0 Members and 1 Guest are viewing this topic.

In case of symmetrical motions, average relative velocity is zero.

Quote from: hamdani yusuf on 12/03/2024 12:11:14In case of symmetrical motions, average relative velocity is zero.?? I don't think you can just claim that separate motions are symmetrical. What do you mean symmetrical? What symmetry?

The asymmetry can then be used to determine absolute frame of reference.

Or, to put it another way, relative velocity. There is no absolute frame of reference.

Quote from: pzkpfw on 20/02/2024 21:06:19(Personally, I think the OP is diving off into Twins' Paradox etc, with ever increasing complexity, when they don't really grasp the significance of the basics of relativity. I think the explanations need to get simpler, not more complex.)Perhaps you can point out which basics of relativity I've missed out in my calculations with symmetrical twin travellings?

(Personally, I think the OP is diving off into Twins' Paradox etc, with ever increasing complexity, when they don't really grasp the significance of the basics of relativity. I think the explanations need to get simpler, not more complex.)

If two clocks have identical histories, you'd expect them to show the same time when they re-unite.

if they accelerate equally in opposite directions and then stop, they would appear to an observer at the starting point to be synchronised to one another, but neither could tell the time on the other's clock (or the observer's) simply by looking at his own.

Quote from: hamdani yusuf on 07/03/2024 03:31:47What if there is another twin travel to the opposite direction with the same speed? And another one in perpendicular direction?The formula for perpendicular velocity addition is:https://en.wikipedia.org/wiki/Velocity-addition_formulaIn our case, vx=vy=0.866c.For convenience, I used spreadsheet to calculate.putting the number to the left formula gives 0.247cputting the number to the right formula gives 1.732cEither result doesn't seem to be correct. It shouldn't be slower than the individual velocity. It shouldn't be higher than c either. Can someone show where the error is?

What if there is another twin travel to the opposite direction with the same speed? And another one in perpendicular direction?

... This messy matrix is not a Lorentz boost. It is, in fact, a combination of a Lorentz boost (corresponding to some velocity in the x−y plane) and a spatial rotation (again in the x−y plane.)However, at least we can get the magnitude of the resulting velocity from this matrix, as the upper left component of the matrix is not affected by the spatial rotation. It is determined purely by the Lorentz boost. A little bit of trivial algebra tells you the magnitude:v=√(vx^2 + vy^2 - vx^2.vy^2/c^2)

How are those numbers calculated?

They should depend on the distance and change of velocity.

The other travelling clock is further than the earth clock during the turn around. But somehow it undergoes less time jump. Where did I miss?

Your numbers in both posts are good except for that bit about 120/4 being 15.Not sure where 120/4 comes from. The 4 was from the perpendicular case, but the case being discussed was not the perpendicular one.

You missed the fact that the Earth clock is inertial between the two events of the 'jump' and the traveling clock is not, so its worldline is half the temporal length that it would have had had it been inertial between its two events.

Is there a formula to calculate the other's clock, when their relative position and motion are known?

Quote from: hamdani yusuf on 18/03/2024 12:57:18Is there a formula to calculate the other's clock, when their relative position and motion are known?Yes. Time dilatation formula. It's sufficiently accurate to allow GPS navigation.

Because it's true. The difference between the two scenarios is "simply by looking at his own."

How does A know what has happened to B, simply by looking at his own clock?