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An 'engineer-friendly' solution to the famous "twin paradox" is given in terms of relativistic Doppler shift. In the previous post (part 2A), the 'paradox' was resolved by means of space-time intervals between events, using this diagram:
Figure 1
This time, let's introduce the sister/brother twins Pam and Jim. 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.
Figure 2 
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.
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.
Figure 2 has shown the situation on a space-time diagram drawn in Jim's reference frame. In the final sub-part, (2c), the same situation will be shown in Pam's reference frame.
You can learn more about relativistic Doppler shift in a free download from this page at the website/eBook Relativity 4 Engineers. (There's also a X-mas special on there, but time is running out...)
Jorrie
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