Thanks for that info @robphy! I didn't know that $k$ was in wide use. I discovered it earlier this year while searching for simple ways to produce rational $(\beta, \gamma)$ pairs. I was quite pleased when I discovered its multiplicative properties, and realised how it's related to rapidity. And I wondered why I hadn't encountered it anywhere, since it seemed pretty useful to me. It's kind of spooky that I also chose the letter $k$ for it. :) I mentioned it in The h Bar physics chatroom, but nobody responded to my comment. So I figured it was either too boring &/or well-known...

@PM 2Ring: thanks for your answer, as well as robphy's addition. I understand everything you wrote, I think. And yet your use of complex numbers to remove minus sign seems artificial. Worse, it seems unnecessary. If (\Delta s)^2=(c\Delta t_2)^2-(\Delta x_2)^2, why not just say delta T is the hypotenuse and delta S and X are the sides? i.e. T being the distance and S and X being orthogonal dimensions?

@Exocytosis We want $\Delta s$ to be the hypotenuse because it's the invariant thing, corresponding to $r$ in the Euclidean equation of $r^2=x^2+y^2$. As I said, when we rotate the coordinate axes we get $r^2=x'^2+y'^2$. The length $r$ is (of course) constant, and it is independent of whatever axes we use to resolve it into $x$ & $y$ components. Exactly the same thing is going on in SR: we have a spacetime measurement $\Delta s$ that's independent of whatever space & time axes we choose. There's an extra twist, though: $(\Delta s)^2$ can be zero or negative, not just positive.

@Exocytosis BTW, the standard derivation of $\Delta s$ and $\gamma$ uses a normal Euclidean right triangle, involving a reflected light beam measured in both the rest frame of the light source and in a frame moving at $v$ relative to that frame.

@PM 2Ring: but delta T can be considered an invariant too if one decides so, is that not an arbitrary choice to choose delta S as such, mathematically speaking? How do you justify substracting space to time otherwise? The radius example may look obvious but I do not see how that makes it right when adding time.

@Exocytosis Which delta T? Remember, there's no privileged frame (although in a given situation 1 or more frames can be convenient or useful). If $(\Delta s)^2>0$ then the interval from A to B is said to be timelike, that is, the worldline of an inertial observer can pass through A & B. In that frame, $\Delta x=0$ (A & B happen at the same location, separated by time but not by space), so their $c\Delta t=\Delta s$ and we call $\Delta t$ the proper time between A & B. The proper time is an invariant, but that's because it's proportional to $\Delta s$.

@Exocytosis There's nothing fishy going on in $(\Delta s)^2=(c\Delta t)^2-(\Delta x)^2$. We're just adding squared distances together; $ct$ has units of distance. Another option is to divide the whole equation by $c^2$ and work in units of time. Actually, it's common in SR to use "natural" units where $c$ is numerically equal to 1, like seconds and light-seconds, or years and light-years. This makes all the equations simpler to remember and to manipulate, although you do have a little bit of fiddly algebra at the end when you need to "decode" stuff back into separate space & time units.

@Exocytosis In SR, 1 second of time has the same magnitude as 1 light-second of distance. However, it's not like the distinction between space & time is lost. For a given spacetime interval AB all observers will agree on whether the interval is timelike ($(\Delta s)^2>0$), lightlike ($(\Delta s)^2=0$), or spacelike ($(\Delta s)^2<0$). But they will disagree on the exact values of the space & time components of the interval. And there is no "true" value for those components, each observer's frame provides a valid view of what's going on.

@PM 2Ring: ok my problem is that there can be a particle staying put while the clock is ticking, but there cannot be a moving particle in zero delay. So that makes space and time dimensions not entirely orthogonal in my opinion. So, unlike X and Y that are independent for 2D coordinates so sqrt(x^2+y^2) gives a distance to the origin, this is not feasible with time against space here. So what I am seeing in t^2-x^2 in that time passes anyway, but if the particle moves, it cannot be counted twice. See where I am going?

@Exocytosis "there cannot be a moving particle in zero delay" True, by definition, motion involves a change in position over time. However, we can have a pure spacelike spacetime interval, where Delta x is nonzero but Delta t is zero.

Eg, if events A, B & C are separated in space in my frame but simultaneous, then the intervals between them are purely spacelike for me. They will be spacelike in other frames, but not simultaneous, and different observers will disagree on the temporal order of those events.

Wikipedia has a nice diagram in en.wikipedia.org/wiki/Relativity_of_simultaneity

> So what I am seeing in t^2-x^2 in that time passes anyway, but if the particle moves, it cannot be counted twice. See where I am going?

Sorry, I'm not sure what you're getting at there.

I've cleaned up some of my comments since they'll be preserved in this room. Unfortunately, we don't have MathJax support in chat. But we can post images, and make directed replies.