TransformInRelativity

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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=1c=1

Suppose the metric solution derived from the General Relativity's Einstein equation is, say,

(dτ) 2=[dt dx dy dz]G[dt dx dy dz](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

(dτ) 2=(dt′) 2−(dx′) 2−(dy′) 2−(dz′) 2(d\tau)^2=(d t')^2-(d x')^2-(d y')^2-(d z')^2

Some transformation is the matrix B such that

[dt′ dx′ dy′ dz′]=B[dt dx dy dz]\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 AA and LL be the matrix satisfy

[1 0 0 0 0 −1 0 0 0 0 −1 0 0 0 0 −1] =L T[1 0 0 0 0 −1 0 0 0 0 −1 0 0 0 0 −1]L G =A T[1 0 0 0 0 −1 0 0 0 0 −1 0 0 0 0 −1]A\begin{aligned}\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\end{aligned}

Then set B=LAB = L A then the transform

[dt′ dx′ dy′ dz′]=LA[dt dx dy dz]\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 result in

(dτ) 2 =[dt′ dx′ dy′ dz′][1 0 0 0 0 −1 0 0 0 0 −1 0 0 0 0 −1][dt′ dx′ dy′ dz′] =[dt dx dy dz]A TL T[1 0 0 0 0 −1 0 0 0 0 −1 0 0 0 0 −1]LA[dt dx dy dz] =[dt dx dy dz]A T[1 0 0 0 0 −1 0 0 0 0 −1 0 0 0 0 −1]A[dt dx dy dz] =[dt dx dy dz]G[dt dx dy dz]\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 −1LAB = A^{- 1} L A and [dt′ dx′ dy′ dz′]=A −1LA[dt dx dy dz]\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

(dτ) 2 =[dt′ dx′ dy′ dz′]G[dt′ dx′ dy′ dz′] =[dt dx dy dz](A −1LA) TG(A −1LA)[dt dx dy dz] =[dt dx dy dz]A TL T[1 0 0 0 0 −1 0 0 0 0 −1 0 0 0 0 −1]LA[dt dx dy dz] =[dt dx dy dz]A T[1 0 0 0 0 −1 0 0 0 0 −1 0 0 0 0 −1]A[dt dx dy dz] =[dt dx dy dz]G[dt dx dy dz] =(dτ) 2\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}

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

[−1 0 0 0 0 1 0 0 0 0 1 0 0 0 0 1]=L T[−1 0 0 0 0 1 0 0 0 0 1 0 0 0 0 1]L\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−n(n+1)2=n(n−1)2n^2 - \frac{n(n + 1)}{2} = \frac{n(n - 1)}{2}

Suppose a movement of a spaceship in GG frame:

G 0 −1(E 1 2+E 2 2+E 3 2)(dτ) 2 =G 0(dt) 2 (G 0 −1E 1 2−13)(dτ) 2 =G 1(dx) 2 (G 0 −1E 2 2−13)(dτ) 2 =G 2(dy) 2 (G 0 −1E 3 2−13)(dτ) 2 =G 3(dz) 2\begin{aligned}G_0^{-1} \left(E_1^2+E_2^2+E_3^2\right) (d \tau)^2&=G_0 (d t)^2\\\left(G_0^{-1}E_1^2 -\frac{1}{3}\right)(d \tau)^2&=G_1(d x)^2\\\left(G_0^{-1}E_2^2-\frac{1}{3}\right)(d \tau)^2&=G_2(d y)^2\\\left(G_0^{-1}E_3^2-\frac{1}{3}\right)(d \tau)^2&=G_3(d z)^2\end{aligned}

where one can verify

G 0(dt) 2−G 1(dx) 2−G 2(dy) 2−G 3(dz) 2\begin{aligned}G_0(d t)^2-G_1(d x)^2-G_2(d y)^2-G_3(d z)^2\end{aligned}

Then with help of LAL A or (L 2A 2) −1L 1A 1(L_2 A_2)^{-1}L_1 A_1 one can see the movement of the spaceship in other frames.

Question 1

how to get a matrix AA satisfying GG ?

