Another Twins Paradox question

[pervect]

Is here is a way to make the distance measured by the Earth at the start of the trip shorter than the distance measured by the ship.

Yes, there is - if the destination is moving away from the Earth, too. Of course, for the ship to catch it, it must be moving away more slowly than the ship.

This applies only to the initial distance measurement, it does not change the situation with respect to the round trip time.

$$\] \unitlength 1mm \begin{picture}(70,110)(0,0) \linethickness{0.3mm} \multiput(10,0)(0.12,0.2){500}{\line(0,1){0.2}} \linethickness{0.3mm} \multiput(30,0)(0.12,0.3){333}{\line(0,1){0.3}} \put(19,2){\makebox(0,0)[cc]{\theta}} \linethickness{0.3mm} \put(10,0){\line(0,1){110}} \put(12,10){\makebox(0,0)[cc]{\theta}} \linethickness{0.3mm} \multiput(10,0)(0.2,0.12){250}{\line(1,0){0.2}} \put(10,-2){\makebox(0,0)[cc]{A}} \put(30,-2){\makebox(0,0)[cc]{B}} \put(38,16){\makebox(0,0)[cc]{C}} \end{picture} \]$$

The line segment AC is the proper distance to the destination in the ship frame, which can be greater than the line segment AB, the proper length in the Earth frame.

However, the returning twin will be younger - in this example, as always. There is no way to make the returining twin older in flat space-time. The inertial path always maximizes the amount of time read on a clock in flat space-time.

This above diagram representing this situation is interesting, though.

We will proceed by finding the x and t coordinates of event C.

Then sqrt(x^2-t^2) = length of AC

The line of simultaneity AC has an equation (using geometric units where velocities are measured as a fraction of c)

t = v x

v is the velocity of the ship.

if AB = K, then the equation of segment BC is

x = K + ut

u is the velocity of the moving target, which must be less than that of the ship if the ship is to ever catch up with it.

Solving these simultaneously for the intersection point, we get

t = vK/(1-uv)
x = K/(1-uv)

The length of AC is just the Lorentz invariant, x^2 - t^2

sqrt(x^2-t^2) = L = K sqrt(1-v^2)/(1-u v)

Plugging in v=.6, u=.4,, we can see that L can be greater than K.

 

[pervect]

Using the approach I used in #19, here is a simple proof that the returning twin, traveling at a constant velocity, will always be older.

On the outbound trip, the traveling twin may decide to turn around when he reaches a specific objective, but he will always note that some time T has elapsed on his watch when he has reached his objective.

It does NOT MATTER what makes the traveling twin decide to turn around, the only thing that's important to solving the problem is that whenever he does so, for WhATEVER reason, his watch will have some definite time reading, which is T.

On the return trip, by symmetry, the same amount of time T will elapse on his watch. Therfore the amount of time for the entire trip for the traveling twin will be 2*T.

Now, let's look at the elapsed time on the Earth. We will compute this by the doppler shift method.

On the outbound trip, the traveling twin will receive a pulse from the Earth emitted at time T/k when he turns around.

When the traveling turn returns to Earth, the total elapsed time by Earth's clocks will be

T/k + k*T

as I argued in post #19

Thus the ratio of the Earth clock to the ship clock will be

(T/k + KT) / 2T = (k + 1/k)/2

A simple graph of this function reveals that it has a minimum when k=1, which means "no doppler shift", possible only if the "travelling twin" did not move.

 

[Al68]

Jesse,

How about if there were no signals sent, and no clocks used. Each twin would calculate both twins' elapsed time by using the known distances and velocity. This could obviously be done correctly by each twin. If we assume SR is correct, we would not need to use physical clocks to measure time, both twins can just calculate their own and the other's elapsed time for the trip.

Each twin would calculate the ship's proper elapsed time as 5 ly/0.866c = ~ 5.75 years.

Each twin would calculate the Earth's proper elapsed time as 10 ly/0.866c = ~ 11.5 years.

Note that in SR, the equation v = d/t is valid as long as we don't mix up the term between frames.

No clocks, no signals, no acceleration, no turnaround. We just calculate how long it would take to get there, from the point of view of each twin.

Pervect,

Thanks for your efforts, but like you said, making the initial distance to the target shorter does not change anything, since that's not the total distance from Earth to the turnaround point if the target is moving away from earth. I was looking for a way to make the total distance traveled by the ship to be specified as a certain distance as measured in the ship's frame.

Here's my reason for that:

In the Twins Paradox, and in the assumptions in every response I have received, the frame in which the distance is specified (or implied by specifying other parameters) is also the inertial frame, while the accelerating frame sees this distance length contracted. This results in two statements that are true in this case.

1. The twin in the inertial frame always ages more.
2. The twin in the frame in which the distance is specified always ages more.

I wanted an example where the frame in which the distance is specified is not the inertial frame.

Notice that the second statement would be true even if distance is not specified explicitly. It could be specified implicitly by specifying other parameters that can be used to calculate distance. Bottom line is, the twin with more mileage ages more, the twin with less mileage ages less.

Thanks,
Alan

 

[JesseM]

Jesse,

How about if there were no signals sent, and no clocks used. Each twin would calculate both twins' elapsed time by using the known distances and velocity. This could obviously be done correctly by each twin. If we assume SR is correct, we would not need to use physical clocks to measure time, both twins can just calculate their own and the other's elapsed time for the trip.

Each twin would calculate the ship's proper elapsed time as 5 ly/0.866c = ~ 5.75 years.

Each twin would calculate the Earth's proper elapsed time as 10 ly/0.866c = ~ 11.5 years.

No, they wouldn't each calculate the Earth's time to be this, unless you specify that they are each calculating the time in the Earth's rest frame. In the twin's rest frame, he sees the Earth moving away from him at 0.866c while the destination is moving towards him at 0.866c, so the time elapsed for him is 5 ly/0.866c = ~5.77 years (again, not 5.75...if you're going to use three significant figures you can't round the last decimal to the nearest 5). But in his frame, he knows using the rules of SR that time on Earth must be passing at half his own rate, so the time elapsed on Earth must be 0.5*5.77 = 2.89 years.

 

[Al68]

No, they wouldn't each calculate the Earth's time to be this, unless you specify that they are each calculating the time in the Earth's rest frame. In the twin's rest frame, he sees the Earth moving away from him at 0.866c while the destination is moving towards him at 0.866c, so the time elapsed for him is 5 ly/0.866c = ~5.77 years (again, not 5.75...if you're going to use three significant figures you can't round the last decimal to the nearest 5). But in his frame, he knows using the rules of SR that time on Earth must be passing at half his own rate, so the time elapsed on Earth must be 0.5*5.77 = 2.89 years.

The aging of the Earth twin would calculated by both using Earth's frame. The aging of the ship's twin would be calculated by both using the ship's frame. You could also say that they each hired their sister to do all the calculations for them ahead of time, so the calculations done by both twins are identical.

I have to disagree with your last statement. The fact that two observers in relative motion will always see the other's clock run slow does NOT mean that each observer will always see a smaller elapsed time on the other's clock after the experiment. The difference in elapsed time is because the event at which each twin stops his clock is not simultaneous in both frames. Total elapsed time is not the same as the rate time passes. Two observers will NOT each see the total elapsed time of the other to be less after the experiment, unless you assume their clocks are stopped simultaneously in both frames, which is impossible. In this case, each will see the other's time pass more slowly than his own.

But in this case, both twins will agree on the following:

1. The event of the ship passing the star, as observed by the ship's twin, occurred at t = 5.77 years, as measured in the ship's frame.
2. The event of the ship passing the star, as observed by the Earth twin, occurred at t = 11.5 years, as measured in the Earth's frame.

Thanks,
Alan

 

[JesseM]

The aging of the Earth twin would calculated by both using Earth's frame.

OK, but that's an arbitrary choice. They could also both calculate things using the traveling twin's rest frame, and in this case they'd conclude the Earth twin aged less. This answer would be just as valid.

I have to disagree with your last statement. The fact that two observers in relative motion will always see the other's clock run slow does NOT mean that each observer will always see a smaller elapsed time on the other's clock after the experiment. The difference in elapsed time is because the event at which each twin stops his clock is not simultaneous in both frames. Total elapsed time is not the same as the rate time passes.

I'm not sure what you mean by this. If I see a clock moving at constant velocity v from one marker A to another marker B, then if the time interval between passing A and passing B is t in my frame, the time elapsed on the clock between passing A and passing B will be $$t*\sqrt{1 - v^2/c^2}$$. Do you disagree?

As an example of this, if the traveling twin in the experiment is towing a rod 5 light years long behind him, then in his frame the event of him reaching the destination and the event of the rod's back end passing the Earth will be simultaneous, and while the time to get from Earth to the destination took 5.77 years in his frame, the event of the rod's back end passing the Earth happened when the Earth's clocks read half this, or 2.89 years. Agreed?

Two observers will NOT each see the total elapsed time of the other to be less after the experiment, unless you assume their clocks are stopped simultaneously in both frames, which is impossible. In this case, each will see the other's time pass more slowly than his own.

Just to be clear, in my thought experiment neither twin changes velocity, "the end of the experiment" is just the moment when the traveling twin reaches the position of the destination, at which point he calculates how much time has elapsed on the Earth clock in the same rest frame he has had throughout the experiment. If you're assuming the twin changes velocity so he's at rest relative to the destination, and that he then figures out the time on the Earth clock in his new rest frame (the same rest frame as the earth), then I agree he will conclude that less time elapsed on his own clock than on the Earth clock. But if the traveling twin is calculating things from the point of view of the same rest frame he had during the outbound leg, then he will conclude that the time elapsed on the Earth's clock is less than the time elapsed in his own frame.

