[ c t ′ x ′ y ′ z ′ ] = [ γ − γ β x − γ β y − γ β z − γ β x 1 + ( γ − 1 ) β x 2 β 2 ( γ − 1 ) β x β y β 2 ( γ − 1 ) β x β z β 2 − γ β y ( γ − 1 ) β y β x β 2 1 + ( γ − 1 ) β y 2 β 2 ( γ − 1 ) β y β z β 2 − γ β z ( γ − 1 ) β z β x β 2 ( γ − 1 ) β z β y β 2 1 + ( γ − 1 ) β z 2 β 2 ] [ c t x y z ] {\displaystyle {\begin{bmatrix}c\,t'\\x'\\y'\\z'\end{bmatrix}}={\begin{bmatrix}\gamma &-\gamma \,\beta _{x}&-\gamma \,\beta _{y}&-\gamma \,\beta _{z}\\-\gamma \,\beta _{x}&1+(\gamma -1){\dfrac {\beta _{x}^{2}}{\beta ^{2}}}&(\gamma -1){\dfrac {\beta _{x}\beta _{y}}{\beta ^{2}}}&(\gamma -1){\dfrac {\beta _{x}\beta _{z}}{\beta ^{2}}}\\-\gamma \,\beta _{y}&(\gamma -1){\dfrac {\beta _{y}\beta _{x}}{\beta ^{2}}}&1+(\gamma -1){\dfrac {\beta _{y}^{2}}{\beta ^{2}}}&(\gamma -1){\dfrac {\beta _{y}\beta _{z}}{\beta ^{2}}}\\-\gamma \,\beta _{z}&(\gamma -1){\dfrac {\beta _{z}\beta _{x}}{\beta ^{2}}}&(\gamma -1){\dfrac {\beta _{z}\beta _{y}}{\beta ^{2}}}&1+(\gamma -1){\dfrac {\beta _{z}^{2}}{\beta ^{2}}}\\\end{bmatrix}}{\begin{bmatrix}c\,t\\x\\y\\z\end{bmatrix}}\,}
Hence:
[ c t 1 ′ x 1 ′ y 1 ′ z 1 ′ ] = [ γ − γ β x − γ β y − γ β z − γ β x 1 + ( γ − 1 ) β x 2 β 2 ( γ − 1 ) β x β y β 2 ( γ − 1 ) β x β z β 2 − γ β y ( γ − 1 ) β y β x β 2 1 + ( γ − 1 ) β y 2 β 2 ( γ − 1 ) β y β z β 2 − γ β z ( γ − 1 ) β z β x β 2 ( γ − 1 ) β z β y β 2 1 + ( γ − 1 ) β z 2 β 2 ] [ c t 1 x 1 0 0 ] = {\displaystyle {\begin{bmatrix}c\,t_{1}'\\x_{1}'\\y_{1}'\\z_{1}'\end{bmatrix}}={\begin{bmatrix}\gamma &-\gamma \,\beta _{x}&-\gamma \,\beta _{y}&-\gamma \,\beta _{z}\\-\gamma \,\beta _{x}&1+(\gamma -1){\dfrac {\beta _{x}^{2}}{\beta ^{2}}}&(\gamma -1){\dfrac {\beta _{x}\beta _{y}}{\beta ^{2}}}&(\gamma -1){\dfrac {\beta _{x}\beta _{z}}{\beta ^{2}}}\\-\gamma \,\beta _{y}&(\gamma -1){\dfrac {\beta _{y}\beta _{x}}{\beta ^{2}}}&1+(\gamma -1){\dfrac {\beta _{y}^{2}}{\beta ^{2}}}&(\gamma -1){\dfrac {\beta _{y}\beta _{z}}{\beta ^{2}}}\\-\gamma \,\beta _{z}&(\gamma -1){\dfrac {\beta _{z}\beta _{x}}{\beta ^{2}}}&(\gamma -1){\dfrac {\beta _{z}\beta _{y}}{\beta ^{2}}}&1+(\gamma -1){\dfrac {\beta _{z}^{2}}{\beta ^{2}}}\\\end{bmatrix}}{\begin{bmatrix}c\,t_{1}\\x_{1}\\0\\0\end{bmatrix}}\,=} [ γ ⋅ c t 1 − x 1 ⋅ γ ⋅ β x − γ ⋅ β x ⋅ c t 1 + x 1 ⋅ ( 1 + ( γ − 1 ) ⋅ β x 2 β 2 ) − γ ⋅ β y ⋅ c t 1 + x 1 ⋅ ( γ − 1 ) ⋅ β x ⋅ β y β 2 − γ ⋅ β z ⋅ c t 1 + x 1 ⋅ ( γ − 1 ) ⋅ β x ⋅ β z β 2 ] {\displaystyle {\begin{bmatrix}\gamma \cdot c\,t_{1}-x_{1}\cdot \gamma \cdot \beta _{x}\\-\gamma \cdot \beta _{x}\cdot c\,t_{1}+x_{1}\cdot (1+(\gamma -1)\cdot {\dfrac {\beta _{x}^{2}}{\beta ^{2}}})\\-\gamma \cdot \beta _{y}\cdot c\,t_{1}+x_{1}\cdot (\gamma -1)\cdot {\dfrac {\beta _{x}\cdot \beta _{y}}{\beta ^{2}}}\\-\gamma \cdot \beta _{z}\cdot c\,t_{1}+x_{1}\cdot (\gamma -1)\cdot {\dfrac {\beta _{x}\cdot \beta _{z}}{\beta ^{2}}}\end{bmatrix}}}
