¿Podemos descomponer un impulso general de Lorentz en una rotación seguida de tres impulsos a lo largo de los ejes de coordenadas?

Question

¿Podemos descomponer un impulso general de Lorentz en una rotación seguida de tres impulsos a lo largo de los ejes de coordenadas?

Física
teoría de grupos
marcos inerciales
lorentz-simetría
sistemas coordinados
relatividad especial

Giuseppe Negro

$\newcommand{\betabold}{\boldsymbol{\beta}} \newcommand{\xbold}{\boldsymbol{x}} \newcommand{\ebold}{\boldsymbol{e}}$ Para $\betabold\in \mathbb R^3$ , con $0<|\betabold|<1$ , denotemos el impulso de Lorentz con velocidad $\betabold$ por

L^{β} (t, X) := (γ t - γ β \cdot X, X_{⊥} + γ X_{∥} - γ β t), dónde γ := \frac{1}{\sqrt{1 - β^{2}}} .

$L^{\betabold}(t, \xbold):=\Big(\gamma t-\gamma \betabold\cdot \xbold, \xbold_\bot +\gamma \xbold_{\parallel}-\gamma\betabold t\Big), \quad \text{where }\ \gamma:=\frac{1}{\sqrt{1-\beta^2}}.$ Aquí

x_{∥} := \frac{x \cdot β}{β^{2}} β

$\xbold_\parallel:=\frac{\xbold\cdot\betabold}{\beta^2}\betabold$ es el componente de

x

$\xbold$ en la dirección de

β

$\betabold$ , y

x_{⊥} := x - x_{∥}

$\xbold_\bot:=\xbold-\xbold_\parallel$ . (La velocidad de la luz se normaliza a

1

$1$ ).

¿Existe una rotación espacial? $R$ y $\alpha_1, \alpha_2, \alpha_3\in(-1, 1)$ tal que
$L^{β} = L^{α_{1} {mi}_{1}} L^{α_{2} {mi}_{2}} L^{α_{3} {mi}_{3}} R ?$ $L^\betabold = L^{\alpha_1\ebold_1}L^{\alpha_2\ebold_2}L^{\alpha_3\ebold_3}R\quad ?$

Dwagg

Una rotación puede mover cualquier vector unitario para que coincida con cualquier eje. Entonces existe una rotación que gira

β

$\mathbf \beta$ hacia

e_{3}

$\mathbf e_3$ eje. Entonces sí; simplemente busque tal rotación y luego elija

α_{3} = | β |

$\alpha_3 =|\mathbf \beta|$ y

α_{2} = α_{1} = 0

$\alpha_2=\alpha_1 = 0$ .

Giuseppe Negro

@Dwagg: ¿No necesita una segunda rotación después del impulso?

usuario178876

El conmutador de dos generadores elevadores es un generador de rotación. Y tu tienes

L^{\vec{φ}} = \exp (i \vec{φ} \cdot \vec{B})

$L^{\vec\varphi}=\exp(\mathrm{i}\vec\varphi\cdot \vec B)$ , dónde

\vec{B} = (B_{x}, B_{y}, B_{z})

$\vec B=(B_x,B_y,B_z)$ es una tupla de 3 de generadores de refuerzo. Tu pregunta es entonces si

L^{\vec{φ}} = L^{α_{1} e_{1}} L^{α_{2} e_{2}} L^{α_{3} e_{3}} R

$L^{\vec\varphi}=L^{\alpha_1e_1}L^{\alpha_2e_2}L^{\alpha_3e_3}R$ para algunos apropiados

α_{i}

$\alpha_i$ y

R

$R$ . La respuesta es sí debido a la fórmula de Baker-Campbell-Hausdorff.

Frobenius

Relacionado: Matriz general Transformación de Lorentz . Por intuición hay infinitas soluciones.

Giuseppe Negro

¿Puede el votante negativo explicar sus razones, por favor?

Frobenius

Al elaborar una respuesta a su pregunta, los primeros resultados muestran que mi intuición acerca de infinitas soluciones es falsa. Podría probar que por dado

β

$\:\boldsymbol{\beta}\:$ existe una y solo una tríada

(α_{1}, α_{2}, α_{3})

$\:\left(\alpha_1,\alpha_2,\alpha_3\right)\:$ y, en consecuencia, una rotación R que satisface su ecuación.

Frobenius

La elaboración es tediosa y larga. Pero publicaré una respuesta en 2-3 días.

Respuestas (1)

¿Podemos descomponer un impulso general de Lorentz en una rotación seguida de tres impulsos a lo largo de los ejes de coordenadas?

Una rotación puede mover cualquier vector unitario para que coincida con cualquier eje. Entonces existe una rotación que gira $\mathbf \beta$ hacia $\mathbf e_3$ eje. Entonces sí; simplemente busque tal rotación y luego elija $\alpha_3 =|\mathbf \beta|$ y $\alpha_2=\alpha_1 = 0$ .
@Dwagg: ¿No necesita una segunda rotación después del impulso?
El conmutador de dos generadores elevadores es un generador de rotación. Y tu tienes $L^{\vec\varphi}=\exp(\mathrm{i}\vec\varphi\cdot \vec B)$ , dónde $\vec B=(B_x,B_y,B_z)$ es una tupla de 3 de generadores de refuerzo. Tu pregunta es entonces si $L^{\vec\varphi}=L^{\alpha_1e_1}L^{\alpha_2e_2}L^{\alpha_3e_3}R$ para algunos apropiados $\alpha_i$ y $R$ . La respuesta es sí debido a la fórmula de Baker-Campbell-Hausdorff.
Relacionado: Matriz general Transformación de Lorentz . Por intuición hay infinitas soluciones.
¿Puede el votante negativo explicar sus razones, por favor?
Al elaborar una respuesta a su pregunta, los primeros resultados muestran que mi intuición acerca de infinitas soluciones es falsa. Podría probar que por dado $\:\boldsymbol{\beta}\:$ existe una y solo una tríada $\:\left(\alpha_1,\alpha_2,\alpha_3\right)\:$ y, en consecuencia, una rotación R que satisface su ecuación.
La elaboración es tediosa y larga. Pero publicaré una respuesta en 2-3 días.

Frobenius · Answer 1

$\begin{matrix} (a) & L (β) = R^{- 1} \cdot L_{3} (α_{3}) \cdot L_{2} (α_{2}) \cdot L_{1} (α_{1}) \end{matrix}$ $\begin{equation} \mathrm{L}\left(\boldsymbol{\beta}\right) =\mathrm{R}^{\boldsymbol{-1}}\cdot\mathrm{L}_3\left(\alpha_3\right)\cdot\mathrm{L}_2\left(\alpha_2\right)\cdot\mathrm{L}_1\left(\alpha_1\right) \tag{a}\label{a} \end{equation}$ $\begin{matrix} (b) & α_{1} = β_{1}, α_{2} = \frac{β_{2}}{\sqrt{1 - β_{1}^{2}}}, α_{3} = \frac{β_{3}}{\sqrt{1 - (β_{1}^{2} + β_{2}^{2})}} \end{matrix}$ $\begin{equation} \alpha_1 =\beta_1\,, \quad \alpha_2 =\dfrac{\beta_2}{\sqrt{1-\beta^2_1}}\,, \quad \alpha_3 =\dfrac{\beta_3}{\sqrt{1-\left(\beta^2_1+\beta^2_2\right)}} \tag{b}\label{b} \end{equation}$ $\begin{matrix} (C) & R = rotación espacial \end{matrix}$ $\begin{equation} \mathrm{R}=\text{space rotation} \tag{c}\label{c} \end{equation}$ $==================================================$

Figura-01 3D

De la figura 01:

Transformación de Lorentz de $\:\mathrm{S}\boldsymbol{\equiv} \{xyz\omega, \omega\boldsymbol{=}ct\}\:$ a $\:\mathrm{S_1}\boldsymbol{\equiv} \{x_1y_1z_1\omega_1, \omega_1\boldsymbol{=}ct_1\}\:$

\begin{matrix} (01) & [\begin{matrix} X_{1} \\ y_{1} \\ z_{1} \\ ω_{1} \end{matrix}] = [\begin{matrix} aporrear ξ & 0 & 0 & - pecado ξ \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ - pecado ξ & 0 & 0 & aporrear ξ \end{matrix}] [\begin{matrix} X \\ y \\ z \\ ω \end{matrix}], bronceado ξ = α_{1} = \frac{{tu}_{1}}{C} \end{matrix}