But in this case, both twins will agree on the following:

1. The event of the ship passing the star, as observed by the ship's twin, occurred at t = 5.77 years, as measured in the ship's frame.
2. The event of the ship passing the star, as observed by the Earth twin, occurred at t = 11.5 years, as measured in the Earth's frame.

Sure, but they'd also agree on this additional proposition:

3. The event of the Earth's clock reading t = 2.89 years, as observed by the ship's twin, occurred at t = 5.77 years, as measured in the ship's frame (and since t = 5.77 years is also the time the ship passed the star in the ship's frame, these events happened simultaneously in the ship's frame).

 

[Al68]

But in this case, both twins will agree on the following:

1. The event of the ship passing the star, as observed by the ship's twin, occurred at t = 5.77 years, as measured in the ship's frame.
2. The event of the ship passing the star, as observed by the Earth twin, occurred at t = 11.5 years, as measured in the Earth's frame.

Sure, but they'd also agree on this additional proposition:

3. The event of the Earth's clock reading t = 2.89 years, as observed by the ship's twin, occurred at t = 5.77 years, as measured in the ship's frame (and since t = 5.77 years is also the time the ship passed the star in the ship's frame, these events happened simultaneously in the ship's frame).

I agree that your #3 would also be correct. But that only means that the ship's twin will see the Earth twin's clock read 2.89 years. The ship's twin will not conclude that the Earth twin has stopped his clock at 2.89 years. In my #1 and #2 above, the times are the total elapsed times of the experiment, after the clocks have stopped. In your #3, the 2.89 years is only a snapshot of a clock while it is running. In my #1 and #2 above, the times are total elapsed proper times, not just a snapshot of someone elses clock while it is running.

I other words, my elasped times are from stopwatches, not from clocks that only show the current time.

The snapshot of a clock while running, taken from someone in another frame, does not represent anyones proper total elapsed time in their own frame. But the total elapsed time recorded on a clock does represent the proper time for its own frame, even if viewed by someone in another frame. The only way a clock reads proper time while running is when seen by someone in the same frame as the clock. If the clock is stopped and shows an elapsed time, it will show the proper elapsed time (between when it was started and stopped) in its own frame, even if viewed by someone in another frame.

Also note in the Twins Paradox, the twins views of each other's clocks while running do not represent the amount each twin actually aged during the trip. For example, if we say after the journey, that the Earth twin aged 12 years and the ship's twin aged 6 years. Then in reality we would assume that the ship's twin aged 3 years on the outbound leg and 3 years on the inboard leg, since the velocity was the same for each leg. This does not correspond to the Earth's snapshots of the ship's twin's clocks.

Also notice that this aging of 3 years during the outbound leg is real, it's not a figment of observation. It exists prior to the observation, so it cannot be actually caused by the observation.

We could also say that in the Twins Paradox, that if both twins accidently lost their clocks right before the reunion, that would only affect their observations. It would not change reality. We could still figure out that the ship's twin is younger.

Thanks,
Alan

 

[JesseM]

I agree that your #3 would also be correct. But that only means that the ship's twin will see the Earth twin's clock read 2.89 years. The ship's twin will not conclude that the Earth twin has stopped his clock at 2.89 years. In my #1 and #2 above, the times are the total elapsed times of the experiment, after the clocks have stopped.

OK, but that's assuming the earth-twin will stop his clock at the moment that is simultaneous with the traveling twin reaching the star in the earth-twin's reference frame. This is an arbitrary choice, it doesn't tell you anything about who was "really" older at the moment the traveling twin reached the star. You could equally well make up the rule that the earth-twin should stop his clock at the moment that is simultaneous with the traveling twin reaching the star in the traveling twin's reference frame.

I other words, my elasped times are from stopwatches, not from clocks that only show the current time.

And it is an arbitrary choice when the earth-twin stops his watch, because if you want him to stop it "at the same moment" that the traveling twin reached the star, this depends on which frame's definition of simultaneity you choose to use.

The snapshot of a clock while running, taken from someone in another frame, does not represent anyones proper total elapsed time in their own frame.

It represents the elapsed time on the clock between the moment they started at the same location (when both clocks read 0) and the present moment, in the frame of the observer. It doesn't represent the elapsed time between the moment they started and the moment they stopped their clocks, but then you have to specify when you want each to stop his own clock. If the earth-twin doesn't stop his clock until the "same time" the traveling twin reaches the destination in the Earth twin's frame, then in the traveling twin's frame, the Earth twin was letting his clock continue to run long after the moment the traveling twin reached his destination and stopped his own clock. If two observers stop their clocks at completely different times, then comparing the reading on each after each one was stopped doesn't tell you anything about who aged less!

But the total elapsed time recorded on a clock does represent the proper time for its own frame, even if viewed by someone in another frame. The only way a clock reads proper time while running is when seen by someone in the same frame as the clock.

I think you're misusing the term "proper time"--by definition, proper time always refers to the time experienced by the clock itself, so even if I see a clock moving relative to me I will say that the time elapsed on the clock between two events on its worldline A and B represents that clock's proper time interval between those two events. Proper time is a frame-invariant concept that always refers to the time along a particular worldline, not the time in any particular frame (so you can also talk about proper time for non-inertial clocks, for example).

If the clock is stopped and shows an elapsed time, it will show the proper elapsed time (between when it was started and stopped) in its own frame, even if viewed by someone in another frame.

Sure.

Also note in the Twins Paradox, the twins views of each other's clocks while running do not represent the amount each twin actually aged during the trip. For example, if we say after the journey, that the Earth twin aged 12 years and the ship's twin aged 6 years. Then in reality we would assume that the ship's twin aged 3 years on the outbound leg and 3 years on the inboard leg, since the velocity was the same for each leg.

In the Earth twin's frame, you mean. In other frames the velocities would be different on the two legs, although they will agree that 3 years elapse on the traveling twin's clock on each leg.

This does not correspond to the Earth's snapshots of the ship's twin's clocks.

What do you mean? The Earth twin will observe 3 years to have passed on the traveling twin's clock at the moment of the turnaround, and will observe 3 more years pass on his clock between the turnaround and returning to earth.

Also notice that this aging of 3 years during the outbound leg is real, it's not a figment of observation. It exists prior to the observation, so it cannot be actually caused by the observation.

Of course. Again, all statements about proper time are frame-invariant.

 

[RandallB]

OK, but that's assuming the earth-twin will stop his clock at the moment that is simultaneous with the traveling twin reaching the star in the earth-twin's reference frame. This is an arbitrary choice, it doesn't tell you anything about who was "really" older at the moment the traveling twin reached the star. You could equally well make up the rule that the earth-twin should stop his clock at the moment that is simultaneous with the traveling twin reaching the star in the traveling twin's reference frame.

Nothing arbitrary about it;
All three of those things are simultaneous. The traveling twin on reaching the star is directly making a comparison to the Earth reference frame and vise-versa locally. And they are both simultaneous with all points of that time in the Earth reference frame including earth; just as that is true for all points of that time in the traveling frame.
That is why an observer at the star watching the traveler go by without turning around can still report back to Earth that the traveler is ageing slower.

What is not simultaneous is the point and time Earth will observe locally going by in the traveling frame – but normally there is nothing there anyway in most examples.
If you don’t get this you won’t get simultaneity or the twins.
Because if you do correctly identify another traveler in the travelers frame passing by Earth at that Earth time, they will be able to correctly report up to the lead traveler that Earth is ageing slower than both of them.

 

[JesseM]

OK, but that's assuming the earth-twin will stop his clock at the moment that is simultaneous with the traveling twin reaching the star in the earth-twin's reference frame. This is an arbitrary choice, it doesn't tell you anything about who was "really" older at the moment the traveling twin reached the star. You could equally well make up the rule that the earth-twin should stop his clock at the moment that is simultaneous with the traveling twin reaching the star in the traveling twin's reference frame.

Nothing arbitrary about it;
All three of those things are simultaneous.

Which 3 things are you talking about? By "things" do you mean 3 events? Presumably the event of the traveling twin reaching the star, with his clock reading t = 5.77 years, would be one of them; are the other two the event of the earth-clock reading t = 2.89 years, and the event of the earth-clock reading t=11.5 years? Obviously these last two events are not simultaneous with each other. The second event is simultaneous with the first event in the frame of the traveling twin, while the third event is simultaneous with the first event in the frame of the earth. The thing I was calling "arbitrary" was the claim Al68 seemed to be making that the first and third event should be used to show that the traveling twin aged less, while ignoring the first and second event, which could just as easily be used to show that the earth-twin aged less. Do you disagree, and think that the traveling twin "really" aged less?

The traveling twin on reaching the star is directly making a comparison to the Earth reference frame and vise-versa locally.

What do you mean by "locally"? Any question of the simultaneity of distant events, like the 3 I listed above, is by definition not a local one. And yes, the traveling twin is free to compare his time with the time on Earth using the earth-frame's definition of simultaneity. He is also free to compare his time with the time on Earth using his own rest frame's definition of simultaneity. It's an arbitrary choice, there's no reason to prefer one frame's definition of simultaneity over another's.

And they are both simultaneous with all points of that time in the Earth reference frame including earth; just as that is true for all points of that time in the traveling frame.

I don't understand what you mean by "all points of that time"...what does "that time" refer to? The time the traveling twin passes by the star? That is only a single point in time.

That is why an observer at the star watching the traveler go by without turning around can still report back to Earth that the traveler is ageing slower.

Assuming the star is at rest relative to the earth, and the star is using its own rest frame, of course. On the other hand, an observer on the ship using his own rest frame will say that the clocks at the star are running slower than those on the ship.

