# Position Matters?

Twins Jack and Jill are initially at rest at x=L/2 in IRF K. Jill travels to the Origin at speed v, and Jack travels to x=L at the same speed. They are still the same age. At some pre-arranged time each twin ignites a rocket and begins accelerating in the positive x direction. Each twin shuts the thruster down at a later, pre-arranged time. They simultaneously come to rest in IRF K’, traveling in the +x direction of K at speed v>0. They decide to come together again, traveling toward one another at the same speed v relative to K’. Relative to K, Jill must travel faster than v and Jack must travel slower than v. When they get together again, Jack has aged less than Jill. Comments?

## Answers and Replies

I don't know what i'm talking about, and i certainly dont know the mathematics to back it up, but i'd like to take a stab to see how my intuition holds up against the knowledgable guys here.

relative to K',
jack and jill did not start accelerating from their respective positions simultaneously, so theyre not the same age when they initially come to rest in K'.

also, wouldn't the final accellerations relative to K play some part on their age difference in K?

DrGreg
Science Advisor
Gold Member
Twins Jack and Jill are initially at rest at x=L/2 in IRF K. Jill travels to the Origin at speed v, and Jack travels to x=L at the same speed. They are still the same age.
Just to clarify, at this point have they both come to rest again relative to K?

Twins Jack and Jill are initially at rest at x=L/2 in IRF K. Jill travels to the Origin at speed v, and Jack travels to x=L at the same speed. They are still the same age.
Assuming they both come to rest in K.
At some pre-arranged time each twin ignites a rocket and begins accelerating in the positive x direction.
Assuming they accelerate at the same rate as measured in frame K or by onboard accelerometers.
Each twin shuts the thruster down at a later, pre-arranged time.
Assuming at the same time, as measured in frame K or the same time as measured by their onboard clocks before resynchronisation.
They simultaneously come to rest in IRF K’, traveling in the +x direction of K at speed v>0.
The above assumptions imply they simultaneously (as measured in frame K) come to rest in IRF K’ . It turns out that it is not important whether you mean simultaneous in frame K or K' but it is important to aware of the difference. Whenever you say "simultaneously" you should specify according to what observer because simultaneity is relative.
They decide to come together again, traveling toward one another at the same speed v relative to K’.
Again, when they decide to come together again, did you mean they start the final phase simultaneously as measured in K, or simultaneously as measured in K' after they have resynchronised their clocks in K'?
Relative to K, Jill must travel faster than v and Jack must travel slower than v. When they get together again, Jack has aged less than Jill. Comments?
I am sure you meant to say that Jack will have aged more than Jill, because as you have already stated, Jill has travelled faster (and further) than Jack in frame K when they get back together again.

Simply put, Jack had a head start of L in the x direction and if they both take off simultaneously as measured in K and travel in the positive x direction and Jill catches up with Jack, then Jill has travelled further than Jack over the same time period (as measured in K) so Jill has travelled faster and aged less.

Last edited:
In the trival case where L = the Origin, Jack and Jill don't part company. All frames will agree on the simultaneity of their actions, because they happen at the same place as each other. Their worldlines will be identical, and so will their ages.

Otherwise... I assume the first pre-arranged time is according to K. And when you say they're the same age as Jill reaches the Origin, and Jack reaches x, I take it their age is measured in K. (In another frame, these events may happen at different times.)

When you say simultaneously come to rest in IRF K', do you mean that this event is simultaneous in K', as well as that they come to rest in K', or do they stop simultaneously in K?

If the simultaneity of their stopping in K' is according to K, then Jill ages less than Jack, as her spacetime trajectory between their parting and meeting is longer.

If the simultaneity of their stopping in K' is according to K', then it's still Jill who ages less. But according to K, Jill stops first, so--if I've got this right--she has to accelerate even more, in comparison to Jack, than in the previous case, Therefore Jill would age even less, compared to Jack, than she did if they stopped in K' at a moment that was simutaneous in K.

GRDixon said:
When they get together again, Jack has aged less than Jill

How do you know this is true ? Can we see your calculation, please ?

Without specifying the acceleration phase it's not possible to work out the proper times.

Draw a space-time diagram, then we'll see what's going on.

They are at equal conditions at start.
Use Jack and Jill own referentials.

As long as they apply the some amount of thrust(energy), during the same time (in their own referential) they will stay equivalent in all aspects.
I think that this can be used as a principle (same action imply the same consequences).

pre-arranged time: It is not impossible the use of an 'instant observer', independent of light travel to do the analisys, where 'pre-arranged time' makes sense. I bet that this step makes everything simpler.
I know that it is not the usual Einstein way of syncronization.

relative to K',
jack and jill did not start accelerating from their respective positions simultaneously, so theyre not the same age when they initially come to rest in K'.

A good point and a legitimate point of view.

Just to clarify, at this point have they both come to rest again relative to K?

Yes, that was the idea. They are both at rest in K when they fire up their thrusters.

