Rivers of Time - The Twin Paradox

Rivers of Time The Twin Paradox There was symmetry in Albert's journey to Alpha Centauri as his spaceship coasted through space at 90% the speed of light, relative to the Earth. Both the spaceship and the Earth possessed uniform motion. Albert observed clocks on Earth running at about 43% the rate of the clocks on his ship, due to the effects of time dilation. This means that Earth clocks would only advance 26 minutes for every hour that passed for Albert. Conversely, an Earth-based observer would see Albert's spaceship clocks running at about 43% the rate of her own clock; again, due to the effects of time dilation. This situation is contradictory, but it is just one of many mysteries that we'll discover about the universe. It is, of course, impossible to peer across the vast distances of space and take readings from a clock that is attached to a fast moving spaceship. It would be feasible, however, to send bursts of light (or radio waves) at one-second intervals from both the Earth's observational platform and the spaceship. Each distant observer could then keep track of the number of pulses received and compare them to the ticks of his or her own clock. Once Albert reaches Alpha Centauri; he will fire his rocket engines, reverse his direction of travel and eventually return to Earth. At the point of turn-around, he will break from uniform motion and, consequently, feel the forces of inertia acting upon his body. He'll quickly return to a speed equal to 90% the speed of light and then coast all the way back to Earth. Will Albert return to Earth and discover that he had aged differently than the friends and family he had left behind? In this section, I'll present the effects of time dilation and spatial contraction for high-speed round trip excursions on board a spaceship. I'll also resolve a famous paradox. RETURNING TO EARTH To simplify the round-trip analysis; consider the impossible situation where Albert travels at 90% the speed of light to Alpha Centauri, makes an instantaneous turn-around, and then travels back to Earth at 90% the speed of light. The Earth-based observer would perceive herself to be at rest while viewing Albert's travels to-and-from the star. She would not experience any inertial forces during the turn-around. For all practical purposes, she is at rest and observing the trip from a completely valid inertial frame. Ignoring the infinite inertial forces that would crush Albert and his spaceship at the moment of turn-around, he would otherwise spend the remaining parts of his trip inside a valid inertial frame. From Albert's prospective, it would be completely reasonable to consider the Earth in motion -racing away from his spaceship at 90% the speed of light- while he and his spaceship remain motionless in space. At some point in time, Albert would see the Earth make an instantaneous turn-around and travel back to him at 90% the speed of light. There appears to be perfect symmetry in the experiences of the Earth-bound observer and Albert. This is the basis of the famous Twin Paradox that is often discussed in the study of Special Relativity. THE TWIN PARADOX One of two identical twins sets off on a journey to the nearest star, traveling in a spaceship at nearly the speed of light. The other twin stays behind on the Earth. The twin located on Earth expects his space-traveling sibling to return to earth looking much younger than himself, due to the effects of time dilation. The twin on the spaceship, however, felt himself at rest (ignoring the moment of his rapid turn-around, of course) and, thus, treats the Earth as a vehicle carrying his sibling on a high-speed round trip journey. The sibling on board the spaceship would -upon his return to Earth- expect to find that his Earth-based sibling looks much younger than himself. Since both prospectives seem to be valid in Special Relativity; there is an apparent paradox when the twins finally meet back on Earth. There really is no paradox, however, because there is no symmetry in motion between the Earth and the spaceship. Due to the turn-around, there was a break in symmetry, and a break from uniform motion. The spaceship actually leaped from one inertial frame to another, each representing a particular direction of travel. I'll provide a qualitative solution to the paradox and then offer the quantitative solution for those needing mathematical proof. RESOLVING THE TWIN PARADOX (A Qualitative Solution) We will assume that the spaceship sends pulses of radio waves toward the earth at one-second intervals. We'll also assume that similar pulses are being sent from the Earth towards the spaceship. These pulses of radio waves represent the