The "Twin Paradox" was born in 1911, after Paul Langevin restated Einstein's time dilation of traveling clocks as an interesting variant: one of two twin brothers undertakes a long, very fast space flight and on his return to Earth, finds his brother quite a bit older than himself.
The 'paradox' came in when skeptics argued that the 'away twin' could just as well have viewed himself as stationary, with the Earth and his twin brother speeding away from him and return to him later. In such a case, the Earth and the 'home twin' must be the younger ones, because they were in the 'moving frame of reference'. When compared to the original postulate, this is obviously paradoxical.
Relativists then say, "remember, the away twin had to suffer 'gforces' to get up to speed, then again to turn around and head back and finally, again to land on Earth. So, the situation is not symmetrical".
Skeptics reply, "so what, we can make the journey arbitrarily long, so that the "gforces" parts become negligible". Further, they say, "we are told that acceleration does not affect the rate of good clocks. So, the paradox stays".
Most relativists will then show how the problem can be broken up into three spacetime intervals, bordered by four events, as shown below, from the home twin's point of view.
Figure 1
In the home twin's reference frame, the turning point is 3 lightyears away and the away twin travels at 60% of c both ways. Hence, the trip takes 5 years each way and home arrival is 10 years after departure.
The spacetime interval applicable to this scenario is expressed as follows:
ds^{2} = c^{2}dt^{2}  dx^{2},
where dx is a space interval, dt a time interval and c=1, due the fact that distance is here expressed in lightyears and time in years (light covers one lightyear in one year).
Let's calculate the first spacetime interval (between events 1 and 2) in the home twin's frame of reference:
ds^{2} = 5^{2}  3^{2 }= 16 square lightyears.
Now, as has been confirmed theoretically and practically many times, spacetime intervals are reference frame independent. In more technical terms, spacetime intervals are invariant under coordinate transformations.
Therefore, the away twin must get the same interval than what the home twin calculated. Since the away twin was present at both events 1 and 2, his space interval dx' = 0. So his dt' must be 4 years. In symbolic form:
ds'^{2} = dt'^{2}  dx'^{2} = 16.
But dx' = 0, so dt'^{2} = 16.
This means that the away twin will reach the turnaround point after 4 years of his own elapsed time and not the 5 years that it will take in the home twin's reference frame. This underlines an important principle of special relativity: an inertial observer present at two events will always observe a shorter time interval between the events than what an observer not present at both events will observe.
If we ignore the short turnaround time and we assume that the inbound flight will take as long as the outbound flight, the away twin's clock and calendar will show that 8 years have elapsed when he arrives on Earth again. The home twin, however, would have recorded 10 years.
Engineers are not a very skeptical lot, but, after nodding their heads and saying, yep, that's what the theory and the calculations show, most will still not feel comfortable with this relativistic explanation.
In the part 2B of this miniseries, a more 'engineeringlike' view, using relativistic Doppler shift, will be discussed. Watch this space!
You can learn more about spacetime intervals in a free download from this page at the website/eBook Relativity 4 Engineers. (There's also a xmas special on there...)


Re: 'Paradoxes' of Relativity Part 2A: The Twin Paradox