I was deriving the matrix form of Lorentz boosts and I came up with a doubt. I don't think I quite understand hyperbolic rotations.
The stadard basis of the Minkowski space is given by $\{e_0, e_1,e_2,e_3\}$, where the square norm of any vector $x=x_0 e_0 + x_1 e_1 + x_2 e_2 + x_3 e_3$ is $|x|^2 = x_0 ^2 - x_1 ^2 - x_2 ^2 -x_3 ^2$. Note: $x_0$ is the time component ($e_0$ indicates time axis), $x_1, x_2, x_3$ is the spatial component ($e_1 , e_2, e_3$ indicate the $x, y,$ and $z$ axis).
Let $B_i$ be a Lorentz boost in the ith direction. This boost will only modify the time component and the $ith$ component, and like any other lorentz transformation, it will preserve the norm of any vector. Consider $B_i e_0 = a e_0 +b e_i = e_0 '$. Then,
$(B_i e_0)^2 = e_0 ^2 = 1$
$a^2 - b^2 = 1$
The solutions of this lie on a hyperbola along the $e_0$ axis. We can parametrize it and obtain:
$a=cosh \theta$ and $b=sinh\theta$
I do the same procedure for $B_i e_i = a e_0 +b e_i = e_i '$. $a$ and $b$ satisfy the equation: $b^2 -a^2 = 1$ which is a hyperbola along the $e_i$ axis.
My intuition: I imagine the $e_0$ as the horizontal axis in the x-y plane and the $e_i$ as the vertical axis. If I shift $e_0$ by and angle $\theta$ (counterclockwise) over the hyperbola, the resulting vector $e_0'$ will fall on the first quadrant. So $e_0 ' = cosh \theta e_0 + sinh \theta e_i $. Now I need to shift $e_i$ by the same angle (counterclockwise) over the corresponding hyperbola so that I can keep $e_i$ and $e_0$ orthogonal to each other. The resulting vector $e_i'$ will fall over the 2nd quadrant. So $e_i ' = -sinh \theta e_0 + cosh \theta e_i $
For $i=1$, I end up with a boost $B_i$ of the form: $$B_i= \left( \begin{matrix} \cosh \theta & -sinh \theta & 0 & 0\\ sinh \theta & cosh \theta & 0 & 0\\ 0 & 0 & 1 & 0\\ 0 & 0 & 0 & 1\\ \end{matrix}\right) $$ when I should end up with:
$$B_i=
\left(
\begin{matrix}
\cosh \theta & sinh \theta & 0 & 0\\
sinh \theta & cosh \theta & 0 & 0\\
0 & 0 & 1 & 0\\
0 & 0 & 0 & 1\\
\end{matrix}\right)
$$
Edit: The columns of this matrix are correct because the matrix satisfies Lorentz transformation condition $\eta = B_1 ^T \eta B_1$ (where $\eta$ is the minkowski metric tensor) while my boost matrix doesn't.
Question : Since taking the square might make me lose/add a negative sign, is there a better way to obtain $e_0'$ and $e_1'$ with all the necessary negative signs in its components? Right now I would just be using trial an error, until the boost matrix I obtain satisfies $\eta = B_1 ^T \eta B_1$
I appreciate any insights, thanks!