$\begin{equation} \begin{bmatrix} x_1\\ y_1\\ z_1\\ \omega_1 \end{bmatrix} \boldsymbol{=} \begin{bmatrix} \hphantom{-}\cosh\!\xi & \hphantom{-}0 & \hphantom{-}0 &-\sinh\!\xi \\ 0 & \hphantom{-}1 & \hphantom{-} 0 & 0\\ 0 & \hphantom{-} 0 & \hphantom{-} 1 & 0\\ -\sinh\!\xi & \hphantom{-} 0 & \hphantom{-} 0 &\hphantom{-}\cosh\!\xi \end{bmatrix} \begin{bmatrix} x\\ y\\ z\\ \omega \end{bmatrix} \,, \quad \tanh\!\xi\boldsymbol{=}\alpha_1\boldsymbol{=}\dfrac{u_1}{c} \tag{01}\label{01} \end{equation}$ o

\begin{matrix} (02) & W_{1} = L_{1} W, L_{1} = [\begin{matrix} aporrear ξ & 0 & 0 & - pecado ξ \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ - pecado ξ & 0 & 0 & aporrear ξ \end{matrix}] \end{matrix}

$\begin{equation} \mathbf{W}_1\boldsymbol{=}\mathrm{L_1}\mathbf{W}\,, \qquad \mathrm{L_1}\boldsymbol{=} \begin{bmatrix} \hphantom{-}\cosh\!\xi & \hphantom{-}0 & \hphantom{-}0 &-\sinh\!\xi \\ 0 & \hphantom{-}1 & \hphantom{-} 0 & 0\\ 0 & \hphantom{-} 0 & \hphantom{-} 1 & 0\\ -\sinh\!\xi & \hphantom{-} 0 & \hphantom{-} 0 &\hphantom{-}\cosh\!\xi \end{bmatrix} \tag{02}\label{02} \end{equation}$

Transformación de Lorentz de $\:\mathrm{S_1}\boldsymbol{\equiv} \{x_1y_1z_1\omega_1, \omega_1\boldsymbol{=}ct_1\}\:$ a $\:\mathrm{S_2}\boldsymbol{\equiv} \{x_2y_2z_2\omega_2, \omega_2\boldsymbol{=}ct_2\}\:$

\begin{matrix} (03) & [\begin{matrix} X_{2} \\ y_{2} \\ z_{2} \\ ω_{2} \end{matrix}] = [\begin{matrix} 1 & 0 & 0 & 0 \\ 0 & aporrear η & 0 & - pecado η \\ 0 & 0 & 1 & 0 \\ 0 & - pecado η & 0 & aporrear η \end{matrix}] [\begin{matrix} X_{1} \\ y_{1} \\ z_{1} \\ ω_{1} \end{matrix}], bronceado η = α_{2} = \frac{{tu}_{2}}{C} \end{matrix}

$\begin{equation} \begin{bmatrix} x_2\\ y_2\\ z_2\\ \omega_2 \end{bmatrix} \boldsymbol{=} \begin{bmatrix} 1 & \hphantom{-}0 & \hphantom{-}0 &\hphantom{-}0 \\ 0 & \hphantom{-}\cosh\!\eta & \hphantom{-} 0 & -\sinh\!\eta\\ 0 & \hphantom{-} 0 & \hphantom{-} 1 & 0\\ 0 & -\sinh\!\eta & \hphantom{-} 0 &\hphantom{-}\cosh\!\eta \end{bmatrix} \begin{bmatrix} x_1\\ y_1\\ z_1\\ \omega_1 \end{bmatrix} \,, \quad \tanh\!\eta\boldsymbol{=}\alpha_2\boldsymbol{=}\dfrac{u_2}{c} \tag{03}\label{03} \end{equation}$ o

\begin{matrix} (04) & W_{2} = L_{2} W_{1}, L_{2} = [\begin{matrix} 1 & 0 & 0 & 0 \\ 0 & aporrear η & 0 & - pecado η \\ 0 & 0 & 1 & 0 \\ 0 & - pecado η & 0 & aporrear η \end{matrix}] \end{matrix}

$\begin{equation} \mathbf{W}_2\boldsymbol{=}\mathrm{L_2}\mathbf{W}_1\,, \qquad \mathrm{L_2}\boldsymbol{=} \begin{bmatrix} 1 & \hphantom{-}0 & \hphantom{-}0 &\hphantom{-}0 \\ 0 & \hphantom{-}\cosh\!\eta & \hphantom{-} 0 & -\sinh\!\eta\\ 0 & \hphantom{-} 0 & \hphantom{-} 1 & 0\\ 0 & -\sinh\!\eta & \hphantom{-} 0 &\hphantom{-}\cosh\!\eta \end{bmatrix} \tag{04}\label{04} \end{equation}$

Transformación de Lorentz de $\:\mathrm{S_2}\boldsymbol{\equiv} \{x_2y_2z_2\omega_2, \omega_2\boldsymbol{=}ct_2\}\:$ a $\:\mathrm{S_3}\boldsymbol{\equiv}\{x_3y_3z_3\omega_3, \omega_3\boldsymbol{=}ct_3\}\:$

\begin{matrix} (05) & [\begin{matrix} X_{3} \\ y_{3} \\ z_{3} \\ ω_{3} \end{matrix}] = [\begin{matrix} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & aporrear ζ & - pecado ζ \\ 0 & 0 & - pecado ζ & aporrear ζ \end{matrix}] [\begin{matrix} X_{2} \\ y_{2} \\ z_{2} \\ ω_{2} \end{matrix}], bronceado ζ = α_{3} = \frac{{tu}_{3}}{C} \end{matrix}

$\begin{equation} \begin{bmatrix} x_3\\ y_3\\ z_3\\ \omega_3 \end{bmatrix} \boldsymbol{=} \begin{bmatrix} 1 & \hphantom{-}0 & \hphantom{-}0 &\hphantom{-}0 \\ 0 & \hphantom{-}1 & \hphantom{-} 0 & \hphantom{-}0\\ 0 & \hphantom{-}0 & \hphantom{-}\cosh\!\zeta & -\sinh\!\zeta\\ 0 & \hphantom{-}0 & -\sinh\!\zeta &\hphantom{-}\cosh\!\zeta \end{bmatrix} \begin{bmatrix} x_2\\ y_2\\ z_2\\ \omega_2 \end{bmatrix} \,, \quad \tanh\!\zeta\boldsymbol{=}\alpha_3\boldsymbol{=}\dfrac{u_3}{c} \tag{05}\label{05} \end{equation}$ o

\begin{matrix} (06) & W_{3} = L_{3} W_{2}, L_{3} = [\begin{matrix} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & aporrear ζ & - pecado ζ \\ 0 & 0 & - pecado ζ & aporrear ζ \end{matrix}] \end{matrix}

$\begin{equation} \mathbf{W}_3\boldsymbol{=}\mathrm{L_3}\mathbf{W}_2\,, \qquad \mathrm{L_3}\boldsymbol{=} \begin{bmatrix} 1 & \hphantom{-}0 & \hphantom{-}0 &\hphantom{-}0 \\ 0 & \hphantom{-}1 & \hphantom{-} 0 & \hphantom{-}0\\ 0 & \hphantom{-}0 & \hphantom{-}\cosh\!\zeta & -\sinh\!\zeta\\ 0 & \hphantom{-}0 & -\sinh\!\zeta &\hphantom{-}\cosh\!\zeta \end{bmatrix} \tag{06}\label{06} \end{equation}$

Tenga en cuenta que, debido a las configuraciones estándar, las matrices $\:\mathrm{L_1}, \mathrm{L_2},\mathrm{L_3}\:$ son realmente simétricos.

