Editing Twin paradox (section)

==Calculation of elapsed time from the Doppler diagram==
The twin on the ship sees low frequency (red) images for 3 years. During that time, he would see the Earth twin in the image grow older by {{nowrap|3/3 {{=}} 1 year}}. He then sees high frequency (blue) images during the back trip of 3&nbsp;years. During that time, he would see the Earth twin in the image grow older by {{nowrap|3 × 3 {{=}} 9 years.}} When the journey is finished, the image of the Earth twin has aged by {{nowrap|1 + 9 {{=}} 10 years.}}

The Earth twin sees 9 years of slow (red) images of the ship twin, during which the ship twin ages (in the image) by {{nowrap|9/3 {{=}} 3 years.}} He then sees fast (blue) images for the remaining 1&nbsp;year until the ship returns. In the fast images, the ship twin ages by {{nowrap|1 × 3 {{=}} 3 years.}} The total aging of the ship twin in the images received by Earth is {{nowrap|3 + 3 {{=}} 6 years}}, so the ship twin returns younger (6&nbsp;years as opposed to 10&nbsp;years on Earth).

===The distinction between what they see and what they calculate===
To avoid confusion, note the distinction between what each twin sees and what each would calculate. Each sees an image of his twin which he knows originated at a previous time and which he knows is Doppler shifted. He does not take the elapsed time in the image as the age of his twin now.
*If he wants to calculate when his twin was the age shown in the image (''i.e.'' how old he himself was then), he has to determine how far away his twin was when the signal was emitted—in other words, he has to consider simultaneity for a distant event.
*If he wants to calculate how fast his twin was aging when the image was transmitted, he adjusts for the Doppler shift. For example, when he receives high frequency images (showing his twin aging rapidly) with frequency <math>\scriptstyle{f_\mathrm{rest}\sqrt{\left({1 + v/c}\right)/\left({1 - v/c}\right)}}</math>, he does not conclude that the twin was aging that rapidly when the image was generated, any more than he concludes that the siren of an ambulance is emitting the frequency he hears. He knows that the [[Doppler effect]] has increased the image frequency by the factor 1 / (1 − ''v''/''c''). Therefore, he calculates that his twin was aging at the rate of
:<math>f_\mathrm{rest}\sqrt{\left({1 + v/c}\right)/\left({1 - v/c}\right)}\times \left(1 - v/c\right) = f_\mathrm{rest}\sqrt{1 - v^2/c^2}\equiv\epsilon f_\mathrm{rest}</math>
when the image was emitted. A similar calculation reveals that his twin was aging at the same reduced rate of ''εf''<sub>rest</sub> in all low frequency images.

===Simultaneity in the Doppler shift calculation===
It may be difficult to see where simultaneity came into the Doppler shift calculation, and indeed the calculation is often preferred because one does not have to worry about simultaneity. As seen above, the ship twin can convert his received Doppler-shifted rate to a slower rate of the clock of the distant clock for both red and blue images. If he ignores simultaneity, he might say his twin was aging at the reduced rate throughout the journey and therefore should be younger than he is. He is now back to square one, and has to take into account the change in his notion of simultaneity at the turnaround. The rate he can calculate for the image (corrected for Doppler effect) is the rate of the Earth twin's clock at the moment it was sent, not at the moment it was received. Since he receives an unequal number of red and blue shifted images, he should realize that the red and blue shifted emissions were not emitted over equal time periods for the Earth twin, and therefore he must account for simultaneity at a distance.