Why Does the Traveling Twin's Clock Show Less Time in the Twin Paradox?

In summary, the twin paradox involves two twins, A and B, where B travels at 60% the speed of light to a star one light year away and back, causing time dilation and a difference in their clocks. However, in a scenario where only half the journey occurs, there is still a discrepancy in their clocks due to the relativity of simultaneity. Twin B, who is traveling, will see a difference in the clocks of Twin A and the person at the star, despite them being at rest relative to each other. This is due to the fact that Twin B's frame of reference has changed, causing a difference in perceived time. In a situation where Twin B does not decelerate at the star, there is
  • #1
timpeac
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Hi, I can see that this topic has been much discussed, but I haven't seen a thread on it with the particular spin I want to give it (just the journey out from Earth).

I understand the traditional view of the twin paradox (I think!): Two twins a and b are on Earth and each has a clock. The clocks are synchronised and one twin travels at, say, 60% the speed of light to a star exactly one light year away, stops and immediately returns. Due to time contraction, during the journey each twin judges that his brother’s clock is running slow – each only showing 0.8 seconds for each second of his own clock - each of which seems, to the holder, to be running normally. On return to Earth when the clocks are compared it is shown that it is the twin who has traveled away and come back whose clock shows less time than his brother’s.

This is the paradox – if you could equally view each twin as the one “at rest” while his brother travels away and back to him, and so both are correct in considering his brother’s clock to be running slow, then how can one of them be shown to really have the slow clock? The answer is that the twin traveling away from Earth has undergone four accelerations, albeit very quick, one when he set off from Earth, one when he arrived at the destination and stopped, one when he accelerated to start back to Earth and one when he decelerated to rest on arrival. This changed his frame of reference and thus changed which times can be considered as simultaneous in each twin’s frame of reference. So far so good – I have no problem with that.

However, my question involves a situation where only half the journey occurs. I can think of two situations (the first of which I can think of explanations for but the second I’d like some help to understand).

a) The first twin sets off as before. At the same time a signal of light is sent from the Earth to the star 1 light year away where someone is waiting to receive it. Since the Earth and the star are at rest with each other they share a frame of reference and can therefore agree on the simultaneity of the journey start. The twin then arrives at the star. Now, for him both the Earth and the star have been moving relative to him at 0.6c and so their clocks have been running slow at 0.8s. Similarly for both the twin on Earth and the person at the star the second twin has been traveling at 0.6c relative to them and they consider that his clock has been running slow (the person waiting at the star just has to remember to add 1 year on to the journey time to get simultaneity because of the time taken for the light signal to arrive).

Now, when the twin has performed this first half of his journey and compares times with the person at the star they can’t both then agree that each other’s clock is slow. On comparing they will see that it is again the twin’s clock which has been running slow. Now, I can understand that because the twin has still undergone two accelerations whereas the person waiting at the star has not. However – in scenario b) below there is no acceleration:

b) Say twin b is not on Earth but traveling from some earlier point at a constant 0.6c past the Earth. As he passes the Earth twin a who is on the Earth sends the light signal towards the star 1 light year away that twin b is also heading directly towards. When twin b approaches the star he doesn’t decelerate to stop but carries on past at the constant 0.6c. However, as he passes the star he and the person their show each other their clocks as they pass. In this situation there has been no acceleration, and again each party can consider the other moving and therefore to have the slow clock. Who will be correct? My suspicion is that again the twin in the spaceship will have the slower clock on comparison – but why? This time there has been no acceleration to change the frame of reference.
Any thoughts? Many thanks.
 
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  • #2
Relativity of Simultaneity. While twin A and the person at the star agree that their clocks read the same time, Twin B does not. According to him, the clock at the Star is actually ahead of Twin A's clock.

