Here I present an alternative thinking to the frame transformation. Consider the four dimension space time the two reference frames, t-x-y-z and the other t'-x'-y'-z'. For math clean, assume 1 meter is defined as the length of light traveling for 1 second aka $c=1$

Suppose the metric solution derived from the General Relativity's Einstein equation is, $(d \tau)^2= \begin{bmatrix}d t & d x & d y & d z\end{bmatrix} G\begin{bmatrix}d t \\d x \\d y \\d z\end{bmatrix}$

This metric, the real entity in universe, for the flat spacetime is

$(d\tau)^2=(d t')^2-(d x')^2-(d y')^2-(d z')^2$

Some transformation is the matrix B such that $\begin{bmatrix}d t' \\d x' \\d y' \\d z'\end{bmatrix}=B\begin{bmatrix}d t \\d x \\d y \\d z\end{bmatrix}$

To keep the metric the same in the two frames. Therefore, let A and L be the matrix satisfy $\begin{bmatrix}1&0&0&0\\0&-1&0&0\\0&0&-1&0\\0&0&0&-1\end{bmatrix}= L^T \begin{bmatrix}1&0&0&0\\0&-1&0&0\\0&0&-1&0\\0&0&0&-1\end{bmatrix} L$

$G=A^T \begin{bmatrix}1&0&0&0\\0&-1&0&0\\0&0&-1&0\\0&0&0&-1\end{bmatrix} A$

Then set $B = L A$ then the transform $\begin{bmatrix}d t' \\d x' \\d y' \\d z'\end{bmatrix}=L A \begin{bmatrix}d t \\d x \\d y \\d z\end{bmatrix}$

will have

$\begin{aligned}\\(d \tau)^2 = \begin{bmatrix}d t'&d x'&d y'&d z'\end{bmatrix} \begin{bmatrix}1&0&0&0\\0&-1&0&0\\0&0&-1&0\\0&0&0&-1\end{bmatrix} \begin{bmatrix}d t' \\d x' \\d y' \\d z'\end{bmatrix} = \begin{bmatrix}d t&d x&d y&d z\end{bmatrix} A^T L^T \begin{bmatrix}1&0&0&0\\0&-1&0&0\\0&0&-1&0\\0&0&0&-1\end{bmatrix} L A \begin{bmatrix}d t\\d x\\d y\\d z\end{bmatrix}\\=\begin{bmatrix}d t&d x&d y&d z\end{bmatrix} A^T \begin{bmatrix}1&0&0&0\\0&-1&0&0\\0&0&-1&0\\0&0&0&-1\end{bmatrix} A \begin{bmatrix}d t\\d x\\d y\\d z\end{bmatrix}=\begin{bmatrix}d t&d x&d y&d z\end{bmatrix} G \begin{bmatrix}d t\\d x\\d y\\d z\end{bmatrix}\end{aligned}$

Also, if $B = A^{- 1} L A$ and $\begin{bmatrix}d t' \\d x' \\d y' \\d z'\end{bmatrix}=A^{- 1} L A \begin{bmatrix}d t \\d x \\d y \\d z\end{bmatrix}$ then

$\begin{aligned}(d \tau)^2=\begin{bmatrix}d t' & d x' & d y' & d z'\end{bmatrix} G\begin{bmatrix}d t' \\d x' \\d y' \\d z'\end{bmatrix}\\=\begin{bmatrix}d t& d x& d y& d z\end{bmatrix} (A^{-1} L A)^T G (A^{-1} L A) \begin{bmatrix}d t\\d x\\d y\\d z\end{bmatrix}\\=\begin{bmatrix}d t& d x& d y& d z\end{bmatrix} A^T L^T \begin{bmatrix}1&0&0&0\\0&-1&0&0\\0&0&-1&0\\0&0&0&-1\end{bmatrix} L A \begin{bmatrix}d t\\d x\\d y\\d z\end{bmatrix}\\=\begin{bmatrix}d t& d x& d y& d z\end{bmatrix} A^T \begin{bmatrix}1&0&0&0\\0&-1&0&0\\0&0&-1&0\\0&0&0&-1\end{bmatrix} A \begin{bmatrix}d t\\d x\\d y\\d z\end{bmatrix}\\=\begin{bmatrix}d t& d x& d y& d z\end{bmatrix}G \begin{bmatrix}d t\\d x\\d y\\d z\end{bmatrix}=(d\tau)^2\end{aligned}$

