## Rivers of Time |

Relativistic Doppler FormulaReferring to figure 3... As Albert travels at speed v _{s} toward Alpha Centauri; a radio signal pulse is sent from the spaceship at time t_{A} and then another at time t_{B}, relative to Earth-time. The first radio signal requires time t_{C} to reach Earth, and the second radio signal requires time t_{D} to reach the Earth. Notice that t_{D} will be slightly greater than t_{C} due to the ship being further away from Earth when the second radio signal is emitted.s _{A} = v_{S} t_{A}s _{A} is the distance the spaceship travels in time t_{A}.t _{C} = s_{A}/ct _{C} is the time required for the radio signal to reach Earth.T _{A} = t_{A} + t_{C}T _{A} is the time the radio signal is received on Earth.T _{A} = t_{A} + s_{A}/c = t_{A} + (v_{S}/c) t_{A} = (1 + v_{S}/c) t_{A}s _{B} = v_{S} t_{B}s _{B} is the distance the spaceship travels in time t_{B}.t _{D} = s_{B}/ct _{D} is the time required for the radio signal to reach Earth.T _{B} = t_{B} + t_{D}T _{B} is the time the radio signal is received on Earth.T _{B} = t_{B} + s_{B}/c = t_{B} + (v_{S}/c) t_{B} = (1 + v_{S}/c) t_{B}T _{B} - T_{A} = (1 + v_{S}/c) t_{A} - (1 + v_{S}/c) t_{B} = (1 + v_{S}/c) (t_{B} - t_{A})∆T = (1 + v _{S}/c) ∆tTime intervals ∆T and ∆t are measured by Earth clocks. If the time between each pulse had been synchronized to a clock located in the moving frame of reference (on board the spaceship, for example), a time dilation correction would be needed. Let ∆t _{S} represent the time interval between the two pulses as measured by a clock located in the frame carrying the radio source. Based on the time dilation formula previously derived:∆t _{S} = [1 - v_{S}^{2}/c^{2}]^{½} ∆t∆t _{S} is the time dilated interval as measured on Earth.Thus: ∆t = [1 - v _{S}^{2}/c^{2}]^{-½} ∆t_{S}Rewriting: ∆T = (1 + v _{S}/c) ∆t = (1 + v_{S}/c) / [1 - v_{S}^{2}/c^{2}]^{½} ∆t_{S}[1 - v _{S}^{2}/c^{2}]^{½} can be rewritten as [(1 + v_{S}/c)(1 - v_{S}/c)]^{½}Then: ∆T = (1 + v _{S}/c) ∆t_{S} / [(1 + v_{S}/c)(1 - v_{S}/c)]^{½}∆T = [(1 + v _{S}/c) / (1 - v_{S}/c)]^{½} ∆t_{S}∆T / ∆t _{S} represents the Relativistic Doppler Shift. A positive speed, +v_{S}, implies that the source is moving away from the receiver (∆T > ∆t_{S}). A negative speed, -v_{S}, implies that the source is moving toward the receiver (∆T < ∆t_{S}).Radio signal pulses are sent, from both the Earth and the spaceship, at one-second intervals. The difference in the time intervals, as measured in the source and observer frames, is due to the effects of Doppler shift and Relativistic time dilation. The time dilation effect can only be extrapolated by counting pulses over the entire duration of the round trip and then comparing the result to a local (reference) clock. Albert is traveling at 90% the speed of light, to-and-from Alpha Centauri. For the outward journey: ∆T / ∆t _{S} = [(1 + .90)/(1 - .90)]^{½} = 4.359Assuming that the spaceship sends radio signal pulses at one-second intervals, relative to ship time; the Earth-based observer will receive one radio pulse every 4.36 seconds. The effects of Doppler shift and Relativistic time dilation caused an apparent ship time to be slowed to 23% of Earth time. (This is red-shifting.) For the return journey: ∆T / ∆t _{S} = [(1 - .90)/(1 + .90)]^{½} = .229Again, assuming that the spaceship sends radio signal pulses at one-second intervals, relative to ship time; the Earth-base observer will receive 4.36 radio pulses every second. The effects of Doppler shift and Relativistic time dilation caused an apparent ship time to be accelerated to 436% of Earth time. (This is blue-shifting.) This is just part of the puzzle... Asymmetry in ViewpointsThe key to resolving the twin paradox is to show that the experiences of the Earth-based observer and space traveler are different (asymmetric) over the course of the journey. Referring to figure 4... The spaceship will pass any point, s, twice during the round-trip; first, on the outbound journey, A, and then on the return journey, B. The Earth-based observer must first wait for the spaceship to reach a given point, and then wait for the radio signal to propagate