What is not simultaneous is the point and time Earth will observe locally going by in the traveling frame

Again, I'm not understanding your language. Do you mean "the point in time"? And if so, what specific event is the Earth observing the point in time of? I have no idea what it means for the Earth to "observe locally going by in the traveling frame", maybe you could give me an example with some specific numbers.

Because if you do correctly identify another traveler in the travelers frame passing by Earth at that Earth time, they will be able to correctly report up to the lead traveler that Earth is ageing slower than both of them.

Why are you introducing a second traveler here? If you're just saying that in the traveler's rest frame, the Earth's clocks are running slow, then of course I agree, but you don't need a second traveler to point that out.

 

[Al68]

OK, but that's assuming the earth-twin will stop his clock at the moment that is simultaneous with the traveling twin reaching the star in the earth-twin's reference frame. This is an arbitrary choice, it doesn't tell you anything about who was "really" older at the moment the traveling twin reached the star.

This is not an arbitrary choice. The fact that the Earth twin stops his clock at the moment that is simultaneous with the traveling twin reaching the star in the earth-twin's reference frame is important. Since this represents how much the Earth twin aged during the experiment.

So, the elapsed time in my #1 represents how much the ships twin aged during the trip, as timed and observed by the ship's twin.

And, the elasped time in my #2 represents how much the Earth twin aged during the trip, as timed and observed by the Earth twin.

The times for stopping the clocks in my example are not arbitrary, they are by design. So that each twin records the time of the "event" in their own respective frames. Not to compare their clocks to each other. Of course the event is not simultaneous to each twin. The event is simultaneous with t = 5.77 years for the ship's twin, and simultaneous with t = 11.5 years for the Earth twin.

Thanks,
Alan

 

[JesseM]

This is not an arbitrary choice. The fact that the Earth twin stops his clock at the moment that is simultaneous with the traveling twin reaching the star in the earth-twin's reference frame is important. Since this represents how much the Earth twin aged during the experiment.

No, it represents how much the Earth twin aged during the experiment in the earth-twin's frame. When you say "during the experiment", don't you just mean "from the time the experiment began (ship leaving earth) to the time the experiment ended (ship arriving at star)"? And don't you agree that the question of what time on Earth corresponds to the time of the ship arriving at the star is one that depends on your choice of reference frame? In the frame of the ship, don't you agree that the event of the ship arriving at the star is simultaneous with the event of the Earth's clock reading t = 2.89, and that therefore that is "the time the experiment ended" on earth, in the ship's own reference frame?

Perhaps the problem is one of language--when we ask "how much time went by on earth" during the experiment, that could be taken to mean how much time went by on the Earth's clock for the duration of the experiment in whatever frame we choose, or it could mean how much time went by in the Earth's own frame. These two meanings are distinct, and I am thinking in terms of the first meaning of the phrase, not the second.

Conceptually, it might also help to make the experiment more symmetrical. Suppose instead of the ship traveling to a star 10 light years from the earth, the ship is traveling along a measuring rod which has one end at the Earth and is at rest relative to earth, and is 10 light years long in the Earth's frame. Now suppose the ship itself is also attached to one end of a measuring rod that's at rest relative to the ship, and extends in the opposite direction as the Earth's measuring rod, and is 10 light years long in the ship's frame. So in this way, during a single trip we can be doing two separate but symmetrical experiments, one where we see how long it takes the ship to reach the far end of the Earth's measuring-rod, and another to see how long it takes the Earth to reach the far end of the ship's measuring rod. In this case the answers for the local readings will be the same--at the time the ship reaches the end of the Earth's measuring rod, the ship's own clock reads t = 5.77 years, and at the time the Earth reaches the end of the ship's measuring rod, the Earth's own clock reads t = 5.77 years. But then for each experiment, we could ask a question analogous to the one you ask in your experiment, namely:

1. How much time passes on Earth during the experiment of the ship traveling from one end of the Earth's measuring-rod to the other?

2. How much time passes on the ship during the experiment of the Earth traveling from one end of the ship's measuring-rod to the other?

How would you answer these questions?

 

[pervect]

Pervect,

Thanks for your efforts, but like you said, making the initial distance to the target shorter does not change anything, since that's not the total distance from Earth to the turnaround point if the target is moving away from earth. I was looking for a way to make the total distance traveled by the ship to be specified as a certain distance as measured in the ship's frame.

Alan

Unless I'm misunderstanding your question, what you are trying to do is impossible, and you are simply not accepting that fact.

[add]The impossibility is a consequence of the "triangle inequality" as it applies to the Minkowski space of special relativity.

I feel like we are going around in circles here. While I think I could improve on my impossibility proof for rigor and completeness, I don't think that would "break the circle", so I'm just going to leave things where they stand, and take a break from the thread.

 

[RandallB]

...What is not simultaneous is the point and time Earth will observe locally going by in the traveling frame ...

Again, I'm not understanding your language. Do you mean "the point in time"? And if so, what specific event is the Earth observing the point in time of? I have no idea what it means for the Earth to "observe locally going by in the traveling frame", maybe you could give me an example with some specific numbers.

O come on don’t play dumb this is a simple SR problem there are only Two reference frames Earth/Star Frame (t) and Travelers Frame (t’). And not that many variables so this shouldn’t be as hard as this thread makes it out to be.
When Earth clock reads t = 0 at Earth it is simultaneous with all other clocks that read t=0 in the earth/star frame including the one observed at the star.
Likewise, when the traveler reaches the star the Earth and the star will simultaneously read t= T, the traveler now at the star can look over at the star clock and see that same t=T but on his own clock he will see t’=T’. Both traveler and star observer can see that the star observer has aged more.
This event is simultaneous with all locations in the Earth frame only at the time t=T for each of those locations including when Earth reaches t=T.
And it is simultaneous with all points in the traveler frame when the reach t’=T’.
BUT between the two frames there is one and only one point that in each frame local see t=T and t’=T’ together.

So now ALAN you know three things are simultaneous with no need to turn the spaceship around.
1) t’=T’ when traveler reaches the star
2) t=T when star sees traveler arrive & go by.
3) t=T at some known distance back at Earth measured in the Earth frame

Don’t you think it would help to figure out exactly WHERE and WHEN the Earth is as measured in the traveler frame when the Earth time reads t=T. It is not impossible to find and can only have one answer. It’s just straight forward SR transforms/math and might help you see what going on.

 

[JesseM]

O come on don’t play dumb this is a simple SR problem there are only Two reference frames Earth/Star Frame (t) and Travelers Frame (t’). And not that many variables so this shouldn’t be as hard as this thread makes it out to be.

I'm not "playing dumb", and I understand the scenario described here just fine, but the phrase "What is not simultaneous is the point and time Earth will observe locally going by in the traveling frame" sounds like gibberish to me. What is it that the Earth is observing locally going by, exactly? The object of the verb "observe" appears to be the "point", but how can a point in time be "going by"? What does that even mean?

When Earth clock reads t = 0 at Earth it is simultaneous with all other clocks that read t=0 in the earth/star frame including the one observed at the star.

Yes, assuming the clock at the star is synchronized with the earth-clock in their common rest frame, then in this rest frame, the event of the earth-clock reading 0 will be simultaneous with the event of the star-clock reading 0. However, in the traveler's frame the event of the earth-clock reading 0 is not simultaneous with the event of the star-clock reading 0; in this frame, at the moment the earth-clock read 0, the star-clock already read 8.66 years.

Likewise, when the traveler reaches the star the Earth and the star will simultaneously read t= T,

They will both read the same time in the earth/star frame, of course. In this example T would be 11.55 years. But the two clocks will not read 11.55 years simultaneously in the traveller's own rest frame.

the traveler now at the star can look over at the star clock and see that same t=T but on his own clock he will see t’=T’.

Yes, when he arrives at the star the star-clock reads 11.55 years, while his clock reads 5.77 years.

Both traveler and star observer can see that the star observer has aged more.

No, because the traveller disagrees that at the moment he left earth, the star-clock read t=0 years. Again, in the traveller's frame the star-clock already read t = 8.66 years at the moment he left earth, and since it read 11.55 years at the moment he arrived at the star, he will say the star-clock only advanced forward by 11.55 - 8.66 = 2.89 years from the time he left Earth to the time he arrived, while his own clock advanced by 5.77 years between these events.

This event is simultaneous with all locations in the Earth frame only at the time t=T for each of those locations including when Earth reaches t=T.

Sure, assuming you're saying that every event which has a time-coordinate of 11.55 years in the earth-frame would be simultaneous with the event of the traveller reaching the star in the earth-frame.

And it is simultaneous with all points in the traveler frame when the reach t’=T’.

Again, if I'm understanding you correctly, I agree--every event that has a time-coordinate of 5.77 years in the traveller's frame would be simultaneous with the event of the traveller reaching the star in the traveller's frame.

BUT between the two frames there is one and only one point that in each frame local see t=T and t’=T’ together.

"In each frame local" is a very weird way of phrasing it to me, but I think I understand what you're saying--there is only one point in spacetime that is assigned a time-coordinate of 11.55 years in the earth-frame and a time-coordinate of 5.77 years in the traveller's frame (and that's the point that corresponds to the traveller arriving at the star).

None of this really helps me in understanding what you meant when you said "What is not simultaneous is the point and time Earth will observe locally going by in the traveling frame", though.

