Professors Mohazzbi and Luo [1] published “Despite several attempts have been made to explain the twin paradox … none of the explanations … resolved the paradox. If the paradox can be ever resolved, it requires a much deeper understanding … of the theory of relativity”. The deeper understanding of resolving the paradox is by applying more explicit definitions of proper time interval, Lorentz transform, time dilation, and aging time.
Professors Mohazzbi and Luo documented the failures to resolve the twin paradox [
The Lorentz transforms frame A, ΔtA, ΔxA coordinates to ΔtB, ΔxB coordinates of frame B, given the velocity (vAB) from frame A to frame B.
Δt time interval units light-seconds.
Δx length interval units light-seconds.
Speed of light c = 1.
(Length interval light travels in one light-second)/(time interval light travels in one light-seconds).
When Δ x 0 = 0 and Δt0, frame 0 has absolute velocity zero (see Equations (4) and (5)).
v 0 A = Δ x A / Δ t A is absolute velocity frame 0 to frame A. (1)
v 0 B = ∇ x B / Δ t B is absolute velocity frame 0 to frame B. (2)
v A B = ( v 0 B − v 0 A ) / ( 1 − v 0 B v 0 A ) is Lorentz velocities (3)
The Minkowski metric relates proper time interval (Δτ) [
Δ τ 2 = Δ t 0 2 = Δ t n 2 − Δ x n 2 , n = 1 , 2 , 3 , ⋯ (4)
Δtn and Δxn are frame n coordinates where the velocity from frame 0 to frame n is
v 0 n = Δ x n / Δ t n (There are many v0n values representing the same Δτ value.) (5)
When Δx0 = 0 and Δt0, frame 0 has absolute velocity zero. By inspection of (4), a frame at rest must have its Δx = 0. Another way to tell if a frame is at rest in space, is an object in the frame has no kinetic energy, it is at rest mass.
Δt0 (Δτ equals Δt0) is an invariant of the Lorentz transform. A given a value of Δτg restricts the transformed coordinates allowed in a transformed Lorentz reference frame. When coordinates are transformed from frame A to frame B using any constant velocity between the frames, the coordinates (ΔtA and ΔxA) and coordinates (ΔtB and ΔtB) are restricted by the Minkowski metric.
Δ τ g 2 = Δ t A 2 − Δ x A 2 = Δ t B 2 − Δ x B 2 (6)
Proper time interval is defined as the Δt0 (clock time interval between two events) observed in your frame (0) when the two events happen in frame (0). If the two events did not happen in your frame (0), Δt0 can be established in your frame (0), Event 1 flashes a green light, Event 2 flashed a red light, and the distances to the two events is known. If the light sources are moving there will be a frequency shift of the red and green flashes.
Consideration of the transit/dynamic nature of the Lorentz transform: consider two rest frames A (ΔxA equals 0) and B (ΔxB equals 0) that are ΔxD distance from frame A to frame B. Let a clock move from frame A to frame B at a constant velocity v (Event 1), stops (Event 2), and return to frame A at a velocity—v (Event 3). Upon return to frame A, it stops (Event 4).
Start
Δ t A = 0 Δ x A = 0 Δ t B = 0 Δ x B = 0
Event 1
Δ t A = Δ x D / v Δ x A = 0 Δ t B = ( Δ x D / v ) / 1 − v 2 Δ x B = Δ x D
Event 2
Δ t A = Δ x D / v Δ x A = 0 Δ t B = Δ x D / v Δ x B = 0
Event 3
Δ t A = ( ( − Δ x D ) / − v ) / 1 − ( − v ) 2 Δ x A = − Δ x D Δ t B = − Δ x D / − v Δ x B = 0
+ Δ x D / v + Δ x D + Δ x D / v
Event 4 Δ t A = 2 Δ x D / v Δ x A = 0 Δ t B = 2 Δ x D / v Δ x B = 0
Total proper time = Event 1 (ΔxD/v) + E 2 (0) + E 3 (ΔxD/v) + E 4 (0) = 2ΔxD/v.
This illustrates, snap shots of coordinates as the clock moves from one stationary Lorentz reference to another and then returns. The key considerations are Lorentz frames, standing still in space with their Δx equal to zero. The going out Δτ equals the return Δτ. The proper time interval Δτ (between Start and Event 4) is 2ΔxD/v. Time dilation of an interval is observable only for an instant.
Consider the relationship between proper time interval Δτ and aging interval. An example of an aging interval is the time between the birth (Event 1) and death of a day fly (Event 2), which is represented by Δτ = 24 hours, which is the life span of a day fly. Proper time interval (Δτ) can represent the aging interval, which is an interval between two events.
This all means astronaut returning to Earth ages the same as his twin that stayed on Earth.
The Schwarzschild Metric models an astronaut accelerating away from Earth and then revering his acceleration, returning to Earth. It is assumed the acceleration and velocity in no way affects his heath. The version of the metric is:
Δ τ 2 = α F Δ t n 2 − α F − 1 Δ x n 2 . (7)
n identifies the reference frame. n = 1 , 2 , 3 , ⋯ .
Δtn is time interval in frame n.
Δxn is length interval in frame n.
αF is parameter representing the constant gravity force moving an object Δxn distance in Δtn time. αF can have different values in a different frame.
(α = 1, represents no force and the Schwarzschild metric becomes the Minkowski metric.)
(α = 0, represents infinite force.)
Δτ, proper time interval, is invariant of Δtk and Δxk coordinates in frame (k) where αF can have different values in different frames representing a different gravity force in that frame.
Consider astronaut rockets from Earth at constant acceleration αA for a proper time interval of 1/4 (ΔτT) and reveres his acceleration for 1/4 (ΔτT) returning to zero velocity. Then keeping acceleration toward Earth for 1/4(ΔτT) and again revering his acceleration away from Earth for 1/4(ΔτT), returning to Earth with zero velocity. The total proper time interval (ΔτT), experienced by the astronaut is ΔτT. His twin has stayed on Earth for the same proper time interval (ΔτT). Both twins will have aged at the same proper time interval.
The astronaut returning to Earth ages the same as his twin that stayed on Earth. This analysis did not address the effects of mass changes caused by an object’s absolute velocity or acceleration.
The author declares no conflicts of interest regarding the publication of this paper.
Lem, D. (2024) Twin Paradox and Proper Time. Journal of Applied Mathematics and Physics, 12, 12-15. https://doi.org/10.4236/jamp.2024.121002