# Boost transformation [closed]

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.

• Do you know how the exponential of a matrix is defined? Nov 19 at 3:17
• Lorentz transformation Nov 19 at 4:52
• Check the identity for $K_1$ and next use the action of rotations on the boost vectors. Nov 19 at 7:24
• If you have some practice with Pauli matrices, you can use the fact that $\sigma_1\oplus 0 =K_1$ and $\sigma_1^2=I$…. Nov 19 at 7:33
• 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 $\:(\alpha \hat{n} \cdot \vec{K})^m=\alpha^m(\hat{n} \cdot \vec{K})^m\:$. Nov 19 at 10:54