De ecuaciones $\eqref{02}$ , $\eqref{04}$ y $\eqref{06}$ tenemos

\begin{matrix} (07) & W_{3} = L_{3} W_{2} = L_{3} L_{2} W_{1} = L_{3} L_{2} L_{1} W ⟹ W_{3} = Λ W \end{matrix}

$\begin{equation} \mathbf{W}_3\boldsymbol{=}\mathrm{L}_3\mathbf{W}_2\boldsymbol{=}\mathrm{L}_3\mathrm{L}_2\mathbf{W}_1\boldsymbol{=}\mathrm{L}_3\mathrm{L}_2\mathrm{L}_1\mathbf{W} \quad \boldsymbol{\Longrightarrow} \quad \mathbf{W}_3\boldsymbol{=}\Lambda\,\mathbf{W} \tag{07}\label{07} \end{equation}$ dónde

Λ

$\:\Lambda\:$ la composición de las tres Transformaciones de Lorentz

L_{1}, L_{2}, L_{3}

$\:\mathrm{L_1}, \mathrm{L_2},\mathrm{L_3}\:$ ^{$\begin{aligned} Λ = L_{3} L_{2} L_{1} = \\ (08) & [\begin{matrix} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & aporrear ζ & - pecado ζ \\ 0 & 0 & - pecado ζ & aporrear ζ \end{matrix}] [\begin{matrix} 1 & 0 & 0 & 0 \\ 0 & aporrear η & 0 & - pecado η \\ 0 & 0 & 1 & 0 \\ 0 & - pecado η & 0 & aporrear η \end{matrix}] [\begin{matrix} aporrear ξ & 0 & 0 & - pecado ξ \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ - pecado ξ & 0 & 0 & aporrear ξ \end{matrix}] \end{aligned}$ $\begin{align} &\Lambda\boldsymbol{=}\mathrm{L}_3\mathrm{L}_2\mathrm{L}_1\boldsymbol{=} \nonumber\\ &\begin{bmatrix} 1 & \hphantom{-}0 & \hphantom{-}0 &\hphantom{-}0 \\ 0 & \hphantom{-}1 & \hphantom{-} 0 & \hphantom{-}0\\ 0 & \hphantom{-}0 & \hphantom{-}\cosh\!\zeta & -\sinh\!\zeta\\ 0 & \hphantom{-}0 & -\sinh\!\zeta &\hphantom{-}\cosh\!\zeta \end{bmatrix} \begin{bmatrix} 1 & \hphantom{-}0 & \hphantom{-}0 &\hphantom{-}0 \\ 0 & \hphantom{-}\cosh\!\eta & \hphantom{-} 0 & -\sinh\!\eta\\ 0 & \hphantom{-} 0 & \hphantom{-} 1 & 0\\ 0 & -\sinh\!\eta & \hphantom{-} 0 &\hphantom{-}\cosh\!\eta \end{bmatrix} \begin{bmatrix} \hphantom{-}\cosh\!\xi & \hphantom{-}0 & \hphantom{-}0 &-\sinh\!\xi \\ 0 & \hphantom{-}1 & \hphantom{-} 0 & 0\\ 0 & \hphantom{-} 0 & \hphantom{-} 1 & 0\\ -\sinh\!\xi & \hphantom{-} 0 & \hphantom{-} 0 &\hphantom{-}\cosh\!\xi \end{bmatrix} \tag{08}\label{08} \end{align}$} ^{eso es
$\begin{matrix} (09) & Λ = L_{3} L_{2} L_{1} = [\begin{matrix} aporrear ξ & 0 & 0 & - pecado ξ \\ pecado η pecado ξ & aporrear η & 0 & - pecado η aporrear ξ \\ pecado ζ aporrear η pecado ξ & pecado ζ pecado η & aporrear ζ & - pecado ζ aporrear η aporrear ξ \\ - aporrear ζ aporrear η pecado ξ & - aporrear ζ pecado η & - pecado ζ & aporrear ζ aporrear η aporrear ξ \end{matrix}] \end{matrix}$ $\begin{equation} \Lambda\boldsymbol{=}\mathrm{L}_3\mathrm{L}_2\mathrm{L}_1\boldsymbol{=} \begin{bmatrix} \hphantom{-}\cosh\!\xi & \hphantom{-}0 & \hphantom{-}0 &-\sinh\!\xi\\ \hphantom{-}\sinh\!\eta\sinh\!\xi & \hphantom{-}\cosh\!\eta & \hphantom{-}0 & -\sinh\!\eta\cosh\!\xi\\ \hphantom{-}\sinh\!\zeta\cosh\!\eta\sinh\!\xi & \hphantom{-}\sinh\!\zeta\sinh\!\eta &\hphantom{-} \cosh\!\zeta & -\sinh\!\zeta\cosh\!\eta\cosh\!\xi\\ -\cosh\!\zeta\cosh\!\eta\sinh\!\xi & -\cosh\!\zeta\sinh\!\eta & -\sinh\!\zeta &\hphantom{-}\cosh\!\zeta\cosh\!\eta\cosh\!\xi \end{bmatrix} \tag{09}\label{09} \end{equation}$}La matriz de transformación de Lorentz

Λ

$\:\Lambda\:$ no es simétrico, por lo que los sistemas

S, S_{3}

$\:\mathrm{S},\mathrm{S_3}\:$ no están en la configuración estándar. Pero es razonable suponer que

\begin{matrix} (10) & Λ = R \cdot L \end{matrix}

$\begin{equation} \Lambda=\mathrm{R}\cdot\mathrm{L} \tag{10}\label{10} \end{equation}$ dónde

L

$\:\mathrm{L}\:$ es la matriz de transformación de Lorentz simétrica de

S

$\:\mathrm{S}\:$ a un sistema intermedio

{S^{'}}_{3}

$\:\mathrm{S'}_3\:$ en configuración estándar a él y co-moviéndose con

S_{3}

$\:\mathrm{S}_3\:$ , mientras

R

$\:\mathrm{R}\:$ es una transformación puramente espacial entre

{S^{'}}_{3}

$\:\mathrm{S'}_3\:$ y

S_{3}

$\:\mathrm{S}_3$ .

Ahora, nuestro objetivo sería expresar la matriz de transformación de Lorentz simétricaen cuanto a la rapidezya que desde $\eqref{10}$

\begin{matrix} (11) & R = Λ \cdot L^{- 1} \end{matrix}

$\begin{equation} \mathrm{R}\boldsymbol{=}\Lambda\cdot\mathrm{L}^{\boldsymbol{-1}} \tag{11}\label{11} \end{equation}$

La matriz de transformación de Lorentz $\:\mathrm{L}\:$ , de $\:\mathrm{S}\:$ al sistema intermedio $\:\mathrm{S'_3}\:$ en la configuración estándar, es:

\begin{matrix} (12) & L (υ) = [\begin{matrix} 1 + (γ - 1) {norte}_{X}^{2} & (γ - 1) {norte}_{X} {norte}_{y} & (γ - 1) {norte}_{X} {norte}_{z} & - \frac{γ υ_{X}}{C} \\ (γ - 1) {norte}_{y} {norte}_{X} & 1 + (γ - 1) {norte}_{y}^{2} & (γ - 1) {norte}_{y} {norte}_{z} & - \frac{γ υ_{y}}{C} \\ (γ - 1) {norte}_{z} {norte}_{X} & (γ - 1) {norte}_{z} {norte}_{y} & 1 + (γ - 1) {norte}_{z}^{2} & - \frac{γ υ_{z}}{C} \\ - \frac{γ υ_{X}}{C} & - \frac{γ υ_{y}}{C} & - \frac{γ υ_{z}}{C} & γ \end{matrix}] \end{matrix}

$\begin{equation} \mathrm{L}\left(\boldsymbol{\upsilon} \right)\boldsymbol{=} \begin{bmatrix} 1\!+\!\left(\gamma\!-\!1\right)\!\mathrm{n}^{2}_{x} & \left(\gamma\!-\!1\right)\!\mathrm{n}_{x}\mathrm{n}_{y} & \left(\gamma\!-\!1\right)\!\mathrm{n}_{x}\mathrm{n}_{z} & \!-\dfrac{\gamma\upsilon_{x}}{c} \vphantom{\dfrac{\dfrac{}{}}{\tfrac{}{}}}\\ \left(\gamma\!-\!1\right)\!\mathrm{n}_{y}\mathrm{n}_{x} & 1\!+\!\left(\gamma\!-\!1\right)\!\mathrm{n}^{2}_{y} &\left(\gamma\!-\!1\right)\!\mathrm{n}_{y}\mathrm{n}_{z} & \!-\dfrac{\gamma\upsilon_{y}}{c} \vphantom{\dfrac{\dfrac{}{}}{\tfrac{}{}}}\\ \left(\gamma\!-\!1\right)\!\mathrm{n}_{z}\mathrm{n}_{x} & \left(\gamma\!-\!1\right)\!\mathrm{n}_{z}\mathrm{n}_{y} & 1\!+\!\left(\gamma\!-\!1\right)\!\mathrm{n}^{2}_{z} & \!-\dfrac{\gamma\upsilon_{z}}{c} \vphantom{\dfrac{\dfrac{}{}}{\tfrac{}{}}}\\ \!-\dfrac{\gamma\upsilon_{x}}{c} & \!-\dfrac{\gamma\upsilon_{y}}{c} &\!-\dfrac{\gamma\upsilon_{z}}{c} & \gamma \vphantom{\dfrac{\dfrac{}{}}{\tfrac{}{}}} \end{bmatrix} \tag{12}\label{12} \end{equation}$ En