So While Twin B measures 1 1/3 years as having passed on his clock during the trip, he will measure 1.06667 years as passing on Twin A's and the clock at at the star. However according to him, the clock at the star reads 0.6 years ahead of Twin A's clock. Thus if Clock A reads 0 at the moment he passes Earth, the star's clock will read .6 years according to him. When he passes the star, his clock reads 1 1/3 years, and the Star's clock will have advanced 1.06667 years to read 1 2/3 years.
 
  • #3
timpeac said:
b) Say twin b is not on Earth but traveling from some earlier point at a constant 0.6c past the Earth. As he passes the Earth twin a who is on the Earth sends the light signal towards the star 1 light year away that twin b is also heading directly towards. When twin b approaches the star he doesn’t decelerate to stop but carries on past at the constant 0.6c. However, as he passes the star he and the person their show each other their clocks as they pass. In this situation there has been no acceleration, and again each party can consider the other moving and therefore to have the slow clock. Who will be correct? My suspicion is that again the twin in the spaceship will have the slower clock on comparison – but why? This time there has been no acceleration to change the frame of reference.
Any thoughts? Many thanks.

In the earth/star rest frame frame, the ship takes 1.67 years to travel one light year, so the star clock shows this time when the ship arrives. The ship clock is moving, and so running slow; it shows 1.33 years when it arrives at the star.

In the ship's rest frame, the distance between the Earth and the star is only 0.8 light years, so it only takes 1.33 years to arrive. The Earth clock and the star clock are synchronised in the earth/star rest frame, but they're out of synch in the ship's rest frame. In the ship's rest frame, when the ship passes earth, the star clock is ahead of the Earth clock, due to the relativity of simultaneity, by 0.6 years. In this frame, it's the star's clock that's moving at 0.6c, and so running slow; in the time it takes the ship clock to record 1.33 years, the star clock only records 1.07 years. But the star clock has a head start of 0.6 years when the ship's clock passes Earth and is synchronised with the Earth clock, and so when the ship reaches the star, the star clock will show 0.6 + 1.07 = 1.67 years. What a coincidence!
 
  • #4
According to traveling twin b, the star clock is ahead of the Earth clock by a factor L*v/c^2 at the start of the journey, while according to Earth twin a, the Earth clock and the star clock are sycronised. L is the distance between the Earth and the star in the twin a's frame. The Earth/star system syncronises its clocks by sending a signal from Earth at time t1 = 0 to the star and then setting the star clock to t2 = L/c to allow for the signal travel time L/c. According to the traveling twin's observations the distance between the Earth and star is L/y due to length contraction, where y is the gamma factor 1/sqrt(1-v^2/c^2). The time for the signal to travel from the Earth to the star in twin b's frame is L/(y(c-v)) and taking time dilation into account this equates to L/(y^2(c-v)) for how much time the Earth clock advances during the signal travel time. So when the star clock t2 is set to L/c the Earth clock t1 reads L/(y^2(c-v)) and the difference between the two readings (t2 - t1) is L*v/c^2. It is this lack of simultaneity of a's clocks from b's point of view, that allows b to reason that a's clocks are running slower but read a greater total ellapsed time due the "head start" programmed into the star clock t2 during the synchronisation process. This difference of opinion between observers as to what is simultaneous, is known as the "relativity of simultaneity".
 
  • #5
Rasalhague said:
... The ship clock is moving, and so running slow; it shows 1.33 years when it arrives at the star.

I think it is not strictly correct to say this. In a purely inertial scenario, there is no "ship clock that is moving, and and so running slow". In this purely inertial "half-twin-paradox", no measurement can tell which clock "runs slow". In order to make a definite prediction (and measurement), at least one of the clocks must be accelerated, so that they can at least pass each other for a second time.
 
  • #6
timpeac said:
b)...My suspicion is that again the twin in the spaceship will have the slower clock on comparison – but why?
The traveling twin will have less elapsed time between the events for the exact same reason as the standard twins paradox.

The second event is defined a priori as 1 ly from Earth in Earth's frame. If you were to instead define the second event as 1 ly from the ship in the ship's frame (buoy trailing the ship?), you would get the opposite result, ie the elapsed time between the events would be less on Earth's clock than the ship clock.
 