So now ALAN you know three things are simultaneous with no need to turn the spaceship around.
1) t’=T’ when traveler reaches the star
2) t=T when star sees traveler arrive & go by.
3) t=T at some known distance back at Earth measured in the Earth frame

This is a little confusing because it doesn't make sense to say time-coordinates are "simultaneous", only events can be simultaneous in relativity...but I assume you mean the following 3 events are simultaneous:

1) the event of the traveller's clock reading T', in this case 5.77 years
2) the event of the star's clock reading T, in this case 11.55 years
3) the event of the Earth's clock reading T, in this case 11.55 years

These events are all simultaneous in the earth-frame, but not in other frames like the traveller's frame. That's what I was saying was "arbitrary", because Alan was using the earth-frame's definition of simultaneity to apparently claim the traveller aged less in some absolute sense, when it would be just as valid to analyze things in the traveller's frame and say the event of the traveller reaching the star was simultaneous with the event of the Earth's clock reading 2.89 years.

 

[Al68]

No, it represents how much the Earth twin aged during the experiment in the earth-twin's frame.

You're right here. But isn't that the definition of aging? How much time passes for a person in their own frame. Someone does't grow old according to time elapsed in a different frame. Someone grows old at the rate time elapses in their own frame. I'm talking about real aging here.

Also, my evaluation here is the same as the first half of the Twins Paradox. Is the differential aging not real unless the twins reunite? Sure, it is more convenient to have the twins reunite for observational purposes, but observation (in the macroscopic world) does not cause reality. This differential aging is a real phenomenom that occurs. It is not a figment of observation, and therefore exists in reality, even if it is not convenient to measure.

Maybe I should state this point more clearly. Actual aging of a person between two events, as measured in their own frame, is equal to the time elapsed between those events, as measured in their own frame. Is this not a valid definition of actual aging?

Also, in your resonse to RandallB, you say that "the traveller disagrees that at the moment he left earth, the star-clock read t=0 years..." Why would the traveler ever conclude that the star-clock would not read exactly the same time as the Earth clock, if the Earth and star are in the same inertial frame. If you say that at any event is simultaneous with t=0 in a particular frame, then the event is simultaneous with t=0 for any location in that frame. What if I specified that the star-clock reads t=0 when the ship leaves earth? And all three clocks are synchronized at the start. Why didn't I think of that?

How about my example where neither twin even has a clock? That made things simple, I thought. Your objection was based on the fact that a clock on Earth would appear to run slow to the ship's twin. But, that assumes that they have clocks. I was referring to total elapsed time for each twin in their own respective frame, between the time the ship left Earth and the time it reached the star, again as observed in their own frame, respectively, and they don't need clocks to figure that out. As in the Twins Paradox, each twin will always see the other's clock run slow. That's different than the elapsed time recorded on a clock. As in the Twin's paradox, the total elapsed time recorded on a clock will not always be observed to be less by a different reference frame, like the rate of time passing is.

And pervect, if you're still around, I agree that we have been going around in circles here. But now you acknowedge my original point of trying to have the distance specified in the frame of the twin on the ship. Even though you think this is not possible. That's a lot better than discussing whether or nor rigid rods exist, which is pointless and my own fault, I admit.

Thanks,
Alan

 

[JesseM]

You're right here. But isn't that the definition of aging? How much time passes for a person in their own frame. Someone does't grow old according to time elapsed in a different frame. Someone grows old at the rate time elapses in their own frame. I'm talking about real aging here.

But you're not talking about an individual twin's age, you're comparing the ages of two different twins, and to do that you must pick a definition of simultaneity. It is of course really true that when the Earth twin's clock reads 11.55 years, he has aged 11.55 years since the traveling twin departed. It is not "really true" (in some objective frame-invariant sense) that when the earth-twin's clock reads 11.55 years, he is older than the traveling twin is at the same moment. In the traveling twin's frame, the event of the earth-twin's clock reading 11.55 years is simultaneous with the event of the traveling twin's clock reading 23.09 years. And when the traveling twin's clock reads 23.09 years, he has really aged by 23.09 years since departing the earth-twin. Likewise, in the traveling twin's frame, when his clock reads 5.77 years, that is simultaneous with the event of the earth-twin's clock reading 2.89 years. And when that event occurs, the earth-twin has really aged by 2.89 years. So why is it any less valid to take the traveling twin's definition of which events are simultaneous rather than the Earth twin's?

Also, my evaluation here is the same as the first half of the Twins Paradox. Is the differential aging not real unless the twins reunite?

No, it isn't, not if by "differential aging" you mean an objective truth about which one has aged less before either one accelerates. Of course there is a definite truth about who has aged less in any given frame--and in the frame where the traveling twin was at rest and the Earth was moving, it is the earth-twin who aged less during the outbound leg. Only if they depart and the reunite can there be an objective truth about which one aged more, and by how much.

Sure, it is more convenient to have the twins reunite for observational purposes, but observation (in the macroscopic world) does not cause reality.

This isn't about "observation causing reality", it's about some quantities not being objective features of reality at all, but instead depending on your choice of coordinate system. Velocity is an example of this. If two objects are moving apart, in one frame the first object may be moving faster than the second, while in another frame the second may be moving faster than the first--would you say there is an objective truth about which one is moving faster? Likewise, with two different spatial coordinate systems, in one coordinate system an object may be at position x=2 and in another it may be at position x=5...would you say there is an objective truth about what its "true" x-coordinate is? Assuming your answer to both these questions is "no", why is it so hard for you to accept that the question of which of two spatially separated events (like the earth-twin's clock reading 5.77 years vs. the traveling twin's clock reading 5.77 years) happened earlier and which happened later? In relativity this is every bit as coordinate-dependent as the question of which of two objects has a larger velocity, or which has a larger x-coordinate.

Maybe I should state this point more clearly. Actual aging of a person between two events, as measured in their own frame, is equal to the time elapsed between those events, as measured in their own frame. Is this not a valid definition of actual aging?

This definition is fine for inertial observers, although it's better to just use the proper time along their wordline between the two events, because that will work for non-inertial observers who don't have a single frame of their own. In any case, even for inertial observers this definition tells you nothing about which of two spatially separated observers has aged more without making some assumption about simultaneity.

Also, in your resonse to RandallB, you say that "the traveller disagrees that at the moment he left earth, the star-clock read t=0 years..." Why would the traveler ever conclude that the star-clock would not read exactly the same time as the Earth clock, if the Earth and star are in the same inertial frame.

This question makes me think you don't understand the way simultaneity works in relativity. Two clocks which are synchronized in their own mutual rest frame will always be out-of-sync in other frames--were you unaware of this? This is just a consequence of Einstein's clock synchronization convention. His idea was that each observer should define space and time coordinates of events using local readings on a network of clocks and measuring-rods which are at rest relative to himself, and with all the clocks "synchronized" using the assumption that light travels at a constant speed in all directions in that observer's rest frame. This necessarily implies that different frames must disagree about simultaneity, as I discussed on this thread:

Suppose I'm on a rocket, and I have two clocks, one at the front of the rocket and one at the back, that I want to synchronize. If I assume that light travels at the same speed in all directions in the rocket's rest frame, then I can set off a flash at the midpoint of the two clocks, and set them to both read the same time at the moment the light reaches them. But now imagine you are in another frame, one in which the rocket is moving forward with some positive velocity. In your frame, the back of the rocket will be moving towards the point in space where the flash was set off, while the front of the rocket will be moving away from it; therefore if you assume light travels at the same speed in both directions in your frame, you will naturally conclude the light must catch up with the clock at the back at an earlier time than it catches up with the clock at the front, since both were at equal distances from the midpoint of the rocket when the flash was set off there. This means that you will judge my two clocks to be out-of-sync if I use the above synchronization procedure, with the back clock ahead of the front clock in your frame.

More technically, you can show using the Lorentz transformation that if two events are simultaneous and 10 light years apart in one frame, then in another frame which is moving at 0.866c relative to the first, one of the events must have happened 8.66 years after the other. Are you familiar with how to use the Lorentz transformation?

If you say that at any event is simultaneous with t=0 in a particular frame, then the event is simultaneous with t=0 for any location in that frame. What if I specified that the star-clock reads t=0 when the ship leaves earth? And all three clocks are synchronized at the start. Why didn't I think of that?

They can only all be synchronized in one frame. In every other frame, the star-clock must either read an earlier or later time than the earth-clock at the moment the earth-clock reads t=0.

How about my example where neither twin even has a clock? That made things simple, I thought. Your objection was based on the fact that a clock on Earth would appear to run slow to the ship's twin. But, that assumes that they have clocks. I was referring to total elapsed time for each twin in their own respective frame, and they don't need clocks to figure that out.

But I don't disagree about the "total elapsed time for each twin in their own respective frame", I just disagree that this tells you anything about which twin aged more in an objective, frame-independent way. When 11.5 years have passed for the earth-twin, then in his frame this is the "same moment" that the traveling twin is reaching the star, but in other frames the traveling twin either has yet to reach the star or already reached it long ago.

To make sense of your claims about who has aged more, it would really help if you would answer this question:

Conceptually, it might also help to make the experiment more symmetrical. Suppose instead of the ship traveling to a star 10 light years from the earth, the ship is traveling along a measuring rod which has one end at the Earth and is at rest relative to earth, and is 10 light years long in the Earth's frame. Now suppose the ship itself is also attached to one end of a measuring rod that's at rest relative to the ship, and extends in the opposite direction as the Earth's measuring rod, and is 10 light years long in the ship's frame. So in this way, during a single trip we can be doing two separate but symmetrical experiments, one where we see how long it takes the ship to reach the far end of the Earth's measuring-rod, and another to see how long it takes the Earth to reach the far end of the ship's measuring rod. In this case the answers for the local readings will be the same--at the time the ship reaches the end of the Earth's measuring rod, the ship's own clock reads t = 5.77 years, and at the time the Earth reaches the end of the ship's measuring rod, the Earth's own clock reads t = 5.77 years. But then for each experiment, we could ask a question analogous to the one you ask in your experiment, namely:

1. How much time passes on Earth during the experiment of the ship traveling from one end of the Earth's measuring-rod to the other?