ticks of a clock, where one clock is fixed to the spaceship and the other to the Earth's observational platform. Since the Earth and the spaceship are moving relative to one another; Doppler shift will affect the rate at which these pulses are received by each observer. Perhaps you've heard a train whistle as it approached and then passed you by. The pitch of the approaching whistle is higher, initially, but falls to its natural pitch at the moment the train is upon you, and then continues to fall as the train moves away. This is the Doppler effect. It occurs with sound, light and radio waves. The pulses of radio waves generated by the spaceship are synchronized to a clock located on-board. When the effects of Relativistic time dilation slows the on-board clock, relative to an Earth clock; the rate at which the pulses are sent toward Earth will be reduced. The same situation holds true for the pulses of radio waves sent from the Earth and received by the spaceship. The effects of time dilation are evidenced by a reduction in the rate by which these pulses are sent towards the distant observer. There is an equation called the Relativistic Doppler formula that calculates normal Doppler shift, but also accounts for the effects of time dilation. This formula will be derived within the quantitative solution. When the source is moving, relatively, toward the observer; the pulses will be spaced more closely along the direction of travel, causing more to be received within a given period of time. This is referred to as Doppler blue shifting. When the source is moving, relatively, away from the observer; the pulses will be spaced further apart along the direction travel, causing fewer to be received within a given period of time. This is referred to as Doppler red shifting. The pulse rates increase and decrease, just like the pitch of the train whistle, due to relative motion between the source and observer. The pulse rates will be further decreased by the effects of time dilation. Imagine that we had placed markers along the path of travel, between Earth and Alpha Centauri. Every time the spaceship passes one of these markers, it would trigger some type of an event at the marker location... perhaps a powerful laser pulse that is directed toward Earth. The laser pulse received by the Earth-based observer will have been delayed. This is due to the fact that light travels at a finite speed c, thus limiting how quickly she could learn about the marker event. As the spaceship travels further and further away from the Earth, more of this delay is introduced. By the time the spaceship reaches Alpha Centauri, it will take light 4.37 years to relay that fact to Earth. The spaceship will spend (4.37 LY)(c)/(.90 c) = 4.86 years traveling to the turn-around point, relative to Earth time. Additionally, 4.37 years will pass before the Earth-based observer receives that pulse of light signifying the turn-around event. From the prospective of an observer on Earth, the spaceship will be traveling away from the Earth for 9.23 years. The Earth-based observer will see a transition from red-shifted pulses to blue-shifted pulses, validating that Albert had begun his journey back toward Earth. The total round-trip time relative to Earth clocks is (2)(4.86 yrs) = 9.72 years, assuming an instantaneous turn-around. The fact that the Earth-based observer didn't learn about the turn-around event for 9.23 years, which is nearly the end of the spaceship's journey, wasn't at-all intuitive for me. It seemed even stranger that the return trip, as perceived by the Earth-based observer, would be compressed into a period of (9.72 yrs - 9.23 yrs) = .49 years! In fact -- I didn't believe it, myself; so I worked out the mathematics required to track the marker events and account for the delay of light propagating back to Earth. To my surprise, the calculations predicted exactly that result! The derivations will be included in the quantitative solution. Albert's experience will be very different, however. From his prospective, there is immediate evidence of his rapid turn-around. The turn-around event happens, literally, under his feet and at exactly the mid-point of his journey! You might ask... What evidence would be at Albert's disposal to validate his turn-around? The earth had been transmitting radio wave pulses since his journey began. During the first half of the trip, the Earth and spaceship were moving, relatively, further and further apart. This would cause the pulses sent from Earth to be red-shifted. The very instant the spaceship turned around and began closing the distance to Earth, those radio wave pulses would become blue-shifted. This instantaneous transition from