3Times Paradox Lorentz Transformation makes Different 3Time Areas (1 is Border)
黒月樹人（KULOTSUKI Kinohito）

Back to Chimera Meam page(Japanese) 

Moving System Time slows down
The most famous element when Special Relativity is used in science fiction movies is to slow down time. The twin paradox is taken up in many Special Relativity.
This has been expanded and interpreted quite freely.
This is because they did not properly understand the meaning of mathematical expression when Einstein leads to it.
Einstein Special Relativity Paper [1] → T. KINEMATICAL PART → §4. Physical Meaning of the Equations Obtained in Respect to Moving Rigid Bodies and Moving Clocks → (Latter half) Moving System Time slows down.
Let me show by a quote a bit after this.
Symbols such as (1−1) for expressions are included for explanation.
Between the quantities x, t, and τ, which refer to the position of the clock, we have, evidently, x=vt (11) and
τ=[1/{1−v^{2}/c^{2} }^{1/2}](t−vx/c^{2}) (1−2)
Therefore,
τ=t{1−v^{2}/c^{2}}^{1/2} (1−3)
=t−(1−{1−v^{2}/c^{2}}^{1/2})t (1−4) … [1]
The formula of (1−4) has been modified for the explanation hereafter.
In (1−3), since (1−5) is always smaller than 1, it leads (1−6).
{1−v^{2}/c^{2}}^{1/2} (1−5)
(1−6)
Einstein uses the relational expression (1−1), when he leads these formulas.
The meaning of (1−1) is that vt is specified as the value of the x coordinate.
Since the equation of the X axis value ξ of the motion system by the formula of Lorentz transformation is (1−7), if (1−1) is substituted into this (1−7), it becomes (1−9).
ξ=β(x−vt) (1−7)
where
β=1/{1−v^{2}/c^{2}}^{1/2} (1−8)
ξ=0 (1−9)
(1−6) and (1−3) are on the origin (ξ = 0) of the moving system.

Looking at the static system K from the moving system k
When the moving system k is translating with the velocity of v for the static system, the Lorentz transformation performs transformations on these coordinate systems.
As translational motion is relative, looking at the static system K from the moving system k, it can be thought that the static system is exercising at the speed of −v.
For the formula of Lorentz transformation, replacing x, y, z, t with ξ, η, ζ, τ, and replacing the speed v with −v, we can derive the inverse transformation equation.

Fig 1 Lorentz transformation and inverse transformation (X axis and time T only)

In this inverse transformation, if the position of (2−1) is designated, x = 0 and (2−2).
ξ=−vτ (2−1)
t=β(τ−v^{2}τ/c^{2})
=β(1−v^{2}/c^{2})τ
=τ[1−v^{2}/c^{2}]^{1/2} (2−2)
Therefore
(2−3)
From above, at the origin ξ = 0 of the moving system k, and at the origin x=0 of the static system K.
There must be a position where between these two origins.

There is a position where the two times coincide with each other at t = τ
Einstein used x = vt as (1−1).
I calculated the t=τ position as the intermediate point is x=(1/2)vt, but this point did not satisfy the condition of t=τ.
I decided to take the opposite way of thinking.
In (3−1), I decided α as the undetermined coefficient.
At this time, τ = t is used as a condition.
x=αvt (3−1)
Substitute this equation into (1−3).
τ=β[1−αv^{2}/c^{2}]t (3−2)
Since τ = t at this time, these variables can be erased.
1=β[1−αv^{2}/c^{2}] (3−3)
Modify this equation.
β^{−1}=1−αv^{2}/c^{2}
αv^{2}/c^{2}=1−β^{−1}
(3−4)
At this time I tried to find x in (3−1), but I cannot decide as it is.
Decide as follows and do numerical calculation.
v=0.8c (3−5)
Let's find α.
α={1/(0.8)^{2}}[1−{1−(0.8)^{2}}^{1/2}]
=(1/0.64)[1−{0.36}^{1/2}]
=0.4/0.64=0.625 (3−6)
Substitute α into (3−1).
x=0.625vt (3−7)
When v = 0.8 c, this position is τ = t.
This is the location of 〇 drawn in light blue in Figure 2.
On (3−8), if γ is obtained under the condition of τ = t, the position of red ○ is determined.
ξ=−γvτ (3−8)
At this time γ becomes the same value as α.
In the numerical calculation with v = 0.8 c, it is (3−9).
ξ=−0.625vt (3−9)
It seems that the position of the light blue ◯ and the position of the red ◯ are far apart.
But when these variables are replaced according to the Lorentz transformation, they are in the same position.

Fig 2 3−Times

Consideration
Previously, it was known that becoming at the origin of the moving system, and conversely, it becomes at the origin of the static system.
In the moving system, it was considered to be in all areas, and in the static system, it was considered to be in all areas.
This was a mistake.
These relationships are established only at the origin of each.
From this, I noticed that there was a position to be between them.
Then, from the time the moving system starts moving relative to the static system, the relationship between these positions and times is divided into the following three.
(1) (area),
(2) (border),
(3) (area).
If there were observer who stayed at position (2)、how can the observer understand what happens in the world of (1) and/or (3) ?
（Written by KLOTSUKI Kinohito, Aug 12, 2018）

Reference (Appendix)
[1] Einstein, Albert (1905d), "On the Electrodynamics of Moving Bodies", Annalen der Physik 17: 89921.
https://www.fourmilab.ch/etexts/einstein/specrel/www/
http://hermes.ffn.ub.es/luisnavarro/nuevo_maletin/Einstein_1905_relativity.pdf
http://www.fourmilab.ch/etexts/einstein/specrel/specrel.pdf

Back to Chimera Meam page(Japanese) 