2. How much time passes on the ship during the experiment of the Earth traveling from one end of the ship's measuring-rod to the other?

How would you answer these questions?

If you used the same logic as you use above, it seems to me you'd have to conclude that the traveling twin aged less than the earth-twin in #1, and that the earth-twin aged less than the traveling twin in #2, despite the fact that both these experiments can be carried out in a single journey, with each twin moving alongside the other twin's measuring rod at the same time. Is that indeed what you'd say?

 

[Al68]

You're right here. But isn't that the definition of aging? How much time passes for a person in their own frame. Someone does't grow old according to time elapsed in a different frame. Someone grows old at the rate time elapses in their own frame. I'm talking about real aging here.

But you're not talking about an individual twin's age, you're comparing the ages of two different twins, and to do that you must pick a definition of simultaneity.

I'm not talking about either one. Of course you can never compare the ages of people in different rest frames at any given point in time. Not in any objective way, as you point out. My #1 and #2 statements earlier were not comparing their ages at any given point in time. I was stating how much each twin aged during the experiment.

It is of course really true that when the Earth twin's clock reads 11.55 years, he has aged 11.55 years since the traveling twin departed. It is not "really true" (in some objective frame-invariant sense) that when the earth-twin's clock reads 11.55 years, he is older than the traveling twin is at the same moment. In the traveling twin's frame, the event of the earth-twin's clock reading 11.55 years is simultaneous with the event of the traveling twin's clock reading 23.09 years. And when the traveling twin's clock reads 23.09 years, he has really aged by 23.09 years since departing the earth-twin. Likewise, in the traveling twin's frame, when his clock reads 5.77 years, that is simultaneous with the event of the earth-twin's clock reading 2.89 years. And when that event occurs, the earth-twin has really aged by 2.89 years. So why is it any less valid to take the traveling twin's definition of which events are simultaneous rather than the Earth twin's?.

I agree with all of your statements here. I wasn't choosing either twin's version of which events were simultaneous. I never said that the event of the Earth twin t=11.55 years was simultaneous with the event of the ship's twin's t=5.77 years. These two events are not simultaneous.

How about my example where neither twin even has a clock? That made things simple, I thought. Your objection was based on the fact that a clock on Earth would appear to run slow to the ship's twin. But, that assumes that they have clocks. I was referring to total elapsed time for each twin in their own respective frame, and they don't need clocks to figure that out.

But I don't disagree about the "total elapsed time for each twin in their own respective frame", I just disagree that this tells you anything about which twin aged more in an objective, frame-independent way.

It tells you a lot if we are talking about how much each twin aged during the experiment. It tells us nothing if we are trying to figure out how much each twin has aged at a particular moment in time, since that moment would not be simultaneous in both frames. Again, I am not claiming anything about how old either twin is at any particular time.

As far as the star clock, the ship's twin would agree that it read t=0 when he left earth, since he was in the same frame when they were synchronized. That doesn't mean that the clocks (on Earth and the star) will always read the same as seen by the ship. I didn't mean to imply that. I was just pointing out that all three clocks were at rest relative to each other when t=0. At least that's the way I read RandallB's post. And I did not mean to imply that I agreed with the rest of his post, either.

I just reread my earlier post, so I need to edit this part. I should have said that the ship would agree that the star-clock read t=0 the moment prior to leaving earth, but not the moment after. If you assume instaneous acceleration, the exact time the ship left Earth could be interpreted as either of these moments, since the ship is both at rest and in relative motion to Earth at t=0. So, you were right, since you were referring to t=0 after the ship was moving relative to earth. This t=0 will no longer be simultaneous at Earth and the star after the ship starts moving.

To make sense of your claims about who has aged more, it would really help if you would answer this question:
Quote:
Conceptually, it might also help to make the experiment more symmetrical. Suppose instead of the ship traveling to a star 10 light years from the earth, the ship is traveling along a measuring rod which has one end at the Earth and is at rest relative to earth, and is 10 light years long in the Earth's frame. Now suppose the ship itself is also attached to one end of a measuring rod that's at rest relative to the ship, and extends in the opposite direction as the Earth's measuring rod, and is 10 light years long in the ship's frame. So in this way, during a single trip we can be doing two separate but symmetrical experiments, one where we see how long it takes the ship to reach the far end of the Earth's measuring-rod, and another to see how long it takes the Earth to reach the far end of the ship's measuring rod. In this case the answers for the local readings will be the same--at the time the ship reaches the end of the Earth's measuring rod, the ship's own clock reads t = 5.77 years, and at the time the Earth reaches the end of the ship's measuring rod, the Earth's own clock reads t = 5.77 years. But then for each experiment, we could ask a question analogous to the one you ask in your experiment, namely:

1. How much time passes on Earth during the experiment of the ship traveling from one end of the Earth's measuring-rod to the other?

2. How much time passes on the ship during the experiment of the Earth traveling from one end of the ship's measuring-rod to the other?

How would you answer these questions?
If you used the same logic as you use above, it seems to me you'd have to conclude that the traveling twin aged less than the earth-twin in #1, and that the earth-twin aged less than the traveling twin in #2, despite the fact that both these experiments can be carried out in a single journey, with each twin moving alongside the other twin's measuring rod at the same time. Is that indeed what you'd say?

Yes, that is what I'd say. Except the part about "each twin moving alongside the other twin's measuring rod at the same time". These two experiments would not end simultaneously in either frame. So they wouldn't be "at the same time". Notice that the Earth twin would see the end of the ship's measuring rod pass Earth at t=5.77 years, and the ship reach the end of the Earth's measuring rod at t=11.5 years. And the ship's twin would see the ship reach the end of the Earth's rod at t=5.77 years, and the Earth reach the end of the ship's rod at t=11.5 years. In other words, your two experiments would not be simultaneous, but would be symmetrical.

And we have to notice that the conclusions you made (on my behalf), and that I agree with, are not contradictary at all. The events that define the ends of your two experiments are not simultaneous in either frame. Yes, the ship's twin will age less than Earth's twin during one of the experiments, and the Earth's twin will age less than the ship's twin during the other. You could not conclude that was contradictary unless you say the experiments are both symmetric and simultaneous. These two experiments would not occur simultaneously in any frame. And notice that the conclusions here are symmetrical.

And notice that I am not saying (in my experiment) that the event of t=5.77 on the ship clock is simultaneous with the event of t=11.5 years on the Earth clock. These clock readings represent how much each twin aged during the experiment, as observed by each twin. These clock readings do not represent represent moments in time that are simultaneous.

And I agree that both twins would have to come to relative rest to ever say one was older than the other at a particular time. Otherwise you could never define this particular time. But they would not have to come to rest to say that one aged more than the other during the experiment.

Thanks,
Alan

 

[RandallB]

They will both read the same time in the earth/star frame, of course. In this example T would be 11.55 years. But the two clocks will not read 11.55 years simultaneously in the traveller's own rest frame.

Well duh, that's the point of simultaneity – what I don’t get is why you refuse to or are unable to identify exactly WHERE and WHEN each of those two Earth-Star frame clocks can be seen in the traveler frame as 11.55.

No, because the traveller disagrees that at the moment he left earth, the star-clock read t=0 years. Again, in the traveller's frame the star-clock already read t = 8.66 years at the moment he left earth, and since it read 11.55 years at the moment he arrived at the star, he will say the star-clock only advanced forward by 11.55 - 8.66 = 2.89 years from the time he left Earth to the time he arrived, while his own clock advanced by 5.77 years between these events.

Balderdash, now that is gibberish and an unfounded arbitrary assumption within the stated problem. By what observation in this example did the traveler SEE the star-clock is at 8.6 the only clock to be seen in the earth-star frame by the traveler is the one there at Earth t=0 just like his t’=0. But what the traveler can do is send a light/radio signal back to all the other travelers behind him – stationary is his frame – to advise them of what t & t’ where when he passed (or Earth passed by him) both “0”.

None of this really helps me in understanding what you meant when you said "What is not simultaneous is the point and time Earth will observe locally going by in the traveling frame", though.

Let's see, how many different places can an Earth bound traveler see in a two reference frame problem? The Earth observer knows he is at location “0” and t= 11.55 in the Earth frame so what else is there to look at! HOW ABOUT THE TIME AND LOCATION OBSERVABLE IN THE OTHER FRAME – the traveler frame! At least put a clock over there, connected by a long measuring rod to the traveler if not another traveler – WHERE and WHEN is it. (Extended translation: What time does a traveler frame clock synchronized with the traveler within their frame read and how far away is it from the traveler as measured within the traveler frame viewable from Earth at Earth t=11.55). There is only room for one clock there, it must have a specific time and location in the traveler frame! So ALAN what is the time and location seen there? What conclusions will be forwarded up to the lead traveler based on same observation that can be seen from either frame?

 

[JesseM]

Well duh, that's the point of simultaneity – what I don’t get is why you refuse to or are unable to identify exactly WHERE and WHEN each of those two Earth-Star frame clocks can be seen in the traveler frame as 11.55.

What are you talking about? I didn't "refuse" to answer this question, you never asked it to me. The answer is that when the traveller's clock reads 11.55, since the traveller observes the Earth clock to be ticking at half the rate of his own clock, he will observe the earth-clock to read 5.77 years. And since the traveller observed the star-clock to start out at 8.66 years, and to be ticking at the same rate as the earth-clock, he will observe the star-clock to read 8.66 + 5.77 = 14.43 years.

