Editing Special relativity (section)

=== Twin paradox ===
{{See also|Twin paradox}}
The reciprocity of time dilation between two observers in separate inertial frames leads to the so-called [[twin paradox]], articulated in its present form by Langevin in 1911.<ref name="Langevin_1911">{{cite journal |last1=Langevin |first1=Paul |title=L'Évolution de l'espace et du temps |journal=Scientia |date=1911 |volume=10 |pages=31–54 |url=https://en.wikisource.org/wiki/Translation:The_Evolution_of_Space_and_Time |access-date=20 June 2023}}</ref> Langevin imagined an adventurer wishing to explore the future of the Earth. This traveler boards a projectile capable of traveling at 99.995% of the speed of light. After making a round-trip journey to and from a nearby star lasting only two years of his own life, he returns to an Earth that is two hundred years older.

This result appears puzzling because both the traveler and an Earthbound observer would see the other as moving, and so, because of the reciprocity of time dilation, one might initially expect that each should have found the other to have aged less. In reality, there is no paradox at all, because in order for the two observers to perform side-by-side comparisons of their elapsed proper times, the symmetry of the situation must be broken: At least one of the two observers must change their state of motion to match that of the other.<ref name='Debs_Redhead'>{{cite journal |author1=Debs, Talal A. |author2=Redhead, Michael L.G. |title=The twin "paradox" and the conventionality of simultaneity |journal=American Journal of Physics |volume=64 |issue=4 |year=1996 |pages=384–392 |doi=10.1119/1.18252 |bibcode=1996AmJPh..64..384D}}</ref>

[[File:Twin paradox Doppler analysis.svg|thumb|Figure 4-4. Doppler analysis of twin paradox]]
Knowing the general resolution of the paradox, however, does not immediately yield the ability to calculate correct quantitative results. Many solutions to this puzzle have been provided in the literature and have been reviewed in the [[Twin paradox]] article. We will examine in the following one such solution to the paradox.

Our basic aim will be to demonstrate that, after the trip, both twins are in perfect agreement about who aged by how much, regardless of their different experiences. {{nowrap|Fig 4-4}} illustrates a scenario where the traveling twin flies at {{nowrap|0.6 c}} to and from a star {{nowrap|3 ly}} distant. During the trip, each twin sends yearly time signals (measured in their own proper times) to the other. After the trip, the cumulative counts are compared. On the outward phase of the trip, each twin receives the other's signals at the lowered rate of {{tmath|1= \textstyle f' = f \sqrt{(1-\beta)/(1+\beta)} }}. Initially, the situation is perfectly symmetric: note that each twin receives the other's one-year signal at two years measured on their own clock. The symmetry is broken when the traveling twin turns around at the four-year mark as measured by her clock. During the remaining four years of her trip, she receives signals at the enhanced rate of {{tmath|1= \textstyle f' ' = f \sqrt{(1+\beta)/(1-\beta)} }}. The situation is quite different with the stationary twin. Because of light-speed delay, he does not see his sister turn around until eight years have passed on his own clock. Thus, he receives enhanced-rate signals from his sister for only a relatively brief period. Although the twins disagree in their respective measures of total time, we see in the following table, as well as by simple observation of the Minkowski diagram, that each twin is in total agreement with the other as to the total number of signals sent from one to the other. There is hence no paradox.<ref name="French_1968"/>{{rp|152–159}}

{| class="wikitable"
|-
! Item!! Measured by the{{br}}stay-at-home !! Fig 4-4 !! Measured by{{br}}the traveler !! Fig 4-4
|-
| Total time of trip
| <math>T = \frac{2L}{v}</math>
| {{nowrap|10 yr}}
| <math>T' =  \frac{2L}{\gamma v} </math>
| {{nowrap|8 yr}}
|-
| Total number of pulses sent
| <math>fT = \frac{2fL}{v}</math>
| 10
| <math>fT' = \frac{2fL}{\gamma v}</math>
| 8
|-
| Time when traveler's turnaround is '''detected'''
| <math>t_1 = \frac{L}{v} + \frac{L}{c}</math>
| {{nowrap|8 yr}}
| <math>t_1' = \frac{L}{\gamma v}</math>
| {{nowrap|4 yr}}
|-
| Number of pulses received at initial <math>f'</math> rate
| <math>f't_1</math> <math>= \frac{fL}{v}(1 + \beta)\left(\frac{1-\beta}{1+\beta}\right)^{1/2}</math>{{br}}<math>= \frac{fL}{v}(1 - \beta ^2)^{1/2}</math>
| 4
| <math>f't_1'</math> <math>= \frac{fL}{v}(1 - \beta^2)^{1/2}\left(\frac{1-\beta}{1+\beta}\right)^{1/2}</math>{{br}}<math>= \frac{fL}{v}(1 - \beta )</math>
| 2
|-
| Time for remainder of trip
| <math>t_2 = \frac{L}{v} - \frac{L}{c}</math>
| {{nowrap|2 yr}}
| <math>t_2' = \frac{L}{\gamma v}</math>
| {{nowrap|4 yr}}
|-
| Number of signals received at final <math>f''</math> rate
| <math>f''t_2</math> <math>= \frac{fL}{v}(1 - \beta)\left( \frac{1 + \beta}{1 - \beta} \right)^{1/2}</math> <math>= \frac{fL}{v}(1 - \beta ^2)^{1/2}</math>
| 4
| <math>f''t_2'</math> <math>= \frac{fL}{v}(1 - \beta^2)^{1/2}\left( \frac{1 + \beta}{1 - \beta} \right)^{1/2}</math> <math>= \frac{fL}{v}(1 + \beta)</math>
| 8
|-
| Total number of received pulses
| <math>\frac{2fL}{v}(1 - \beta ^2)^{1/2}</math> <math>= \frac{2fL}{\gamma v}</math>
| 8
| <math>\frac{2fL}{v}</math>
| 10
|-
| Twin's calculation as to how much the '''''other''''' twin should have aged
| <math>T' = \frac{2L}{\gamma v}</math>
| {{nowrap|8 yr}}
| <math>T = \frac{2L}{v}</math>
| {{nowrap|10 yr}}
|}