G≡[G 0 0 0 0 −G 13 0 0 0 0 −G 13 0 0 0 0 −G 13]=A T[1 0 0 0 0 −1 0 0 0 0 −1 0 0 0 0 −1]AG \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. GG, being symmetric, can be column-row operations to become [1 0 0 0 0 −1 0 0 0 0 −1 0 0 0 0 −1]\begin{bmatrix}1&0&0&0\\0&-1&0&0\\0&0&-1&0\\0&0&0&-1\end{bmatrix}

Let A=[G 12 0 0 0 0 G 16 0 0 0 0 G 16 0 0 0 0 G 16]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} which leads to G=A T[1 0 0 0 0 −1 0 0 0 0 −1 0 0 0 0 −1]AG = 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

[dt′ dx′ dy′ dz′]=LA[dt dx dy dz]\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 lead to

(dτ) 2=(dt′) 2−(dx′) 2−(dy′) 2−(dz′) 2=G(dt) 2−G −13(dx) 2−G −13(dy) 2−G −13(dz) 2(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 1L_1 and L 2L_2 are again a Lorentz transform.

(L 1L 2) T[−1 0 0 0 0 1 0 0 0 0 1 0 0 0 0 1]L 1L 2 =L 2 TL 1 T[−1 0 0 0 0 1 0 0 0 0 1 0 0 0 0 1]L 1L 2 =L 2 T[−1 0 0 0 0 1 0 0 0 0 1 0 0 0 0 1]L 2 =[−1 0 0 0 0 1 0 0 0 0 1 0 0 0 0 1]\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

[* * * * * * 0 0 * 0 * 0 * 0 0 *]\begin{bmatrix}*&*&*&*\\ *&*&0&0\\ *&0&*&0\\ *&0&0&*\end{bmatrix}

Consider the 2nd column whose entries are L 01,L 11L_{01},L_{11} then 1=−L 01 2+L 11 21 = - L_{01}^2 + L_{11}^2 so L 11=1+L 01 2L_{11} = \sqrt{ 1 + L_{01}^2}. Consider the 3rd column whose entries are L 02,L 22L_{02},L_{22} then 0=−L 01L 020 = - L_{01} L_{02} so L 02=0L_{02} = 0 then 1=−L 02 2+L 22 21 = - L_{02}^2 + L_{22}^2 so L 22=1L_{22} = 1 . Similarly, L 03=0L_{03} = 0 and L 33=1L_{33} = 1

Consider the 1st column whose entries are L 00,L 10,L 20,L 30L_{00}, L_{10}, L_{20}, L_{30} and the orthogonal equation with 2nd column 0=−L 00L 01+L 10L 110 = - L_{00} L_{01} + L_{10} L_{11} then L 10=L 01L 00L 11L_{10} = \frac{L_{01} L_{00}}{L_{11}}

also orthogonal with 3rd column 0=L 200 = L_{20} similarly 0=L 300 = L_{30} then −1=−L 00 2+L 10 2=(−1+L 01 21+L 01 2)L 10 2- 1 = - L_{00}^2 + L_{10}^2 = ( - 1 + \frac{L_{01}^2}{1 + L_{01}^2} ) L_{10}^2 therefore L 10=L 01L_{10} = L_{01} and L 00=1+L 01 2L_{00} = \sqrt{ 1 + L_{01}^2}

To conclude, it is of the form:

[1+L 01 2 L 01 0 0 L 01 1+L 01 2 0 0 0 0 1 0 0 0 0 1]\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 01dt+1+L 01 2dx0 = L_{01}d t +\sqrt{1 + L_{01}^2}d x aka dxdt=−L 011+L 01 2\frac{d x}{d t} = - \frac{L_{01}}{\sqrt{1 + L_{01}^2}} then dx′=0d x' = 0

so the '-frame which is moving in XX-axis direction at speed of v≡−L 011+L 01 2v \equiv - \frac{L_{01}}{\sqrt{ 1 + L_{01}^2}}

so v 2(1+L 01) 2=L 01 2v^2 (1 + L_{01})^2 = L_{01}^2 aka L 01 2=v 21−v 2L_{01}^2 = \frac{v^2}{1 - v^2} and 1+L 01 2=11−v 21 + L_{01}^2 = \frac{1}{1 - v^2}

Then define γ≡11−v 2≥1\gamma \equiv \frac{1}{\sqrt{1 - v^2}} \geq 1, the form is:

[γ −γv 0 0 −γv γ 0 0 0 0 1 0 0 0 0 1]\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 YY or ZZ-axis only

Question 4

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

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

[1 0 0 K] T[−1 0 0 0 0 1 0 0 0 0 1 0 0 0 0 1][1 0 0 K] =[1 0 0 K T][−1 0 0 0 0 1 0 0 0 0 1 0 0 0 0 1][1 0 0 K] =[1 0 0 K T][−1 0 0 K] =[−1 0 0 I] =[−1 0 0 0 0 1 0 0 0 0 1 0 0 0 0 1]\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}

Also, K=[X Y Z]K = \begin{bmatrix} X&Y&Z\end{bmatrix}, then

I=K TK=[X T Y T Z T][X Y Z]=[X Y Z][X T Y T Z T]=XX T+YY T+ZZ TI = 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=[v x v y v z]v = \begin{bmatrix} v_x\\v_y\\v_z\end{bmatrix} . Euclidean-rotation first so that in the '-frame 1|v|v\frac{1}{|v|} v is the X-axis direction vector, freely choose some YY for the YY-axis direction vector and some ZZ for the ZZ-axis direction.

[dx dy dz]=[1|v|v Y Z][dx′ dy′ dz′]\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≡[1|v|v Y Z]K \equiv \begin{bmatrix} \frac{1}{|v|} v&Y&Z\end{bmatrix} then [dx dy dz]=K[dx′ dy′ dz′]\begin{bmatrix} d x\\d y\\d z\end{bmatrix} = K \begin{bmatrix} d x'\\d y'\\d z'\end{bmatrix}. Because I=K TK,K −1=K TI = K^T K, K^{-1} = K^T

[dx′ dy′ dz′]=K −1[dx dy dz]=K T[dx dy dz]=[1|v|v T Y T Z T][dx dy dz]\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

[dt ‴ dx ‴ dy ‴ dz ‴] =[1 0 0 0 0 1|v|v Y Z][γ −γ|v| 0 0 −γ|v| γ 0 0 0 0 1 0 0 0 0 1][1 0 0 1|v|v T 0 Y T 0 Z T][dt dx dy dz] =[1 0 0 0 0 1|v|v Y Z][γ −γv T −γ|v| γ|v|v T 0 Y T 0 Z T][dt dx dy dz] =[γ −γv T −γv γ|v| 2vv T+YY T+ZZ T][dt dx dy dz] =[γ −γv T −γv γ−1|v| 2vv T+1|v| 2vv T+YY T+ZZ T][dt dx dy dz] =[γ −γv T −γv γ−1|v| 2vv T+I][dt dx dy dz]\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 [γ −γv T −γv γ−1|v| 2vv T+I]\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-v:

[γ −γv T −γv γ−1|v| 2vv T+I][γ γv T γv γ−1|v| 2vv T+I] =[γ 2(1−v Tv) γ 2v T−γ(γ−1)|v| 2v Tvv T−γv T −γ 2v+γ(γ−1)|v| 2vv Tv+γv −γ 2vv T+(γ−1) 2|v| 4vv Tvv T+2(γ−1)|v| 2vv T+I] =[γ 2(1−|v| 2) γ 2v T−γ(γ−1)v T−γv T −γ 2v+γ(γ−1)v+γv −γ 2vv T+(γ−1) 2|v| 2vv T+2(γ−1)|v| 2vv T+I] =[1 0 0 −γ 2|v| 2+γ 2−2γ+1+2γ−2)|v| 2vv T+I] =[1 0 0 γ 2(1−|v| 2)−1|v| 2vv T+I] =[1 0 0 1]\begin{aligned}&\begin{bmatrix} \gamma&-\gamma v^T\\-\gamma v&\frac{\gamma-1}{|v|^2} v v^T + I\end{bmatrix} \begin{bmatrix} \gamma&\gamma v^T\\\gamma v&\frac{\gamma-1}{|v|^2} v v^T + 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,

[dt dx dy dz] =[1 v]dt [γ −γv T −γv γ−1|v| 2vv T+I][dt dx dy dz] =[γ −γv T −γv γ−1|v| 2vv T+I][1 v]dt =[1γ−γv Tv −γv+γ−1|v| 2vv Tv+v]dt =[γ−γ|v| 2 −γv+(γ−1)v+v]dt =[γ −1 0]dt [γ −1 0] =[γ −γv T −γv γ−1|v| 2vv T+I][1 v]\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:

[γ −γv T −γv γ−1|v| 2vv T+I] −1[1 0]=[γ γv T γv γ−1|v| 2vv T+I][1 0]=[γ γv]\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}

dt=γdt′d t = \gamma d t'

[dx dy dz]=γvdt′\begin{bmatrix}d x\\d y\\d z\end{bmatrix}= \gamma v d t'

For idle-object of orig-frame:

[γ −γv T −γv γ−1|v| 2vv T+I][1 0]=[γ −γv]\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}

dt′=γdtd t' = \gamma d t

[dx′ dy′ dz′]=−γvdt\begin{bmatrix}d x'\\d y'\\d z'\end{bmatrix}= - \gamma v d t

For relationship with classic transform, put back the light speed cc, it is [γ −γc 2v T −γv γ−1|v| 2vv T+I]\begin{bmatrix} \gamma&-\frac{\gamma}{c^2} v^T\\-\gamma v&\frac{\gamma-1}{|v|^2} v v^T + I\end{bmatrix} where γ≡11−|v| 2c 2\gamma \equiv \frac{1}{\sqrt{1-\frac{|v|^2}{c^2}}}. When cc is infinite, it becomes [1 0 −v I]\begin{bmatrix} 1&0\\-v&I\end{bmatrix} classically.

Question 6

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

(dτ) 2=[dt dx dy dz][1 0 0 0 0 −1 0 0 0 0 −1 0 0 0 0 −1][dt dx dy dz](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 dτd \tau becomes

(dτ) 2=[dt′ dx′ dy′ dz′]G[dt′ dx′ dy′ dz′](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 GG and the frame-transform is therefore

[dt dx dy dz]=[γ γv T γv γ−1|v| 2vv T+I]A[dt′ dx′ dy′ dz′]\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,Av, A where

G=A T[1 0 0 0 0 −1 0 0 0 0 −1 0 0 0 0 −1]AG = 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=[G 0 0 0 0 0 −G 1 0 0 0 0 −G 2 0 0 0 0 −G 3]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=[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]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

[dt′ dx′ dy′ dz′]=[G 0 −0.5dτ 0 0 0]\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

[dt dx dy dz]=[γ γv T γv γ−1|v| 2vv T+I]A[G 0 −0.5dτ 0 0 0]=[γ γv T γv γ−1|v| 2vv T+I][dτ 0 0 0]=[1 v]γdτ\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 τ\tau. The G 0,G 1,G 2,G 3,vG_0,G_1,G_2,G_3,v can be seen as a function of τ\tau. Then:

T SpaceShip′ =∫ 0 S SpaceShipG 0 −0.5dτ T SpaceShip =∫ 0 S SpaceShipγdτ\begin{aligned}T_{SpaceShip}'&= \int_0^{S_{SpaceShip}} G_0^{-0.5} d \tau\\T_{SpaceShip} &= \int_0^{S_{SpaceShip}} \gamma d \tau\end{aligned}

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

0=∫ 0 S SpaceShipvγdτ0 = \int_0^{S_{SpaceShip}} v \gamma d \tau

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

[dt dx dy dz]=[dτ 0 0 0]\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

[dt′ dx′ dy′ dz′]=A −1[γ −γv T −γv γ−1|v| 2vv T+I][dτ 0 0 0]=[G 0 −0.5 −G 1 −0.5v 1 −G 2 −0.5v 2 −G 3 −0.5v 3]γdτ\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′ =∫ 0 S EarthγG 0 −0.5dτ T Earth =∫ 0 S Earthdτ\begin{aligned}T_{Earth}' &= \int_0^{S_{Earth}} \gamma G_0^{-0.5} d \tau\\T_{Earth} &= \int_0^{S_{Earth}} d \tau\end{aligned}

for some different S EarthS_{Earth} such that

0=∫ 0 S Earth[−G 1 −0.5v 1 −G 2 −0.5v 2 −G 3 −0.5v 3]γdτ0 = \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 that object's aging is caused by the accumulated dτd \tau of the object, TT is confusing unless G 0=1G_0=1 and movement in other 3 dimensions are idle.

