by Mohammed_I
P: 8,430
 Quote by Al68 If we did use the 4.5 yrs, we would have to add 8 yrs to that when the ship came to rest to get 12.5 yrs. The ship would see the earth clock "jump ahead" by 8 yrs when the ship comes to rest with earth. Only after the ship comes to rest with earth is a one way trip half of a two way trip.
Based on what? You certainly aren't using a single inertial frame here. If we allow arbitrary non-inertial coordinate systems, there's no reason that the coordinate system's definition of simultaneity at the moment the ship comes to rest relative to the Earth would have to match up with the definition of simultaneity in the inertial frame where the ship is at rest at that moment.
Mentor
P: 15,620
 Quote by neopolitan Now, for the purposes of explaining to the student, I would have thought we could say that we know that the square of the proper time in a single inertial frame which is taken to be at rest is equal to the sum of the squares of the proper time and coordinate distance in another inertial frame (or set of inertial frames) - between any two simultaneous events (simultaneous as defined by Einstein). Is that true (even if clumsily worded)?
I think that you are thinking the right thing and just writing it wrong (clumsily), but no, what you wrote is not correct. First, proper times are frame invariant, so it doesn't make much sense to say "the proper time in a single inertial frame". Second, simultaneity is frame variant, so you need to specify which frame. What you have described here only works in the rest frame of one of the worldlines, in any other rest frame you will have to consider the distance traveled by both worldlines. Third, I don't know if this is general for arbitrary paths or if it only applies for straight worldlines. Finally, it sounds like you are considering scenarios where the twins do not start together and reunite at some other time. If so, then there are 4 events of interest, the two events of each worldline starting and the two events of each worldline ending. Those two starting events must be simultaneous with each other in the rest frame of one worldline as must the two ending events.
P: 801
 Quote by JesseM But you said it would work even if you didn't use the Earth's frame--so, explain how you could do it "correctly" in the frame where the ship is at rest during the outbound trip and the Earth is moving at 0.8c. Do you deny that according to this frame's definition of simultaneity, when the ship's clock reads 7.5 years the Earth's clock reads 4.5 years?
No. I would just point out that after the ship is at rest with earth, everyone will agree that earth's clock reads 12.5 yrs. The difference in simultaneity between the two frames is 8 yrs. 4.5 + 8 = 12.5.
 If you want to instead use the definition of simultaneity in the Earth frame, then you're admitting that your method only works if you make use of the Earth frame rather than any other frame.
My method in this example is the same method used in the standard resolution. The twins end up in the same frame with the ship's twin 5 yrs younger than earth's twin.

Al
P: 801
 Quote by DaleSpam That's not an inertial frame. Obviously you can make it work there, but also obviously you won't be using the standard intertial frame rules.
I've never heard of a twins paradox variation where the ship stays in a single inertial frame. I'm using the same rules as the standard resolution.

Al
P: 8,430
 Quote by Al68 No. I would just point out that after the ship is at rest with earth, everyone will agree that earth's clock reads 12.5 yrs.
Only if you choose to use the inertial frame where the Earth is at rest, and the traveler is instantaneously at rest at the midpoint of the acceleration. But this was exactly my point, that your method only works if you use the coordinates of a single inertial frame, which you seemed to be disagreeing with when you said "It works in every frame. All frames would agree on the result." If you still maintain it works in every frame, tell me how it works in the inertial frame where the traveler is at rest during the outbound journey before accelerating (which is not the same as a non-inertial frame where the traveler is at rest at every moment, including during the acceleration).
P: 801
 Quote by JesseM You didn't add this until after I had already responded: Based on what? You certainly aren't using a single inertial frame here. If we allow arbitrary non-inertial coordinate systems, there's no reason that the coordinate system's definition of simultaneity at the moment the ship comes to rest relative to the Earth would have to match up with the definition of simultaneity in the inertial frame where the ship is at rest at that moment.
Well if the ship is at rest with earth, they're in the same frame by definition. t=12.5 is simultaneous with the ship's clock reading 7.5 yrs after the ship comes to rest at the end of the trip.

Al
P: 8,430
 Quote by Al68 Well if the ship is at rest with earth, they're in the same frame by definition. t=12.5 is simultaneous with the ship's clock reading 7.5 yrs after the ship comes to rest at the end of the trip.
You're not answering my question about other frames though--it's not like in relativity you're forced to only use frames that objects are at rest in, you can use any inertial frame you want! Are you agreeing that your method can only work in the frame where the Earth is at rest, that it can't be generalized to an inertial frame where the Earth is in motion?
P: 801
 Quote by JesseM Only if you choose to use the inertial frame where the Earth is at rest, and the traveler is instantaneously at rest at the midpoint of the acceleration. But this was exactly my point, that your method only works if you use the coordinates of a single inertial frame, which you seemed to be disagreeing with when you said "It works in every frame. All frames would agree on the result." If you still maintain it works in every frame, tell me how it works in the inertial frame where the traveler is at rest during the outbound journey before accelerating (which is not the same as a non-inertial frame where the traveler is at rest at every moment, including during the acceleration).
It works the same in that frame as in the standard resolutions. The ship's twin realizes that if the earth sends a signal at t=4.5, and if his ship broke so he could never accelerate, when he receives this signal and subtracts the time it had to take for the signal to reach him at c, he would conclude that the signal was sent at t'= 7.5 yrs in his frame. And he would realize that if he had decelerated to come to rest with earth, then earth's clock would read 12.5 yrs simultaneously with the ship's clock reading 7.5 yrs.