No, because the traveller disagrees that at the moment he left earth, the star-clock read t=0 years. Again, in the traveller's frame the star-clock already read t = 8.66 years at the moment he left earth, and since it read 11.55 years at the moment he arrived at the star, he will say the star-clock only advanced forward by 11.55 - 8.66 = 2.89 years from the time he left Earth to the time he arrived, while his own clock advanced by 5.77 years between these events.

Balderdash, now that is gibberish and an unfounded arbitrary assumption within the stated problem. By what observation in this example did the traveler SEE the star-clock is at 8.6 the only clock to be seen in the earth-star frame by the traveler is the one there at Earth t=0 just like his t’=0.

I wasn't talking about what the traveller SEES if he looks around in his immediate vicinity--I try to avoid using the word "sees", instead using the word "observes" which in relativity is usually understood to just mean what is true about what's happening in that observer's rest frame (with 'frame' understood as a coordinate system filling all of space and time), not what he's actually seeing using light-signals...for example, the rate that I "see" a moving clock ticking is different from the rate I "observe" it ticking, because what I "see" is affected by doppler shift as well as time dilation. In this particular quote which you're responding to, I didn't use the word "see" or "observe", I just talked about what is true "in the traveller's frame" so there would be no ambiguity. And it is true that in the traveller's frame, the event of the star-clock reading 8.66 years happens at a time-coordinate of t' = 0 in terms of his frame's time-coordinate. Do you deny this? If so, try doing a Lorentz transformation starting from the coordinates of this event in the earth/star frame (ie x = 10 light years, t = 8.66 years).

But what the traveler can do is send a light/radio signal back to all the other travelers behind him – stationary is his frame – to advise them of what t & t’ where when he passed (or Earth passed by him) both “0”.

Well, likewise, if the traveler has a clock that is 5 light years ahead of him, at rest with respect to him and synchronized with his own clock according to the Einstein clock synchronization convention, then when that clock reads t'=0 years, it will be passing by the star and noting that the star's clock reads t=8.66 years. It could radio that observation to the traveller, and that's one way for the traveller to verify that the star-clock read 8.66 years at the moment he left in his own frame (but he could also figure that out without an actual physical clock in front of him, just by doing calculations based on when he observed the star-clock reading 8.66 years in his telescope, or based on backtracking from the time he observed when he passed the star clock and the rate he observed it ticking).

None of this really helps me in understanding what you meant when you said "What is not simultaneous is the point and time Earth will observe locally going by in the traveling frame", though.

Lets see, how many different places can an Earth bound traveler see in a two reference frame problem? The Earth observer knows he is at location “0” and t= 11.55 in the Earth frame so what else is there to look at! HOW ABOUT THE TIME AND LOCATION OBSERVABLE IN THE OTHER FRAME – the traveler frame! At least put a clock over there, connected by a long measuring rod to the traveler if not another traveler – WHERE and WHEN is it. (Extended translation: What time does a traveler frame clock synchronized with the traveler within their frame read and how far away is it from the traveler as measured within the traveler frame viewable from Earth at Earth t=11.55).

OK, I think I see what you're saying now, you're specifying that all statements about when things happen in a frame must be based on local measurements by clocks that are synchronized in that frame. If that's what you mean, it isn't obvious to me how that original sentence was supposed to imply that, but never mind. Anyway, in principle I agree all statements about times of distant events in a particular frame should be describable this way, although it isn't completely necessary since you can also figure out the times by calculation, based on backtracking from the time you received the light from an event.

As for the question of what a second clock synchronized with the traveller's will read when it passes by the Earth and the earth-clock reads t = 11.55 years, the answer is that it will read 23.09 years.

I am not really sure what the point of this discussion is--can you explain again what you were disagreeing with in the first post of mine that you responded to?

 

[JesseM]

I'm not talking about either one. Of course you can never compare the ages of people in different rest frames at any given point in time. Not in any objective way, as you point out. My #1 and #2 statements earlier were not comparing their ages at any given point in time. I was stating how much each twin aged during the experiment.

But to answer how much the Earth twin aged "during the experiment", you have to identify what point on the Earth twin's worldline corresponds to the end of the experiment, and you can't do that without picking a definition of simultaneity.

I think we may not be having any disagreement on substantive issues here, just over the meaning of the phrase "during the experiment". You seem to feel that the question "how much does the earth-twin age during the experiment" automatically implies we define the end of the experiment in terms of the definition of simultaneity in the earth-twin's own frame. I don't think so, I think it makes perfect sense to ask a question like "how much does the earth-twin age during the experiment in the traveller's frame"--would you say this question is incoherent or unnatural somehow?

You earlier seemed to make the argument that we should define things this way because "isn't that the definition of aging? How much time passes for a person in their own frame. Someone does't grow old according to time elapsed in a different frame. Someone grows old at the rate time elapses in their own frame. I'm talking about real aging here." I would basically agree with this, but for me the question of how much the earth-twin aged "during the experiment" is broken up into two steps:

1. Find the points on the earth-twin's worldline that correspond to the beginning and the end of the experiment

2. Find the proper time along the earth-twin's worldline between these two points

If the earth-twin was moving inertially, then #2 is equivalent to finding the amount of time between these two points in the earth-twin's own rest frame. So while I agree that you need to use the earth-twin's own frame to figure out how much he aged in #2, that doesn't imply that you also have to use the earth-twin's definition of simultaneity to figure out the point on his worldline that's "the end of the experiment" in #1. For example, if someone asked you "how much did the earth-twin age during the experiment in the traveller's frame", that means in #1 you'd use the traveller's definition of simultaneity to figure out which point on the earth-twin's worldline corresponds to the end of the experiment, but then in #2 you'd use the earth-twin's own frame to find out how much he aged between these two points. So your argument "Someone grows old at the rate time elapses in their own frame" is true but as long as we use the earth-twin's own frame in #2 I think it is being satisfied, regardless of whose frame we use to define "the end of the experiment" in #1.

It tells you a lot if we are talking about how much each twin aged during the experiment. It tells us nothing if we are trying to figure out how much each twin has aged at a particular moment in time, since that moment would not be simultaneous in both frames. Again, I am not claiming anything about how old either twin is at any particular time.

But how can you define "during the experiment" without specifying what moment in time on Earth corresponds to the end of the experiment?

As far as the star clock, the ship's twin would agree that it read t=0 when he left earth, since he was in the same frame when they were synchronized.

I thought you said earlier the traveling twin was just passing by the Earth at constant velocity on his way to the star, rather than accelerating from the earth. In post #33 you had said:

A spaceship plans to go to the nearest star system a distance of 10 light years away, at speed v = 0.866c. This spaceship first travels in the opposite direction from this star system, then turns around and passes Earth at speed v = 0.866c and both twins start their clocks at this time. Since they are not separated by any distance in the direction of travel, this will be t=0 for both twins. There will be no acceleration of the spaceship between Earth and the nearest star system.

In any case, a frame cannot accelerate, so I could rephrase that by saying "in the inertial frame where the traveller is at rest during his trip between Earth and the star, the star-clock read 8.66 years at the moment the earth-clock read 0 years".

Yes, that is what I'd say. Except the part about "each twin moving alongside the other twin's measuring rod at the same time". These two experiments would not end simultaneously in either frame.

I agree, I just meant that whichever frame you choose, there will be some duration of time between the event of the ship and Earth initially departing and the event of one of them reaching the end of the other one's measuring-rod, and for that duration they will both be moving alongside each other's measuring-rod "at the same time".

And we have to notice that the conclusions you made (on my behalf), and that I agree with, are not contradictary at all. The events that define the ends of your two experiments are not simultaneous in either frame. Yes, the ship's twin will age less than Earth's twin during one of the experiments, and the Earth's twin will age less than the ship's twin during the other. You could not conclude that was contradictary unless you say the experiments are both symmetric and simultaneous. These two experiments would not occur simultaneously in any frame. And notice that the conclusions here are symmetrical.

And notice that I am not saying (in my experiment) that the event of t=5.77 on the ship clock is simultaneous with the event of t=11.5 years on the Earth clock. These clock readings represent how much each twin aged during the experiment, as observed by each twin. These clock readings do not represent represent moments in time that are simultaneous.

And I agree that both twins would have to come to relative rest to ever say one was older than the other at a particular time. Otherwise you could never define this particular time. But they would not have to come to rest to say that one aged more than the other during the experiment.

OK, as long as you aren't claiming there's a single objective truth about who is aging more slowly in this type of experiment, then like I said I don't think we're really having any substantive disagreements over the physics of SR, just over semantic issues like what it means to ask how much the earth-twin aged "during the experiment".

 

[Al68]

Jesse,

First of all, you are right about the star-clock. There has been so much discussion of different experiments that I lost track of which one RandallB was talking about. He was talking about my example where the ship was already moving relative to earth. So that was my mistake.

Also, I agree that we are not disagreeing about any relevant facts about my example. We only disagree about how relevant each fact is.

Another way to point out the asymmetry in my experiment is this:

The ship's twin will say, "during the experiment, as measured in my frame, I aged 5.77 years and my brother aged 2.89 years."

The Earth twin will say, "during the experiment, as measured in my frame, I aged 11.5 years and my brother aged 5.77 years."

Each twin will agree that the other is telling the truth.

I think these statements incorporate your points and mine, and show that the experiment is still asymmetrical. We could add more true statements, and the asymmetry would still exist. My main point was that the experiment is asymmetrical, even without acceleration.

Do you agree with this?

Thanks,
Alan

 

[JesseM]

Jesse,

First of all, you are right about the star-clock. There has been so much discussion of different experiments that I lost track of which one RandallB was talking about. He was talking about my example where the ship was already moving relative to earth. So that was my mistake.

Also, I agree that we are not disagreeing about any relevant facts about my example. We only disagree about how relevant each fact is.