back to Earth.A) Consider the outbound journey : s(t) = v t t is the travel time of the spaceship, as required to reach any point, s(t), along its journey. t _{d}(t) = s(t)/ct _{d}(t) is the time required for the radio signal to propagate back to Earth.T(t) = t + t _{d} = t + (v/c) tT(t) is the time the radio signal is received by the Earth-based observer. T(t) = (1 + v/c) t (0 ≤ t ≤ s _{O}/v)B) Consider the return journey: Let t' be the elapsed time after the turn-around. s(t') = s _{O} - v t't _{d}(t') = s(t')/c = s_{O}/c - (v/c) t'T(t') = t' + t _{d}(t') = t' + [s_{O}/c - (v/c) t] = s_{O}/c + [t' - (v/c) t']T(t') = s _{O}/c + (1 - v/c) t'Accounting for the time required to reach the turn-around point... t = t' + s _{O}/vThen: t' = t - s _{O}/vT(t) = s _{O}/c + (1 - v/c)(t - s_{O}/v) + s_{O}/vThe last term, s _{O}/v, was tacked-on to account for the time required to reach the turn-around point.Expanding and rewriting: T(t) = 2 (s _{O}/c) + (1 - v/c) t (s_{O}/v < t ≤ 2 [s_{O}/v])For the outbound trip: T(t = 0) = 0 T(t = s _{O}/v) = (1 + v/c)(s_{O}/v) = s_{O}/v + s_{O}/c= (4.37 LY)(c)/(.90 c) + (4.37 LY)(c)/(c) = 4.855 years + 4.370 years = 9.225 years The Earth-based observer will, for a period of 9.23 years, receive radio pulses at 23% the rate at which they were sent by the spaceship. For the return trip: T(t = s _{O}/v) = 2 (s_{O}/v) + (1 - v/c) t = s_{O}/v + s_{O}/c= (4.37 LY)(c)/(.90 c) + (4.37 LY)(c)/(c) = 4.86 years + 4.37 years = 9.225 years T(t = 2 s _{O}/v) = 2 s_{O}/v= (2)(4.37 LY)(c)/(.90 c) = (2)(4.86 years) = 9.711 years The Earth-based observer will, for a period of 9.711 years - 9.225 years = .486 yrs, receive radio pulses at 436% the rate at which they were sent by the spaceship. The measured round-trip time, on Earth, is 9.72 years. Albert, on the other hand, will see the transition from red-shifted pulses to blue-shifted pulses at exactly the halfway point of his journey. His trip will also be shorter than the elapse time on Earth due to time dilation. ∆t _{ship} = [1 - v^{2}/c^{2}]^{½} ∆t_{earth}∆t _{ship} = [1 - .90^{2}]^{½} 4.86 years∆t _{ship} = 2.12 yrsThis is the one way trip time, according to Albert's ship clock. Albert will, for a period of 2.12 years, receive radio pulses at 23% the rate at which they were sent from the Earth. He will then, for a period of 2.12 years, receive radio pulses at 436% the rate at which they were sent from the Earth. The measured round-trip time on board the spaceship is 4.24 years. Concluding RemarksThe Earh-based observer experiences 9.72 years for a complete round trip. Albert, who is riding on the spaceship, experiences 4.24 years for a complete round trip. How many years does the Earth-based observer calculate for ship time? Answer: 23% of normal time for a period of 9.23 years and then 436% of normal time for a period of .49 years: (.229)(9.225) + (4.359)(.49) = 2.11 + 2.13 = 4.24 years This agrees with Albert's on board clock. How many years does Albert calculate for Earth time? Answer: 23% of normal time for a period of 2.12 years and then 436% normal time for a period of 2.12 years: (.229)(2.12) + (4.359)(2.12) = .485 + 9.24 = 9.72 years This agrees with Earth-based observer's clock. There is a slight rounding error that is not shown in these calculations. Everyone is in agreement as to how much Albert has aged relative to people on Earth. Through careful analysis, the paradox does not exist. Albert would have aged by 4.14 years, whereas people on the Earth would have aged by 9.72 years. |

SUMMARY

Time dilation is valid and predictable for spaceships that make round-trips to distant stars. Everyone agrees that the space traveler will age more slowly than those who remained on Earth. Spatial contraction is identical to the one-way analysis presented in the previous section. The direction of travel has no influence on the calculation.

It is completely impractical to make an instantaneous turn-around. We ignored the long-periods of acceleration and deceleration needed for a practical journey to the stars. In the next section, I'll formulate the effects of time dilation and spatial contraction for a uniformly accelerated spaceship.