(12)

$\eqref{12}$

\begin{aligned} (13.1) & \frac{υ}{C} & = (\frac{υ_{X}}{C}, \frac{υ_{y}}{C}, \frac{υ_{z}}{C}) = (bronceado ξ, \frac{bronceado η}{aporrear ξ}, \frac{bronceado ζ}{aporrear ξ aporrear η}) \equiv β = (β_{1}, β_{2}, β_{3},) \\ (13.2) & {(\frac{υ}{C})}^{2} & = {(\frac{υ_{X}}{C})}^{2} + {(\frac{υ_{y}}{C})}^{2} + {(\frac{υ_{z}}{C})}^{2} = 1 - {(\frac{1}{aporrear ξ aporrear η aporrear ζ})}^{2} = \frac{γ^{2} - 1}{γ^{2}} \\ (13.3) & γ & = {(1 - \frac{υ^{2}}{C^{2}})}^{- \frac{1}{2}} = aporrear ξ aporrear η aporrear ζ = γ_{1} γ_{2} γ_{3} \\ (13.4) & norte & = ({norte}_{X}, {norte}_{y}, {norte}_{z}) = \frac{υ / C}{υ / C} = \frac{(pecado ξ aporrear η aporrear ζ, pecado η aporrear ζ, pecado ζ)}{\sqrt{{aporrear}^{2} ξ {aporrear}^{2} η {aporrear}^{2} ζ - 1}} \end{aligned}

$\begin{align} \dfrac{\boldsymbol{\upsilon}}{c} & \boldsymbol{=} \left(\dfrac{\upsilon_{x}}{c},\dfrac{\upsilon_{y}}{c},\dfrac{\upsilon_{z}}{c}\right)\boldsymbol{=}\left(\tanh\!\xi,\dfrac{\tanh\!\eta}{\cosh\!\xi},\dfrac{\tanh\!\zeta}{\cosh\!\xi\cosh\!\eta}\right)\equiv \boldsymbol{\beta}\boldsymbol{=} \left(\beta_1,\beta_2,\beta_3,\right) \tag{13.1}\label{13.1}\\ \left(\dfrac{\upsilon}{c}\right)^{2} & \boldsymbol{=} \left(\dfrac{\upsilon_{x}}{c}\right)^{2}+\left(\dfrac{\upsilon_{y}}{c}\right)^{2}+\left(\dfrac{\upsilon_{z}}{c}\right)^{2} \boldsymbol{=}1 \boldsymbol{-}\left(\dfrac{1}{\cosh\!\xi\cosh\!\eta\cosh\!\zeta}\right)^{2}=\dfrac{\gamma^{2}\!-\!1}{\gamma^{2}} \tag{13.2}\label{13.2}\\ \gamma & \boldsymbol{=} \left(\!1\!-\!\frac{\upsilon^{2}}{c^{2}}\right)^{-\frac12}\boldsymbol{=} \cosh\!\xi\cosh\!\eta\cosh\!\zeta \boldsymbol{=}\gamma_1\gamma_2\gamma_3 \tag{13.3}\label{13.3}\\ \mathbf{n} & = \left(\mathrm{n}_{x},\mathrm{n}_{y},\mathrm{n}_{z}\right) \boldsymbol{=}\dfrac{\boldsymbol{\upsilon}/c}{\upsilon/c}\boldsymbol{=}\dfrac{\left(\sinh\!\xi\cosh\!\eta\cosh\!\zeta ,\sinh\!\eta\cosh\!\zeta,\sinh\!\zeta\right)}{\sqrt{\cosh^2\!\xi\cosh^2\!\eta\cosh^2\!\zeta-1}} \tag{13.4}\label{13.4} \end{align}$ dónde

υ

$\:\boldsymbol{\upsilon}\:$ es el vector velocidad del origen

{O^{'}}_{3} (\equiv O_{3})

$\:\mathrm{O'}_{\!\!3}\left(\equiv \mathrm{O}_{3}\right)\:$ con respecto a

S

$\:\mathrm{S}$ ⁽¹⁾ ,

n

$\:\mathbf{n}\:$ el vector unitario a lo largo

υ

$\:\boldsymbol{\upsilon}\:$ y

γ

$\:\gamma\:$ el correspondiente

γ -

$\:\gamma-$ factor.

Entonces la matrizde ecuación $\eqref{12}$ como función de las rapidezes $\:\xi,\eta,\zeta\:$ es ⁽²⁾

\begin{aligned} L (υ) = L (ξ, η, ζ) = \\ (14) & [\begin{matrix} 1 + \frac{{pecado}^{2} ξ {aporrear}^{2} η {aporrear}^{2} ζ}{1 + aporrear ξ aporrear η aporrear ζ} & \frac{pecado ξ aporrear η pecado η {aporrear}^{2} ζ}{1 + aporrear ξ aporrear η aporrear ζ} & \frac{pecado ξ aporrear η aporrear ζ pecado ζ}{1 + aporrear ξ aporrear η aporrear ζ} & - pecado ξ aporrear η aporrear ζ \\ \frac{pecado ξ aporrear η pecado η {aporrear}^{2} ζ}{1 + aporrear ξ aporrear η aporrear ζ} & 1 + \frac{{pecado}^{2} η {aporrear}^{2} ζ}{1 + aporrear ξ aporrear η aporrear ζ} & \frac{pecado η aporrear ζ pecado ζ}{1 + aporrear ξ aporrear η aporrear ζ} & - pecado η aporrear ζ \\ \frac{pecado ξ aporrear η aporrear ζ pecado ζ}{1 + aporrear ξ aporrear η aporrear ζ} & \frac{pecado η aporrear ζ pecado ζ}{1 + aporrear ξ aporrear η aporrear ζ} & 1 + \frac{{pecado}^{2} ζ}{1 + aporrear ξ aporrear η aporrear ζ} & - pecado ζ \\ - pecado ξ aporrear η aporrear ζ & - pecado η aporrear ζ & - pecado ζ & aporrear ξ aporrear η aporrear ζ \end{matrix}] \end{aligned}