Assuming at the same time, as measured in frame K or the same time as measured by their onboard clocks before resynchronisation.

They shut their thrusters down simultaneously according to the clocks of K.

I am sure you meant to say that Jack will have aged more than Jill, because as you have already stated, Jill has travelled faster (and further) than Jack in frame K when they get back together again.

You're right. My apologies for the mental short circuit.

When you say simultaneously come to rest in IRF K', do you mean that this event is simultaneous in K', as well as that they come to rest in K', or do they stop simultaneously in K?

I meant that they come to rest in simultaneously in K', as determined by the clocks in K.

If the simultaneity of their stopping in K' is according to K, then Jill ages less than Jack, as her spacetime trajectory between their parting and meeting is longer.

Right. I fouled up here. Jack will be older than Jill once they join up again.

What a pathetic thread. It's like arguing about the size of sqrt(2) when you could just calculate the value. Instead of handwaving and guesswork, why not draw the worldlines for the scenario in question and calculate the proper-times ? It's called 'science', some of you may have heard of it.

After my disrepectful remarks , I have no choice but to draw the diagram myself. I think each traveller has 4 phases of which I've labelled A 1, B1 etc. The small triangle and text show how to calculate the proper-time for each leg. Maybe GRDixon can fix the pic to fit his scenario and work out the proper-times. I had some trouble understanding when the changeovers happen after phase B. The purple line is the worldline of someone at L/2, from whose viewpoint the diagram is drawn. ( whoops, the origin is at L/2, makes no difference).

#### Attachments

• jack+Jill.png
2.5 KB · Views: 328
Maybe GRDixon can fix the pic to fit his scenario and work out the proper-times.

I'll have a go, but must confess that I'm just now reading "Relativity Simply Explained" by Martin Gardner. In brief, I'm out of my depth, at times, with some of the bright young minds who express their views on Physics Forums, particularly in GRT. I had a look at your diagram and like it! I'm going to learn how to do this stuff. At 71, I expect it's not going to be as easy as things were 50 years ago (ouch!) at MIT. But I know that it will be as much fun as ever. Speaking of GRT, I was anticipating more comments to my doggerel thread a la the Equivalence Principle. Thanks for commenting. Yours in Physics, GRD.

What a pathetic thread. It's like arguing about the size of sqrt(2) when you could just calculate the value. Instead of handwaving and guesswork, why not draw the worldlines for the scenario in question and calculate the proper-times ? It's called 'science', some of you may have heard of it.

"WHEELER'S FIRST MORAL PRINCIPLE. Never make a calculation until you know the answer" (Spacetime Physics, p. 60).

DrGreg
Science Advisor
Gold Member
The wording is still a little ambiguous, so I've drawn two possible diagrams. They both start the same.

(A) Jack and Jill begin moving apart, each at v relative to K in opposite directions

(B) They come to rest in K, simultaneously in K

(C) They start to accelerate identically, simultaneously in K.

(D) When they each reach a speed of v relative to K, they stop accelerating. This will occur simultaneously in K. At this point, both are at rest in K'.

Up to this point, by considering the symmetry relative to K, the proper time (i.e. time experienced by Jack or Jill) for each stage (A-B, B-C, C-D) is equal for Jack and Jill. So at the events D1 and D2, Jack and Jill are the same age.

This is where I now have to make an assumption.

Assumption 1 (left hand diagram)

(E) Jack and Jill begin moving towards each other, each at v relative to K' in opposite directions simultaneously in K'

(F) Jack and Jill meet and come to rest relative to K'.

Because K and K' have different definitions of simultaneity, Jill ages less from D1 to E1 than Jack does from D2 to E2. Now considering the symmetry relative to K', Jack and Jill age the same amount from E to F. Over the whole journey the only segment where there's an age difference is D to E, where Jack ages the most.

Assumption 2 (right hand diagram)

(E) Jack and Jill begin moving towards each other, each at v relative to K' in opposite directions simultaneously in K

(F) Jack and Jill meet and come to rest relative to K'.

Jack and Jill age the same amount from D to E. From E to F, Jack is moving slower than Jill relative to K, and so ages more. Over the whole journey the only segment where there's an age difference is E to F, where Jack ages the most. So, same outcome as assumption 1.

I suspect the questioner intended assumption 2, and may also have intended that the B-C and D-E segments are all zero duration.

#### Attachments

• Jack & Jill 2.png
42.9 KB · Views: 362
I suspect the questioner intended assumption 2, and may also have intended that the B-C and D-E segments are all zero duration.

That is so fine. I did have Assumption 2 in mind, but both are interesting. Now if I could only tie this in with the Equivalence Principle ... :) Muchas Gracias.

where Jack ages the most. So, same outcome as assumption 1.
I'm glad my intuitive hunch that jack ages the most with either assumption was correct (this time).
That is so fine. I did have Assumption 2 in mind, but both are interesting. Now if I could only tie this in with the Equivalence Principle ... :) Muchas Gracias.
I suspect we might not have addressed the crux of what you were asking in this thread because the title says "position matters?" and we have not really discussed what you meant by that. Also, could you post a link to the EP thread that you asked about earlier?

