Changing inertial reference states and the twin paradox

[Leo_Bloom]

Say I climb to Mt Everest, hang out for a two weeks, and come back down. Lo, I am younger than my twin at sea level.

By two weeks?

Or, two weeks minus the my “age-loss” in each moment (at whatever scale) to ascend?

Same question, with descent.

Same question (adjusted) for up and down. Or is the predicate in the above question false–ie the paradox must always be for up and down?

Another question: what is the minimum “inertial reference-change” step, and are they summed? Is that question even germane to all the above questions?

Questions, questions. Thank you to all who respond.

Leo

[Leo_Bloom]

Sorry, forgot to add two other basic questions: should I have used “flies really fast” instead of "the guy on Mt. Everest?

Can there be some third inertial reference frame where I can watch these events transpire and say, cool, look.at these guys go at it, boy will they be surprised."

[MikeS]

@Leo_Bloom

Say I climb to Mt Everest, hang out for a two weeks, and come back down. Lo, I am younger than my twin at sea level.

Older, actually. See below.

By two weeks?

Or, two weeks minus the my “age-loss” in each moment (at whatever scale) to ascend?

Your age difference is much, much less than two weeks. The best way to state the result is that the ratio of the time elapsed on your clock vs. the time elapsed on your twin’s clock is

(your time elapsed)/(twin’s time elapsed) = 1 + (∆Phi)/c[sup]2[/sup]

where ∆Phi is the difference in the gravitational potential between your location and your twin’s location, and c is the speed of light. Since you have a higher gravitational potential than your twin does when you’re at the top of Mount Everest, ∆Phi is positive, and so you actually age slightly more than your twin does.

However, this ratio is very very close to 1 even if you’re all the way at the top of Mount Everest. If you actually did the experiment you’re describing (and let’s ignore the amount of time you’d need to climb and descend the mountain), then you would age about 1.1 microsecond more than your twin at sea level. If you stayed up there for four weeks, you’d age about 2.2 microseconds extra; six weeks, 3.3 microseconds extra; and so on.

Same question, with descent.

I assume you mean that you go down to the bottom of Death Valley or something like that? By the above logic, if you hang out at a lower gravitational potential, you’ll age less.

Same question (adjusted) for up and down. Or is the predicate in the above question false–ie the paradox must always be for up and down?

Another question: what is the minimum “inertial reference-change” step, and are they summed? Is that question even germane to all the above questions?

I’m not quite clear what you’re asking in the above questions. However, you seem to be a little confused between two effects here: the classic twin paradox, which doesn’t require gravity at all but does require one of the twins to be moving at a substantial fraction of the speed of light; and the phenomenon of gravitational time dilation, which says that clocks run faster when they’re higher in a gravitational well, regardless of how fast they’re moving relative to that gravity field.

[Exapno_Mapcase]

You don’t have to go anywhere or do anything. Your head is a few nanoseconds older than your feet.

[The_Hamster_King]

@MikeS

However, you seem to be a little confused between two effects here: the classic twin paradox, which doesn’t require gravity at all but does require one of the twins to be moving at a substantial fraction of the speed of light; and the phenomenon of gravitational time dilation, which says that clocks run faster when they’re higher in a gravitational well, regardless of how fast they’re moving relative to that gravity field.

To nitpick: The classic twin paradox requires one of the twins to experience significant acceleration. It’s not the trip at near light speed that puts one twin permanently behind the other, it’s the turnaround to come home.