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 $\Lambda^0_0$
1 Answers
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.
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