To avoid math symbols ambiguity and clearly, a {A:B} subscript indicates its a variable in A frame about object B and dr 2d r^2 means (dr) 2(d r)^2. For example in 1 dim movement of the flat spacetime for earth, the metric is

dτ object 2=dt earth:object 2−dt earth:object 2d \tau_{object}^2=d t_{earth:object}^2-d t_{earth:object}^2 and the moving object and the moving Bob

dt earth:object 2 =E earth:object 2dτ object 2 dr earth:object 2 =(E earth:object 2−1)dτ object 2 dt earth:bob 2 =E earth:bob 2dτ bob 2 dr earth:bob 2 =(E earth:bob 2−1)dτ bob 2\begin{aligned}d t_{earth:object}^2&=E_{earth:object}^2 d \tau_{object}^2\\\\d r_{earth:object}^2&=(E_{earth:object}^2-1) d \tau_{object}^2\\\\d t_{earth:bob}^2&=E_{earth:bob}^2 d \tau_{bob}^2\\\\d r_{earth:bob}^2&=(E_{earth:bob}^2-1) d \tau_{bob}^2\end{aligned}

By general relativity, the metric in Bob frame is, assuming Bob moving rightward

α(τ bob) ≡1E earth:bob(τ bob) 2−1dE earth:bob(τ bob)dτ bob dτ object 2 =(1+α(τ bob)r bob:object)dt bob:object 2−dr bob:object 2 G =[1+α(τ bob)r bob:object 0 0 −1] A =[1+α(τ bob)r bob:object 0 0 1] L =[E earth:bob(τ bob) E earth:bob 2(τ bob)−1 E earth:bob 2(τ bob)−1 E earth:bob(τ bob)]\begin{aligned}\alpha(\tau_{bob}) &\equiv \frac{1}{\sqrt{E_{earth:bob}(\tau_{bob})^2-1}} \frac{d E_{earth:bob}(\tau_{bob})}{d \tau_{bob}}\\\\d \tau_{object}^2&=\left(1+\alpha(\tau_{bob}) r_{bob:object}\right) d t_{bob:object}^2- d r_{bob:object}^2\\\\G &=\begin{bmatrix}1+\alpha(\tau_{bob}) r_{bob:object}&0\\0&-1\end{bmatrix}\\\\A&=\begin{bmatrix}\sqrt{1+\alpha(\tau_{bob}) r_{bob:object}}&0\\0&1\end{bmatrix}\\\\L&=\begin{bmatrix}E_{earth:bob}(\tau_{bob})&\sqrt{E_{earth:bob}^2(\tau_{bob})-1}\\\sqrt{E_{earth:bob}^2(\tau_{bob})-1}&E_{earth:bob}(\tau_{bob})\end{bmatrix}\end{aligned}