$\begin{align} &\mathrm{L}\left(\boldsymbol{\upsilon} \right)\boldsymbol{=}\mathrm{L}\left(\xi,\eta,\zeta \right)\boldsymbol{=} \nonumber\\ &\begin{bmatrix} 1\!+\!\dfrac{\sinh^{2}\!\xi\cosh^2\!\eta\cosh^2\!\zeta}{1\!+\!\cosh\!\xi\cosh\!\eta\cosh\!\zeta} &\dfrac{\sinh\!\xi\cosh\!\eta\sinh\!\eta\cosh^2\!\zeta}{1\!+\!\cosh\!\xi\cosh\!\eta\cosh\!\zeta} & \dfrac{\sinh\!\xi\cosh\!\eta\cosh\!\zeta\sinh\!\zeta}{1\!+\!\cosh\!\xi\cosh\!\eta\cosh\!\zeta} & \!\boldsymbol{-} \sinh\!\xi\cosh\!\eta\cosh\!\zeta \vphantom{\dfrac{\dfrac{}{}}{\tfrac{}{}}}\\ \dfrac{\sinh\!\xi\cosh\!\eta\sinh\!\eta\cosh^2\!\zeta}{1\!+\!\cosh\!\xi\cosh\!\eta\cosh\!\zeta} & 1\!+\!\dfrac{\sinh^2\!\eta\cosh^2\!\zeta}{1\!+\!\cosh\!\xi\cosh\!\eta\cosh\!\zeta } &\dfrac{\sinh\!\eta\cosh\!\zeta\sinh\!\zeta}{1\!+\!\cosh\!\xi\cosh\!\eta\cosh\!\zeta} & \!\boldsymbol{-} \sinh\!\eta\cosh\!\zeta \vphantom{\dfrac{\dfrac{}{}}{\tfrac{}{}}}\\ \dfrac{\sinh\!\xi\cosh\!\eta\cosh\!\zeta\sinh\!\zeta}{1\!+\!\cosh\!\xi\cosh\!\eta\cosh\!\zeta} & \dfrac{\sinh\!\eta\cosh\!\zeta\sinh\!\zeta}{1\!+\!\cosh\!\xi\cosh\!\eta\cosh\!\zeta} & 1\!+\!\dfrac{\sinh^2\!\zeta}{1\!+\!\cosh\!\xi\cosh\!\eta\cosh\!\zeta } & \!\boldsymbol{-}\sinh\!\zeta \vphantom{\dfrac{\dfrac{}{}}{\tfrac{}{}}}\\ \!\boldsymbol{-}\sinh\!\xi\cosh\!\eta\cosh\!\zeta & \!\boldsymbol{-}\sinh\!\eta\cosh\!\zeta &\!\boldsymbol{-}\sinh\!\zeta & \cosh\!\xi\cosh\!\eta\cosh\!\zeta \vphantom{\dfrac{\dfrac{}{}}{\tfrac{}{}}} \end{bmatrix} \tag{14}\label{14} \end{align}$ mientras^{$\begin{aligned} L^{- 1} (υ) = L (- υ) = L (- ξ, - η, - ζ) = \\ (15) & [\begin{matrix} 1 + \frac{{pecado}^{2} ξ {aporrear}^{2} η {aporrear}^{2} ζ}{1 + aporrear ξ aporrear η aporrear ζ} & \frac{pecado ξ aporrear η pecado η {aporrear}^{2} ζ}{1 + aporrear ξ aporrear η aporrear ζ} & \frac{pecado ξ aporrear η aporrear ζ pecado ζ}{1 + aporrear ξ aporrear η aporrear ζ} & pecado ξ aporrear η aporrear ζ \\ \frac{pecado ξ aporrear η pecado η {aporrear}^{2} ζ}{1 + aporrear ξ aporrear η aporrear ζ} & 1 + \frac{{pecado}^{2} η {aporrear}^{2} ζ}{1 + aporrear ξ aporrear η aporrear ζ} & \frac{pecado η aporrear ζ pecado ζ}{1 + aporrear ξ aporrear η aporrear ζ} & pecado η aporrear ζ \\ \frac{pecado ξ aporrear η aporrear ζ pecado ζ}{1 + aporrear ξ aporrear η aporrear ζ} & \frac{pecado η aporrear ζ pecado ζ}{1 + aporrear ξ aporrear η aporrear ζ} & 1 + \frac{{pecado}^{2} ζ}{1 + aporrear ξ aporrear η aporrear ζ} & pecado ζ \\ pecado ξ aporrear η aporrear ζ & pecado η aporrear ζ & pecado ζ & aporrear ξ aporrear η aporrear ζ \end{matrix}] \end{aligned}$ $\begin{align} &\mathrm{L}^{\boldsymbol{-1}}\left(\boldsymbol{\upsilon} \right)\boldsymbol{=}\mathrm{L}\left(\boldsymbol{-}\boldsymbol{\upsilon} \right)\boldsymbol{=}\mathrm{L}\left(\boldsymbol{-}\xi,\boldsymbol{-}\eta,\boldsymbol{-}\zeta \right)\boldsymbol{=} \nonumber\\ &\begin{bmatrix} 1\!+\!\dfrac{\sinh^{2}\!\xi\cosh^2\!\eta\cosh^2\!\zeta}{1\!+\!\cosh\!\xi\cosh\!\eta\cosh\!\zeta} &\dfrac{\sinh\!\xi\cosh\!\eta\sinh\!\eta\cosh^2\!\zeta}{1\!+\!\cosh\!\xi\cosh\!\eta\cosh\!\zeta} & \dfrac{\sinh\!\xi\cosh\!\eta\cosh\!\zeta\sinh\!\zeta}{1\!+\!\cosh\!\xi\cosh\!\eta\cosh\!\zeta} & \! \sinh\!\xi\cosh\!\eta\cosh\!\zeta \vphantom{\dfrac{\dfrac{}{}}{\tfrac{}{}}}\\ \dfrac{\sinh\!\xi\cosh\!\eta\sinh\!\eta\cosh^2\!\zeta}{1\!+\!\cosh\!\xi\cosh\!\eta\cosh\!\zeta} & 1\!+\!\dfrac{\sinh^2\!\eta\cosh^2\!\zeta}{1\!+\!\cosh\!\xi\cosh\!\eta\cosh\!\zeta } &\dfrac{\sinh\!\eta\cosh\!\zeta\sinh\!\zeta}{1\!+\!\cosh\!\xi\cosh\!\eta\cosh\!\zeta} & \! \sinh\!\eta\cosh\!\zeta \vphantom{\dfrac{\dfrac{}{}}{\tfrac{}{}}}\\ \dfrac{\sinh\!\xi\cosh\!\eta\cosh\!\zeta\sinh\!\zeta}{1\!+\!\cosh\!\xi\cosh\!\eta\cosh\!\zeta} & \dfrac{\sinh\!\eta\cosh\!\zeta\sinh\!\zeta}{1\!+\!\cosh\!\xi\cosh\!\eta\cosh\!\zeta} & 1\!+\!\dfrac{\sinh^2\!\zeta}{1\!+\!\cosh\!\xi\cosh\!\eta\cosh\!\zeta } & \!\sinh\!\zeta \vphantom{\dfrac{\dfrac{}{}}{\tfrac{}{}}}\\ \!\sinh\!\xi\cosh\!\eta\cosh\!\zeta & \!\sinh\!\eta\cosh\!\zeta &\!\sinh\!\zeta & \cosh\!\xi\cosh\!\eta\cosh\!\zeta \vphantom{\dfrac{\dfrac{}{}}{\tfrac{}{}}} \end{bmatrix} \tag{15}\label{15} \end{align}$}

De ecuaciones $\eqref{09}$ , $\eqref{11}$ y $\eqref{15}$

\begin{aligned} R = Λ \cdot L^{- 1} = \\ [\begin{matrix} aporrear ξ & 0 & 0 & - pecado ξ \\ pecado η pecado ξ & aporrear η & 0 & - pecado η aporrear ξ \\ pecado ζ aporrear η pecado ξ & pecado ζ pecado η & aporrear ζ & - pecado ζ aporrear η aporrear ξ \\ - aporrear ζ aporrear η pecado ξ & - aporrear ζ pecado η & - pecado ζ & aporrear ζ aporrear η aporrear ξ \end{matrix}] \\ [\begin{matrix} 1 + \frac{{pecado}^{2} ξ {aporrear}^{2} η {aporrear}^{2} ζ}{1 + aporrear ξ aporrear η aporrear ζ} & \frac{pecado ξ aporrear η pecado η {aporrear}^{2} ζ}{1 + aporrear ξ aporrear η aporrear ζ} & \frac{pecado ξ aporrear η aporrear ζ pecado ζ}{1 + aporrear ξ aporrear η aporrear ζ} & pecado ξ aporrear η aporrear ζ \\ \frac{pecado ξ aporrear η pecado η {aporrear}^{2} ζ}{1 + aporrear ξ aporrear η aporrear ζ} & 1 + \frac{{pecado}^{2} η {aporrear}^{2} ζ}{1 + aporrear ξ aporrear η aporrear ζ} & \frac{pecado η aporrear ζ pecado ζ}{1 + aporrear ξ aporrear η aporrear ζ} & pecado η aporrear ζ \\ \frac{pecado ξ aporrear η aporrear ζ pecado ζ}{1 + aporrear ξ aporrear η aporrear ζ} & \frac{pecado η aporrear ζ pecado ζ}{1 + aporrear ξ aporrear η aporrear ζ} & 1 + \frac{{pecado}^{2} ζ}{1 + aporrear ξ aporrear η aporrear ζ} & pecado ζ \\ pecado ξ aporrear η aporrear ζ & pecado η aporrear ζ & pecado ζ & aporrear ξ aporrear η aporrear ζ \end{matrix}] = \\ (dieciséis) & [\begin{matrix} \frac{aporrear ξ + aporrear η aporrear ζ}{1 + aporrear ξ aporrear η aporrear ζ} & - \frac{pecado ξ pecado η aporrear ζ}{1 + aporrear ξ aporrear η aporrear ζ} & - \frac{pecado ξ pecado ζ}{1 + aporrear ξ aporrear η aporrear ζ} & 0 \\ \frac{pecado ξ pecado η}{1 + aporrear ξ aporrear η aporrear η} & \frac{aporrear η + aporrear ζ aporrear ξ}{1 + aporrear ξ aporrear η aporrear ζ} & - \frac{aporrear ξ pecado η pecado ζ}{1 + aporrear ξ aporrear η aporrear ζ} & 0 \\ \frac{pecado ξ aporrear η pecado ζ}{1 + aporrear ξ aporrear η aporrear ζ} & \frac{pecado η pecado ζ}{1 + aporrear ξ aporrear η aporrear η} & \frac{aporrear ζ + aporrear ξ aporrear η}{1 + aporrear ξ aporrear η aporrear ζ} & 0 \\ 0 & 0 & 0 & 1 \end{matrix}] \end{aligned}