  • #7
Jorrie said:
I think it is not strictly correct to say this. In a purely inertial scenario, there is no "ship clock that is moving, and and so running slow". In this purely inertial "half-twin-paradox", no measurement can tell which clock "runs slow". In order to make a definite prediction (and measurement), at least one of the clocks must be accelerated, so that they can at least pass each other for a second time.
None of the clocks were accelerated in the scenario presented. But the elapsed time between events is easy to determine and measure in both frames.
 
  • #8
AI68 and I have posted several explanations of the TP in prior years - based upon reducing the problem to a one way trip and doubling the result, The one way trip analysis is the easy way to break the problem down and eliminate accelerations that confound most explanationss - starting with and initial speed (an on the fly reading as the clocks pass to start) and the same on the fly reading on arrival, eliminates any weasel wording that depends upon a false application of a pseudo G field as Einstein proposed in1918 The one way trip is simply an application of the principle of interval invariance in all frames
 
  • #9
yogi said:
The one way trip is simply an application of the principle of interval invariance in all frames

Agreed, but IMO it is false to compare the propertime interval of the inertially 'moving twin' with the coordinate time interval of the inertial 'home twin' for this situation and then declare that the 'moving twin' has aged less than the 'home-twin' during the away trip. To do so would mean giving preference to one inertial frame above another. Only the 'away twin' measures the invariant interval between the departure and arrival events directly.
 
  • #10
Jorrie said:
I think it is not strictly correct to say this. In a purely inertial scenario, there is no "ship clock that is moving, and and so running slow". In this purely inertial "half-twin-paradox", no measurement can tell which clock "runs slow". In order to make a definite prediction (and measurement), at least one of the clocks must be accelerated, so that they can at least pass each other for a second time.

It's the sort of language often used in these contexts, as in the popular motto "a moving clock runs slow", and movement is defined by the frame, so there's no ambiguity (the ship clock is moving in the earth/star rest frame; the Earth and star clocks are moving in the ship's rest frame). But I agree that when these ideas as introduced in the form of a scenario that contrasts a rocket or train or other vehicle with a planet or station or some other thing that our non-relativistic intuition tells us is stationary, it can beguile those instincts into thinking of motion as absolute, and obscure the equivalence. Perhaps the best way, pedagogically, is to introduce the basic idea of relativity using two identical space ships. Then there isn't that distraction for beginners of wondering whether it's significant that one of the frames is centred on a vehicle and the other on something that our non-relativistic instincts tell us is inherently "not moving".
 
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  • #11
Thank you all for your replies - they make perfect sense. I was forgetting that just because both twins agree that passing the Earth is simultaneous, and twin a and the person at the star agree that the event is simultaneous this does not necessarily imply that twin b and the person at the star will also agree the same thing. In twin b's opinion the person at the star has a "head start".
 
  • #12
Rasalhague said:
It's the sort of language often used in these contexts, as in the popular motto "a moving clock runs slow", and movement is defined by the frame, so there's no ambiguity (the ship clock is moving in the earth/star rest frame; the Earth and star clocks are moving in the ship's rest frame). But I agree that when these ideas as introduced in the form of a scenario that contrasts a rocket or train or other vehicle with a planet or station or some other thing that our non-relativistic intuition tells us is stationary, it can beguile those instincts into thinking of motion as absolute, and obscure the equivalence. Perhaps the best way, pedagogically, is to introduce the basic idea of relativity using two identical space ships. Then there isn't that distraction for beginners of wondering whether it's significant that one of the frames is centred on a vehicle and the other on something that our non-relativistic instincts tell us is inherently "not moving".

Yes. I think that this is a problem with the train/embankment scenario when first seen by beginners. It seems obvious that the train is moving and the embankment is not. That's the way the world appears to be.

Matheinste.
 