So

[dt earth:object dr earth:object] =LA[dt bob:object dr bob:object] dτ object 2 =[dt earth:object dr earth:object][1 0 0 −1][dt earth:object dr earth:object] =[dt earth:object dr earth:object]G[dt earth:object dr earth:object] [dt bob:object dr bob:object] =A −1L −1[dt earth:object dr earth:object] =[11+α(τ bob)r bob:object 0 0 1][E earth:bob(τ bob) −E earth:bob 2(τ bob)−1 −E earth:bob 2(τ bob)−1 E earth:bob(τ bob)][E earth:object(τ object) E earth:object 2(τ object)−1]dτ object =[11+α(τ bob)r bob:object(E earth:bob(τ bob)E earth:object(τ object)−E earth:bob 2(τ bob)−1E earth:object 2(τ object)−1) E earth:bob(τ bob)E earth:object 2(τ object)−1−E earth:bob 2(τ bob)−1E earth:object(τ object)]dτ object\begin{aligned}\begin{bmatrix}d t_{earth:object}\\d r_{earth:object}\end{bmatrix}&=L A \begin{bmatrix}d t_{bob:object}\\d r_{bob:object}\end{bmatrix}\\\\d \tau_{object}^2&=\begin{bmatrix}d t_{earth:object}&d r_{earth:object}\end{bmatrix}\begin{bmatrix}1&0\\0&-1\end{bmatrix}\begin{bmatrix}d t_{earth:object}\\d r_{earth:object}\end{bmatrix}\\\\&=\begin{bmatrix}d t_{earth:object}&d r_{earth:object}\end{bmatrix}G\begin{bmatrix}d t_{earth:object}\\d r_{earth:object}\end{bmatrix}\\\\\begin{bmatrix}d t_{bob:object}\\d r_{bob:object}\end{bmatrix}&=A^{-1}L^{-1}\begin{bmatrix}d t_{earth:object}\\d r_{earth:object}\end{bmatrix}\\\\&=\begin{bmatrix}\frac{1}{\sqrt{1+\alpha(\tau_{bob}) r_{bob:object}}}&0\\0&1\end{bmatrix}\begin{bmatrix}E_{earth:bob}(\tau_{bob})&-\sqrt{E_{earth:bob}^2(\tau_{bob})-1}\\-\sqrt{E_{earth:bob}^2(\tau_{bob})-1}&E_{earth:bob}(\tau_{bob})\end{bmatrix}\begin{bmatrix}E_{earth:object}(\tau_{object})\\\sqrt{E_{earth:object}^2(\tau_{object})-1}\end{bmatrix} d \tau_{object}\\\\&=\begin{bmatrix}\frac{1}{\sqrt{1+\alpha(\tau_{bob}) r_{bob:object}}}\left(E_{earth:bob}(\tau_{bob})E_{earth:object}(\tau_{object})-\sqrt{E_{earth:bob}^2(\tau_{bob})-1}\sqrt{E_{earth:object}^2(\tau_{object})-1}\right)\\E_{earth:bob}(\tau_{bob})\sqrt{E_{earth:object}^2(\tau_{object})-1}-\sqrt{E_{earth:bob}^2(\tau_{bob})-1}E_{earth:object}(\tau_{object})\end{bmatrix} d \tau_{object}\end{aligned}

Twin paradox in special relativity

dτ object 2 =dt alice:object 2−dr alice:object 2=dt bob:object 2−dr bob:object 2 γ =11−v 2,v=1−1γ 2 dt bob:object =γdt alice:object−γvdr alice:object dr bob:object =−γvdt alice:object+γdr alice:object\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=dr bob:bob0=d r_{bob:bob} so v=dr alice:bobdt alice:bobv=\frac{d r_{alice:bob}}{d t_{alice:bob}}

Object in Alice-frame, where E alice:objectE_{alice:object} is the energy per unit mass, which may not be a constant, of the object in Alice-frame:

E alice:object 2dτ object 2 =dt alice:object 2 (E alice:object 2−1)dτ object 2 =dr alice:object 2 E alice:objectdτ object =dt alice:object E alice:object 2−1dτ object =±dr alice:object dr alice:objectdt alice:object =E alice:object 2−1E alice:object\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}

As a result,

γ=11−v 2,v=dr alice:bobdt alice:bob=E alice:bob 2−1E alice:bob\gamma=\frac{1}{\sqrt{1-v^2}},v=\frac{d r_{alice:bob}}{d t_{alice:bob}}=\frac{\sqrt{E_{alice:bob}^2-1}}{E_{alice:bob}}

It follows γ=E alice:bob,γv=E alice:bob 2−1\gamma=E_{alice:bob}, \gamma v=\sqrt{E_{alice:bob}^2-1}

Object in Bob-frame:

dt bob:object =E alice:bobdt alice:object−E alice:bob 2−1dr alice:object =(E alice:bobE alice:object−(E alice:bob 2−1)(E alice:object 2−1))dτ object dr bob:object =−E alice:bob 2−1dt alice:object+E alice:bobdr alice:object =(E alice:bobE alice:object 2−1−E alice:objectE alice:bob 2−1)dτ object\begin{aligned}d t_{bob:object}&=E_{alice:bob} d t_{alice:object} -\sqrt{E_{alice:bob}^2-1} d r_{alice:object}\\\\&=\left(E_{alice:bob} E_{alice:object} - \sqrt{(E_{alice:bob}^2-1)(E_{alice:object}^2-1)}\right) d \tau_{object}\\\\d r_{bob:object}&=-\sqrt{E_{alice:bob}^2-1} d t_{alice:object}+E_{alice:bob} d r_{alice:object}\\\\&=\left(E_{alice:bob} \sqrt{E_{alice:object}^2-1}-E_{alice:object}\sqrt{E_{alice:bob}^2-1}\right)d \tau_{object}\end{aligned}

