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## Homework Statement

Given the matrix

$$ \Omega = \begin{pmatrix}

0 & -\psi & 0 & 0 \\

-\psi & 0 & 0 & 0 \\

0 & 0 & 0 & 0 \\

0 & 0 & 0 & 0

\end{pmatrix}$$

show that ## e^{\Omega}## is a Lorentz transformation along the x-axis with ## \beta = tanh(\psi)##

## Homework Equations

During the lesson we derived from the standard Lorentz transformation matrix the following matrix, where ##\psi## is the rapidity:

$$ \Lambda = \begin{pmatrix}

cosh(\psi) & -sinh(\psi) & 0 & 0 \\

-sinh(\psi) & cosh(\psi) & 0 & 0 \\

0 & 0 & 0 & 0 \\

0 & 0 & 0 & 0

\end{pmatrix} $$

Other equations:

##cosh(\psi)=\gamma##

##sinh(\psi)=\gamma \beta##

## The Attempt at a Solution

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From ## \beta = tanh(\psi)##:

## \psi=arctg(\beta) = \ln\sqrt{\frac{1+\beta}{1-\beta}} ##

## e^{-\psi} = \sqrt{\frac{1-\beta}{1+\beta}}##

I think i have to show that the two matrices (##\Lambda## and ##e^{-\Omega}##) are the same, but i can't understand why there are zeros on the diagonal. For the two first zeros on the diagonal ##cosh(\psi)=0##, so ##\psi = \frac \pi 2##.

I think there's an error somewhere, because with the previous formulas it turns out that ##\beta = 1## and ##v=c##.

To be honest, i can't find the right way to solve the problem, maybe it's just algebra? Can you please give me a hint? Thank you!