When trying to find the Lorentz transformation in matrix form in the $x^2+x^3$-direction, I tried simply mapping the Lorentz boost in the $x^2$-direction to the $x+x^3$-direction by rotating it $45°$ around the $x^1$ axis:

$$ R_{x}^{-1}(\theta=45°)\left(\begin{array}{cccc}\gamma & 0 & -\gamma\beta &0\\0&1&0&0\\-\gamma\beta&0&\gamma&0\\0&0&0&1\end{array}\right)R_{x}(\theta=45°) $$ Where $\beta=\dfrac vc$ and $\gamma = \dfrac{1}{\sqrt{1-\beta^2}}$.

However, I am not sure of the proper form of the rotation matrix in 4-dimensional spacetime. Especially since we are using 'mostly minus' conventions. Using 'mostly plus' I would guess:

$$ R_x=\left(\begin{array}{cccc}-1&0&0&0\\0&1&0&0\\0&0&\cos\theta&-\sin\theta\\0&0&\sin\theta&\cos\theta\end{array}\right). $$

How would this translate to mostly minus conventions?

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    $\begingroup$ I believe the 00 component should be $+1$, as a rotation acts as the identity on $x^0$. $\endgroup$ – Dwagg Aug 11 at 20:00
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    $\begingroup$ What you are looking for is a Lorentz boost $B_{\beta}$ in the $x^2$ direction conjugated by a rotation $S_{\theta=\pi/4}$ in the $(x^2,x^3)$ plane, or $S_\theta^{-1} B_\beta S_\theta$, I believe. $\endgroup$ – Dwagg Aug 11 at 20:27

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