Al
P: 8,430
 Quote by Al68 It works the same in that frame as in the standard resolutions. The ship's twin realizes that if the earth sends a signal at t=4.5, and if his ship broke so he could never accelerate, when he receives this signal and subtracts the time it had to take for the signal to reach him at c, he would conclude that the signal was sent at t'= 7.5 yrs in his frame.
Yes, of course. But this 7.5 years doesn't enter into the calculations for the total time elapsed on Earth when the traveler returns, if we use your method.
 Quote by Al68 And he would realize that if he had decelerated to come to rest with earth, then earth's clock would read 12.5 yrs simultaneously with the ship's clock reading 7.5 yrs.
It would read 7.5 years simultaneously with the Earth's clock reading 12.5 years in the frame of the Earth. So what you're telling me has nothing to do with using your method in the other frame where the ship was at rest during the outbound journey--all you're saying is that regardless of what the traveler's current rest frame is, he can calculate that in the frame of the Earth the turnaround happens when the Earth clock reads 12.5 years, and then your method is to double that. Of course any observer, no matter what his physical state of motion, can calculate how things will work in the Earth's rest frame, but when I (and DaleSpam) said that your method only works in the Earth's rest frame, I was talking about calculations, the physical motion of the observer doing the calculations is irrelevant.
P: 801
 Quote by JesseM Yes, of course. But this 7.5 years doesn't enter into the calculations for the total time elapsed on Earth when the traveler returns, if we use your method. It would read 7.5 years simultaneously with the Earth's clock reading 12.5 years in the frame of the Earth. So what you're telling me has nothing to do with using your method in the other frame where the ship was at rest during the outbound journey--all you're saying is that regardless of what the traveler's current rest frame is, he can calculate that in the frame of the Earth the turnaround happens when the Earth clock reads 12.5 years, and then your method is to double that. Of course any observer, no matter what his physical state of motion, can calculate how things will work in the Earth's rest frame, but when I (and DaleSpam) said that your method only works in the Earth's rest frame, I was talking about calculations, the physical motion of the observer doing the calculations is irrelevant.
Well, I must not understand what you're saying. "My" method here is the same method used in the standard resolutions. How would you explain a one way trip where the ship ended up at rest with earth with the ship's twin younger than the earth twin?

Al
P: 8,430
 Quote by Al68 Well, I must not understand what you're saying. "My" method here is the same method used in the standard resolutions. How would you explain a one way trip where the ship ended up at rest with earth with the ship's twin younger than the earth twin?
I never said your method was incorrect! If you look at post #225 I just agreed with DaleSpam that the method of taking the Earth's time at the turnaround and doubling it only works if you're using the Earth's rest frame. If you are, you do get the right answer. But this doesn't really address the question which is the basis for the twin paradox, which is "why can't you consider things from the traveling twin's perspective instead of the Earth-twin's perspective"? And the answer to that is that you can analyze the problem from the perspective a frame which is different from the Earth's frame, but if you pick an inertial frame then the traveling twin will change speed in this frame and for part of the journey his clock will be running slower than the Earth twin's, whereas if you try to use a coordinate system where the traveling twin is at rest throughout the journey, this is a non-inertial coordinate system so you can't assume the usual time dilation formula still works.
 P: 4 You are reffering to time dilation in an inertial reference frame, but the rocket twin is undergoing acceleration and thus is not in an inertial reference frame.
P: 801
 Quote by JesseM I never said your method was incorrect! If you look at post #225 I just agreed with DaleSpam that the method of taking the Earth's time at the turnaround and doubling it only works if you're using the Earth's rest frame. If you are, you do get the right answer. But this doesn't really address the question which is the basis for the twin paradox, which is "why can't you consider things from the traveling twin's perspective instead of the Earth-twin's perspective"? And the answer to that is that you can analyze the problem from the perspective a frame which is different from the Earth's frame, but if you pick an inertial frame then the traveling twin will change speed in this frame and for part of the journey his clock will be running slower than the Earth twin's, whereas if you try to use a coordinate system where the traveling twin is at rest throughout the journey, this is a non-inertial coordinate system so you can't assume the usual time dilation formula still works.
I must have missed something somewhere, I don't see where we disagree. My only point was that the twins paradox could be analyzed by just analyzing two one way trips. Same problem. Same result.

The question "why can't you consider things from the traveling twin's perspective instead of the Earth-twin's perspective?" has the same answer either way.