Verify for any object in Bob's frame,

dt bob:object 2−dr bob:object 2 =(E alice:object 2E alice:bob 2+(E alice:bob 2−1)(E alice:object 2−1)−2E alice:objectE alice:bob(E alice:bob 2−1)(E alice:object 2−1))dτ object 2 −(E alice:object 2(E alice:bob 2−1)+E alice:bob 2(E alice:object 2−1)−2E alice:objectE alice:bob(E alice:bob 2−1)(E alice:object 2−1))dτ object 2 =(E alice:object 2−(E alice:object 2−1))dτ object 2 =dτ object 2\begin{aligned}&d t_{bob:object}^2-d r_{bob:object}^2\\\\&=\left(E_{alice:object}^2 E_{alice:bob}^2+(E_{alice:bob}^2-1)(E_{alice:object}^2-1)-2E_{alice:object} E_{alice:bob} \sqrt{(E_{alice:bob}^2-1)(E_{alice:object}^2-1)}\right)d \tau_{object}^2\\\\&-\left(E_{alice:object}^2(E_{alice:bob}^2-1)+E_{alice:bob}^2 (E_{alice:object}^2-1)-2E_{alice:object}E_{alice:bob}\sqrt{(E_{alice:bob}^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:bobE alice:object−(E alice:bob 2−1)(E alice:object 2−1)E_{bob:object}=E_{alice:bob} E_{alice:object} - \sqrt{(E_{alice:bob}^2-1)(E_{alice:object}^2-1)}

When object is Bob itself:

dt bob:bob=(E alice:bob 2−(E alice:bob 2−1)(E alice:bob 2−1))dτ bob=dτ bob dr bob:bob=(E alice:bobE alice:bob 2−1−E alice:bobE alice:bob 2−1)dτ bob=0\begin{aligned}&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 XX to Alice. At Alice's time TT which is also Alice's proper time as Alice is idle, she receives data the object sends at object's XX proper time, then:

T−E alice:objectX=E alice:object 2−1X X=Sig alice:object(T)≡TE alice:object+E alice:object 2−1\begin{aligned}&T-E_{alice:object} X = \sqrt{E_{alice:object}^2-1} X\\\\&X = Sig_{alice:object}(T) \equiv \frac{T}{E_{alice:object}+\sqrt{E_{alice:object}^2-1}}\end{aligned}

Any info of the object beyond its proper time XX is a speculation of Alice, such as the object ageing now TE\frac{T}{E} by the definition "now" being whatever happens at Alice's TT, as the object might explode at some time after XX and before TE\frac{T}{E}. In this regard, twin paradox is a badly posed problem that Alice and Bob see themselves aging TT and the other aging TE\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. Would Alice and Bob skype with each other continuously, they would be just like seeing old and slow movie of each other and see the other one aging less.

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 by Alice, any formula of XX could serve. But by the commander Alice's speculation "the other submarine Bob sends the info at my time EXE X and at distance E 2−1X\sqrt{E^2-1}X, so, if it is still alive now, it must be aging EX+E 2−1XE X +\sqrt{E^2-1}X which is TT , the same aging as my submarine"; should the calculated number EX+E 2−1XE X +\sqrt{E^2-1}X be smaller, then Alice concludes Bob aging younger 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" even both do not meet again yet.

Initially Alice, Bob, Charlie at the same point, r alice:bob=r alice:charlie=τ allice=τ bob=τ charlie=0r_{alice:bob}=r_{alice:charlie}=\tau_{allice}=\tau_{bob}=\tau_{charlie}=0. At Alice's time τ alice=T\tau_{alice}=T, Alice received Bob's SOS signal saying his τ bob\tau_{bob} and his spaceship stops moving with respect to Alice. Alice then sends Charlie for rescue and bring back to Alice.

