Causality: Invariant Under Lorentz Transformation

I'll begin by stating that I have only studied a very small bit of special relativity: a couple of lectures or so around the end of the Physics course I took, intended just to "give us a taste". This means that my familiarity with is is very, very limited.

We were taught – and that does seem rather intuitive – that two events could have a causal relation if light speed will suffice to travel from one event to the other (that is: if $$\frac{\Delta r}{\Delta t} \leq c$$), since if that is the case, information can be transferred between the two locations "in time". Furthermore, the professor also mentioned almost as a side remark that if two reference frames are related through a Lorentz transformation, then this "property" of the two events will be invariant with respect to those reference frames.

I was hoping to have these two points clarified:

1. Which of these is true: is it that two events are, or simply could be causally related in a reference frame $$\iff$$ those two events are causally related in every reference frame that is related to that one through a Lorentz transformation? What I'm asking, in other words, is: is the property $$\frac{\Delta r}{\Delta t} \leq c$$ invariant, or is it the propety of actually being causally related? (Like the difference between me being close enough to someone to yell and get their attention, and me actually yelling to get their attention.)
2. How might I go about proving this? If the invariant part is $$\frac{\Delta r}{\Delta t} \leq c$$ then I assume the proof is algebric, and if so then I'd really appreciate seeing it. If the property is the actual causal relation then how does it even work? If events are too far to have information transferred between them in a given reference frame, but not in another, and that information does indeed get passed in one then it will get passed in the other as well, but that would mean information gets passed faster than light in the second reference frame... which shouldn't be possible according to special relativity's postulates, to my understanding.

2. Assume that in a given frame the spatial distance $$\Delta r$$ and time difference $$\Delta t$$ are such that $$\left|\frac{\Delta r}{\Delta t}\right| < c.$$ Then it is equally true that $$\left(\frac{\Delta r}{\Delta t}\right)^2 < c^2$$ and it follows $$(\Delta r)^2 < (c\Delta t)^2$$, that is: $$(\Delta r)^2 - (c\Delta t)^2 < 0.$$ Now you recognise that the righthand side is the Lorentz invariant interval. This is the same in every Lorentz frame. So if in a new frame with primed coordinates we also have $$(\Delta r')^2 - (c\Delta t')^2 < 0.$$ Repeating the steps above in the reverse order yields $$\left|\frac{\Delta r'}{\Delta t'}\right| < c.$$
• You are most welcome, @ShyGuy. Lorentz transformations can be quite intimidating the first time you encounter them, with all those $\gamma$s and $c^2$ floating around. In my opinion, the best approaches are the more geometrical ones that emphasise the similarity with the rotations. Also it helps a lot to set $c=1$ in all the formulas. Sep 16, 2021 at 9:04