I'm trying to learn by myself some special relativity. By reading online I've come across the fact that Lorentz transformations are rotations on a 4D spacetime with a Minkowski metric. A rotation $\Lambda$ must preserve lengths, therefore, for a generic 4-vector $\vec{x}$:
$$ || \Lambda \vec{x} || = || \vec{x} || \Rightarrow \left( \Lambda^\alpha_\mu x^\mu \right) \left( \Lambda^\beta_\nu x^\nu \right) \eta_{\alpha \beta} = x^\mu x^\nu \eta_{\mu \nu} \Rightarrow \Lambda^\alpha_\mu \Lambda^\beta_\nu \eta_{\alpha \beta} = \eta_{\mu \nu} \Rightarrow \Lambda^T \eta \Lambda = \eta$$
which $$ \Vert\Lambda \vec{x}\Vert = \Vert\vec{x}\Vert \Rightarrow \left( \Lambda^\alpha_\mu x^\mu \right) \left( \Lambda^\beta_\nu x^\nu \right) \eta_{\alpha \beta} = x^\mu x^\nu \eta_{\mu \nu} \Rightarrow \Lambda^\alpha_\mu \Lambda^\beta_\nu \eta_{\alpha \beta} = \eta_{\mu \nu} \Rightarrow \Lambda^T \eta \Lambda = \eta,$$ which is the condition that allows $\Lambda \in O(1, 3)$. Now it is necessary to exclude spatial reflections, therefore it is necessary to force $\mathrm{det}\Lambda = 1$:
$$ \Lambda \in SO(1, 3) \subset O(1, 3) $$
I've $$ \Lambda \in SO(1, 3) \subset O(1, 3).$$ I've also read that it is necessary to exclude time reflections, which swap the past and the present. The condition which must be forced is $\Lambda^0_0 \geq 0$, the orthochronous condition, so that $\Lambda \in SO^+(1,3)$. \
Now I find it difficult to understand why that condition excludes time reflections. I've tried to interpret the condition like this: applying $\Lambda$ to a 4-vector $\vec{x}$ produces a 4-vector $\vec{x}'$ whose time component should be:
$$ x'^0 = \Lambda^0_\mu x^\mu = \Lambda^0_0 ct + \Lambda^0_1 x + \Lambda^0_1 y + \Lambda^0_1 z $$
I $$ x'^0 = \Lambda^0_\mu x^\mu = \Lambda^0_0 ct + \Lambda^0_1 x + \Lambda^0_1 y + \Lambda^0_1 z,$$ I get that if $\Lambda^0_0 < 0$ the sign in front of $ct$ changes, but it is not guaranteed that the whole $x'^0$ is actually negative... Someone couldCould someone please explain it to me?
Thank you!
