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The Twin Paradox Part 2

Part 2 is an 'engineer-friendly' solution to the famous "twin paradox". It is given in terms of relativistic Doppler shift. In the previous post (part 1), the 'paradox' was resolved by means of space-time intervals between events, using this diagram:

Twin paradox in intervals

This time, let's introduce the sister/brother twins Pam and Jim into the twin paradox. They have agreed to put Einstein to the test, starting on New Year's Day, 2007.

Pam, after launching on New Year's Eve and gathering speed, passes Earth at the stroke of midnight on New Year's Day, with a relative speed of 0.6c. Four years later, on New Year's Day 2011 (according to Pam), having covered three-quarters of the distance to the nearest stellar neighbor to our Sun, she performs a rapid turnaround and heads back home at -0.6c relative to Earth.

To find out how the calendar/clocks of Pam and Jim differ, let Pam and Jim send each other a yearly New Year's greeting during Pam's trip, obviously using their own calendars and clocks. We can use the well-verified relativistic Doppler shift formula to calculate periods for Pam and Jim. The one-way relativistic ratio of periods is given by

Tr / T = sqrt[(c+v)/(c-v)], where Tr is the received period, T the transmitted period, and v an opening velocity between the transmitter and the receiver. As usual, c is the speed of light in vacuum. So, during the outbound trip, with relative velocity v = 0.6c, the Doppler shift ratio is:

Tr / T = sqrt[1.6/0.4] = 2, meaning that a greetings transmitted once a year will reach the receiver every two years.

During Pam's inbound trip, with relative velocity v = -0.6c (a closing velocity), the relativistic ratio of periods is:

Tr / T = sqrt[0.4/1.6] = 0.5, meaning that greetings transmitted once a year will reach the receiver every six months.

These relativistic Doppler shifts are completely symmetrical - it does not matter which twin does the transmitting and which one does the receiving - it only depends on the relative velocity between them. Figure 2 illustrates this in the context of the twin paradox.

Twin paradox with Doppler shift

In the four years that Pam heads away from home, she will receive only two New Year's messages from Jim. This is because Jim here represents the T period of 1 year. Pam is the receiver, with the Tr period of 2 years. On her calendar she will receive "happy New Year 2008" only on Jan 1, 2009, and the next one ("happy New Year 2009") on Jan 1, 2011. Weird, but this is due to the increasing distance between them and the time light takes to cross the gap.

During her return trip, the situation is reversed, so in the last four years, she will receive eight New Year's messages from Jim, one every six months! She will receive the last (tenth) message, on New Year's Day 2015 on her calendar, as she makes a close fly-by of Earth.

Does this solve the twin paradox? Not quite, yet.

Now that was Pam receiving Jim's messages. At what rate will Jim receive New Year's messages from his sister? For the first eight years, he will also have to wait two years for every 'happy New Year' message. This means that it will be 2015 on Earth before Jim gets the message that his sister has sent on New Year's Day 2011, with a note that she has just turned around for the home leg of her trip.

Then, for the last two years, Jim will receive a New Year's message every six months - four of them. Add them up and Jim will receive only eight messages from Pam in the decade that he waited for her return. Conclusion: Pam recorded only eight years during her voyage, while Jim recorded 10 years.

What we have not yet done is to view the twin paradox strictly from Pam's frame of reference. The problem we face is that Pam does not find herself in one inertial frame of reference for her whole trip. If we insist on drawing Pam's two phases (outbound and inbound) on one inertial frame, we are forced to insert a discontinuity in the inertial frame of brother Jim, as illustrated in figure below.

Twin paradox from moving reference frame

The vertical line is Pam's world line, with the bullets showing the 8 years that elapsed. During Pam's acceleration for the turn-around, the lines of simultaneity change drastically in her reference frame. This causes the apparent discontinuity in the home twin's world line, showing that one should rather not represent two different inertial frames as a single line on a space-time diagram.

A much better way to view the twin paradox is to stick to one of Pam's inertial frames as the reference. In figure 2 below, I chose Pam's home bound trip as the inertial reference frame. In this frame, Pam moves at 'double speed' until her turn-around point, meaning we must do a relativistic doubling of her velocity: (0.6c+0.6c)/(1+0.62/c^2) = 0.882c.

Twin paradox from return home frame

At the turnaround point, Pam decelerates until she is stationary in the reference frame and then waits for Jim to catch up. Jim is doing a steady 0.6c in the reference frame and the reunion happens 12.5 years later on the calendars of the chosen reference frame.

However, Pam and Jim again aged 8 and 10 years respectively, as before. The New Year's messages were also received just like in the previous article, because the relative Doppler shift ratios stayed the same.

Readers are urged to carefully look at the last figure. If one understands the reasoning behind these distances and times of the solved twin paradox, one understands special relativity!

Readers can discuss this article at: CR4 Twin Paradox Blog

Return from the Twin Paradox Part 2 page to the Paradoxes page.



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