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I have a boost defined as

$$L = \exp(\alpha \hat{n} \cdot \vec{K}), \quad \hat{n}^2=1, \quad \tanh\alpha = - \frac{v}{c}$$

where

$$K_{1} = \left(\begin{array}{cccc} 0 & 1 & 0 & 0 \\ 1 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ \end{array}\right), \quad K_{2}=\left(\begin{array}{cccc} 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 0 \\ 1 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ \end{array}\right), \quad K_{3}=\left(\begin{array}{cccc} 0 & 0 & 0 & 1 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 1 & 0 & 0 & 0 \\ \end{array}\right).$$

I wanted to show that transformation of every boost could be written as

$$L = 1 + (\hat{n} \cdot \vec{K})\; \sinh \alpha + (\cosh \alpha - 1)(\hat{n} \cdot \vec{K})^2.$$

I don't know how to show this.

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    $\begingroup$ Do you know how the exponential of a matrix is defined? $\endgroup$
    – Ghoster
    Nov 19 at 3:17
  • $\begingroup$ Lorentz transformation $\endgroup$
    – Frobenius
    Nov 19 at 4:52
  • $\begingroup$ Check the identity for $K_1$ and next use the action of rotations on the boost vectors. $\endgroup$ Nov 19 at 7:24
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    $\begingroup$ If you have some practice with Pauli matrices, you can use the fact that $\sigma_1\oplus 0 =K_1$ and $\sigma_1^2=I$…. $\endgroup$ Nov 19 at 7:33
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    $\begingroup$ Hint : Replace $\:x\:$ by $\:(\alpha \hat{n} \cdot \vec{K})\:$ in $$ e^x=\sum\limits_{m=0}^{m=\infty}\dfrac{x^m}{m!}=\dfrac{\cosh x+\sinh x}{2} $$ and try to find what about the powers $\:(\alpha \hat{n} \cdot \vec{K})^m=\alpha^m(\hat{n} \cdot \vec{K})^m\:$. $\endgroup$
    – Frobenius
    Nov 19 at 10:54

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