$\begin{align} &\mathrm{R}\boldsymbol{=}\Lambda\cdot\mathrm{L}^{\boldsymbol{-1}}\boldsymbol{=} \nonumber\\ &\begin{bmatrix} \hphantom{-}\cosh\!\xi & \hphantom{-}0 & \hphantom{-}0 &-\sinh\!\xi\vphantom{\dfrac{a}{b}}\\ \hphantom{-}\sinh\!\eta\sinh\!\xi & \hphantom{-}\cosh\!\eta & \hphantom{-}0 & -\sinh\!\eta\cosh\!\xi\vphantom{\dfrac{a}{b}}\\ \hphantom{-}\sinh\!\zeta\cosh\!\eta\sinh\!\xi & \hphantom{-}\sinh\!\zeta\sinh\!\eta &\hphantom{-} \cosh\!\zeta & -\sinh\!\zeta\cosh\!\eta\cosh\!\xi\vphantom{\dfrac{a}{b}}\\ -\cosh\!\zeta\cosh\!\eta\sinh\!\xi & -\cosh\!\zeta\sinh\!\eta & -\sinh\!\zeta &\hphantom{-}\cosh\!\zeta\cosh\!\eta\cosh\!\xi\vphantom{\dfrac{a}{b}} \end{bmatrix} \nonumber\\ &\begin{bmatrix} 1\!+\!\dfrac{\sinh^{2}\!\xi\cosh^2\!\eta\cosh^2\!\zeta}{1\!+\!\cosh\!\xi\cosh\!\eta\cosh\!\zeta} &\dfrac{\sinh\!\xi\cosh\!\eta\sinh\!\eta\cosh^2\!\zeta}{1\!+\!\cosh\!\xi\cosh\!\eta\cosh\!\zeta} & \dfrac{\sinh\!\xi\cosh\!\eta\cosh\!\zeta\sinh\!\zeta}{1\!+\!\cosh\!\xi\cosh\!\eta\cosh\!\zeta} & \! \sinh\!\xi\cosh\!\eta\cosh\!\zeta \vphantom{\dfrac{\dfrac{}{}}{\tfrac{}{}}}\\ \dfrac{\sinh\!\xi\cosh\!\eta\sinh\!\eta\cosh^2\!\zeta}{1\!+\!\cosh\!\xi\cosh\!\eta\cosh\!\zeta} & 1\!+\!\dfrac{\sinh^2\!\eta\cosh^2\!\zeta}{1\!+\!\cosh\!\xi\cosh\!\eta\cosh\!\zeta } &\dfrac{\sinh\!\eta\cosh\!\zeta\sinh\!\zeta}{1\!+\!\cosh\!\xi\cosh\!\eta\cosh\!\zeta} & \! \sinh\!\eta\cosh\!\zeta \vphantom{\dfrac{\dfrac{}{}}{\tfrac{}{}}}\\ \dfrac{\sinh\!\xi\cosh\!\eta\cosh\!\zeta\sinh\!\zeta}{1\!+\!\cosh\!\xi\cosh\!\eta\cosh\!\zeta} & \dfrac{\sinh\!\eta\cosh\!\zeta\sinh\!\zeta}{1\!+\!\cosh\!\xi\cosh\!\eta\cosh\!\zeta} & 1\!+\!\dfrac{\sinh^2\!\zeta}{1\!+\!\cosh\!\xi\cosh\!\eta\cosh\!\zeta } & \!\sinh\!\zeta \vphantom{\dfrac{\dfrac{}{}}{\tfrac{}{}}}\\ \!\sinh\!\xi\cosh\!\eta\cosh\!\zeta & \!\sinh\!\eta\cosh\!\zeta &\!\sinh\!\zeta & \cosh\!\xi\cosh\!\eta\cosh\!\zeta \vphantom{\dfrac{\dfrac{}{}}{\tfrac{}{}}} \end{bmatrix}\boldsymbol{=} \nonumber\\ &\begin{bmatrix} \dfrac{\cosh\!\xi\!+\!\cosh\!\eta\cosh\!\zeta}{1\!+\!\cosh\!\xi\cosh\!\eta\cosh\!\zeta} & \boldsymbol{-}\dfrac{\sinh\!\xi\sinh\!\eta\cosh\!\zeta}{1\!+\!\cosh\!\xi\cosh\!\eta\cosh\!\zeta} & \boldsymbol{-}\dfrac{\sinh\!\xi\sinh\!\zeta}{1\!+\!\cosh\!\xi\cosh\!\eta\cosh\!\zeta} & 0 \vphantom{\dfrac{\dfrac{}{}}{\tfrac{}{}}}\\ \dfrac{\sinh\!\xi\sinh\!\eta}{1\!+\!\cosh\!\xi\cosh\!\eta\cosh\!\eta} & \hphantom{\boldsymbol{-}}\dfrac{\cosh\!\eta\!+\!\cosh\!\zeta\cosh\!\xi}{1\!+\!\cosh\!\xi\cosh\!\eta\cosh\!\zeta} & \boldsymbol{-}\dfrac{\cosh\!\xi\sinh\!\eta\sinh\!\zeta}{1\!+\!\cosh\!\xi\cosh\!\eta\cosh\!\zeta} & 0 \vphantom{\dfrac{\dfrac{}{}}{\tfrac{}{}}}\\ \dfrac{\sinh\!\xi\cosh\!\eta\sinh\!\zeta}{1\!+\!\cosh\!\xi\cosh\!\eta\cosh\!\zeta} & \dfrac{\sinh\!\eta\sinh\!\zeta}{1\!+\!\cosh\!\xi\cosh\!\eta\cosh\!\eta} & \dfrac{\cosh\!\zeta\!+\!\cosh\!\xi\cosh\!\eta}{1\!+\!\cosh\!\xi\cosh\!\eta\cosh\!\zeta} & 0 \vphantom{\dfrac{\dfrac{}{}}{\tfrac{}{}}}\\ 0 & 0 & 0 & 1 \vphantom{\dfrac{\dfrac{}{}}{\tfrac{}{}}} \end{bmatrix} \tag{16}\label{16} \end{align}$ ^{$\begin{matrix} (17) & R = [\begin{matrix} \frac{aporrear ξ + aporrear η aporrear ζ}{1 + aporrear ξ aporrear η aporrear ζ} & - \frac{pecado ξ pecado η aporrear ζ}{1 + aporrear ξ aporrear η aporrear ζ} & - \frac{pecado ξ pecado ζ}{1 + aporrear ξ aporrear η aporrear ζ} \\ \frac{pecado ξ pecado η}{1 + aporrear ξ aporrear η aporrear η} & \frac{aporrear η + aporrear ζ aporrear ξ}{1 + aporrear ξ aporrear η aporrear ζ} & - \frac{aporrear ξ pecado η pecado ζ}{1 + aporrear ξ aporrear η aporrear ζ} \\ \frac{pecado ξ aporrear η pecado ζ}{1 + aporrear ξ aporrear η aporrear ζ} & \frac{pecado η pecado ζ}{1 + aporrear ξ aporrear η aporrear η} & \frac{aporrear ζ + aporrear ξ aporrear η}{1 + aporrear ξ aporrear η aporrear ζ} \end{matrix}] \end{matrix}$ $\begin{equation} \mathcal R= \begin{bmatrix} \dfrac{\cosh\!\xi\!+\!\cosh\!\eta\cosh\!\zeta}{1\!+\!\cosh\!\xi\cosh\!\eta\cosh\!\zeta} & \boldsymbol{-}\dfrac{\sinh\!\xi\sinh\!\eta\cosh\!\zeta}{1\!+\!\cosh\!\xi\cosh\!\eta\cosh\!\zeta} & \boldsymbol{-}\dfrac{\sinh\!\xi\sinh\!\zeta}{1\!+\!\cosh\!\xi\cosh\!\eta\cosh\!\zeta} \vphantom{\dfrac{\dfrac{}{}}{\tfrac{}{}}}\\ \dfrac{\sinh\!\xi\sinh\!\eta}{1\!+\!\cosh\!\xi\cosh\!\eta\cosh\!\eta} & \hphantom{\boldsymbol{-}}\dfrac{\cosh\!\eta\!+\!\cosh\!\zeta\cosh\!\xi}{1\!+\!\cosh\!\xi\cosh\!\eta\cosh\!\zeta} & \boldsymbol{-}\dfrac{\cosh\!\xi\sinh\!\eta\sinh\!\zeta}{1\!+\!\cosh\!\xi\cosh\!\eta\cosh\!\zeta} \vphantom{\dfrac{\dfrac{}{}}{\tfrac{}{}}}\\ \dfrac{\sinh\!\xi\cosh\!\eta\sinh\!\zeta}{1\!+\!\cosh\!\xi\cosh\!\eta\cosh\!\zeta} & \dfrac{\sinh\!\eta\sinh\!\zeta}{1\!+\!\cosh\!\xi\cosh\!\eta\cosh\!\eta} & \dfrac{\cosh\!\zeta\!+\!\cosh\!\xi\cosh\!\eta}{1\!+\!\cosh\!\xi\cosh\!\eta\cosh\!\zeta} \vphantom{\dfrac{\dfrac{}{}}{\tfrac{}{}}} \end{bmatrix} \tag{17}\label{17} \end{equation}$} ^{$\begin{matrix} (18) & R = [\begin{matrix} porque θ + (1 - porque θ) {metro}_{X}^{2} & (1 - porque θ) {metro}_{X} {metro}_{y} + pecado θ {metro}_{z} & (1 - porque θ) {metro}_{X} {metro}_{z} - pecado θ {metro}_{y} \\ (1 - porque θ) {metro}_{y} {metro}_{X} - pecado θ {metro}_{z} & porque θ + (1 - porque θ) {metro}_{y}^{2} & (1 - porque θ) {metro}_{y} {metro}_{z} + pecado θ {metro}_{X} \\ (1 - porque θ) {metro}_{z} {metro}_{X} + pecado θ {metro}_{y} & (1 - porque θ) {metro}_{z} {metro}_{y} - pecado θ {metro}_{X} & porque θ + (1 - porque θ) {metro}_{z}^{2} \end{matrix}] \end{matrix}$ $\begin{equation} \mathcal R= \begin{bmatrix} \cos\theta+(1-\cos\theta)\mathrm{m}_{x}^2 & (1-\cos\theta)\mathrm{m}_{x}\mathrm{m}_{y}+\sin\theta\mathrm{m}_{z} & (1-\cos\theta)\mathrm{m}_{x}\mathrm{m}_{z}-\sin\theta\mathrm{m}_{y}\vphantom{\dfrac{\dfrac{}{}}{\tfrac{}{}}}\\ (1-\cos\theta)\mathrm{m}_{y}\mathrm{m}_{x}-\sin\theta\mathrm{m}_{z}& \cos\theta+(1-\cos\theta)\mathrm{m}_{y}^2 & (1-\cos\theta)\mathrm{m}_{y}\mathrm{m}_{z}+\sin\theta\mathrm{m}_{x}\vphantom{\dfrac{\dfrac{}{}}{\tfrac{}{}}}\\ (1-\cos\theta)\mathrm{m}_{z}\mathrm{m}_{x}+\sin\theta\mathrm{m}_{y} & (1-\cos\theta)\mathrm{m}_{z}\mathrm{m}_{y}-\sin\theta\mathrm{m}_{x} & \cos\theta+(1-\cos\theta)\mathrm{m}_{z}^2\vphantom{\dfrac{\dfrac{}{}}{\tfrac{}{}}} \end{bmatrix} \tag{18}\label{18} \end{equation}$} ^{$\begin{aligned} 2 porque θ + 1 = t r a C mi (R) = \\ \frac{(aporrear ξ + aporrear η + aporrear ζ) + (aporrear ξ aporrear η + aporrear η aporrear ζ + aporrear ζ aporrear ξ)}{(1 + aporrear ξ aporrear η aporrear ζ)} \\ (19) & \frac{(1 + aporrear ξ) (1 + aporrear η) (1 + aporrear ζ) - (1 + aporrear ξ aporrear η aporrear ζ)}{(1 + aporrear ξ aporrear η aporrear ζ)} \end{aligned}$ $\begin{align} & 2\cos\theta+1=\mathrm{trace}(\mathcal R)= \nonumber\\ &\dfrac{\left(\cosh\!