  • #13
timpeac said:
Thank you all for your replies - they make perfect sense. I was forgetting that just because both twins agree that passing the Earth is simultaneous, and twin a and the person at the star agree that the event is simultaneous this does not necessarily imply that twin b and the person at the star will also agree the same thing. In twin b's opinion the person at the star has a "head start".

Good to see you grasped the concept so quickly. :smile: These sort of threads normally go on for hundreds of posts.

It might be worth mentioning in conclusion, that when the calculations are done correctly as shown for example by Janus and Rasalhague, then it is true that there is no way to actually determine which clock if any is "really" running slower than the other in a one way non-accelerating experiment.
 
  • #14
kev said:
Good to see you grasped the concept so quickly. :smile: These sort of threads normally go on for hundreds of posts.

It might be worth mentioning in conclusion, that when the calculations are done correctly as shown for example by Janus and Rasalhague, then it is true that there is no way to actually determine which clock if any is "really" running slower than the other in a one way non-accelerating experiment.

But doesn't the situation I described in b) in my original post show that both clocks are indeed running slow in relation to each other without acceleration taking place? Twin b's clock is indeed slower than the person at the star's on comparison when he arrives and the person at the star's clock - while being ahead of twin b's in absolute terms - is still less than it would have been if time hadn't contracted from twin b's point of view (he had a 0.6 year head start when twin a passed the Earth and this has reduced to 1/3 years by the time twin b arrives).
 
  • #15
timpeac said:
But doesn't the situation I described in b) in my original post show that both clocks are indeed running slow in relation to each other without acceleration taking place? Twin b's clock is indeed slower than the person at the star's on comparison when he arrives and the person at the star's clock - while being ahead of twin b's in absolute terms - is still less than it would have been if time hadn't contracted from twin b's point of view (he had a 0.6 year head start when twin a passed the Earth and this has reduced to 1/3 years by the time twin b arrives).

Yes, both observer's measure the other observer's clocks to be running slower than their own clocks and in the "hard core relativistic" or traditional interpretation that means both clocks are "really" running slower than each other. Personally I believe it is physically impossible and logically inconsistent for two clocks to physically run slower than each other and I prefer to think that in this situation the physical reality is undetermined and we can not say in an absolute sense which clock is running slower than the other, or if they are running at the same rate. It is only when one observer turns around and returns, that proper elapsed times can be compared in a way that all observers can agree on. That is my personal view, but it probably does not represent the formal accepted view.
 
  • #16
kev said:
... Personally I believe it is physically impossible and logically inconsistent for two clocks to physically run slower than each other, or if they are running at the same rate. It is only when one observer turns around and returns, that proper elapsed times can be compared in a way that all observers can agree on. That is my personal view, but it probably does not represent the formal accepted view.

For the all-inertial case, I agree. There is however an interesting twist in the logic when we consider an all-inertial, three-clock (ABC) experiment, where a third inertial clock (C) does the return trip at -v relative to A. By this setup and Lorentz invariance:

[tex]\Delta\tau_B = \Delta\tau_C = \gamma\frac{\Delta\tau_A}{2}[/tex]

where the [itex]\Delta\tau[/itex] are the propertime intervals between the three events making up the Minkowski triangle. Although this does not prove anything, it is extremely suggestive that clocks B and C must record time 'slower' than clock A. However, the fact that all three are inertial observers, makes the conclusion invalid, not so?
 
  • #17
Jorrie said:
Agreed, but IMO it is false to compare the propertime interval of the inertially 'moving twin' with the coordinate time interval of the inertial 'home twin' for this situation and then declare that the 'moving twin' has aged less than the 'home-twin' during the away trip. To do so would mean giving preference to one inertial frame above another. Only the 'away twin' measures the invariant interval between the departure and arrival events directly.

