In the twins paradox of relativity, is this an accurate non-mathematical description of what happens in the the traveling twin's frame of reference?

Question

In the twins "paradox" thought experiment of special relativity, in the frame of the rocket, upon accelerating quickly to almost the speed of light, the clocks attached to the destination star ten light years away "jump" from displaying "zero" to "ten years", due to time desynchronization. Also, in the frame of the rocket, the star jumps from ten light years away, to a few light hours away, due to the Lorentz contraction.

Over the next few hours, the star travels towards the rocket at just under the speed of light, covering the few light hours in a few hours, and when it reaches the rocket, the time on the star clock is still ten years (plus a few hours) because of time dilation.

Then the rocket accelerates towards the earth, stopping at the star momentarily and then whizzing off towards earth at just under the speed of light. During this acceleration the clocks of earth jump from displaying zero to ten years momentarily, and then to displaying 20 years, due to the old time desychronization effect getting replaced with first nothing, and then a new and opposite one.

The earth jumps to being 10 light years away for that instant when the rocket is motionless during its reversal, and at that time the earth's clocks run at a normal speed, and then jumps to being a few light hours from rocket, due to the Lorentz contraction, when the rocket starts moving at almost the speed of light towards it, and the clocks slow down to a glacial speed again due to time dilation.

The earth covers the few light hours in a few hours and when it reaches the rocket, the earth clocks still display 20 years (plus a double helping of a few hours) because of time dilation, while the rocket clock shows a double helping of a few hours, since it shows a few hours more than it did when it was at the star.

The rocket then stops, which has no effect on the time displayed on the rocket clock nor the earth clock, but does cause the earth clock and the star clock to start running at a normal rate.

It also causes the star clock to jump from displaying 10 years (plus a few hours) to displaying 20 years (plus a double helping of a few hours) by removing the time desychronization.

Thus at the end, like at the start, the earth and star clocks display the exact same time in the rocket frame (and in the earth frame). The rocket clock is the odd one out with just a double helping of a few hours displayed on it, while all three sets of clocks are now running at a normal rate.

And, of course, the two twins are again both aging at a normal rate in the rocket frame, but the earth twin is twenty years older while the traveling twin is only a double helping of a few hours older.

Would that be accurate?

Edit: Willo's answer pointed out an obvious mistake I made, which was to forget that whenever the rocket gets a new velocity (or do I mean a new acceleration --I am confused now) the frame of reference attached to it strictly speaking becomes a new frame of reference. So the question should have asked about the frames of reference of the rocket, I guess. I'm adding this because I don't want every answer to repeat what Willo has pointed out.

Edit: I meant that when the rocket "jumps" from one velocity to another (because the acceleration lasts only a moment, or a second) which means that there is a new frame of reference, the clock "jumps" (actually taking as long to change to the new value as the time that acceleration lasts). I guess the idea the clocks jumping is connected to the idea that the accelerations last a negligible time, and so the time it takes for the clock to change from displaying "zero" to "ten years" is likewise negligible. I was using the word "jump" in the layman's sense of a sudden or rapid change followed by a period of no change. I didn't mean to suggest that there was any discontinuity. It's easier to visualize a clock jumping from one value to another than to visualize it changing over time until it gets to the new value. Likewise, it's easier to picture a clock as frozen rather than moving so slowly that it looks frozen. It's also simpler and more memorable, as well as more attention-grabbing, to talk of a star (or the earth) jumping from far away to nearby. I tend to favor simple ideas because it is easier for everyone to use a simple idea, and remember it, including me. Of course, I would never want to oversimplify, but it's a fine line I guess (in this case I think it's worth it. BTW I didn't come up with this memorable way of putting it. I read a comment the other day by the user called "Max", where he used it. I still think it is an awesome way of explaining it in an introduction for a layman.

Answer 1

I think your description of clocks 'jumping' is problematic, although the gist of what you suggest is broadly right if you put the jumping clocks aside.

In the explanation below, let's assume the clocks on Earth and on the distant star are set to Earth time, and that the Earth's frame can be treated as being inertial (ie we can ignore its rotation etc).

When the travelling twin (TT) and the remaining twin (RT) are initially at rest (let's say that's on January 1st 2023), for both twins the 'current' time on the distant star is the same as the current time on Earth, ie it is January 1st 2023.

However, as soon as the TT has instantaneously accelerated to near c, the TT's plane of simultaneity tilts upwards through time in the direction in which the TT is travelling, as a consequence of which the 'current' time on the distant star in the TT's outbound frame is ten years ahead of the 'current' time on the distant star in the RT's frame- ie it is already January 1st 2033 on the distant star when the RT starts coasting towards it. I think that is what you meant when you talked about clocks on the distant star 'jumping forward ten years'.

A day later, the TT reaches the distant star on January 2nd 2033 Earth time. In the TTs outbound frame, owing to the relativity of simultaneity, the 'current' time on Earth is still January 2nd 2023, so the clocks on the Earth and on the star seem 10 years out of sync to the TT.

If the TT now decelerates instantaneously to be at rest with the distant star, the TT's plane of simultaneity rotates back to being level with the RTs. That rotation causes the 'current' time on Earth for the TT to be 2nd January 2033.

So you see that it is the acceleration/deceleration of the RT that causes the RT to shift between inertial reference frames, and those shifts cause tilts in the planes of simultaneity of the RT, and it is those tilts which cause big changes to the 'current' time in faraway places.

The TT's return journey is just a repeat of the outward one, so there is nothing new to be gained by examining it.

The explanation so far has focussed on the effect the relativity of simultaneity has on timings, so let's talk now about the effect on distances.

When you ask how far away an object is, what you mean is how far away is it 'now', because clearly the position of an object can change with time. That immediately highlights the impact of simultaneity, because if two observers are in motion relative to one another, the two observers will disagree where the distant object is 'now', because they each have a different view of what time 'now' is.

To go back to your example, at the outset of the coasting part of the TT's outbound trip, the TT thinks 'now' on the distant star is ten years ahead of 'now' according to the RT. The TT therefore thinks the distance star is much closer 'now' because the TT is considering the position of the star ten years later than the RT is, during which time the distant star has moved much closer in the TTs outbound frame of reference.

Answer 2

Whether this (or anything) is accurate depends on the meanings you choose to assign to various words, but I do not recommend the meanings you are choosing.

There is no (orthogonal) frame in which any properly functioning clock ever jumps ahead. Initially, the rocket shares a frame with the earth and the distant star. (Even this is a slight abuse of language, because it identifies a frame at one event with a frame at another, but it should be clear enough what it means.) After the rocket accelerates, it is attached to a new frame.

The event when the rocket accelerates (which we can take to be instantaneous) is simultaneous with the distant clock showing (say) noon in the old frame and simultaneous with the distant clock showing (say) 3PM in the new frame. The spatial distance to that star is one thing in the old frame and a different thing in the new frame. (Note that this is distance to the star, not to any particular event on that star).

In any given frame, all clocks tick at the same rate and never jump.