I Choice of signature important for superluminal 4-velocity? (Minkowski)

[PeterDonis]

I would think that parallel means "same sign".

Same sign of what?

A point that is stationary

I didn't describe a point, I described a curve. If you want to view it as a curve describing the "motion" of something, that something is certainly not staying at the same "point" (it goes from $x = - \infty$ to $x = \infty$), so "stationary" does not seem like a good word to describe it.

the x-axis is literally the position correct?

It's a coordinate line describing the range of possible positions in the $x$ direction.

time stays the same while position changes.

More precisely, coordinate time stays the same while position changes.

I don't know how to interpret that.

Well, we're talking about hypothetical objects that can "move" on spacelike trajectories. This particular spacelike trajectory covers every possible position on the $x$ axis in zero coordinate time. What does that suggest? (For example, what coordinate "speed" would you say such an object has?)

I don't see any other way how $\frac{dt}{ds}$ can change with frames without considering relative motion.

First, what is $dt / ds$ for the curve I described, where the only coordinate that changes as a function of $s$ is $x$?

Second, what does that same curve look like if we Lorentz transform to a different frame? I.e., what will the four functions $t'(s)$, $x'(s)$, $y'(s)$, and $z'(s)$ look like? Assume we are doing a Lorentz boost in the $x$ direction (which means $y' = y$ and $z' = z$, so only the $t'(s)$ and $x'(s)$ functions are of interest).

Third, given the answers above, will $dt / ds$ stay the same when we transform to the new frame, or will it change?

 

[PeterDonis]

Hopefully that definition is not limited to time like vectors.

No, it's valid in general since Rindler is taking the absolute value, which means this process discards the information about whether the vector is timelike or spacelike. (Note, however, that his definition as you give cannot possibly apply to null vectors, since their magnitude is zero.)

However, the fact that this definition throws away the information about whether the vector is timelike or spacelike means it can't possibly be used to determine a different formula for spacelike vectors than for timelike vectors. But that is what you are trying to use it to do.