As in this case x 1 = − c t 1 {\displaystyle x_{1}=-c\,t_{1}} one gets:
c t 1 ′ = γ ⋅ c t 1 + c t 1 ⋅ γ ⋅ β x = γ c t 1 ( 1 + β x ) {\displaystyle c\,t_{1}'=\gamma \cdot c\,t_{1}+c\,t_{1}\cdot \gamma \cdot \beta _{x}=\gamma \,c\,t_{1}(1+\beta _{x})}
x 1 ′ = − γ ⋅ β x ⋅ c t 1 − c t 1 ⋅ ( 1 + ( γ − 1 ) ⋅ β x 2 β 2 ) = − c t 1 ( 1 + β x γ ( 1 + γ − 1 γ β 2 ⋅ β x ) ) = − c t 1 ( 1 + β x γ ( 1 + 1 − 1 − β 2 β 2 ⋅ β x ) ) {\displaystyle x_{1}'=-\gamma \cdot \beta _{x}\cdot c\,t_{1}-c\,t_{1}\cdot (1+(\gamma -1)\cdot {\dfrac {\beta _{x}^{2}}{\beta ^{2}}})=-c\,t_{1}\left(1+\beta _{x}\gamma \left(1+{\dfrac {\gamma -1}{\gamma \beta ^{2}}}\cdot \beta _{x}\right)\right)=-c\,t_{1}\left(1+\beta _{x}\gamma \left(1+{\dfrac {1-{\sqrt {1-\beta ^{2}}}}{\beta ^{2}}}\cdot \beta _{x}\right)\right)}
y 1 ′ = − γ ⋅ β y ⋅ c t 1 − c t 1 ⋅ ( γ − 1 ) ⋅ β x ⋅ β y β 2 = − c t 1 β y γ ( 1 + γ − 1 γ β 2 ⋅ β x ) = − c t 1 β y γ ( 1 + 1 − 1 − β 2 β 2 ⋅ β x ) {\displaystyle y_{1}'=-\gamma \cdot \beta _{y}\cdot c\,t_{1}-c\,t_{1}\cdot (\gamma -1)\cdot {\dfrac {\beta _{x}\cdot \beta _{y}}{\beta ^{2}}}=-c\,t_{1}\beta _{y}\gamma \left(1+{\dfrac {\gamma -1}{\gamma \beta ^{2}}}\cdot \beta _{x}\right)=-c\,t_{1}\beta _{y}\gamma \left(1+{\dfrac {1-{\sqrt {1-\beta ^{2}}}}{\beta ^{2}}}\cdot \beta _{x}\right)}
z 1 ′ = − γ ⋅ β z ⋅ c t 1 − c t 1 ⋅ ( γ − 1 ) ⋅ β x ⋅ β z β 2 = − c t 1 β z γ ( 1 + γ − 1 γ β 2 ⋅ β x ) = − c t 1 β z γ ( 1 + 1 − 1 − β 2 β 2 ⋅ β x ) {\displaystyle z_{1}'=-\gamma \cdot \beta _{z}\cdot c\,t_{1}-c\,t_{1}\cdot (\gamma -1)\cdot {\dfrac {\beta _{x}\cdot \beta _{z}}{\beta ^{2}}}=-c\,t_{1}\beta _{z}\gamma \left(1+{\dfrac {\gamma -1}{\gamma \beta ^{2}}}\cdot \beta _{x}\right)=-c\,t_{1}\beta _{z}\gamma \left(1+{\dfrac {1-{\sqrt {1-\beta ^{2}}}}{\beta ^{2}}}\cdot \beta _{x}\right)}
whereby 0.5 + 0.125 β 2 ≤ 1 − 1 − β 2 β 2 ≤ 0.5 + r β 2 {\displaystyle 0.5+0.125\beta ^{2}\leq \ {\dfrac {1-{\sqrt {1-\beta ^{2}}}}{\beta ^{2}}}\leq 0.5+r\beta ^{2}} with r = 0.5 for |β| ≤ 1 and r = 0.144 for |β| ≤ 0.5 and r = 0.126 for |β| ≤ 0.125
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one gets:
with r = 0.5 for |β| ≤ 1 and r = 0.144 for |β| ≤ 0.5 and r = 0.126 for |β| ≤ 0.125