I'm glad my intuitive hunch that jack ages the most with either assumption was correct (this time).

I suspect we might not have addressed the crux of what you were asking in this thread because the title says "position matters?" and we have not really discussed what you meant by that. Also, could you post a link to the EP thread that you asked about earlier?

I am too. Even though Professor Einstein cautioned us about overly relying on our intuition, I listen to mine all the time (er-r-r, almost all the time).

What I had in mind, when I titled the thread "position matters?" was the idea that Jill ages less than Jack does by opting to ride in the back of the bus, so to speak. If their being at rest in a gravitational field is equivalent, then "how deep" one is in the field ... even a uniform field ... would determine whether one aged faster or slower than someone at a different depth. That to me (particularly in the case of uniform fields) is one of the profound, new insights in GRT. The more I work into it, the more excited I become.

DrGreg
Science Advisor
Gold Member
What I had in mind, when I titled the thread "position matters?" was the idea that Jill ages less than Jack does by opting to ride in the back of the bus, so to speak. If their being at rest in a gravitational field is equivalent, then "how deep" one is in the field ... even a uniform field ... would determine whether one aged faster or slower than someone at a different depth. That to me (particularly in the case of uniform fields) is one of the profound, new insights in GRT. The more I work into it, the more excited I become.
Actually the example you gave isn't quite equivalent to being at rest in a gravitational field. At the C events, in my diagram, Jack and Jill are a distance L apart, measured in K. Because of the specific way you specified the acceleration, they are still a distance L apart, measured in K, at the end of the acceleration, the D events. Because this length must be contracted relative to the length measured in K', Jack and Jill are at a greater distance apart, measured in K. From their own point of view, they move apart during the acceleration phase, and do not consider each other to be at rest. For more about this, look up "Bell Paradox".

To make the experiment equivalent to the gravitational case we need to modify it slightly.

(A) Jack and Jill begin moving apart, each at v relative to K in opposite directions

(B) They come to rest in K, simultaneously in K

Assumption 3

(C) They start to accelerate in such a way that they regard themselves to be at rest relative to each other. This is called "Born rigid" acceleration. To prevent the moving apart that occurred under the other assumptions, Jill must accelerate at a higher rate than Jack. As measured by K, Jack and Jill get closer to each other.

(D) When they each reach a speed of v relative to K, they stop accelerating. This will occur simultaneously in K'. At this point, both are at rest in K'. Because Jack's acceleration was lower, it took him a longer time to reach v along C2-D2 than it did for Jill along C1-D1.

(E) Jack and Jill begin moving towards each other, each at v relative to K' in opposite directions simultaneously in K'

(F) Jack and Jill meet and come to rest relative to K'.

Now considering the symmetry relative to K', Jack and Jill age the same amount from D to E to F. Over the whole journey the only segment where there's an age difference is C to D, where Jack ages the most. Again, the same result as assumptions 1 and 2, but under assumption 3 Jack and Jill consider themselves to be at rest relative to each other from B to E.

#### Attachments

• Jack & Jill 3.png
54 KB · Views: 369
Actually the example you gave isn't quite equivalent to being at rest in a gravitational field. At the C events, in my diagram, Jack and Jill are a distance L apart, measured in K. Because of the specific way you specified the acceleration, they are still a distance L apart, measured in K, at the end of the acceleration, the D events. Because this length must be contracted relative to the length measured in K', Jack and Jill are at a greater distance apart, measured in K. From their own point of view, they move apart during the acceleration phase, and do not consider each other to be at rest. For more about this, look up "Bell Paradox".

To make the experiment equivalent to the gravitational case we need to modify it slightly.

(A) Jack and Jill begin moving apart, each at v relative to K in opposite directions

(B) They come to rest in K, simultaneously in K

Assumption 3

(C) They start to accelerate in such a way that they regard themselves to be at rest relative to each other. This is called "Born rigid" acceleration. To prevent the moving apart that occurred under the other assumptions, Jill must accelerate at a higher rate than Jack. As measured by K, Jack and Jill get closer to each other.

(D) When they each reach a speed of v relative to K, they stop accelerating. This will occur simultaneously in K'. At this point, both are at rest in K'. Because Jack's acceleration was lower, it took him a longer time to reach v along C2-D2 than it did for Jill along C1-D1.

(E) Jack and Jill begin moving towards each other, each at v relative to K' in opposite directions simultaneously in K'

(F) Jack and Jill meet and come to rest relative to K'.

Now considering the symmetry relative to K', Jack and Jill age the same amount from D to E to F. Over the whole journey the only segment where there's an age difference is C to D, where Jack ages the most. Again, the same result as assumptions 1 and 2, but under assumption 3 Jack and Jill consider themselves to be at rest relative to each other from B to E.

Many thanks. I have printed this one out and will ponder it.