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 $T_{Earth}'$

Twin paradox in special relativity

$\begin{aligned}d \tau_{object}^2&=d t_{alice:object}^2-d r_{alice:object}^2=d t_{bob:object}^2-d r_{bob:object}^2\\\\\gamma&=\frac{1}{\sqrt{1-v^2}},v=\sqrt{1-\frac{1}{\gamma^2}}\\\\d t_{bob:object}&=\gamma d t_{alice:object}-\gamma v d r_{alice:object}\\\\d r_{bob:object}&=-\gamma v d t_{alice:object}+\gamma d r_{alice:object}\end{aligned}$

When object is Bob, $0=d r_{bob:bob}$ so $v=\frac{d r_{alice:bob}}{d t_{alice:bob}}$

Object in Alice-frame, where $E_{alice:object}$ is the energy, which may not be a constant, of the object in Alice-frame:

$\begin{aligned}E_{alice:object}^2 d \tau_{object}^2&=d t_{alice:object}^2\\\\(E_{alice:object}^2-1) d \tau_{object}^2&=d r_{alice:object}^2\\\\E_{alice:object} d \tau_{object}&=d t_{alice:object}\\\\\sqrt{E_{alice:object}^2-1} d \tau_{object}&=\pm d r_{alice:object}\\\\\frac{d r_{alice:object}}{d t_{alice:object}}&=\frac{\sqrt{E_{alice:object}^2-1}}{E_{alice:object}}\end{aligned}$

Object, assuming moving rightward, in Bob-frame:

$\begin{aligned}d t_{bob:object}&=\gamma d t_{alice:object} -\gamma v d r_{alice:object}\\\\&=E_{alice:object} \gamma d \tau_{object}-\gamma v \sqrt{E_{alice:object}^2-1}d \tau_{object}\\\\&=\left(E_{alice:object} \gamma - \sqrt{(\gamma^2-1)(E_{alice:object}^2-1)}\right) d \tau_{object}\\\\d r_{bob:object}&=-\sqrt{\gamma^2-1} d t_{alice:object}+\gamma d r_{alice:object}\\\\&=-E_{alice:object}\sqrt{\gamma^2-1} d \tau_{object}+\gamma \sqrt{E_{alice:object}^2-1}d \tau_{object}\\\\&=\left(\gamma \sqrt{E_{alice:object}^2-1}-E_{alice:object}\sqrt{\gamma^2-1}\right)d \tau_{object}\end{aligned}$

Verify for any object in Bob's frame,

$\begin{aligned}d t_{bob:object}^2-d r_{bob:object}^2&=\left(E_{alice:object}^2 \gamma^2+(\gamma^2-1)(E_{alice:object}^2-1)-2E_{alice:object} \gamma \sqrt{(\gamma^2-1)(E_{alice:object}^2-1)}\right)d \tau_{object}^2\\\\&-\left(E_{alice:object}^2(\gamma^2-1)+\gamma^2 (E_{alice:object}^2-1)-2E_{alice:object}\gamma\sqrt{(\gamma^2-1)(E_{alice:object}^2-1)}\right)d \tau_{object}^2\\\\&=\left(E_{alice:object}^2-(E_{alice:object}^2-1)\right)d \tau_{object}^2\\\\&=d \tau_{object}^2\end{aligned}$

Also,

$E_{bob:object}=E_{alice:object} \gamma - \sqrt{(\gamma^2-1)(E_{alice:object}^2-1)}$

When object is Bob itself:

$\begin{aligned}&v=\frac{d r_{alice:bob}}{d t_{alice:bob}}=\frac{\sqrt{E_{alice:bob}^2-1}}{E_{alice:bob}},E_{alice:bob}=\frac{1}{\sqrt{1-v^2}}\\\\&E_{alice:bob}=\gamma\\\\&d t_{bob:bob}=\left(E_{alice:bob}^2-\sqrt{(E_{alice:bob}^2-1)(E_{alice:bob}^2-1)}\right) d \tau_{bob}=d \tau_{bob}\\\\&d r_{bob:bob}=\left(E_{alice:bob}\sqrt{E_{alice:bob}^2-1}-E_{alice:bob}\sqrt{E_{alice:bob}^2-1}\right)d \tau_{bob}=0\end{aligned}$

The object, moving at constant speed, continuously sends its proper time $X$ to Alice. At Alice's time $T$ which is also Alice's proper time as Alice is idle, she receives data the object sends at object's $X$ proper time, then:

$\begin{aligned}&T-E_{alice:object} X = \sqrt{E_{alice:object}^2-1} X\\\\&X = Sig(T) \equiv \frac{T}{E_{alice:object}+\sqrt{E_{alice:object}^2-1}}\end{aligned}$

Any info of the object beyond its proper time $X$ is a speculation of Alice, such as the object ageing now $\frac{T}{E}$ by the definition "now" being whatever happens at Alice's $T$, as the object might explode at some time after $X$ and before $\frac{T}{E}$. In this regard, twin paradox is a badly posed problem that both Alice and Bob see themselves aging $T$ and the other aging $\frac{T}{E}$ and wonder who is older. In fact, the light cone of the traditional "now" has only one single point intersection at any moment. No submarine commanders have twin paradox when they measure aging of the other submarine by info transmitted at the speed of sound in the sea.

Only the object itself knows object's real aging. At Alice or Bob's current time they might not see the real aging of the object due to the object's space-time point not in Alice or Bob's light cone yet. Suppose Bob is the object and a speculation of Bob's proper time is needed, any formula of $X$ could serve. But by the commander Alice's speculation "the other submarine Bob sends the info at my time $E X$ and at distance $\sqrt{E^2-1}X$, so, if it is still alive now, it must be aging $E X +\sqrt{E^2-1}X$ which is $T$ , the same aging as my submarine"; should the calculated number $E X +\sqrt{E^2-1}X$ be larger, then Alice concludes Bob aging older than Alice and sends a query to Bob. Bob, perhaps years later, replies Alice's query by saying "yes, I passed an unexpected black hole nearby".