# Show that a matrix is a Lorentz transformation

## 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!

## Answers and Replies

PeroK
Science Advisor
Homework Helper
Gold Member
2020 Award
What is the definition of ##e^{\Omega}##?

Can you calculate ##\Omega^n##?

##\Omega^n## can be obtained multiplying the matrix n times. I underestimated the definition of ##e^{\Omega}##, i guess i can't obtain it taking the exponential of each element... now i'm reading the definition on wikipedia.

nrqed
Science Advisor
Homework Helper
Gold Member
##\Omega^n## can be obtained multiplying the matrix n times. I underestimated the definition of ##e^{\Omega}##, i guess i can't obtain it taking the exponential of each element... now i'm reading the definition on wikipedia.
The exponential of a matrix is defined as in quantum mechanics, through its Taylor expansion (here around ##\psi=0##).