Show that a matrix is a Lorentz transformation

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SUMMARY

The discussion focuses on demonstrating that the matrix $$ \Omega = \begin{pmatrix} 0 & -\psi & 0 & 0 \\ -\psi & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \end{pmatrix}$$ leads to a Lorentz transformation when exponentiated, specifically $$ e^{\Omega}$$. The rapidity $$\psi$$ is related to the velocity parameter $$\beta$$ through the equation $$\beta = tanh(\psi)$$. The standard Lorentz transformation matrix $$ \Lambda $$ is given as $$ \begin{pmatrix} cosh(\psi) & -sinh(\psi) & 0 & 0 \\ -sinh(\psi) & cosh(\psi) & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \end{pmatrix}$$. The key to solving the problem lies in understanding the matrix exponential and its Taylor series expansion.

PREREQUISITES
  • Matrix exponentiation and its applications in physics
  • Understanding of rapidity and its relation to velocity in special relativity
  • Familiarity with hyperbolic functions: cosh and sinh
  • Basic knowledge of Taylor series expansions
NEXT STEPS
  • Study the properties of matrix exponentiation in quantum mechanics
  • Learn about the derivation of Lorentz transformations from rapidity
  • Explore the Taylor series expansion of matrices
  • Investigate the relationship between hyperbolic functions and Lorentz transformations
USEFUL FOR

Students of physics, particularly those studying special relativity, mathematicians interested in linear algebra, and anyone looking to deepen their understanding of matrix operations in the context of transformations.

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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