# Deriving hyperbolic form of Lorentz transform

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?

• The determinant has to be 1. Mar 2, 2021 at 1:51
• @robphy That's the very next question, so it seems highly unlikely we're expected to know that for this question. Mar 2, 2021 at 1:52
• The determinant of a rotation is 1, and the determinant of a Galilean transformation is 1. Mar 2, 2021 at 1:55
• @robphy That's well and good, but we haven't shown that the Lorentz transform is a rotation. Mar 2, 2021 at 1:56
• 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 2, 2021 at 1:59

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.

• Thank you. That works quite nicely. The only question is: why do we ignore the possibility of $\det \Lambda = -1$? Mar 2, 2021 at 2:25
• 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. Mar 2, 2021 at 3:00
• Ah right, thanks for clarifying. I suppose that the formula $\det \Lambda = \gamma^2 (1 - \beta^2)$ immediately rules out any negative values. Mar 2, 2021 at 4:45
• 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.) Mar 2, 2021 at 4:52
• 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. Mar 2, 2021 at 5:42