The one way experiment is easily done in a lab using pions -a hi speed pion is generated and passes a detector with a clock #1 attached - a few feet away is a second clock with a detector #2 positioned at the point where the particle disintegrates. These two clocks are in the same room and they are always in sync - but the difference between the time measured by the two clocks is the proper time difference between the two events - the pion carries its own clock in the form of a decay period - this pion clock is not going to read the same as the difference between #1 clock and #2 clock

There is no acceleration - the pion is up to its travel speed at the time it reaches #1 clock.
Its important to keep your eye on the experiment - there is a space interval in the pions frame and space interval in the lab frame - there is a time difference logged by the pion's clock and there is a time difference logged by the difference between #1 and #2 clocks
The space time interval must be the same in both frames
 
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  • #18
yogi said:
Its important to keep your eye on the experiment - there is a space interval in the pions frame and space interval in the lab frame - there is a time difference logged by the pion's clock and there is a time difference logged by the difference between #1 and #2 clocks

Yes, but the pion's space interval is zero, that's why its "own clock" observes the propertime interval, which then equals the spacetime interval. The lab clock does not measure the propertime interval, but rather the larger lab coordinate time interval. Therefore we should not say that the pion clock is "running slower" than the lab clock (which is the reason behind the current discussion).

I do not quite understand what you mean by "... there is a time difference logged by the difference between #1 and #2 clocks".
 
  • #19
Jorrie said:
Yes, but the pion's space interval is zero, that's why its "own clock" observes the propertime interval, which then equals the spacetime interval. The lab clock does not measure the propertime interval, but rather the larger lab coordinate time interval. Therefore we should not say that the pion clock is "running slower" than the lab clock (which is the reason behind the current discussion).
The current discussion is about which frame has less elapsed time between the same two events, not which clock ran "slower".

yogi's scenario is analogous to the OP's in that the pion clock shows less elapsed coordinate time than the lab frame clocks between the two defined events.
 
  • #20
According to the theory of relativity only force free objects can be considered true reference frames. In the case of the spaceship moving at a constant velocity, it would have certainly accelerated at some point in the past so as to achieve its current motion, which renders it unsuitable as a reference frame.

Hence, it is the stationary twin and the stationary star who are true reference frames and therefore it is the traveling twin whose watch runs slower.

Fairly new to this subject and hence ask for suggestions in case you find the above out of place.

Thanks
 
  • #21
shyamalshukla, that's not what relativity says. It says that all inertial frames are equivalent, not that some are more at rest than others.
 
  • #22
shyamalshukla said:
According to the theory of relativity only force free objects can be considered true reference frames. In the case of the spaceship moving at a constant velocity, it would have certainly accelerated at some point in the past so as to achieve its current motion, which renders it unsuitable as a reference frame.

Hence, it is the stationary twin and the stationary star who are true reference frames and therefore it is the traveling twin whose watch runs slower.

Fairly new to this subject and hence ask for suggestions in case you find the above out of place.

Thanks
In the scenario being discussed, there is no acceleration involved. Whether or not the ship accelerated in the past is irrelevant, it is "force free" at all times during the scenario.

The reference frame in which the ship is at rest is a perfectly valid inertial frame, and is not "rendered unsuitable" because of any previous acceleration of the ship. In that inertial frame there is less elapsed time between the defined events than in Earth's inertial rest frame, and there was no acceleration whatsoever involved. That was the point.
 
  • #23
shyamalshukla said:
In the case of the spaceship moving at a constant velocity, it would have certainly accelerated at some point in the past so as to achieve its current motion, which renders it unsuitable as a reference frame.


Thanks

Hello shyamalshukla.

In special relativity any object moving without acceleration can be considered at rest or moving as suits us for ease of problem solving or illustration. Only relative motion has any meaning. But, as you have said, any such object may, in the past, have been accelerated, and in most cases we cannot know or do not care about its complete history. For most cases, for purposes of illustration or instruction, this past history is ignored or considered irrelevant and it is tacitly assumed that the object has always been in its present state of rest or motion. The acceleration history, although it may have other effects, does not disqualify such objcts from being considered inertial.

