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

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SUMMARY

The discussion centers on the significance of signature choice in the context of superluminal 4-velocities within Minkowski spacetime. The original poster successfully computed the tangent vector to a spacelike curve using the signature (-,+,+,+) but encountered issues with (+,-,-,-). It was concluded that the term "superluminal 4-velocities" is misleading, as the focus should be on tangent vectors to spacelike curves, where the squared norm's sign is determined by the signature used. The choice of signature does not affect the underlying physics or mathematics, provided that consistent definitions are applied throughout.

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  • #61
PhDeezNutz said:
I would think that parallel means "same sign".

Same sign of what?

PhDeezNutz said:
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.

PhDeezNutz said:
the x-axis is literally the position correct?

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

PhDeezNutz said:
time stays the same while position changes.

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

PhDeezNutz said:
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?)

PhDeezNutz said:
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?
 
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  • #62
PhDeezNutz said:
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.
 
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