# The physical interpretation of the Minkowski spacetime diagram

Discussion in 'Physics & Math' started by geordief, Dec 21, 2017.

1. ### Neddy BateValued Senior Member

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Yes, assuming the turnaround point has a clock Einstein synced with the start point, the traveler clock reads 20 seconds and the turnaround clock reads 40 seconds.

I am also trying to figure out how to reconcile the traveling-twin's "telescope view" of the home-twin's clock with the two different times he must calculate for the home-twin's clock, before and after the turn around point. I remember doing a calculation like that a long time ago, but I can't remember what I did. I am currently working on it, and will post when I figure it out.

For now, I may be wrong, but I think the time he sees in the telescope would be 40-34.64.=5.36. When I changed my scenario to match the scenario on Mike_Fontenot's website, the distance between the start and turnaround became 34.64 in the home-twin's frame, and 17.32 in the traveling twin's frame. Sorry, but that is all I have so far.

Confused2 likes this.

3. ### Confused2Registered Senior Member

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Sorry, I plucked 17.32 light seconds in the home twin frame without checking - 34.64 light seconds in that frame looks consistent with the scenario of 0.866c and 40 seconds travel time in the home twin frame. The figures in my post will reflect that failure to check and are wrong.

5. ### Mike_FontenotRegistered Member

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That's right. And when the CADO equation is used, the same result is obtained. At the instants in the traveler's life immediately before and immediately after the instantaneous turnaround (which are denoted as t = 20- and t = 20+, respectively), CADO_T(20-) = 10, and CADO_T(20+) = 70. So using the CADO equation directly at those two instants in the traveler's life gives the same CUMULATIVE times as the Lorentz equation does, and we both have to subtract those two times to get the ageing of the home twin during the instantaneous turnaround. So my only objection to your wording of your results is that you don't at that point in the analysis do the subtraction and state that the home twin gets 60 years older during the instantaneous turnaround, so that the three components of the home twin's ageing are kept track of separately. The fact that the home twin suddenly ages by a large amount during the turnaround is what is important in the resolution of the twin "paradox", and it deserves to be clearly emphasized.

As explained on the webpage, since nothing on the right-hand-side of the CADO equation changes during the instantaneous turnaround except for the velocity, it's possible to get the ageing of the home twin during the turnaround directly and very simply. We just define the "delta CADO equation" as:

The change in v during the turnaround is

delta(v) = v(20+) - v(20-) = (-0.866) - (0.866) = - 1.732.

(The relative velocity is taken as positive when the twins are moving apart, so on the outbound leg, v is positive, and on the inbound leg, v is negative.)

So we have

delta_CADO_T(20) = - L(20) * [ v(20+) - v(20-) ]
= -L(20) * delta(v)
= -34.64 * ( - 1.732 )
= 60 years.

So, to directly answer your question, the -1.732 number is the change in velocity at the instantaneous turnaround.

The "delta CADO equation" is just doing the subtraction that you and I needed to do manually when using either the Lorentz equations or the CADO equation. So it's not conceptually a big deal, but it just simplifies and speeds up the calculation of what happens at the turnaround, when you are doing lots of them.

7. ### Neddy BateValued Senior Member

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Okay, I think I'm starting to understand your method a little better now. I'll have to play around with it a bit to see if I can get the hang of it. Thanks!

Last edited: Dec 31, 2017
8. ### Neddy BateValued Senior Member

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Okay, I've been thinking more about this, and it seems to me that the traveling-twin's "telescope view" of the home-twin's clock is probably not going to be of much use. The problem is that the situation is perfectly reciprocal prior to the turnaround. Just as we can give the traveling twin a telescope to look at the home twin's clock, we can also give a telescope to the home twin to look at the traveling twin's clock.

Assuming our logic was correct in calculating that the time the traveling twin would see in the telescope would be 40-34.64.=5.36, we can use the same logic to calculate that the time the home twin would see in the telescope would be 20-17.32.=2.68. If we find it hard to see the reciprocality in that, then we can imagine letting the journey go twice as long in length, then it would be 40-34.64.=5.36 which is exactly the same.

Another way to see that the situation is perfectly reciprocal prior to the turnaround is to provide another Einstein-synchronised clock to be the reciprocal of the one that Confused2 added. This new Einstein-synchronised clock would be at rest in the traveling twin's frame, and strategically located so that it is co-located with the home twin simultaneously (in the traveling twin reference frame) with the turnaround. That means the time on both the traveling twin's clock and the newly added clock would be t'=20, when the home twin's clock displays t=10. Again, this shows the reciporcality of the situation.

To break the symmetry we can consider a different reference frame where a doppelganger of the traveling twin has been traveling from deep space, toward the home twin instead of away from him. We can imagine that at the moment the doppelganger and the original traveling twin are co-located, both of their clocks display t'=20. We provide the doppelganger frame its own Einstein-synchronised clock at rest in the doppelganger frame, and strategically located so that it is co-located with the home twin simultaneously (in the doppelganger frame) with t'=20. That means the time on both the doppelganger clocks would be t'=20, but the home twin's clock displays t=70 in that case. This finally breaks the symmetry, and represents what happens when the original traveling twin turns around.

9. ### phytiRegistered Senior Member

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317

When the reversal is abstracted to an unreal instantaneous event (left), the change in position of the axis of simultaneity (green) jumps from Bt=10 to Bt=70. A sees B age 60 units instantly!

If A performs a more realistic deceleration (right), the axis of simultaneity sweeps thru the angle with the events increasing in frequency for the 60 unit interval.

This is not observing aging, but doppler effected clock rates.

A receives 10/20 (1/2) while outbound and 70/20 (7/2) while inbound.

B receives 20/70 (2/7) and 20/10 (2) units.

The aging (accumulated time) requires a comparison of clocks at a common location.

10. ### phytiRegistered Senior Member

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317
Neddy Bate #52

I would think an infinite amount of force would be required. Maybe A would not be aware of any change, but that would be because he would be dead from the force smashing his body to a bloody pulp.

The force f is proportional to 1/r*r, and would increase without limit as r decreases. The problem with the discontinuous path, it's indeterminate at the break, and r=0. Many will misinterpret this as 1/0, or 'infinity'. There is no such value, it's a relation, literally without a boundary. If light travels instantaneously from A to B, it requires zero time. You can't have a=dv/dt if t=0. At the end of the outbound path, a=0.

11. ### phytiRegistered Senior Member

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317
Neddy Bate #52
The force f is proportional to 1/r*r, and would increase without limit as r decreases. The problem with the discontinuous path, it's indeterminate at the break, and r=0. Many will misinterpret this as 1/0, or 'infinity'. There is no such value, it's a relation, literally without a boundary. If light travels instantaneously from A to B, it requires zero time. You can't have a=dv/dt if t=0. At the end of the outbound path, a=0.

12. ### Neddy BateValued Senior Member

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Okay, but I think we can say that in the limit as the time-duration of a non-zero velocity change approaches zero, the required force approaches infinity?