$G$, being symmetric, can be column-row operations to become $\begin{bmatrix}1&0&0&0\\0&-1&0&0\\0&0&-1&0\\0&0&0&-1\end{bmatrix}$

A frame transformation matrix $L$ satisfies the above property is named Lorentz Transform which can also be defined by

$\begin{bmatrix}-1&0&0&0\\0&1&0&0\\0&0&1&0\\0&0&0&1\end{bmatrix} = L^T \begin{bmatrix}-1&0&0&0\\0&1&0&0\\0&0&1&0\\0&0&0&1\end{bmatrix} L$

Total 10 equations and 16 variables so, assuming no degeneration, 6 variables will be selected as parameters in the process of "finding normal basis algorithm". For Lorentz of dimension n space-time, the number of parameters are $n^2 - \frac{n(n + 1)}{2} = \frac{n(n - 1)}{2}$

Question 1

how to get a matrix $A$ satisfying

$G \equiv \begin{bmatrix}G&0&0&0\\0&-G^{\frac{1}{3}}&0&0\\0&0&-G^{\frac{1}{3}}&0\\0&0&0&-G^{\frac{1}{3}}\end{bmatrix}= A^T \begin{bmatrix}1&0&0&0\\0&-1&0&0\\0&0&-1&0\\0&0&0&-1\end{bmatrix} A$

Ans. let $A = \begin{bmatrix}G^{\frac{1}{2}}&0&0&0\\0&-G^{\frac{1}{6}}&0&0\\0&0&-G^{\frac{1}{6}}&0\\0&0&0&-G^{\frac{1}{6}}\end{bmatrix}$ is a solution of $G = A^T \begin{bmatrix}1&0&0&0\\0&-1&0&0\\0&0&-1&0\\0&0&0&-1\end{bmatrix} A$

Then

$\begin{bmatrix}d t' \\d x' \\d y' \\d z'\end{bmatrix}= L A \begin{bmatrix}d t\\d x\\d y\\d z\end{bmatrix}$

will have

$(d \tau)^2 = (d t')^2 - (d x')^2 - (d y')^2 - (d z')^2= G (d t)^2 - G^{-\frac{1}{3}} (d x)^2 - G^{-\frac{1}{3}} (d y)^2 - G^{-\frac{1}{3}} (d z)^2$

Question 2

Show the product of two Lorentz transforms $L_1$ and $L_2$ are again a Lorentz transform.

$\begin{aligned}(L_1 L_2)^T \begin{bmatrix}-1&0&0&0\\0&1&0&0\\0&0&1&0\\0&0&0&1\end{bmatrix} L_1 L_2 = L_2^T L_1^T \begin{bmatrix}-1&0&0&0\\0&1&0&0\\0&0&1&0\\0&0&0&1\end{bmatrix} L_1 L_2 \\= L_2^T \begin{bmatrix}-1&0&0&0\\0&1&0&0\\0&0&1&0\\0&0&0&1\end{bmatrix} L_2 = \begin{bmatrix}-1&0&0&0\\0&1&0&0\\0&0&1&0\\0&0&0&1\end{bmatrix}\end{aligned}$

Question 3

What is the Lorentz transform who is of the form

$\begin{bmatrix}*&*&*&*\\ *&*&0&0\\ *&0&*&0\\ *&0&0&*\end{bmatrix}$

Consider the 2nd column whose entries are $L_{01},L_{11}$ then $1 = - L_{01}^2 + L_{11}^2$ so $L_{11} = \sqrt{ 1 + L_{01}^2}$. Consider the 3rd column whose entries are $L_{02},L_{22}$ then $0 = - L_{01} L_{02}$ so $L_{02} = 0$ then $1 = - L_{02}^2 + L_{22}^2$ so $L_{22} = 1$ . Similarly, $L_{03} = 0$ and $L_{33} = 1$

Consider the 1st column whose entries are $L_{00}, L_{10}, L_{20}, L_{30}$ and the orthogonal equation with 2nd column $0 = - L_{00} L_{01} + L_{10} L_{11}$ then $L_{10} = \frac{L_{01} L_{00}}{L_{11}}$

also orthogonal with 3rd column $0 = L_{20}$ similarly $0 = L_{30}$ then $- 1 = - L_{00}^2 + L_{10}^2 = ( - 1 + \frac{L_{01}^2}{1 + L_{01}^2} ) L_{10}^2$ therefore $L_{10} = L_{01}$ and $L_{00} = \sqrt{ 1 + L_{01}^2}$

