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I am given the question:

Using the determinant, show that the 1+1d Lorentz transformation matrix $\Lambda$ can be written in terms of hyperbolic trig functions, $$ \Lambda = \begin{pmatrix} \cosh u & -\sinh u \\ -\sinh u & \cosh u \end{pmatrix} . $$

This seems like a pretty paltry hint. Sure, the determinant of the usual 1+1d Lorentz matrix is $1 - \beta^2$, but how does that even remotely help?

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  • $\begingroup$ The determinant has to be 1. $\endgroup$
    – robphy
    Mar 2, 2021 at 1:51
  • $\begingroup$ @robphy That's the very next question, so it seems highly unlikely we're expected to know that for this question. $\endgroup$
    – Noldorin
    Mar 2, 2021 at 1:52
  • $\begingroup$ The determinant of a rotation is 1, and the determinant of a Galilean transformation is 1. $\endgroup$
    – robphy
    Mar 2, 2021 at 1:55
  • $\begingroup$ @robphy That's well and good, but we haven't shown that the Lorentz transform is a rotation. $\endgroup$
    – Noldorin
    Mar 2, 2021 at 1:56
  • $\begingroup$ Any Lorentz transformation obeys $\Lambda^T \eta \Lambda = \eta$ by definition. If you take the determinant of this equation you find immediately that any Lorentz transformation has $\det \Lambda = \pm 1$. $\endgroup$
    – Gold
    Mar 2, 2021 at 1:59

1 Answer 1

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Suppose that

$$ \Lambda = \begin{pmatrix} \gamma & -\beta\gamma \\ -\beta\gamma & \gamma \end{pmatrix} $$

where $\beta=\frac{v}{c}$ and $\gamma = (1-\beta^2)^{-\frac{1}{2}}$. Note that $\beta \in (-1, 1)$, so we can define $u$ as the unique real number such that $\beta = \tanh u$. Now, $\det \Lambda = \gamma^2 - \beta^2\gamma^2 = \gamma^2(1 - \beta^2) = 1$ so

$$ \gamma^2 = \frac{1}{1 - \tanh^2 u} = \cosh^2 u, $$

but $\cosh$ and $\gamma$ are positive so $\gamma = \cosh u$. Therefore,

$$ \Lambda = \begin{pmatrix} \cosh u & -\sinh u \\ -\sinh u & \cosh u \end{pmatrix} $$

as expected.

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  • $\begingroup$ Thank you. That works quite nicely. The only question is: why do we ignore the possibility of $\det \Lambda = -1$? $\endgroup$
    – Noldorin
    Mar 2, 2021 at 2:25
  • $\begingroup$ Lorentz transformations can be classified into proper (those with $\det\Lambda = +1$) and improper (those with $\det\Lambda = -1$). Examples of the former are rotations and Lorentz boosts and examples of the latter are inversion and time reversal. I think the question only applies to Lorentz boosts since it's easy to show that the determinant of the matrix given in the question is always $+1$ (rotations are trivial in $1+1$ space). Consequently, improper Lorentz transformations cannot be put into the form requested. $\endgroup$ Mar 2, 2021 at 3:00
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    $\begingroup$ Ah right, thanks for clarifying. I suppose that the formula $\det \Lambda = \gamma^2 (1 - \beta^2)$ immediately rules out any negative values. $\endgroup$
    – Noldorin
    Mar 2, 2021 at 4:45
  • $\begingroup$ My only remaining query is why $\Lambda^T \eta \Lambda = \eta$. (The answer in the comments didn't make sense to me, I'm afraid.) $\endgroup$
    – Noldorin
    Mar 2, 2021 at 4:52
  • $\begingroup$ It is the definition: a linear transformation $\Lambda$ is called a Lorentz transformation if $\Lambda^T\eta\Lambda = \eta$. It's similar to how a linear transformation $A$ is said to be orthogonal if $O^T O = O^T I O = I$. Intuitively, the equation $\Lambda^T\eta\Lambda = \eta$ says that $\Lambda$ preserves the spacetime interval, just like $O^T O = I$ says that an orthogonal transformation preserves Euclidean distance. $\endgroup$ Mar 2, 2021 at 5:42

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