Twin-Paradox Events and Transformations

Note 3 inertial reference frames:

S = The frame of the "stay at home" twin — uses coordinates (t,x),
S' = The frame of the "outbound part of the trip" — uses coordinates (t',x'),
S" = The frame of the "inbound part of the trip" — uses coordinates (t",x").

1. The Lorentz transformation S → S'
   S' recedes from S with velocity v along their x- and x'-axes. They synchronize their clocks at the Start event E with coordinates: E: ( t, x ) = ( 0, 0 ) = ( t', x' ). Letting γ = 1/√(1 - v²/c²),

   S → S'

   t' = γ(t - vx/c2), x' = γ(x - vt).

   Summary of Reference Frames

   t(worldline of S) [R] R: Return event (t,x) = (2T,0) (t",x")=(2T/γ,0) t"(worldline of inbound S") t'(worldline of x" outbound S') [A] A: Turnaround event (t,x) = (T,vT) (t',x')=(T/γ,0) (t",x")=(T/γ,0) x' x [E] E: Start event (t,x)=(0,0)=(t',x')

2. The coordinates of the Turnaround event A
   The outbound twin (now in frame S') has been travelling during a time T (according to frame S) and decides to turn back to her brother. Let's call this event A. According to frame S this event has the coordinates:
   A: ( t, x ) = ( T, vT ).
   To find the coordinates according to the travelling twin who is in frame S', we apply the Lorentz transformation S → S':
   A: ( t', x' ) = ( γ{T - v[vT]/c²}, γ{vT - vT} ) = ( γT/γ², 0 ) = ( T/γ, 0 ).
   We see that the travelling twin's clock shows T/γ elapsed time at this event. She is now ready to (carefully!) jump on frame S" and return to her brother. They synchronize their clocks at event A with coordinates:
   A: ( t", x" ) = ( T/γ, 0 ).
   Here we have carried over the already elapsed clock time of the travelling twin since we are interested in the time the twin's clock will show when they reunite again.

   Coordinate Translation

   P: (x',y') = (x-2,y-1)

3. The Lorentz transformation S → S"
   Frame S" [translated from S by ( T, vT )] approaches S with velocity v (recedes with -v) along their x- and x"-axes.

   S →S"

   t" - T/γ = γ( [t - T] + v[x - vT]/c2 ), x" = γ( [x - vT] + v[t - T] ).

4. The coordinates of the Return event R
   When the twin (now in frame S") has been approaching S during a time T (again according to frame S), she will reunite with her brother. Let's call this event R. According to frame S this event has coordinates:
   R: ( t, x ) = ( 2T, 0 ).
   Applying the Lorentz transformation frame S → S":
   R: ( t" - T/γ, x" ) = ( γ{T - v[vT]/c²}, γ{-vT + vT} ) = ( T/γ, 0 ) → R: ( t", x" ) = ( 2T/γ, 0 ).

5. Conclusion
   At this event R, both clocks of frames S and S" are at the same place denoted by x = x" = 0, but the clock of S shows an elapsed time t = 2T and the clock of S" shows an elapsed time t" = 2T/γ. So if v = 0.8c and T = 1.25 years (travels vT = 0.8 × 1.25 = 1 light-year & return), then γ = 1/√(1 - 0.8²) = 1.66667 and the "stay at home" twin will have aged 2.5 years while his travelling twin sister will have aged 2.5/1.66667 = 1.5 years.

• Twin Paradox Animated Spacetime Diagram

Food for Thought: Simultaneous — but to whom?

S' → S  x = γ(x' + vt'), t = γ(t' + vx'/c2).

A rod whose resting length L₀= 10 meters is moving (parallel) to the right along the positive x-axis of the S-frame with velocity v = 0.8c. A rider on the rod (i.e. in the S'-frame) sets off two explosions simultaneously — so he says.

• How far apart are the 2 explosions in the S-frame?
  The two explosions occur simultaneously in the S'-frame which implies that Δt' = 0. Applying the Lorentz reverse transformation S' → S, and using Δx' = L₀,
  Δx = γ(Δx' + vΔt') = γL₀= 1.667(10m) = 16.67m apart.
• What is the time interval between the two explosions as measured in the S-frame? Which explosion occurred first in the S-frame?
  Again, using Δt' = 0, Δx' = L₀ as above,
  Δt = γ(Δt' + vΔx'/c²) = γ(0.8)L₀/c = 1.3336(10m)/300m/μsec =0.0445μsec.
  Since Δt = t_R - t_L> 0, the explosion on the left occurred first in the S-frame.
• What is the length of the rod as measured in the S-frame?
  The distance between the 2 explosions (non-simultaneous in the S-frame) is not the same as the length of the rod as measured in the S-frame, which corresponds to marking both ends of the rod simultaneously in the S-frame (i.e. Δt = 0). Applying the Lorentz transformation S → S', and using Δx = L,
  Δx' = L₀ = γ(Δx - vΔt) = γL → L = L₀/γ = (10m)/1.667 = 6m.