"Triplet Paradox" involving a single one-way trip

[mathyou9]

The "triplet paradox" that comes up in my Googling and search results involve one triplet that stays at rest and the other two venture out and back again, but in opposite directions. Essentially two "twin paradoxes" occurring together.

But what about a scenario in which triplets A and B are together and triplet C is waiting at some far-off location many light years away [maybe not triplets, per se, but three individuals born at the same time in the same reference frame.] A and C remain at rest. Let's say B departs A at 0.8c heading toward C. Since A and C remain in the same reference frame, albeit many light years apart, obviously they share simultaneity planes and age "together at the same rate," right? But how does B's age compare to A and C when he gets to C's location?

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I'm not sure what I'm missing, but I can't seem to wrap my head around it. Maybe I just need some very-detailed Minkowski diagrams (I really don't know.) Thanks. :-)

 

[1977ub]

If A B & C are the same age, and [at rest] in the same inertial frame, and B travels between A & C at relativistic speed, B will have aged much less than A & C have during the trip.

 

[mathyou9]

Ahhh, I got it now. My hangup was that each frame measures clocks in the other frame as going slow. I completely forgot (and I don't know why) to take into account B's "slanted" simultaneity plane relative to A & C's simultaneity plane. The other clock may run slow to you (and yours to it) but *when* your measurement starts and ends *in* the other frame is the key. Brainfart. :-)

[edited for clarity.]

 

[ghwellsjr]

The "triplet paradox" that comes up in my Googling and search results involve one triplet that stays at rest and the other two venture out and back again, but in opposite directions. Essentially two "twin paradoxes" occurring together.

But what about a scenario in which triplets A and B are together and triplet C is waiting at some far-off location many light years away [maybe not triplets, per se, but three individuals born at the same time in the same reference frame.] A and C remain at rest. Let's say B departs A at 0.8c heading toward C. Since A and C remain in the same reference frame, albeit many light years apart, obviously they share simultaneity planes and age "together at the same rate," right? But how does B's age compare to A and C when he gets to C's location?

---

I'm not sure what I'm missing, but I can't seem to wrap my head around it. Maybe I just need some very-detailed Minkowski diagrams (I really don't know.) Thanks. :-)

In the rest frame of A and C, B's clock runs at 60% of the coordinate time. If we assume that A and C are separated by 4 light years, it will take B 5 years of coordinate time to make the trip but his clock will show 3 years of elapsed time:

 

[mathyou9]

In the rest frame of A and C, B's clock runs at 60% of the coordinate time. If we assume that A and C are separated by 4 light years, it will take B 5 years of coordinate time to make the trip but his clock will show 3 years of elapsed time:

View attachment 83237​

I got myself (over)thinking this. Since the scenario involves one-way trip(s), symmetry between both frames of reference remains intact. So from B's frame of reference, he's the rest frame and A and C are the ones moving with slower clocks. ?

:

 

[Ibix]

Almost. You've forgotten the relativity of simultaneity, which means that (in this frame) the red clock starts ticking well before the traveller starts out, and finishes when he gets there. The blue clock starts ticking when the traveller sets out, but doesn't finish ticking until well after the traveller has arrived. That's how they manage to fit in six slow ticks - they either start early or finish late.

 

[mathyou9]

Ahhh, yes! Like this, right?

I totally understand simultaneity planes, but not seeing it in a proper Minkowski diagram, I guess I keep giving myself brainfarts. Haha! Thanks for being the "Gas X" for my brain. :-)

 

[ghwellsjr]

I got myself (over)thinking this. Since the scenario involves one-way trip(s), symmetry between both frames of reference remains intact. So from B's frame of reference, he's the rest frame and A and C are the ones moving with slower clocks. ?

View attachment 83424 :

There's a couple things wrong with your diagram. First, you've got black B aging 5 years when he actually ages 3 years and second, you have A and C aging only 3 years when you should show them aging 5 years:

 

[Ibix]

Ahhh, yes! Like this, right?
View attachment 83425
I totally understand simultaneity planes, but not seeing it in a proper Minkowski diagram, I guess I keep giving myself brainfarts. Haha! Thanks for being the "Gas X" for my brain. :-)

Again - almost. If you look at the red and blue tick marks in ghwellsjr's version of the diagram you'll see that they are considerably further apart than yours. Yours actually tick at the same rate as the black clock - George's tick slower.

 

[mathyou9]

Yeah...I made my diagram using MS Paint. But you can rest assured I envisioned ghwellsjr's diagram as I made mine. :-)

 

[ghwellsjr]

Yeah...I made my diagram using MS Paint. But you can rest assured I envisioned ghwellsjr's diagram as I made mine. :-)

You can avoid mistakes by using the Lorentz Transformation equations to get from one diagram to the next. You don't have to transform the coordinates of all the dots since equally spaced dots along one worldline will be equally spaced along their transformed worldline. And it helps to pick dots that evaluate to integers like the two dots at the intersections and the dots on the blue and red worldlines that are three dots away from the intersections. I recommend that you perform the evaluations on at least those four dots just to assure yourself that the process works and to give yourself confidence to try it that way the next time.