\xi\!+\!\cosh\!\eta\!+\!\cosh\!\zeta\right)\!+\!\left(\cosh\!\xi\cosh\!\eta\!+\!\cosh\!\eta\!\cosh\!\zeta\!+\!\cosh\!\zeta\cosh\!\xi\right)}{\left(1\!+\!\cosh\!\xi\cosh\!\eta\cosh\!\zeta\right)} \nonumber\\ &\dfrac{\left(1\!+\!\cosh\!\xi\right)\left(1\!+\!\cosh\!\eta\right)\left(1\!+\!\cosh\!\zeta\right)-\left(1\!+\!\cosh\!\xi\cosh\!\eta\cosh\!\zeta\right)}{\left(1\!+\!\cosh\!\xi\cosh\!\eta\cosh\!\zeta\right)} \tag{19}\label{19} \end{align}$
$\begin{matrix} (20) & porque θ = \frac{(1 + aporrear ξ) (1 + aporrear η) (1 + aporrear ζ) - 2 (1 + aporrear ξ aporrear η aporrear ζ)}{2 (1 + aporrear ξ aporrear η aporrear ζ)} \end{matrix}$ $\begin{equation} \cos\theta=\dfrac{\left(1\!+\!\cosh\!\xi\right)\left(1\!+\!\cosh\!\eta\right)\left(1\!+\!\cosh\!\zeta\right)-2\left(1\!+\!\cosh\!\xi\cosh\!\eta\cosh\!\zeta\right)}{2\left(1\!+\!\cosh\!\xi\cosh\!\eta\cosh\!\zeta\right)} \tag{20}\label{20} \end{equation}$
$\begin{matrix} (21) & porque θ = \frac{(1 + γ_{1}) (1 + γ_{2}) (1 + γ_{3}) - 2 (1 + γ_{1} γ_{2} γ_{3})}{2 (1 + γ_{1} γ_{2} γ_{3})}, γ_{ȷ} = {(1 - α_{ȷ})}^{- \frac{1}{2}} \end{matrix}$ $\begin{equation} \cos\theta=\dfrac{\left(1\!+\!\gamma_1\right)\left(1\!+\!\gamma_2\right)\left(1\!+\!\gamma_3\right)-2\left(1\!+\!\gamma_1\gamma_2\gamma_3\right)}{2\left(1\!+\!\gamma_1\gamma_2\gamma_3\right)}\,,\quad \gamma_{\jmath}=\left(1\!-\!\alpha_{\jmath}\right)^{\boldsymbol{-\frac12}}\:\: \vphantom{\dfrac{\dfrac{a}{b}}{\dfrac{c}{d}}} \tag{21}\label{21} \end{equation}$
$\begin{aligned} (22.1) & pecado θ {metro}_{X} & = - \frac{(1 + aporrear ξ) pecado η pecado ζ}{2 (1 + aporrear ξ aporrear η aporrear ζ)} \\ (22.2) & pecado θ {metro}_{y} & = + \frac{(1 + aporrear η) pecado ζ pecado ξ}{2 (1 + aporrear ξ aporrear η aporrear ζ)} \\ (22.3) & pecado θ {metro}_{z} & = - \frac{(1 + aporrear ζ) pecado η pecado ξ}{2 (1 + aporrear ξ aporrear η aporrear ζ)} \end{aligned}$ $\begin{align} \sin\theta\,\mathrm{m}_{x} & = \boldsymbol{-}\dfrac{\left(1\!+\!\cosh\!\xi\right)\sinh\!\eta\sinh\!\zeta}{2\left(1\!+\!\cosh\!\xi\cosh\!\eta\cosh\!\zeta\right)} \tag{22.1}\label{22.1}\\ \sin\theta\,\mathrm{m}_{y} & = \boldsymbol{+}\dfrac{\left(1\!+\!\cosh\!\eta\right)\sinh\!\zeta\sinh\!\xi}{2\left(1\!+\!\cosh\!\xi\cosh\!\eta\cosh\!\zeta\right)} \tag{22.2}\label{22.2}\\ \sin\theta\,\mathrm{m}_{z} & = \boldsymbol{-}\dfrac{\left(1\!+\!\cosh\!\zeta\right)\sinh\!\eta\sinh\!\xi}{2\left(1\!+\!\cosh\!\xi\cosh\!\eta\cosh\!\zeta\right)} \tag{22.3}\label{22.3} \end{align}$
$\begin{aligned} pecado θ & = \frac{\sqrt{{(1 + γ_{1})}^{2} (γ_{2}^{2} - 1) (γ_{3}^{2} - 1) + {(1 + γ_{2})}^{2} (γ_{3}^{2} - 1) (γ_{1}^{2} - 1) + {(1 + γ_{3})}^{2} (γ_{1}^{2} - 1) (γ_{2}^{2} - 1)}}{2 (1 + γ_{1} γ_{2} γ_{3})} \\ (23) & γ_{ȷ} & = {(1 - α_{ȷ})}^{- \frac{1}{2}}, θ \in [0, π] \end{aligned}$ $\begin{align} \sin\theta & =\dfrac{\sqrt{\left(1\!+\!\gamma_1\right)^2\left(\gamma_2^2\!-\!1\right)\left(\gamma_3^2\!-\!1\right)\!+\!\left(1\!+\!\gamma_2\right)^2\left(\gamma_3^2\!-\!1\right)\left(\gamma_1^2\!-\!1\right)\!+\!\left(1\!+\!\gamma_3\right)^2\left(\gamma_1^2\!-\!1\right)\left(\gamma_2^2\!-\!1\right)}}{2\left(1\!+\!\gamma_1\gamma_2\gamma_3\right)} \nonumber\\ \gamma_{\jmath} & =\left(1\!-\!\alpha_{\jmath}\right)^{\boldsymbol{-\frac12}}\,,\quad \theta\in [0,\pi] \tag{23}\label{23} \end{align}$
$\begin{aligned} broncearse θ & = \frac{\sqrt{{(1 + γ_{1})}^{2} (γ_{2}^{2} - 1) (γ_{3}^{2} - 1) + {(1 + γ_{2})}^{2} (γ_{3}^{2} - 1) (γ_{1}^{2} - 1) + {(1 + γ_{3})}^{2} (γ_{1}^{2} - 1) (γ_{2}^{2} - 1)}}{(1 + γ_{1}) (1 + γ_{2}) (1 + γ_{3}) - 2 (1 + γ_{1} γ_{2} γ_{3})} \\ (24) & γ_{ȷ} & = {(1 - α_{ȷ})}^{- \frac{1}{2}}, θ \in [0, π] \end{aligned}$ $\begin{align} \tan\theta & =\dfrac{\sqrt{\left(1\!+\!\gamma_1\right)^2\left(\gamma_2^2\!-\!1\right)\left(\gamma_3^2\!-\!1\right)\!+\!\left(1\!+\!\gamma_2\right)^2\left(\gamma_3^2\!-\!1\right)\left(\gamma_1^2\!-\!1\right)\!+\!\left(1\!+\!\gamma_3\right)^2\left(\gamma_1^2\!-\!1\right)\left(\gamma_2^2\!-\!1\right)}}{\left(1\!+\!\gamma_1\right)\left(1\!+\!\gamma_2\right)\left(1\!+\!\gamma_3\right)-2\left(1\!+\!\gamma_1\gamma_2\gamma_3\right)} \nonumber\\ \gamma_{\jmath} & =\left(1\!-\!\alpha_{\jmath}\right)^{\boldsymbol{-\frac12}}\,,\quad \theta\in [0,\pi] \tag{24}\label{24} \end{align}$
$\begin{aligned} metro & = \frac{[(1 + γ_{1}) {(γ_{2}^{2} - 1)}^{\frac{1}{2}} {(γ_{3}^{2} - 1)}^{\frac{1}{2}}, (1 + γ_{2}) {(γ_{3}^{2} - 1)}^{\frac{1}{2}} {(γ_{1}^{2} - 1)}^{\frac{1}{2}}, {(1 + γ_{3})}^{2} {(γ_{1}^{2} - 1)}^{\frac{1}{2}} {(γ_{2}^{2} - 1)}^{\frac{1}{2}}]}{\sqrt{{(1 + γ_{1})}^{2} (γ_{2}^{2} - 1) (γ_{3}^{2} - 1) + {(1 + γ_{2})}^{2} (γ_{3}^{2} - 1) (γ_{1}^{2} - 1) + {(1 + γ_{3})}^{2} (γ_{1}^{2} - 1) (γ_{2}^{2} - 1)}} \\ (25) & γ_{ȷ} & = {(1 - α_{ȷ})}^{- \frac{1}{2}}, θ \in [0, π] \end{aligned}$ $\begin{align} \mathbf{m} & =\dfrac{\biggl[\left(1\!+\!\gamma_1\right)\left(\gamma_2^2\!-\!1\right)^{\boldsymbol{\frac12}}\left(\gamma_3^2\!-\!1\right)^{\boldsymbol{\frac12}}\:\boldsymbol{,}\:\left(1\!+\!\gamma_2\right)\left(\gamma_3^2\!-\!1\right)^{\boldsymbol{\frac12}}\left(\gamma_1^2\!-\!1\right)^{\boldsymbol{\frac12}}\:\boldsymbol{,}\:\left(1\!+\!\gamma_3\right)^2\left(\gamma_1^2\!-\!1\right)^{\boldsymbol{\frac12}}\left(\gamma_2^2\!-\!1\right)^{\boldsymbol{\frac12}}\biggr]}{\sqrt{\left(1\!+\!\gamma_1\right)^2\left(\gamma_2^2\!-\!1\right)\left(\gamma_3^2\!-\!1\right)\!+\!\left(1\!+\!\gamma_2\right)^2\left(\gamma_3^2\!-\!1\right)\left(\gamma_1^2\!-\!1\right)\!+\!\left(1\!+\!\gamma_3\right)^2\left(\gamma_1^2\!-\!1\right)\left(\gamma_2^2\!-\!1\right)}} \nonumber\\ \gamma_{\jmath} & =\left(1\!-\!\alpha_{\jmath}\right)^{\boldsymbol{-\frac12}}\,,\quad \theta\in [0,\pi] \tag{25}\label{25} \end{align}$}

