# How do time-like separated points preserve temporal ordering under orthochronous Lorentz Transformations?

How do time-like separated points preserve temporal ordering under orthochronous Lorentz Transformations? This question has already been asked in this Phys.SE post but I want to derive this result in another way which I suspect should be possible but I think I'm missing something. So I work in the $$(-,+,+,+)$$ signature. I have $$\Lambda^{\mu}_{\rho} \Lambda^{\nu}_{\sigma} \eta^{\rho \sigma}=\eta^{\mu \nu}$$ and say the difference of the two space time points in a frame S is given by $$z^{\mu}=(x^{\mu}-y^{\mu})$$. Now for time like separated points I have $$-(z^0)^2 + \sum_{i}(z^i)^2<0$$, I choose $$z^0$$ so that it is positive in S which means $$(z^0)^2 > \sum_{i}(z^i)^2$$. Now the $$0,0$$ Lorentz equation gives me ($$\Lambda^0_{0})^2=1+ \Lambda^0_{i}\Lambda^0_{i}$$ If we work with proper orthochronous transformations then this implies that $$\Lambda^0_0$$ is positive and greater than $$\Lambda^0_i \Lambda^0_{i}$$. Now Lorentz transforming $$z^0$$ we have $$(z')^0 = \Lambda^0_0 z^0 + \Lambda^0_i z^i$$. Now this is where I am kind of stuck. How exactly do I use the above inequalities to show that $$(z')^0$$ is positive? Because taking the square root of the inequalities I have $$z^0> \sqrt{\sum_{i}(z^i)^2}$$ and similarly for $$\Lambda^0_0$$

So I have found an answer by Qmechanic here this question is even more elaborate. I shall just translate the answer to my question in the notation I have used. So what I need to show is that $$(z')^0= \Lambda^0_0 z^0 + \Lambda^0_i z^i>0$$. Now first note that we we write $$\bar{\Lambda}^0_i= [\Lambda^0_1, \Lambda^0_2...]$$ and even the $$\bar{z}^i=[z^1,z^2...]$$ We now have the inequalities $$0\leq \Big(\frac{ \bar{\Lambda}^0_i}{\Lambda^0_0} + \frac{\bar{z}^i}{z^0}\Big)^2$$ which implies $$-2 \frac{{\Lambda}^0_iz^i}{z^0 \Lambda^0_0}\leq \Big(\frac{ \bar{\Lambda}^0_i}{\Lambda^0_0}\Big)^2 + \Big(\frac{\bar{z}^i}{z^0}\Big)^2 < \frac{(\Lambda^0_0)^2-1}{(\Lambda^0_0)^2}+1 <2$$. Thus we can now see $$0 < 2\Big(1+ \frac{{\Lambda}^0_iz^i}{z^0 \Lambda^0_0}\Big)$$ which is what we need.