red-shifted pulses to blue-shifted pulses convinces Albert that the Earth just performed a rapid turn-around, relative to his spaceship. Notice the asymmetry between the experiences of the Earth-based observer and Albert. From a prospective on the Earth, pulses received from the spaceship will be red-shifted for (9.23 yrs)/(9.72 yrs) = .95 (95% of the total trip time). From a prospective on board the spaceship, pulses received from the Earth will be red-shifted for exactly half the trip and blue-shifted for the other half of the trip. The red-shifting phenomenon implies fewer pulses received in a given time period, meaning less time passed at the source. The Earth-based observer received red-shifted pulses for 95% of the trip, whereas Albert received red-shifted pulses for only 50% of the trip. When comparing clock ticks, from either prospective; both would agree that Albert's spaceship clock lost time, relative to Earth-based clocks over the entire trip. When Albert returned to Earth, he discovered that he had aged significantly less than his friends and family back on Earth! The paradox is resolved! RESOLVING THE TWIN PARADOX (A Quantitative Solution) The highlighted area below provides a mathematical treatment of high-speed round-trip excursion on board a spaceship. This section can be skipped without loss of comprehension. Relativistic Doppler Formula Referring to figure 3... As Albert travels at speed vs toward Alpha Centauri; a radio signal pulse is sent from the spaceship at time tA and then another at time tB, relative to Earth-time. The first radio signal requires time tC to reach Earth, and the second radio signal requires time tD to reach the Earth. Notice that tD will be slightly greater than tC due to the ship being further away from Earth when the second radio signal is emitted. sA = vS tA sA is the distance the spaceship travels in time tA. tC = sA/c tC is the time required for the radio signal to reach Earth. TA = tA + tC TA is the time the radio signal is received on Earth. TA = tA + sA/c = tA + (vS/c) tA = (1 + vS/c) tA sB = vS tB sB is the distance the spaceship travels in time tB. tD = sB/c tD is the time required for the radio signal to reach Earth. TB = tB + tD TB is the time the radio signal is received on Earth. TB = tB + sB/c = tB + (vS/c) tB = (1 + vS/c) tB TB - TA = (1 + vS/c) tA - (1 + vS/c) tB = (1 + vS/c) (tB - tA) ∆T = (1 + vS/c) ∆t Time 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 ∆tS 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: ∆tS = [1 - vS2/c2]½ ∆t ∆tS is the time dilated interval as measured on Earth. Thus: ∆t = [1 - vS2/c2]-½ ∆tS Rewriting: ∆T = (1 + vS/c) ∆t = (1 + vS/c) / [1 - vS2/c2]½ ∆tS [1 - vS2/c2]½ can be rewritten as [(1 + vS/c)(1 - vS/c)]½ Then: ∆T = (1 + vS/c) ∆tS / [(1 + vS/c)(1 - vS/c)]½ ∆T = [(1 + vS/c) / (1 - vS/c)]½ ∆tS ∆T / ∆tS represents the Relativistic Doppler Shift. A positive speed, +vS, implies that the source is moving away from the receiver (∆T > ∆tS). A negative speed, -vS, implies that the source is moving toward the receiver (∆T < ∆tS). 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 / ∆tS = [(1 + .90)/(1 - .90)]½ = 4.359 Assuming 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 / ∆tS = [(1 - .90)/(1 + .90)]½ = .229 Again, 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 Viewpoints The 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. td(t) = s(t)/c td(t) is the time required for the radio signal to propagate back to Earth. T(t) = t + td = t + (v/c) t T(t) is the time the radio signal is received by the Earth-based observer. T(t) = (1 + v/c) t   (0 ≤ t ≤ sO/v) B) Consider the return journey: Let t' be the elapsed time after the turn-around. s(t') = sO - v t' td(t') = s(t')/c = sO/c - (v/c) t' T(t') = t' + td(t') = t' + [sO/c - (v/c) t] = sO/c + [t' - (v/c) t'] T(t') = sO/c + (1 - v/c) t' Accounting for the time required to reach the turn-around point... t = t' + sO/v Then: t' = t - sO/v T(t) = sO/c + (1 - v/c)(t - sO/v) + sO/v The last term, sO/v, was tacked-on to account for the time required to reach the turn-around point. Expanding and rewriting: T(t) = 2 (sO/c) + (1 - v/c) t   (sO/v < t ≤ 2 [sO/v]) For the outbound trip: T(t = 0) = 0 T(t = sO/v) = (1 + v/c)(sO/v) = sO/v + sO/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 = sO/v) = 2 (sO/v) + (1 - v/c) t = sO/v + sO/c = (4.37 LY)(c)/(.90 c) + (4.37 LY)(c)/(c) = 4.86 years + 4.37 years = 9.225 years T(t = 2 sO/v) = 2 sO/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. ∆tship = [1 - v2/c2]½ ∆tearth ∆tship = [1 - .902]½ 4.86 years ∆tship = 2.12 yrs This 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 Remarks The 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.