. 