Another way to point out the asymmetry in my experiment is this:

The ship's twin will say, "during the experiment, as measured in my frame, I aged 5.77 years and my brother aged 2.89 years."

The Earth twin will say, "during the experiment, as measured in my frame, I aged 11.5 years and my brother aged 5.77 years."

Each twin will agree that the other is telling the truth.

I think these statements incorporate your points and mine, and show that the experiment is still asymmetrical. We could add more true statements, and the asymmetry would still exist. My main point was that the experiment is asymmetrical, even without acceleration.

Do you agree with this?

Thanks,
Alan

Yup, I agree it is asymmetrical in this sense. The asymmetry here can be understood as a consequence of the fact that the distance each sees the other move cannot be symmetrical, since the traveling twin sees the distance between the Earth and the star Lorentz-contracted. If instead each looked at the time for the Earth to pass by a buoy which was behind the ship and at rest relative to it, then the asymmetry would be in the opposite direction--the ship twin's value for how much he aged would be twice the Earth twin's value.

 

[Al68]

Yup, I agree it is asymmetrical in this sense. The asymmetry here can be understood as a consequence of the fact that the distance each sees the other move cannot be symmetrical, since the traveling twin sees the distance between the Earth and the star Lorentz-contracted. If instead each looked at the time for the Earth to pass by a buoy which was behind the ship and at rest relative to it, then the asymmetry would be in the opposite direction--the ship twin's value for how much he aged would be twice the Earth twin's value.

Thanks, Jesse. What you just stated was exactly what my initial point was. Maybe I should have worded it the way you did.

That being said, it's not even that important. If I had known it would take up 4 forum pages, I would have never brought it up.

I only wanted to make this point because this asymmetry is usually ignored in the Twins Paradox.

Thanks,
Alan

 

[RandallB]

The answer is that when the traveller's clock reads 11.55, since the traveller observes the Earth clock to be ticking at half the rate of his own clock, he will observe the earth-clock to read 5.77 years. And since the traveller observed the star-clock to start out at 8.66 years, and to be ticking at the same rate as the earth-clock, he will observe the star-clock to read 8.66 + 5.77 = 14.43 years.

I was specific in pointing the need for knowing WHERE and WHEN in the traveling frame for BOTH clocks in the Earth Star Frame. Again you only work the star clock. I’m only complaining about leaving out the easy details such as where and when in the traveling frame is the EARTH CLOCK at t = 11.55.

'Yada – Yada
Yada – Yada - Yada'
As for the question of what a second clock synchronized with the traveler’s will read when it passes by the Earth and the earth-clock reads t = 11.55 years, the answer is that it will read 23.09 years.

Well finally at least you identified the WHEN; but what about the WHERE in the traveler frame for this clock that reads 23.09 years. We know it’s at distance “0” in the Earth Frame.

Alan are you working these numbers, would this distance be the length of the measuring rod you were referring to earlier?

At least there is enough here to see that this means the Earth frame can clearly report that t’ = 23.09 years on the traveler clock near Earth is simultaneous with t’ = 5.77 years on the traveler clock near the star. OBVIOUSLY the traveling frame has screwed up synchronization – right?.

BUT Alan, work out the details on that rod length your were thinking about. Where is the Earth in relation to the traveler when t’= 5.77 years?? Once you know where the Earth is at you can have another observer in traveler frame there with a clock at t’=5.77 look over to the Earth to see WHEN that is in the Earth frame.
You shouldn’t be surprised to see that now it is the Earth frame that has lost touch with keeping things synchronized.
An “Asymmetrical disagreement” if you like.
BUT that is not the point.
And nether is which twin in the one way trip here is “really” younger or older!
Einstein’s point here is that what you perceive to be simultaneous between things separated by any distance, even just across the room from your is only within your own frame of reference and is not “real” and does not need to be “real”. IF you demand that some frame be a reference that is best described as a preferred frame of reference as in Lorentz Relativity "LR" not SR.

TO get a better ‘feel’ for these I recommend detailing out accurate “When” and “Where” information in every frame you use for everything used in an exsample. IMO using the ‘ability’ to SEE or OBSERVE clocks at great distances can to easily cause problems in understanding and doesn’t really make sense anyway. With any distance you must have a delayed observation, which is in effect local observer sending a report to you.

Keep working the details and it will get clearer.

 

[Al68]

Alan are you working these numbers, would this distance be the length of the measuring rod you were referring to earlier?

At least there is enough here to see that this means the Earth frame can clearly report that t’ = 23.09 years on the traveler clock near Earth is simultaneous with t’ = 5.77 years on the traveler clock near the star. OBVIOUSLY the traveling frame has screwed up synchronization – right?.

BUT Alan, work out the details on that rod length your were thinking about. Where is the Earth in relation to the traveler when t’= 5.77 years?? Once you know where the Earth is at you can have another observer in traveler frame there with a clock at t’=5.77 look over to the Earth to see WHEN that is in the Earth frame.
You shouldn’t be surprised to see that now it is the Earth frame that has lost touch with keeping things synchronized.
An “Asymmetrical disagreement” if you like.
BUT that is not the point.
And nether is which twin in the one way trip here is “really” younger or older!
Einstein’s point here is that what you perceive to be simultaneous between things separated by any distance, even just across the room from your is only within your own frame of reference and is not “real” and does not need to be “real”. IF you demand that some frame be a reference that is best described as a preferred frame of reference as in Lorentz Relativity "LR" not SR.

TO get a better ‘feel’ for these I recommend detailing out accurate “When” and “Where” information in every frame you use for everything used in an exsample. IMO using the ‘ability’ to SEE or OBSERVE clocks at great distances can to easily cause problems in understanding and doesn’t really make sense anyway. With any distance you must have a delayed observation, which is in effect local observer sending a report to you.

Keep working the details and it will get clearer.

Randall,

Although it looks like we've got my point resolved, you seem to be hinting at something else. Is there some other important point to be made here?

Just because I kept trying to redirect the discussion to my point doesn't mean that I'm not aware of other aspects of SR and the Twins Paradox that are more important.

I just didn't want to get sidetracked discussing points that have been addressed extensively in discussions of the Twins Paradox and SR. Are you referring to something else here?

Thanks,
Alan

 

[pervect]

I only wanted to make this point because this asymmetry is usually ignored in the Twins Paradox.

Alan

That's because there isn't any.

 

[Al68]

I only wanted to make this point because this asymmetry is usually ignored in the Twins Paradox.

Alan

That's because there isn't any.

pervect,

You do not think it's asymmetric that, in the Twins Paradox, one twin travels a longer distance than the other (as measureed in each twin's respective frame)?

Or do you just think that this point is irrelevant or unimportant?

Just curious.

Thanks,
Alan

 

[JesseM]

pervect,

You do not think it's asymmetric that, in the Twins Paradox, one twin travels a longer distance than the other (as measureed in each twin's respective frame)?

Or do you just think that this point is irrelevant or unimportant?

Just curious.

Thanks,
Alan

I think this point depends on what you and pervect mean by "asymmetry". Usual when physicists talk about symmetries they are talking about the laws of physics, and there is no asymmetry in how the laws of physics work in the two frames. But if you like you can say there is an "asymmetry" in the details of how this particular physical setup appears in the two frames; you don't even really have to get into the times to see this, you can just note that in one frame there's two parts of the system moving and the one in the middle at rest, while in the other frame two parts are at rest and one is moving between them (this 'asymmetry' would be present in the Newtonian version of the problem too).

 

[RandallB]

Randall
Although it looks like we've got my point resolved, you seem to be hinting at something else. Is there some other important point to be made here?
Are you referring to something else here?

My main concern was that IMO you were correct in looking for additional details related to the twins such as what amounted to the when and where the Earth was when the star saw the traveler arrive. When you were told that some of that information was arbitrary or unknowable I just disagreed and didn’t want you to be sidetracked from following a productive oath of your own choice. For any location and time in a reference frame there is one and only one time and place for it in another defined reference frame. There is no reason you cannot determine that kind helpful information exactly and there is nothing arbitrary in it.

As to “something else here” – that depends on where you want to go beyond the twins.
You can already see from looking at simultaneous events in the earth-star frame you find from complete SR detail it reveals t’= 5.77 and t’= 23.09 are. A paradox you can duplicate from the t’ frame when you looking at the earth-star t frame.
I don’t think your asymmetry point is the key.
Rather you might ask the question, can you trust your frame to correctly tell you if in reality two events separated by distance are simultaneous or not??

Based on what you’ve learned from the Twins, IMO I think you must say NO, and this was Einstein’s point.
Thus moving beyond the twins you might ask: ‘Can I define a "preferred frame" where I can determine if two separated events are REALLY simultaneous?’

Myself, I’m committed to SR and would say NO again, as it would take something very solid for me to reconsider LR “Lorentz Relativity” and the preferred reference frame used there.
So this could be “something else” for you to consider / work on / think about; beyond the twins, maybe open another post after you consider and research it a bit if you like.

You seem to have the twins issue fairly well in hand at this point,
Look on Twins/SR as a tool.
RB

 

[Al68]

RandallB,

OK, I thought you might be hinting at something else. I had other reasons to pursue my asymmetry point. Which is why I kind of ignored the other points presented. I just wanted to return to your points because it seemed like you were hinting at something else that hasn't been mentioned.

Thanks,
Alan

 

[Physical_Anarchist]

The reason why many people can't understand the Twins Paradox is simply because it doesn't make sense.
Never fear, I am here to cut through all the cr4p and tell it like it is.