Al
P: 645
 Quote by DaleSpam I don't know if you are thinking the right thing and just writing it wrong, but no, what you wrote is not correct. First, proper times are frame invariant, so it doesn't make much sense to say "the proper time in a single inertial frame". Second, simultaneity is frame variant, so you need to specify which frame. What you have described here only works in the rest frame of one of the worldlines, in any other rest frame you will have to consider the distance traveled by both worldlines. Third, I don't know if this is general for arbitrary paths or if it only applies for straight worldlines. Finally, it sounds like you are considering scenarios where the twins do not start together and reunite at some other time. If so, then there are 4 events of interest, the two events of each worldline starting and the two events of each worldline ending. Those two starting events must be simultaneous with each other in the rest frame of one worldline as must the two ending events.

 Quote by neopolitan Now, for the purposes of explaining to the student, I would have thought we could say that we know that the square of the proper time in a single inertial frame which is taken to be at rest is equal to the sum of the squares of the proper time and coordinate distance in another inertial frame (or set of inertial frames) - between any two simultaneous events (simultaneous as defined by Einstein).
Initially, I thought about making the red "a" the definite article "the", to keep it to the twin paradox scenario, but then I thought that in reality it doesn't have to be.

As I read my paragraph again, I see I left ambiguity. I did indeed mean four simultaneous events, in two pairs. I should have written "two pairs of simultaneous events". My error. Also, to remove less obvious ambiguity, I should point out that the coordinate distance in the other inertial frame is according to the first mentioned "taken to be at rest" frame, and the simultaneous events are simultaneous in the "taken to be at rest frame". I took these latter two to be obvious, but I shouldn't do that.

So, rephrasing:

 Quote by neopolitan, rephrasing Given an inertial frame as our reference, a frame which is taken to be at rest (thus having a coordinate distance of zero), and another inertial frame, or set of contiguous inertial frames, if we consider two pairs of simultaneous events (each pair of events has one event to each frame) and if we use values according to an observer in our reference frame then: we know that the square of the proper time in former inertial frame which is taken to be at rest is equal to the sum of the squares of the proper time and coordinate distance in the other inertial frame (or set of inertial frames).
Then this would work for the twin paradox scenario, with two pairs of events which are actually just two events (departure and return) or pairs of events which are not collocated in spacetime, or a mix (like Al68 has sort of been discussing when breaking the twin paradox into two legs).

Is this less ambiguous and, being less ambiguous, correct?

If correct, could we be less exacting in order to convey the concept to the student, and then slowly build up the understanding of the conditions under which the equation is correct? (Sort of how one would explain a rainbow in steps, without at first talking about suspension of droplets of water of the correct size to refract rays of light, or the relative placement of those droplets, or how the image of the rainbow is formed in the eye and is not something external to our perceptions of it.)

cheers,

neopolitan
 Mentor P: 15,620 I'm too tired to untangle your rephrasing. Given two (straight) worldlines in some inertial coordinate system the spacetime interval of those lines can be written (using my convention above and units where c=1): St² = Tt² - Xt² Sh² = Th² - Xh² If Tt = Th (e.g. the coordinate time between two pairs of simultaneous events) then St² - Sh² = Xh² - Xt² or St² + Xt² = Sh² + Xh² That is all. Just a little algebra on the spacetime interval formula in the special case where the coordinate times are equal for two straight worldlines. So what? What is the value? How is it in any way preferable to the more general spacetime interval formula?
P: 645
 Quote by DaleSpam I'm too tired to untangle your rephrasing. Given two (straight) worldlines in some inertial coordinate system the spacetime interval of those lines can be written (using my convention above and units where c=1): St² = Tt² - Xt² Sh² = Th² - Xh² If Tt = Th (e.g. the coordinate time between two pairs of simultaneous events) then St² - Sh² = Xh² - Xt² or St² + Xt² = Sh² + Xh² That is all. Just a little algebra on the spacetime interval formula in the special case where the coordinate times are equal for two straight worldlines. So what? What is the value? How is it in any way preferable to the more general spacetime interval formula?
Basically all I am saying is in the first equation you raised.

St² = Tt² - Xt² (where Xh is zero and Tt = Th, which is all my convoluted phrasing said)

And, yes, it is in the third and fourth equations as well. The point is, I guess, that we are talking about spacetime intervals which are the same, so long as you use a consistent inertial frame. Which gets us back to the original point, if you don't use a consistent inertial frame it doesn't work.

This seems to be the one factor common to all approaches for explaining the twin paradox. If you use a consistent inertial frame it works. If you don't, you need to do something extra (which can be to consider simultaneity, or to note that there is a missing amount of time equal in magnitude to the disagreed distance travelled, which can be calculated as shown in an earlier post, or you can show that the line is bent in a spacetime vector diagram).

I don't think we are getting any further than that, and I don't really think there is any further to go than that.

cheers,

neopolitan
Mentor
P: 15,620
 Quote by neopolitan This seems to be the one factor common to all approaches for explaining the twin paradox. If you use a consistent inertial frame it works. If you don't, you need to do something extra ... I don't think we are getting any further than that, and I don't really think there is any further to go than that.
I agree 100% with that. And really, that is the whole point of the relativity postulate: if you use any inertial frame everything works the same. (it even rhymes )

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