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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?

Qmechanic
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Noldorin
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  • The determinant has to be 1. – robphy Mar 02 '21 at 01:51
  • @robphy That's the very next question, so it seems highly unlikely we're expected to know that for this question. – Noldorin Mar 02 '21 at 01:52
  • The determinant of a rotation is 1, and the determinant of a Galilean transformation is 1. – robphy Mar 02 '21 at 01:55
  • @robphy That's well and good, but we haven't shown that the Lorentz transform is a rotation. – Noldorin Mar 02 '21 at 01:56
  • 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 $\det \Lambda = \pm 1$. – Gold Mar 02 '21 at 01:59
  • Yeah, that's the hint in the next question. Whoever set this question seems to be getting at something else... as if there's a simpler way. – Noldorin Mar 02 '21 at 02:00
  • IMHO there's hardly a simpler way than just taking $\det$ of the defining equation of a Lorentz transformation. – Gold Mar 02 '21 at 02:01
  • Gold, robphy, Okay, I'm going to assume for now that the lecturer accidentally put the questions the wrong way round, and go that way. Thanks. – Noldorin Mar 02 '21 at 02:03
  • @Gold By the way, is there any obvious reason that $\Lambda^T \eta \Lambda = \eta $ holds other than just working through the algebra? – Noldorin Mar 02 '21 at 02:12
  • A Lorentz transformation is by definition a linear transformation preserving the Minkowski metric. Because of this they obey $$\Lambda^\mu_{\phantom\mu \alpha}\Lambda^\nu_{\phantom\nu\beta}\eta_{\mu\nu}=\eta_{\alpha\beta}.$$ If you know how tensor components transform, the LHS is the transformation of the metric tensor under $\Lambda$ and you are just saying that it is kept invariant. The equation $\Lambda^T\eta\Lambda = \eta$ is just this exact same equation in matrix notation. – Gold Mar 02 '21 at 02:15
  • @Gold That's certainly not how we defined the Lorentz transform. (We derived it from first principles.) I don't know anything about tensors, by the way. – Noldorin Mar 02 '21 at 02:22
  • You say the determinant of the Lorentz transformation matrix you defined in class is $1-\beta^2$, but this is wrong - as robphy says, it should be 1. Is that a typo? – J. Murray Mar 02 '21 at 03:53
  • @J.Murray Sorry, I accidentally left out the $\gamma^2$ factor. This is the determinant simply computed from the usual form of the Lorentz matrix. – Noldorin Mar 02 '21 at 04:41

1 Answers1

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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.

Adam Zalcman
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  • Thank you. That works quite nicely. The only question is: why do we ignore the possibility of $\det \Lambda = -1$? – Noldorin Mar 02 '21 at 02:25
  • 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. – Adam Zalcman Mar 02 '21 at 03:00
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    Ah right, thanks for clarifying. I suppose that the formula $\det \Lambda = \gamma^2 (1 - \beta^2)$ immediately rules out any negative values. – Noldorin Mar 02 '21 at 04:45
  • 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.) – Noldorin Mar 02 '21 at 04:52
  • 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. – Adam Zalcman Mar 02 '21 at 05:42
  • While I'm sure that's a valid way to define a Lorentz transformation, we defined it as a transformation between two reference frames that preserves homogeneity and isotropy of space and keeps the laws of physics the same (including the constant speed of light). I don't immediately see how these are related. – Noldorin Mar 02 '21 at 16:34
  • We see that the laws of physics should not change just because we may have chosen a different coordinate system to describe them. Let's consider the consequences of this constraint. Some laws of physics are expressed in terms of the spacetime interval. Thus, we see that the spacetime interval - determined by $\eta$ - must not change when we apply a legal coordinate system transformation $\Lambda$. But, this just means that legal coordinate transformations obey $\Lambda^T\eta\Lambda = \eta$. – Adam Zalcman Mar 02 '21 at 18:12
  • Could you kindly clarify why your last sentence is true? – Noldorin Mar 02 '21 at 18:40
  • I'm happy to! Would you mind submitting a separate question about the equivalence of the two definitions? It's interesting in its own right and goes beyond the hyperbolic representation the current post concerns. Also, new question will benefit from more scrutiny and attention. In any case, the system indicates that our discussion here is getting too long :-) – Adam Zalcman Mar 02 '21 at 21:09
  • Sure, that sounds reasonable, will do! – Noldorin Mar 02 '21 at 22:22
  • Here you go: https://physics.stackexchange.com/questions/618222/prove-lambdat-eta-lambda-eta – Noldorin Mar 02 '21 at 23:22