Matheinste.
 
  • #24
Al68 said:
The current discussion is about which frame has less elapsed time between the same two events, not which clock ran "slower".

No, sorry, the current discussion (since post #15) is mainly about "which clock is running slower". My problem is that experts tend to 'brush this under the carpet' a bit. IMO, Kev's view is the better one: you simply cannot tell which clock is 'running slow' in a purely inertial scenario.

Yes, every inertial observer obtain different coordinate time and space intervals between events and after Lorentz transformation, they all agree on the spacetime interval between the events. It still tells them nothing about their relative clock rates, however.

From a teaching POV, IMO, it confuses novices if we tell them that yes, each clock can be 'running slow', depending on who measures what, the relativity of simultaneity, etc. All they need to know is that the spacetime interval is invariant and how to use that.
 
  • #25
Jorrie said:
No, sorry, the current discussion (since post #15) is mainly about "which clock is running slower".
It's been discussed, but as a side issue, incidental to the subject of the thread.
My problem is that experts tend to 'brush this under the carpet' a bit. IMO, Kev's view is the better one: you simply cannot tell which clock is 'running slow' in a purely inertial scenario.
Sure I can. In each inertial frame, the clock in motion runs slower than the clock at rest in that frame. Which clock runs slow is frame dependent, that is very different than "you cannot tell". And it hasn't been brushed under the carpet in this thread because as far as I can see, that wasn't the question being asked.
Yes, every inertial observer obtain different coordinate time and space intervals between events and after Lorentz transformation, they all agree on the spacetime interval between the events. It still tells them nothing about their relative clock rates, however.
Who was asking about relative clock rates? The OP was asking about the elapsed time between events in each frame.
 
  • #26
Al68 said:
In each inertial frame, the clock in motion runs slower than the clock at rest in that frame. Which clock runs slow is frame dependent, that is very different than "you cannot tell".

And it hasn't been brushed under the carpet in this thread because as far as I can see, that wasn't the question being asked. Who was asking about relative clock rates? The OP was asking about the elapsed time between events in each frame.

I do not want to be confrontational, but you seem to have missed some of the OP's questions, e.g.
timpeac said:
b)Say twin b is not on Earth but traveling from some earlier point at a constant 0.6c past the Earth. ... In this situation there has been no acceleration, and again each party can consider the other moving and therefore to have the slow clock. Who will be correct? My suspicion is that again the twin in the spaceship will have the slower clock on comparison – but why?

timpeac said:
But doesn't the situation I described in b) in my original post show that both clocks are indeed running slow in relation to each other without acceleration taking place? Twin b's clock is indeed slower than the person at the star's on comparison when he arrives and the person at the star's clock - while being ahead of twin b's in absolute terms - is still less than it would have been if time hadn't contracted from twin b's point of view (he had a 0.6 year head start when twin a passed the Earth and this has reduced to 1/3 years by the time twin b arrives).

That said, I realize that 'running slow' is a subjective concept that depends which frame does the measurements. My point is that unless you can make two co-located measurements, you do not know which inertial clock recorded less propertime interval. Coordinate time intervals are 'real' as measurements, but, as Kev has pointed out in post #15 (in reply to timpeac's last question):

"Personally I believe it is physically impossible and logically inconsistent for two clocks to physically run slower than each other and I prefer to think that in this situation the physical reality is undetermined and we can not say in an absolute sense which clock is running slower than the other, or if they are running at the same rate."

AFAIK, this is actually the mainstream position, but I would like to hear some moderator's opinion on it.
 
  • #27
Jorrie said:
My point is that unless you can make two co-located measurements, you do not know which inertial clock recorded less propertime interval. Coordinate time intervals are 'real' as measurements, but, as Kev has pointed out in post #15 (in reply to timpeac's last question):

"Personally I believe it is physically impossible and logically inconsistent for two clocks to physically run slower than each other and I prefer to think that in this situation the physical reality is undetermined and we can not say in an absolute sense which clock is running slower than the other, or if they are running at the same rate."

