Shinoteki is right—moving slower than light is timelike, while moving faster than light is spacelike. No relativistic change of reference frame will interchange those.
IIRC, movement in spacetime is the same no matter which axis you designate as being time.
No. The metric treats time differently from space even as they are all on a single manifold. The Minkowski metric has three spacial dimensions with a +, and time gets a -. This is why space and time are different. Thinking of spacetime as R^4 is misleading because one doesn’t have the Euclidean metric on it.
It shouldn’t. Moving from B to A slower than light is possible*, moving from A to B faster than light isn’t, and you can’t change whether something is possible by changing reference frames.
What I’m trying to get at is, What does a physicist mean when she says she saw X move from A to B faster than light? The measurement is made from a single point; say A. So the physicist is at A, sees X leave at time tX, sends a photon to B at time t0, and gets a photon back from B at time t1, which shows X at B at some time tB. I’m tempted to set tB = (t0+t1)/2, but I don’t think relativity lets me do that, except within a particular reference frame.
“X travelled faster than light” only means that tX < t1. The FTL interpretation is t0 < tX < tB < t1: The photon left at t0, then X left at tX, and both met at B at time tB, X travelling faster than light.
Is there a mundane interpretation under which tB < tX < t1? The photon left A at t0, met X at B at tB, causing X to travel back to A and arrive there at tX.
The answer appears to be No, because X would need to travel faster than light on the return trip. And this also explains that Owen’s original answer was correct: You can say that X travelled from A to B faster than light, or from B to A faster than light.
moving from A to B faster than the speed of light in one reference frame is equivalent to moving from B to A faster than the speed of light in another reference frame
or
moving from A to B faster than the speed of light in one reference frame is equivalent to moving from B to A slower than the speed of light in another reference frame
I meant the first one: faster than light in both directions.
You can think of it this way: if any reference frame perceived travel from B to A slower than light, then so would every reference frame. The only way for two observers to disagree about whether the object is at A or B first, is for both to observe the motion as being faster than light.
Second ‘faster’ should be ‘slower’, I think.
Shinoteki is right—moving slower than light is timelike, while moving faster than light is spacelike. No relativistic change of reference frame will interchange those.
What do you mean by “spacelike”?
IIRC, movement in spacetime is the same no matter which axis you designate as being time.
No. The metric treats time differently from space even as they are all on a single manifold. The Minkowski metric has three spacial dimensions with a +, and time gets a -. This is why space and time are different. Thinking of spacetime as R^4 is misleading because one doesn’t have the Euclidean metric on it.
It shouldn’t. Moving from B to A slower than light is possible*, moving from A to B faster than light isn’t, and you can’t change whether something is possible by changing reference frames.
*(Under special relativity without tachyons)
What I’m trying to get at is, What does a physicist mean when she says she saw X move from A to B faster than light? The measurement is made from a single point; say A. So the physicist is at A, sees X leave at time tX, sends a photon to B at time t0, and gets a photon back from B at time t1, which shows X at B at some time tB. I’m tempted to set tB = (t0+t1)/2, but I don’t think relativity lets me do that, except within a particular reference frame.
“X travelled faster than light” only means that tX < t1. The FTL interpretation is t0 < tX < tB < t1: The photon left at t0, then X left at tX, and both met at B at time tB, X travelling faster than light.
Is there a mundane interpretation under which tB < tX < t1? The photon left A at t0, met X at B at tB, causing X to travel back to A and arrive there at tX.
The answer appears to be No, because X would need to travel faster than light on the return trip. And this also explains that Owen’s original answer was correct: You can say that X travelled from A to B faster than light, or from B to A faster than light.
An interpretation putting t1<tX seems to have the photon moving faster than light backwards in time to get from B back to A
My question is whether he meant to say
moving from A to B faster than the speed of light in one reference frame is equivalent to moving from B to A faster than the speed of light in another reference frame
or
moving from A to B faster than the speed of light in one reference frame is equivalent to moving from B to A slower than the speed of light in another reference frame
both of which involve moving faster than light.
I meant the first one: faster than light in both directions.
You can think of it this way: if any reference frame perceived travel from B to A slower than light, then so would every reference frame. The only way for two observers to disagree about whether the object is at A or B first, is for both to observe the motion as being faster than light.
I know Owen was not talking about impossibility, I brought up impossibility to show that what you thought Owen meant could not be true.
Moving from B to A slower than the speed of light does not involve moving faster than light.