Show that a matrix is a Lorentz transformation

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Homework Help Overview

The discussion revolves around demonstrating that the matrix \( e^{\Omega} \) represents a Lorentz transformation along the x-axis, where \( \Omega \) is defined as a specific matrix involving a parameter \( \psi \) related to rapidity. The context includes the relationship between rapidity and the Lorentz transformation matrix.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants explore the definition of the matrix exponential \( e^{\Omega} \) and its calculation. There are attempts to relate the matrix \( \Omega \) to the standard Lorentz transformation matrix \( \Lambda \) and questions about the presence of zeros on the diagonal of \( \Omega \). Some participants express confusion regarding the implications of the equations involving rapidity and velocity.

Discussion Status

Participants are actively discussing the properties of the matrix \( \Omega \) and its exponential. Some have noted the need to clarify the definition of \( e^{\Omega} \) and its computation, while others are considering the implications of the derived relationships. There is no explicit consensus yet, but the conversation is progressing with various interpretations being explored.

Contextual Notes

Participants are grappling with the definitions and properties of matrix exponentials, as well as the implications of rapidity in the context of Lorentz transformations. There is mention of potential errors in understanding the relationship between the matrices and the rapidity parameter.

fineTuner
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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


[/B]
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!
 
Physics news on Phys.org
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.
 
fineTuner said:
##\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##).
 

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