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Relativity and Cosmology

This is a Blog on relativity and cosmology for engineers and the like. You are welcome to comment upon or question anything said on my website (http://www.relativity-4-engineers.com), in the eBook or in the snippets I post here.

Comments/questions of a general nature should preferably be posted to the FAQ section of this Blog (http://cr4.globalspec.com/blogentry/316/Relativity-Cosmology-FAQ).

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Regards, Jorrie

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2 comments

Paradoxes of Relativity Part 2c: Twin Paradox (concluded)

Posted December 24, 2006 11:00 PM by Jorrie
Pathfinder Tags: special relativity Twin Paradox

In this 'engineer-friendly' mini-series on the famous "twin paradox" of relativity, where Pam sets off on a long, fast space journey and on return finds herself younger than her twin brother, we have viewed the situation in various ways. What we have not done yet is to view the situation 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 1 below.

Figure 1:

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 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/c2) = 0.882c.

Figure 2:

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 Fig. 2. If one understands the reasoning behind these distances and times, you understand special relativity! You can learn more about the twin paradox in a free download from this page at the website/eBook Relativity 4 Engineers. (There's still a X-mas special on there, but time is running out...)

In the next installment on relativistic paradoxes, the "long ladder paradox" will be discussed.


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#1

Re: Paradoxes of Relativity Part 2c: Twin Paradox (concluded)

01/05/2007 2:41 AM

Jorrie, fascinating stuff!

I have a doubt however: after your Fig. 2 you said "... and the reunion happens 12.5 years later on the calendars of the chosen reference frame" as is also shown on the figure.

Are we supposed to believe that an observer sitting stationary in your "chosen" reference frame would have aged 12.5 years, compared to the 10 and 8 years of your twins?

Guru
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#2
In reply to #1

Re: Paradoxes of Relativity Part 2c: Twin Paradox (concluded)

01/05/2007 7:51 AM

Guest wrote: "Are we supposed to believe that an observer sitting stationary in your "chosen" reference frame would have aged 12.5 years, compared to the 10 and 8 years of your twins?"

No, we are absolutely not supposed to believe that! No observer in the chosen reference frame (fig. 2) can be present at both Pam's departure event and the reunion event, so we cannot say how any observer stationary in the reference frame would have aged during Pam's journey.

Say we had two stationary observers with clocks synchronized by Einstein's method, Jack at the origin and Jill at 7.5 l.y. away. Jack is present at the start of Pam's journey and Jill is present at the reunion.

If Jack and Jill exchange messages afterwards (waiting 7.5 years for the signals to cover the distance) and compare the date and time of departure and reunion, they will calculate a time interval of 12.5 years.

However, Jim, who made the 0.6c journey between the two events will not agree with Jack and Jill's calculation, because his clock and calendar tells him that the journey took 10 years. Since Jim was present at both events without being accelerated, we better believe him!

Hope it clears the uncertainty!

Jorrie

__________________
"Curiosity has its own reason for existence" -- Albert Einstein
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