To conclude, it is of the form:

$\begin{bmatrix} \sqrt{1+L_{01}^2}&L_{01}&0&0\\ L_{01}&\sqrt{1+L_{01}^2}&0&0\\ 0&0&1&0\\ 0&0&0&1\end{bmatrix}$

When $0 = L_{01}d t +\sqrt{1 + L_{01}^2}d x$

aka $\frac{d x}{d t} = - \frac{L_{01}}{\sqrt{1 + L_{01}^2}}$

then $d x' = 0$

so the '-frame which is moving in $X$-axis direction at speed of $v \equiv - \frac{L_{01}}{\sqrt{ 1 + L_{01}^2}}$ so $v^2 (1 + L_{01})^2 = L_{01}^2$

aka $L_{01}^2 = \frac{v^2}{1 - v^2}$

and $1 + L_{01}^2 = \frac{1}{1 - v^2}$

Then define $\gamma \equiv \frac{1}{\sqrt{1 - v^2}} \geq 1$, the form is:

$\begin{bmatrix} \gamma&-\gamma v&0&0\\ -\gamma v&\gamma&0&0\\ 0&0&1&0\\ 0&0&0&1\end{bmatrix}$

similar for moving at $Y$ or $Z$-axis only

Question 4

What is the Lorentz transform who is zero-moving speed?

It is the Euclidean rotation matrix. Let $L = \begin{bmatrix} 1&0\\ 0&K\end{bmatrix}$ where the bottom-right 3-by-3 matrix $K$ is the Euclidean rotation matrix. Since $I = K^T K$ where $I$ for now denote the 3-by-3 identity matrix and 0 is for a 1-by-3 or 3-by-1 zero matrix,

$\begin{aligned}\begin{bmatrix} 1&0\\ 0&K\end{bmatrix}^T \begin{bmatrix}-1&0&0&0\\0&1&0&0\\0&0&1&0\\0&0&0&1\end{bmatrix} \begin{bmatrix} 1&0\\ 0&K\end{bmatrix} = \begin{bmatrix} 1&0\\ 0&K^T\end{bmatrix} \begin{bmatrix}-1&0&0&0\\0&1&0&0\\0&0&1&0\\0&0&0&1\end{bmatrix} \begin{bmatrix} 1&0\\ 0&K\end{bmatrix} \\= \begin{bmatrix} 1&0\\ 0&K^T\end{bmatrix} \begin{bmatrix} -1&0\\ 0&K\end{bmatrix} = \begin{bmatrix} -1&0\\ 0&I\end{bmatrix}=\begin{bmatrix}-1&0&0&0\\0&1&0&0\\0&0&1&0\\0&0&0&1\end{bmatrix}\end{aligned}$

Let $K = \begin{bmatrix} X&Y&Z\end{bmatrix}$, then $I = K^T K = \begin{bmatrix} X^T\\Y^T\\Z^T\end{bmatrix} \begin{bmatrix} X&Y&Z\end{bmatrix} = \begin{bmatrix} X&Y&Z\end{bmatrix} \begin{bmatrix} X^T\\Y^T\\Z^T\end{bmatrix} = X X^T + Y Y^T + Z Z^T$

Question 5

With axis-parallel frame, what is the Lorentz transform of a 3 dimension v-moving frame?

Let the moving is at $v = \begin{bmatrix} v_x\\v_y\\v_z\end{bmatrix}$ . Euclidean-rotation first so that in the '-frame $\frac{1}{|v|} v$ is the X-axis direction vector, freely choose some $Y$ for the $Y$-axis direction vector and some $Z$ for the $Z$-axis direction.

