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The Twins Paradox
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 space-ships.  
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 space-ship 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 stay-at-home 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 tA be the total time according to A  
Let tB 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.  
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