- #1
Buckethead
Gold Member
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First off, let me just say I love this forum and all the people in it! The world is a better place because of all of you who take the time to post here and I thank you all!
The most popular explanation to resolve the twin paradox seems to be the asymmetry that arises because the ship undergoes acceleration. I have to admit I have never been able to find comfort in that explanation even though I know it is correct. It also led me to wonder about a triplet situation whereby a non accelerating ship (A) passes Earth, syncs its clock at that time and heads out to space where it passes a non accelerating inbound ship (B), at which time B records A’s time. B eventually does a flyby past Earth where Earth and Ship B compare times. With no acceleration involved in that scenario I then had to study more explanations still with no real comfort.
But finally, in post #18 of the thread entitled “A Geometrical View of Time Dilation and the Twin Paradox – Comments” my non acceleration scenario was laid out perfectly along with a beautiful resolution.
https://www.physicsforums.com/threa...me-dilation-real-or-perspective-based.911615/
My hat is off to the_emi_guy for writing that post and I can now die in peace! I encourage anyone that is still scratching their head on the twin paradox to read it. It certainly helped me and most importantly, it’s really simple. As explained in the post it comes down the fact that although ship A sees Earth’s clock as moving more slowly than its own, the clock on ship B is moving more slowly still by virtue of the fact that the relative velocity between the two ships is greater than the relative velocity between ship A and the Earth. Hence the clock on ship B is moving slowest of all relative to ship A. When ship B arrives back at Earth, its clock will show less accumulated time than Earth and everyone is happy. Wonderful!
My reason for this post though is because it led me to a very interesting question.
A few years ago I asked what the clock on a twin paradox ship is actually doing while it is accelerating away from Earth and before the turnaround. Was it moving faster or slower or undefined. I seem to recall the answers left me unsatisfied and I don’t think anyone could say for sure. So here is my question with some backstory to reveal why I’m asking it.
I had a thought that there might be a definitive answer not only to my question above but to the question of what all the clocks are doing in the triplet scenario. I began by recognizing that the clocks on ship A and B, before they left Earth were fitted with identical clocks and calibrated to match the clock on Earth. So far so good. But then necessarily, ship B accelerated to begin its voyage to deep space, turned around, and headed back. Likewise ship A also accelerated in the opposite direction, turned around and headed back to do its initial Earth flyby. So both ship A and B had acceleration in their history and Earth did not.
I thought back on all non gravitational situations involving time dilation and realized that every single one of them involved an acceleration history that resulted in a differential velocity. Even a ship orbiting a low gravity asteroid using retro-rocket would first have to accelerate (for example) off the asteroid to begin its orbit or have, at some point in its history, accelerated relative to the asteroid. In all cases the clocks that underwent acceleration had slower running clocks when they were compared to their non accelerated counterparts.
This led me to think that in the triplet scenario, could it be possible that upon acceleration (and the resulting velocity differential) clock A runs slower than the clock on Earth not in a relative fashion but in an absolute fashion and the same for clock B. In addition, since clock A and clock B are moving at let’s say .6c relative to the Earth and since both went through identical acceleration curves, could it be both clock A and clock B are ticking at the same rate in an absolute, not a relative sense. This feels satisfying to me because it works in the triplet scenario and also in the twins scenario. In the twins paradox, what if when the ship accelerates, its clock slows, then when it decelerates and stops its clock speeds up again to match the rate of the clock on Earth. Then when it accelerates again, its clock slows again, then when it reaches Earth and comes to a stop, its clock again matches the tic rate of the clock on Earth but has accumulated less time overall.
So my question is, is it possible that this could be true? That when a clock accelerates from a zero differential velocity to a non-zero differential velocity, its clock slows in an absolute fashion and when it decelerates back to a zero differential velocity its clock returns to match the tic rate of its counterpart?
The most popular explanation to resolve the twin paradox seems to be the asymmetry that arises because the ship undergoes acceleration. I have to admit I have never been able to find comfort in that explanation even though I know it is correct. It also led me to wonder about a triplet situation whereby a non accelerating ship (A) passes Earth, syncs its clock at that time and heads out to space where it passes a non accelerating inbound ship (B), at which time B records A’s time. B eventually does a flyby past Earth where Earth and Ship B compare times. With no acceleration involved in that scenario I then had to study more explanations still with no real comfort.
But finally, in post #18 of the thread entitled “A Geometrical View of Time Dilation and the Twin Paradox – Comments” my non acceleration scenario was laid out perfectly along with a beautiful resolution.
https://www.physicsforums.com/threa...me-dilation-real-or-perspective-based.911615/
My hat is off to the_emi_guy for writing that post and I can now die in peace! I encourage anyone that is still scratching their head on the twin paradox to read it. It certainly helped me and most importantly, it’s really simple. As explained in the post it comes down the fact that although ship A sees Earth’s clock as moving more slowly than its own, the clock on ship B is moving more slowly still by virtue of the fact that the relative velocity between the two ships is greater than the relative velocity between ship A and the Earth. Hence the clock on ship B is moving slowest of all relative to ship A. When ship B arrives back at Earth, its clock will show less accumulated time than Earth and everyone is happy. Wonderful!
My reason for this post though is because it led me to a very interesting question.
A few years ago I asked what the clock on a twin paradox ship is actually doing while it is accelerating away from Earth and before the turnaround. Was it moving faster or slower or undefined. I seem to recall the answers left me unsatisfied and I don’t think anyone could say for sure. So here is my question with some backstory to reveal why I’m asking it.
I had a thought that there might be a definitive answer not only to my question above but to the question of what all the clocks are doing in the triplet scenario. I began by recognizing that the clocks on ship A and B, before they left Earth were fitted with identical clocks and calibrated to match the clock on Earth. So far so good. But then necessarily, ship B accelerated to begin its voyage to deep space, turned around, and headed back. Likewise ship A also accelerated in the opposite direction, turned around and headed back to do its initial Earth flyby. So both ship A and B had acceleration in their history and Earth did not.
I thought back on all non gravitational situations involving time dilation and realized that every single one of them involved an acceleration history that resulted in a differential velocity. Even a ship orbiting a low gravity asteroid using retro-rocket would first have to accelerate (for example) off the asteroid to begin its orbit or have, at some point in its history, accelerated relative to the asteroid. In all cases the clocks that underwent acceleration had slower running clocks when they were compared to their non accelerated counterparts.
This led me to think that in the triplet scenario, could it be possible that upon acceleration (and the resulting velocity differential) clock A runs slower than the clock on Earth not in a relative fashion but in an absolute fashion and the same for clock B. In addition, since clock A and clock B are moving at let’s say .6c relative to the Earth and since both went through identical acceleration curves, could it be both clock A and clock B are ticking at the same rate in an absolute, not a relative sense. This feels satisfying to me because it works in the triplet scenario and also in the twins scenario. In the twins paradox, what if when the ship accelerates, its clock slows, then when it decelerates and stops its clock speeds up again to match the rate of the clock on Earth. Then when it accelerates again, its clock slows again, then when it reaches Earth and comes to a stop, its clock again matches the tic rate of the clock on Earth but has accumulated less time overall.
So my question is, is it possible that this could be true? That when a clock accelerates from a zero differential velocity to a non-zero differential velocity, its clock slows in an absolute fashion and when it decelerates back to a zero differential velocity its clock returns to match the tic rate of its counterpart?