Time dilation triplet paradox

Question

Here is the twin paradox with a twist.

Scenario 1: An observer (A) leaves from the equator of the earth and travels with an acceleration of $9.8\,\mathrm{m/s^2}$ in a north direction ( i.e. in the direction of the pole star) and continues to accelerate at the same rate for 5 earth years, then decelerates for 5 earth years ( -$9.8\,\mathrm{m/s^2}$) comes to a halt and immediately starts returning to the earth accelerating for 5 earth years at (-) $9.8\,\mathrm{m/s^2}$ and then decelerating at $9.8\,\mathrm{m/s^2}$ and arrives back after a total of 20 earth years. A second observer (B) stays at the equator where (A) left and is still there when (A) returns - assuming B lives so long....

From what I understand (A) has had a time dilation so although (B)'s clock says 20 years have passed, (A)'s clock will say less than 20 years have past

i.e. the typical twin paradox..

Scenario 2: A third observer (C) leaves from the same point on earth's equator at the same time as (A), but goes due to south (i.e. in the direction of the southern cross) and travels with the opposite accelerations and decelerations as (A) for the same Earth time, so that (C) arrives back when (B)'s Clock says 20 years have passed. i.e. at the same time as (A).

So from what I understand (C) has had a time dilation so although (B)'s clock says 20 years have passed, (C)'s clock will say less than 20 years have past.

In fact as the direction doesn't make a difference, (A) and (C)'s clocks should show identical times, even though (A) and (C)'s relative velocity was at all times twice the relative velocity of (A) and (B)'s and twice the relative velocity (B) and (C)'s.

Therefore one would guess that it is not the relative velocity itself that causes the different times, but the acceleration.

Now the question is, if an accelerating frame and a gravitational frame are equivalent and as (A), (B), and (C) are all subject to a acceleration of $9.8\,\mathrm{m/s^2}$ for the entire Earth 20 years, why would (B)'s clock be different to that of (A) and (C)?

Is there a difference between an accelerational field and forces caused by gravity and accelerational field and forces caused by a rocket? (ie in a gravitational accelerational field the time dilation is less than in rocket powered accelerational field)?

If there is a difference then why is there no difference between inertial mass and gravitational mass (surely if one is subject to a time dilation and the other not, there is also mass dilation in the one and not in the other?)

Is the answer that the direction of the acceleration make a difference? (What does that say about australian and english time?)

Answer 1

Therefore one would guess that it is not the relative velocity itself that causes the different times, but the acceleration.

It isn't acceleration either, there is a paper that gives example of motions with the same accelerations that have different time clock readings.

Clocks measure the proper time of worldlines. So given your path, you measure the proper time of the corresponding worldline. The reason the stay at home triplet measures the most time is because they follow a straight line, and straight lines have the greatest proper time of all the paths between two events. Similar to how straight lines have the least distance of all paths between two points in Euclidean space.

Now the question is, if an accelerating frame and a gravitational frame are equivalent and as (A), (B), and (C) are all subject to a acceleration of  for the entire Earth 20 years, why would (B)'s clock be different to that of (A) and (C)?

Acceleration is only locally like gravity. And it's actually the opposite. Real gravity causes tidal forces, what you are used to calling gravity is the effect of acceleration. The inertial frames are the freely falling frames (and only defined locally) and the person at home is being accelerated by the pressure of the ground beneath them so is not moving inertially.

In general relativity, since the inertial frames are infinitesimally small, you simply learn to work in an arbitrary frame by computing things locally. And you find out that you take the path and break it into a bunch of infinitesimal pieces and compute the proper time of each piece using your coordinate system and your metric. But there is still a path that gives the most proper time, and it isn't the one on the ground, it is the one that orbits the earth during the whole 20 years. Throw something up with a speed and angle to have it orbit the earth for the entire 20 years to crash down just in time.

So in Newtonian mechanics you are used to measuring gravity in a frame that isn't actually an inertial frame and most of the things you call gravity are an artifact of being in a non inertial (accelerating) frame.

And the real effects of gravity are real and are still there. They are time forces and gravitational time dilation which has an effect to make things orbit.

Is there a difference between an accelerational field and forces caused by gravity

In general relativity there are no forces caused by gravity. The inertial frame is the frame you like to say is falling. You think there is a force proportional to mass because you aren't computing in the inertial frame.

and accelerational field and forces caused by a rocket? (ie in a gravitational accelerational field the time dilation is less than in rocket powered accelerational field)?

There is a metric, caused by the real gravitational effect. It tells you how to measure proper time differences of small curves based on the coordinate changes in that infinitesimally small region. Every triplet had to use that metric. And every triplet had to take their curve and break it into pieces and use the metric to find the proper time of each one.

The most time would be measured by a freely falling person, which is none of them. Two triplets have rockets firing and the other one has the ground pushing on them. All of them are accelerating, so not moving inertially, so aging less than they could have.

So there isn't a force of gravity in general relativity. So your question doesn't make sense and we can't compare acceleration to something that doesn't exist.

An example would to imagine putting strings on a globe. The smallest path is the great circle. You can compute it by breaking it into pieces, finding the change of latitude and longitude, and using the metric to find the length of that piece. And if you wrapped a string in a not tight way it could be longer and you'd do the same thing. It would be longer and you could still find the length by breaking it into pieces, finding the change of latitude and longitude, and using the metric to find the length of that piece, and adding up the length of each piece.

In your case you have a worldline, you can break it into pieces, find the change in coordinates of each piece, use the metric to get a little bit of proper time for that piece, and then add them up and get the total proper time, which is what the clock reading will tell you. All the paths of the three triplets will be shorter in time than they could have been since none of them was the geodesic path between the events.