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Figura-02 3D

^{(1) ver APÉNDICE C - Suma relativista de velocidades}

^{(2) ver ANEXO B - La matriz L}

^{(3) Construcción $\:\boldsymbol{\alpha}\:$ de $\:\boldsymbol{\beta}\:$}

Impresionante. Eso es mucho trabajo. Me tomará un tiempo digerirlo, pero muchas gracias por su tiempo.
@Frobenius Genial! Una pequeña pregunta: ¿Qué software/paquete está utilizando para el gráfico?
@JC: GeoGebra. Una de sus ventajas es la inserción de ecuaciones en $\LaTeX$ .

¿Podemos descomponer un impulso general de Lorentz en una rotación seguida de tres impulsos a lo largo de los ejes de coordenadas?

Giuseppe Negro

Dwagg

Giuseppe Negro

usuario178876

Frobenius

Giuseppe Negro

Frobenius

Frobenius

Respuestas (1)

Frobenius

Giuseppe Negro

JC

Frobenius

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Prueba de unicidad de transformación entre marcos relativistas

Transformación de Lorentz, problema con la derivación

Derivación de la transformación de Lorentz

Diferencia horaria como resultado del impulso de Lorentz

Transformación de Lorentz

¿No hay ordenamiento espacial en el espacio-tiempo plano de Minkowski 1+1D?

Matriz general Transformación de Lorentz

Invariancia del intervalo de espacio-tiempo directamente del postulado

Homogeneidad e isotropía y derivación de las transformaciones de Lorentz