dt charlie:bob =(E alice:charlieE alice:bob−(E alice:charlie 2−1)(E alice:bob 2−1))dτ bob =E alice:charliedτ bob dr charlie:bob =(E alice:charlieE alice:bob 2−1−E alice:bobE alice:charlie 2−1)dτ bob =−E alice:charlie 2−1dτ bob\begin{aligned}d t_{charlie:bob}&=\left(E_{alice:charlie} E_{alice:bob} - \sqrt{(E_{alice:charlie}^2-1)(E_{alice:bob}^2-1)}\right) d \tau_{bob}\\\\&=E_{alice:charlie} d \tau_{bob}\\\\d r_{charlie:bob}&=\left(E_{alice:charlie} \sqrt{E_{alice:bob}^2-1}-E_{alice:bob}\sqrt{E_{alice:charlie}^2-1}\right)d \tau_{bob}\\\\&=-\sqrt{E_{alice:charlie}^2-1} d \tau_{bob}\end{aligned}

Any frame can be used for calculation of aging but here Alice-frame is used because it is flat and mathematically simpler. Total aging are:

τ alice =T+2E alice:bob 2−1Sig alice:bob(T)E alice:charlieE alice:charlie 2−1 =T+E alice:bob 2−1Sig alice:bob(T)E alice:charlie 2−1⋅2E alice:charlie τ bob =Sig alice:bob(T)+E alice:bob 2−1Sig alice:bob(T)+E alice:bob 2−1Sig alice:bob(T)(E alice:charlie+1)E alice:charlie 2−1 =1+E alice:bob 2−1E alice:bob+E alice:bob 2−1T+E alice:bob 2−1Sig alice:bob(T)E alice:charlie 2−1⋅(E alice:charlie+1) τ charlie =T+E alice:bob 2−1Sig alice:bob(T)E alice:charlie 2−1⋅2\begin{aligned}\tau_{alice}&=T+ 2 \frac{\sqrt{E_{alice:bob}^2-1} Sig_{alice:bob}(T) E_{alice:charlie}}{\sqrt{E_{alice:charlie}^2-1}}\\\\&=T+\frac{\sqrt{E_{alice:bob}^2-1} Sig_{alice:bob}(T)}{\sqrt{E_{alice:charlie}^2-1}} \cdot 2E_{alice:charlie}\\\tau_{bob}&=Sig_{alice:bob}(T)+ \sqrt{E_{alice:bob}^2-1} Sig_{alice:bob}(T)+\frac{\sqrt{E_{alice:bob}^2-1} Sig_{alice:bob}(T) \left(E_{alice:charlie}+1\right)}{\sqrt{E_{alice:charlie}^2-1}}\\\\&=\frac{1+\sqrt{E_{alice:bob}^2-1}}{E_{alice:bob}+\sqrt{E_{alice:bob}^2-1}}T+\frac{\sqrt{E_{alice:bob}^2-1} Sig_{alice:bob}(T) }{\sqrt{E_{alice:charlie}^2-1}} \cdot \left(E_{alice:charlie}+1\right)\\\\\tau_{charlie}&=T+ \frac{\sqrt{E_{alice:bob}^2-1} Sig_{alice:bob}(T)}{\sqrt{E_{alice:charlie}^2-1}} \cdot 2\end{aligned}

If Charlie is so-called beam-me-up method, then

E alice:charlie =∞ τ alice =T+E alice:bob 2−1Sig alice:bob(T)⋅2 τ bob =1+E alice:bob 2−1E alice:bob+E alice:bob 2−1T+E alice:bob 2−1Sig alice:bob(T) τ charlie =T\begin{aligned}E_{alice:charlie}&=\infty\\\\\tau_{alice}&=T+\sqrt{E_{alice:bob}^2-1} Sig_{alice:bob}(T) \cdot 2\\\\\tau_{bob}&=\frac{1+\sqrt{E_{alice:bob}^2-1}}{E_{alice:bob}+\sqrt{E_{alice:bob}^2-1}}T+\sqrt{E_{alice:bob}^2-1} Sig_{alice:bob}(T)\\\\\tau_{charlie}&=T\end{aligned}

If Bob is also light except the idle period, then as light itself has no clock it follows: object

E alice:charlie =∞,E alice:bob=∞ τ alice =2T τ bob =T τ charlie =T\begin{aligned}E_{alice:charlie}&=\infty, E_{alice:bob}=\infty\\\\\tau_{alice}&=2T\\\\\tau_{bob}&=T\\\\\tau_{charlie}&=T\end{aligned}