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Relativity

The Twins "Paradox"

An interesting and amusing result predicted by relativity theory is often called the twins "paradox".

We are used to the idea that twins have the same age (plus or minus an hour or two). If one of a pair of twins goes on a long, fast journey and then returns home, it will be found that the twins have aged differently. The fact that the "stay-at-home" twin is always the older of the two caused some people to suggest that Einstein’s theory had internal contradictions (hence the term the twins "paradox"). It is generally agreed today that there is no real paradox but the term is still used when considering such situations.

The situation is actually easier to illustrate if we have three observers, A, B and C. Consider the series of events below.

1...B and C move towards A. Each has the same speed, v, relative to A.

2. ..At the instant when B is next to A, B and A both set their clocks to read zero.

3.

 B sends flashes of light to A at 10 second intervals (the first flash being sent at t = 10 seconds on B's clock).
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).
4.  B meets C at the instant when the tenth flash is sent.

At this instant, C sets his/her clock to read the same as B’s clock.
5. C sends flashes at 10 second intervals which will be seen by A at 5 second intervals (remember, if v is changed to -v, k becomes 1/k). C will, of course, meet A at the instant of sending the tenth flash.

Now consider the time between two events from the sequence above. The two events we will consider are

the meeting between A and B

the meeting between A and C

From A’s clock the time between these two events is

(10 × 20) + (10 × 5) = 250 seconds

Combining the times measured by B and C we have

(10 × 10) + (10 × 10) = 200 seconds

Let us now forget about C. Imagine instead a similar situation but with only A and B, the "twins". This time B goes away from A with speed v, as before, but after a time, B uses some kind of rocket motor to reverse direction and return to A (also at speed v). If the time taken for the reversal is small compared with the total time of the journey then we can apply (at least approximately) the results calculated above.

Conclusion: even though A and B are twins, when B returns to A, A will be older than B!

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).

Putting the calculations in more general terms

Let T = interval of transmission of flashes
Let tA = total time according to A
Let tB = total time according to B (or according to B and C in the original)
Let n = number of flashes sent in each half of the journey

then

tB = 2nT

and

tA = nkT + nT/k = nT(k + 1/k)

So, if k >> 1

tA = (approx) nkT

Therefore, if v = 0·99c which corresponds to k = (approx) 200, we have

tA/tB = nkT/2nT = 100

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

© David Hoult 2008