$\begin{bmatrix} d x\\d y\\d z\end{bmatrix} = \begin{bmatrix} \frac{1}{|v|} v&Y&Z\end{bmatrix} \begin{bmatrix} d x'\\d y'\\d z'\end{bmatrix}$

Let $K \equiv \begin{bmatrix} \frac{1}{|v|} v&Y&Z\end{bmatrix}$ then $\begin{bmatrix} d x\\d y\\d z\end{bmatrix} = K \begin{bmatrix} d x'\\d y'\\d z'\end{bmatrix}$. Because $I = K^T K, K^{-1} = K^T$

$\begin{bmatrix} d x'\\d y'\\d z'\end{bmatrix}= K^{-1} \begin{bmatrix} d x\\d y\\d z\end{bmatrix} = K^T \begin{bmatrix} d x\\d y\\d z\end{bmatrix} = \begin{bmatrix} \frac{1}{|v|} v^T\\Y^T\\Z^T\end{bmatrix} \begin{bmatrix} d x\\d y\\d z\end{bmatrix}$

Then go on $X$-axis-only Lorentz transform to ''-frame then inverse the K rotation to '''-frame, so the over all Lorentz transform is

$\begin{aligned}\begin{bmatrix} d t^{'''}\\d x^{'''}\\d y^{'''}\\d z^{'''}\end{bmatrix} = \begin{bmatrix} 1&0&0&0\\0&\frac{1}{|v|}v&Y&Z\end{bmatrix} \begin{bmatrix} \gamma&-\gamma|v|&0&0\\ -\gamma|v|&\gamma&0&0\\ 0&0&1&0\\ 0&0&0&1\end{bmatrix} \begin{bmatrix} 1&0\\0&\frac{1}{|v|}v^T\\0&Y^T\\0&Z^T\end{bmatrix} \begin{bmatrix} d t\\d x\\d y\\d z\end{bmatrix} = \begin{bmatrix} 1&0&0&0\\0&\frac{1}{|v|}v&Y&Z\end{bmatrix} \begin{bmatrix} \gamma&-\gamma v^T\\-\gamma |v|&\frac{\gamma}{|v|}v^T\\0&Y^T\\0&Z^T\end{bmatrix} \begin{bmatrix} d t\\d x\\d y\\d z\end{bmatrix}\\= \begin{bmatrix} \gamma&-\gamma v^T\\-\gamma v&\frac{\gamma}{|v|^2} v v^T + Y Y^T + Z Z^T\end{bmatrix} \begin{bmatrix} d t\\d x\\d y\\d z\end{bmatrix}= \begin{bmatrix} \gamma&-\gamma v^T\\-\gamma v&\frac{\gamma-1}{|v|^2} v v^T + \frac{1}{|v|^2} v v^T + Y Y^T + Z Z^T\end{bmatrix} \begin{bmatrix} d t\\d x\\d y\\d z\end{bmatrix}= \begin{bmatrix} \gamma&-\gamma v^T\\-\gamma v&\frac{\gamma-1}{|v|^2} v v^T + I\end{bmatrix} \begin{bmatrix} d t\\d x\\d y\\d z\end{bmatrix}\end{aligned}$

So the Lorentz transform is $\begin{bmatrix} \gamma&-\gamma v^T\\-\gamma v&\frac{\gamma-1}{|v|^2} v v^T + I\end{bmatrix}$ whose inverse is the matrix with $-v$:

$\begin{aligned}\begin{bmatrix} \gamma&-\gamma v^T\\-\gamma v&\frac{\gamma-1}{|v|^2} + I\end{bmatrix} \begin{bmatrix} \gamma&\gamma v^T\\\gamma v&\frac{\gamma-1}{|v|^2} + I\end{bmatrix} = \begin{bmatrix} \gamma^2 (1-v^T v)&\gamma^2 v^T-\frac{\gamma(\gamma-1)}{|v|^2} v^T v v^T -\gamma v^T\\-\gamma^2 v+\frac{\gamma(\gamma-1)}{|v|^2} v v^T v+\gamma v&-\gamma^2 v v^T+\frac{(\gamma-1)^2}{|v|^4} v v^T v v^T +\frac{2(\gamma-1)}{|v|^2} v v^T + I\end{bmatrix} \\=\begin{bmatrix} \gamma^2 (1-|v|^2)&\gamma^2 v^T-\gamma(\gamma-1)v^T -\gamma v^T\\-\gamma^2 v+\gamma(\gamma-1) v+\gamma v&-\gamma^2 v v^T+\frac{(\gamma-1)^2}{|v|^2} v v^T +\frac{2(\gamma-1)}{|v|^2} v v^T + I\end{bmatrix}=\begin{bmatrix}1&0\\0&\frac{-\gamma^2 |v|^2 +\gamma^2-2\gamma+1+2\gamma-2)}{|v|^2} v v^T + I\end{bmatrix}\\=\begin{bmatrix}1&0\\0&\frac{\gamma^2 (1- |v|^2)-1}{|v|^2} v v^T + I\end{bmatrix} = \begin{bmatrix}1&0\\0&1\end{bmatrix}\end{aligned}$