No, I'm no physicist or anything, but even I can see that there's no logic to the explanations offered here or anywhere. Take this one for instance:
http://sciam.com/print_version.cfm?articleID=000BA7D8-2FB2-1E6D-A98A809EC5880105
It's clear from the word go that if you use your conclusion in the proof, you will at the end get the conclusion you were looking for. Let me sum up this especially pathetic attempt:
1.The star is 6 light-years away.
2.The trip takes 10 years (to the one staying at home).
3.The trip feels only like 8 years, because of length contraction.
4.Length contraction = Time Dilation
5.We're supposedly trying to prove time dilation!
6.Return trip, 10 years.
7.Again, feels like 8 years for some reason.
8. 8 + 8 = 16 < 20 = 10 + 10
9.Throw in useless Doppler shifts to confuse tired brains...
10.Conclude that you've proven your point.

Seriously, though, let's define the problem before attemping to solve it shall we?
I'm no expert, so feel free to correct me here but the root of the problem arises from some strange property of light: it passes you by at the constant speed of 299 792 458 m/s. Even when you travel at 100 000 000 m/s relative to your friend, the light passes him by at 299 792 458 m/s and it also passes you by at 299 792 458 m/s. The answer then is time dilation. It allows you to become slower even while you're traveling fast, so that you can see light pass you by at the same rate as before. Notice, however, that there is no acceleration in the problem, which means there shouldn't be in the paradox. Time dilation is a function of speed here, not acceleration.
How do we then define the Twin Paradox without needlessly confusing the issue with accelerations? There's a number of ways we can do that.
1.Suppose that the twins are both astronauts. They each embark on a spaceship. They accelerate at the same rate, for the same predetermined length of time. Afterwards, one of them immediately engages his thrusters in reverse, in order to decelerate, and then to accelerate in the opposite direction, and finally decelerate again in order to stop at the point of origin. Meanwhile, the second one has stopped accelerating, so he is cruising at a uniform velocity. One year later, he does the same decelarating and accelerating maneouvres his bro did earlier. One year later, on final approach, he decelerates the same way his twin did and also ends up at the point of origin. Now they are together again, and the only difference is that one has been static for 2 years, while his bro has been in motion at a constant rate of speed. SR tells us that Time Dilation did occur in this scenario and that consequently one of them will be younger than the other. I challenge anyone to prove such a thing happened. (Edit: I guess they could even have started out in opposite directions...)
2.Another way we can frame the question is by having an alien with an atomick clock, onboard a spaceship in motion coincidentally synchronise it with an atomic clock situated somewhere along his flighpath. I am not suggesting causality, so there shouldn't be a problem with the fact that the 2 atomic clocks were reset at the same exact instant. The question then becomes: when the alien gets a snapshot of the second clock while flying by it, will the clocks still be synchronised?

 

[pervect]

The reason why many people can't understand the Twins Paradox is simply because it doesn't make sense.
Never fear, I am here to cut through all the cr4p and tell it like it is.

No, I'm no physicist or anything, but even I can see that there's no logic to the explanations offered here or anywhere.

Take this one for instance:
http://sciam.com/print_version.cfm?articleID=000BA7D8-2FB2-1E6D-A98A809EC5880105
It's clear from the word go that if you use your conclusion in the proof, you will at the end get the conclusion you were looking for.

You should re-read the forum guidelines you signed, espeically the ones about "unfounded attacks on mainstream science will not be tolerated".

But a few things should be obvious even to a non-physicist like yourself.

1) The article does not attempt to "prove" relativity. The article only attempts to explain it.

2) The proof of relativity is not deductive. The proof of relativity is in how well it's theoretical results are confirmed by experiment.

 

[Physical_Anarchist]

What makes my "attack" unfounded? And how is it an attack on mainstream science? I can see it as an attack on lazy defining of a problem and an attack on equally lazy explanations of a theoretical phenomena.
I've also seen many posts from some users who "wish to argue Relativity's validity, or advertise their own personal theories", that didn't earn a warning about forum guidelines. What's so much worse about mine? Does it make too much sense? Scared? Go ahead and delete it if you must... If you want to stifle intelligent debate...

 

[JesseM]

I'm no expert, so feel free to correct me here but the root of the problem arises from some strange property of light: it passes you by at the constant speed of 299 792 458 m/s. Even when you travel at 100 000 000 m/s relative to your friend, the light passes him by at 299 792 458 m/s and it also passes you by at 299 792 458 m/s. The answer then is time dilation.

Not alone, no. The fact that light is measured to travel at the same speed by all observers is a consequence of how Einstein proposed that each observer should define their own coordinate system--using a network of rulers and clocks which are at rest with respect to themselves, with the clocks synchronized using the "Einstein clock synchronization convention", which is based on each observer assuming that light travels at the same speed in all directions in their coordinate system (so two clocks in an observer's system are defined to be synchronized if, when you set off a flash of light at their midpoint, they both read the same time at the moment the light from the flash reaches each one). The rationale for defining each observer's coordinate system this way becomes clear in retrospect, when you see that it is only when you define your coordinates this way that the laws of physics will be observed to work the same way in each observer's coordinate system (this is because the laws of physics have a property known as 'Lorentz-symmetry', the name based on the fact that the different coordinate systems described above will be related to each other by a set of equations known as the 'Lorentz transformation'). You're still free to define your coordinate systems in a different way, but then the laws of physics would take a different form in different observer's frames.

The Einstein clock synchronization is enough to insure that each observer will measure light to travel at a constant speed in all directions, as opposed to faster in some directions than others. But to explain why the magnitude of this constant speed is the same for different observers, you also have to know that each observer will measure the rulers of those moving at velocity v relative to him to shrink by a factor of $$\sqrt{1 - v^2/c^2}$$ and the ticks of clocks expand by a factor of $$1/\sqrt{1 - v^2/c^2}$$ (as long as the laws governing the rulers and the clocks have the property of Lorentz-symmetry, it's guaranteed this will happen). So, the fact that all observers measure the speed of light to be constant in all directions and the same from one observer's frame to another's is really a consequence of three things combined: time dilation, length contraction and "the relativity of simultaneity" (meaning that different observers will disagree whether a given pair of events happened 'at the same time' or not) which is a consequence of Einstein's clock synchronization convention. I posted a simple numerical example of how these three factors interact to insure a constant speed of light in this thread, if you're interested.

It allows you to become slower even while you're traveling fast, so that you can see light pass you by at the same rate as before. Notice, however, that there is no acceleration in the problem, which means there shouldn't be in the paradox. Time dilation is a function of speed here, not acceleration.

Yes, in an inertial frame time dilation is always a function of speed--if a clock is traveling at velocity v at a given moment, its rate of ticking will always be $$\sqrt{1 - v^2/c^2}$$ times the rate of ticking of clocks at rest in that frame at that moment.

How do we then define the Twin Paradox without needlessly confusing the issue with accelerations? There's a number of ways we can do that.
1.Suppose that the twins are both astronauts. They each embark on a spaceship. They accelerate at the same rate, for the same predetermined length of time.

It is usually convenient in statements of the twin paradox to just assume the acceleration period is instantaneously brief, so that the twin switches from one velocity to another instantaneously.

Afterwards, one of them immediately engages his thrusters in reverse, in order to decelerate, and then to accelerate in the opposite direction, and finally decelerate again in order to stop at the point of origin. Meanwhile, the second one has stopped accelerating, so he is cruising at a uniform velocity. One year later, he does the same decelarating and accelerating maneouvres his bro did earlier. One year later, on final approach, he decelerates the same way his twin did and also ends up at the point of origin. Now they are together again, and the only difference is that one has been static for 2 years, while his bro has been in motion at a constant rate of speed. SR tells us that Time Dilation did occur in this scenario and that consequently one of them will be younger than the other. I challenge anyone to prove such a thing happened.

Simple, just analyze the problem from the point of view of the inertial frame of the spot where they both departed and later reunited (we can assume it's the earth, say). In this frame, one twin spent only a brief time moving at high velocity (suppose he instantaneously accelerated to 0.8c moving away from the earth, then after 0.01 years instaneously accelerated to 0.8c moving back towards it, then after another 0.1 years he reached Earth again and instantaneously accelerated so he was at rest on earth), while the other spent a whole year moving at high velocity. The first twin's clock was only ticking slow in this frame during the time he was moving relative to the earth, while the other twin's clock was ticking slow during the entire year, so the second twin's clock will have elapsed less time. Using the Lorentz transform, we could analyze this same situation from the point of view of any other inertial frame, and we'd always get the same answer to what the two clocks read when they reunited--I could show you the math if you want.

2.Another way we can frame the question is by having an alien with an atomick clock, onboard a spaceship in motion coincidentally synchronise it with an atomic clock situated somewhere along his flighpath. I am not suggesting causality, so there shouldn't be a problem with the fact that the 2 atomic clocks were reset at the same exact instant. The question then becomes: when the alien gets a snapshot of the second clock while flying by it, will the clocks still be synchronised?

Just to be clear, are there 2 different clocks in the alien's flightpath, as well as a third atomic clock on the alien's ship? And you're saying the alien's clock reads the same time as the first clock in his path at the moment he passes it, and then you want to know what will happen as he passes the second clock in his path and compares it with his own clock? In this case the answer will depend on which frame the two clocks were synchronized, because again, the "relativity of simultaneity" means that different frames disagree on whether two events (such as two different clocks ticking 12 o clock) happened at the same time or different times. If the two clocks are at rest with respect to each other and synchronized in their own rest frame, then in the alien's rest frame the first clock he passes will be ahead of the second one by a constant amount, and this explains why, even though both clocks are running slower than his, his clock still reads a smaller time than that of the second clock he passes (in the clocks' own frame, this is because the alien's clock was running slow). Again, I could show you a numerical example to explain why both frames make the same prediction about what the clocks read at the moment they pass despite disagreeing about which clock was running slow and whether or not the two clocks in his path were synchronized.