AFAIK, this is actually the mainstream position, but I would like to hear some moderator's opinion on it.
Yes, the mainstream position is that each clock runs slow relative to the other in the other's rest frame, in a reciprocal relationship. But there is a big difference between "which clock runs slower than the other" and "which clock shows less elapsed time between events".

As far as knowing "which inertial clock recorded less propertime interval", that's easy, because every clock records its own proper time. And the proper time interval recorded by a clock equals the coordinate time elapsed between two events if the clock was started and stopped simultaneous with the events, whether they were local to the clock or not.

kev's post was referring to the frame dependent nature of which clock runs slower, not the frame independent result of which frame has less elapsed time between the events.
 
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  • #28
Al68 said:
As far as knowing "which inertial clock recorded less propertime interval", that's easy, because every clock records its own proper time. And the proper time interval recorded by a clock equals the coordinate time elapsed between two events if the clock was started and stopped simultaneous with the events, whether they were local to the clock or not.

OK, I think we have different definitions of the propertime interval between events. I'm used to propertime interval in terms of the clock differential of a clock co-located with both events. I suppose one can more generally view that as the spacetime interval and it happens to be also the coordinate time (and propertime) interval recorded by such a clock.

By analogy, you then also name the coordinate time interval measured by any clock not co-located with both events the propertime interval between the events, as measured in that clock's frame.

Hmm... Not sure I'm happy with it, but I'll give it a think...
 
  • #29
Jorrie said:
OK, I think we have different definitions of the propertime interval between events. I'm used to propertime interval in terms of the clock differential of a clock co-located with both events. I suppose one can more generally view that as the spacetime interval and it happens to be also the coordinate time (and propertime) interval recorded by such a clock.

By analogy, you then also name the coordinate time interval measured by any clock not co-located with both events the propertime interval between the events, as measured in that clock's frame.
No, I didn't say that at all. I'm using the same definition you are. I'm not calling the coordinate time between non-local events a proper time interval, I'm only saying the two quantities are equal in this specific case.

If a clock measures the proper time interval between two events, A and B, both local to the clock, and (non-local) event C is simultaneous with event B in that clock's rest frame, then the coordinate time elapsed between events A and C is equal to the proper time interval measured by the clock between events A and B.
 
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  • #30
Al68 said:
If a clock measures the proper time interval between two events, A and B, both local to the clock, and (non-local) event C is simultaneous with event B in that clock's rest frame, then the coordinate time elapsed between events A and C is equal to the proper time interval measured by the clock between events A and B.

Tx, I now understand what you said and I agree. I think some of the difficulties in people's (including the OP's) minds arise from what you said earlier:
Al68 said:
... there is a big difference between "which clock runs slower than the other" and "which clock shows less elapsed time between events".

Trying to equate the two causes many of the apparent paradoxes.
 
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Related to Why Does the Traveling Twin's Clock Show Less Time in the Twin Paradox?

1. What is the "Half the Twin Paradox"?

The "Half the Twin Paradox" is a thought experiment that explores the concept of time dilation in the context of a journey through space. It involves two identical twins, one of whom stays on Earth while the other travels into space at high speeds. When the traveling twin returns to Earth, they will have aged less than their twin who remained on Earth. This paradox challenges our understanding of time and relativity.

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4. What are the implications of the "Half the Twin Paradox"?

The "Half the Twin Paradox" has significant implications for our understanding of time, space, and the universe. It challenges our perception of time as a constant and shows that it can be affected by various factors. It also has practical applications in fields such as space travel and GPS technology, where precise timing is crucial.

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The "Half the Twin Paradox" highlights the complex and mysterious nature of the universe. It shows that our understanding of time and space is limited and constantly evolving. It also raises questions about the nature of reality and the fabric of the universe. Further exploration and research into this paradox can lead to new discoveries and advancements in our understanding of the universe.

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