For moving-frame verification,

$\begin{aligned}\begin{bmatrix}d t\\d x\\d y\\d z\end{bmatrix} = \begin{bmatrix}1\\v\end{bmatrix} d t\\\begin{bmatrix} \gamma&-\gamma v^T\\-\gamma v&\frac{\gamma-1}{|v|^2} v v^T + I\end{bmatrix} \begin{bmatrix}d t\\d x\\d y\\d z\end{bmatrix}= \begin{bmatrix} \gamma&-\gamma v^T\\-\gamma v&\frac{\gamma-1}{|v|^2} v v^T + I\end{bmatrix} \begin{bmatrix}1\\v\end{bmatrix} d t = \begin{bmatrix}{1}\gamma-\gamma v^T v\\-\gamma v + \frac{\gamma-1}{|v|^2} v v^T v + v\end{bmatrix} d t\\= \begin{bmatrix}\gamma-\gamma |v|^2\\-\gamma v + (\gamma-1) v + v\end{bmatrix} d t= \begin{bmatrix}\gamma^{-1}\\0\end{bmatrix} d t\\\begin{bmatrix}\gamma^{-1}\\0\end{bmatrix}=\begin{bmatrix} \gamma&-\gamma v^T\\-\gamma v&\frac{\gamma-1}{|v|^2} v v^T + I\end{bmatrix} \begin{bmatrix}1\\v\end{bmatrix}\end{aligned}$

For idle-object of '-frame:

$\begin{bmatrix} \gamma&-\gamma v^T\\-\gamma v&\frac{\gamma-1}{|v|^2} v v^T + I\end{bmatrix}^{-1} \begin{bmatrix}1\\0\end{bmatrix}=\begin{bmatrix} \gamma&\gamma v^T\\\gamma v&\frac{\gamma-1}{|v|^2} v v^T + I\end{bmatrix} \begin{bmatrix}1\\0\end{bmatrix}=\begin{bmatrix}\gamma\\\gamma v\end{bmatrix}$

$d t = \gamma d t'$

$\begin{bmatrix}d x\\d y\\d z\end{bmatrix}= \gamma v d t'$

For idle-object of orig-frame:

$\begin{bmatrix} \gamma&-\gamma v^T\\-\gamma v&\frac{\gamma-1}{|v|^2} v v^T + I\end{bmatrix} \begin{bmatrix}1\\0\end{bmatrix} = \begin{bmatrix}\gamma\\-\gamma v\end{bmatrix}$

$d t' = \gamma d t$

$\begin{bmatrix}d x'\\d y'\\d z'\end{bmatrix}= - \gamma v d t$

For relationship with classic transform, put back the light speed $c$, it is

$\begin{bmatrix} \gamma&-\frac{\gamma}{c^2} v^T\\-\gamma v&\frac{\gamma-1}{|v|^2} v v^T + I\end{bmatrix}$

where

$\gamma \equiv \frac{1}{\sqrt{1-\frac{|v|^2}{c^2}}} \geq 1$

When $c$ is infinite, it becomes $\begin{bmatrix} 1&0\\-v&I\end{bmatrix}$ classically.

Question 6

Twin paradox. Suppose the orig-frame, ignoring Earth's gravity, is flat so its metric is

$(d \tau)^2=\begin{bmatrix}d t&d x&d y&d z\end{bmatrix} \begin{bmatrix}1&0&0&0\\0&-1&0&0\\0&0&-1&0\\0&0&0&-1\end{bmatrix} \begin{bmatrix}d t\\d x\\d y\\d z\end{bmatrix}$

When '-frame is some huge-accelerating and decelerating space ship, its metric is no longer flat and by general relativity theory the universal metric $d \tau$ becomes

