Accelerated motion behaves in a similar way duting initial stage although velocity ... same time interval t: 0⤠t ⤠tN. = [ ... With the same initial conditions:.
Numerical Example Illustrating relative simultaneity During previous discussions it has been suggested that the two points in transverse motion need to accelerate to have a realistic case. Rather than going into complex analysis I present mathematical model of the case and give an example of numerical computations: As it is seen from seen from analysis and plots attached below, for both - accelerated and non-accelerated motions, the distance between moving points 1 and 2 naturally changes when the point 2 is at rest while number 1 is moving. This is direct consequence of relative simultaneity. In the case of uniform motion the distance is increasing and stabilisses after the poin 2 moves out. Accelerated motion behaves in a similar way duting initial stage although velocity profile is naturally different. Quite unexpected result is the fact that in accelerated motion case the distance between both points will be increasing indefinitely (unless we request the stop at y’=0 ) which is not the case in the stationary system. This is yet another confirmation of the illusion associated with the relative simultaneity concept. The distance plots for each case from t’=0 until the first point (1) hits line Y’=0, are below,for the following parametwrs: ,c = 1, v = 05c, 𝑣𝑦 = 0.5𝑐, 𝐷𝑦 = 1, 𝐷𝑥 = 1, 𝑎 = 1, 𝛾 = 1/√1 − 0.52 Two cases were analysed in detail 1. A set of two distinct trajectories of two objects in K moving during the same time interval t : 0≤ t ≤ tN 𝒄𝒕 𝒄𝒕 −𝑫𝒙 𝟎 𝑻𝟏 = [ ] 𝑎𝑛𝑑 𝑻𝟐 = [ ] −𝑫𝒚 + 𝒗𝒚 𝒕 −𝑫𝒚 + 𝒗𝒚 𝒕 𝟎 𝟎
(1)
With initial conditions: P01 ≡ [0,0, -Dy, 0], P02 ≡ [0,-Dx, -Dy, 0], Dy>0,vy,>0,Dx>0. Transformation of the trajectories in accordance with equations: 𝑻′𝟏 = 𝑳𝑻𝟏 𝑎𝑛𝑑 𝑻𝟐 ′ = 𝑳𝑻𝟐 With
(2)
𝛾 L=
−𝑣
𝛾 0 [ 0 𝑐
−𝑣 𝑐
𝛾
𝛾 0 0
0 0 0 0 1 0 0 1]
yields: 𝒄𝒕′ −𝒗𝒕′ 𝑻′𝟏 = [ ], −𝑫𝒚 + 𝒗𝒚 𝒕′ /𝜸 𝟎 𝒄𝒕′ 𝒗𝟐 𝜸𝑫𝒙 ⁄𝒄𝟐 − 𝜸𝑫𝒙 − 𝒗𝒕′ 𝑻′𝟐 = [ ] −𝑫𝒚 − 𝒗𝒗𝒚 𝑫𝒙 ⁄𝒄𝟐 + 𝒗𝒚 𝒕′ /𝜸 𝟎
(3)
2. A set of two distinct trajectories of two objects in K moving during the same time interval t : 0≤ t ≤ tN 𝒄𝒕 𝒄𝒕 −𝑫𝒙 𝟎 𝑻𝟏 = [ ] 𝑎𝑛𝑑 𝑻𝟐 = [ ] 𝟐 −𝑫𝒚 + 𝒂𝒕 /𝟐 −𝑫𝒚 + 𝒂𝒕𝟐 /𝟐 𝟎 𝟎
(4)
With the same initial conditions: P01 ≡ [0,0, -Dy, 0], P02 ≡ [0,-Dx, -Dy, 0], Dy>0,vy,>0,Dx>0. Transformation of the trajectories in accordance with equations: 𝑻′𝟏 = 𝑳𝑻𝟏 𝑎𝑛𝑑 𝑻𝟐 ′ = 𝑳𝑻𝟐 𝒄𝒕′ −𝒗𝜸𝒕′ − 𝜸𝑫𝒙
𝒄𝒕′ −𝒗𝜸𝒕′
𝑻′ 𝟏 =
−𝑫𝒚 + [
𝟎
𝒂𝒕′
, 𝑻′𝟐 =
𝟐
𝟐𝜸𝟐
]
(5)
−𝑫𝒚 + [
𝒂𝒗𝟐 𝑫𝒙 𝟐 𝒕′ 𝟐𝒄𝟒
−
𝟎
𝒂𝒗𝑫𝒙 𝒕′ 𝜸𝒄𝟐
+
𝒂𝒕′
𝟐
𝟐𝜸𝟐
]
The the distance between two points of the two sets of trajectories has been calculated using the Euclidean norm in 3 dimensions x,y,z. The trsnsformstion of initial condidtions indicates that while the starting point of point 1 are 𝑡 ′ = 0, 𝑥 ′ = 0 𝑦 ′ = −𝑫𝒚 , the point 2 initial conditions are: 𝑡 ′ = 𝑣𝛾𝐷𝑥 ⁄𝑐 2 , 𝑥 ′ = 𝛾𝐷𝑥 𝑦 ′ = −𝑫𝒚
The point 2 between t’=0 and 𝑡 ′ = 𝑣𝛾𝐷𝑥 ⁄𝑐 2 is at rest due to relative simultaneity while point 1 is moving in uniform or accelerated motion from t’=0.