$(d \tau)^2 = \begin{bmatrix}d t'& d x'&d y'&d z'\end{bmatrix} G \begin{bmatrix}d t'\\ d x'\\d y'\\d z'\end{bmatrix}$

for some $G$ and the frame-transform is therefore

$\begin{bmatrix}d t\\d x\\d y\\d z\end{bmatrix}=\begin{bmatrix} \gamma&\gamma v^T\\\gamma v&\frac{\gamma-1}{|v|^2} v v^T + I\end{bmatrix} A \begin{bmatrix}d t'\\ d x'\\d y'\\d z'\end{bmatrix}$

for some $v, A$ where

$G = A^T \begin{bmatrix}1&0&0&0\\0&-1&0&0\\0&0&-1&0\\0&0&0&-1\end{bmatrix} A$

Suppose an example. $G = \begin{bmatrix}G_0&0&0&0\\0&-G_1&0&0\\0&0&-G_2&0\\0&0&0&-G_3\end{bmatrix}$

and

$A = \begin{bmatrix}G_0^{0.5}&0&0&0\\0&G_1^{0.5}&0&0\\0&0&G_2^{0.5}&0\\0&0&0&G_3^{0.5}\end{bmatrix}$

So the space-ship twin object in '-frame is

$\begin{bmatrix}d t'\\ d x'\\d y'\\d z'\end{bmatrix}=\begin{bmatrix}G_0^{-0.5}d \tau\\0\\0\\0\end{bmatrix}$

and the same space-ship twin object in earth-frame is

$\begin{bmatrix}d t\\d x\\d y\\d z\end{bmatrix}=\begin{bmatrix} \gamma&\gamma v^T\\\gamma v&\frac{\gamma-1}{|v|^2} v v^T + I\end{bmatrix} A \begin{bmatrix}G_0^{-0.5}d \tau\\0\\0\\0\end{bmatrix}=\begin{bmatrix} \gamma&\gamma v^T\\\gamma v&\frac{\gamma-1}{|v|^2} v v^T + I\end{bmatrix} \begin{bmatrix}d \tau\\0\\0\\0\end{bmatrix}= \begin{bmatrix}1\\v\end{bmatrix} \gamma d \tau$

While movement of different objects seen in the same frame can have different space-time length, The movement of an object in the space time have the same space-time length in every frame and can be parameterized by $\tau$. The $G_0,G_1,G_2,G_3,v$ can be seen as a function of $\tau$. Then:

$T_{SpaceShip}'= \int_0^{S_{SpaceShip}} G_0^{-0.5} d \tau$

$T_{SpaceShip} = \int_0^{S_{SpaceShip}} \gamma d \tau$

where $S_{SpaceShip}$ is such that for the loop, seeing "the space ship away and back":

$0 = \int_0^{S_{SpaceShip}} v \gamma d \tau$

Similarly, for the Earth-twin object in earth-frame is

$\begin{bmatrix}d t\\ d x\\d y\\d z\end{bmatrix}=\begin{bmatrix}d \tau\\0\\0\\0\end{bmatrix}$

who in '-frame is

$\begin{bmatrix}d t'\\ d x'\\d y'\\d z'\end{bmatrix} = A^{-1} \begin{bmatrix} \gamma&-\gamma v^T\\-\gamma v&\frac{\gamma-1}{|v|^2} v v^T + I\end{bmatrix} \begin{bmatrix}d \tau\\0\\0\\0\end{bmatrix} = \begin{bmatrix}G_0^{-0.5}\\-G_1^{-0.5} v_1\\-G_2^{-0.5} v_2\\-G_3^{-0.5} v_3\end{bmatrix} \gamma d \tau$

For a close loop, seeing "the Earth away and back",

$T_{Earth}' = \int_0^{S_{Earth}} \gamma G_0^{-0.5} d \tau$

$T_{Earth} = \int_0^{S_{Earth}} d \tau$

for some different $S_{Earth}$ such that

$0 = \int_0^{S_{Earth}} \begin{bmatrix}-G_1^{-0.5} v_1\\-G_2^{-0.5} v_2\\-G_3^{-0.5} v_3\end{bmatrix} \gamma d \tau$

Mathematically, the paradox is:

• space-ship twin thinks his own aging $T_{SpaceShip}'$ and Earth twin thinks his space-ship twin aging $T_{SpaceShip}$
• Earth twin thinks his own aging $T_{Earth}$ and space-ship twin thinks his Earth twin aging $T_{Earth}'$