Why Does the Ship's Twin Observe Earth's Clock Jump in the Twin Paradox?

[Al68]

In the standard "twins paradox", we have a turnaround 8 ly from earth, v = 0.8c. Omitting the math, the ship's twin observes Earth's clock to "jump time" from 3.6 yrs to 16.4 yrs (adjusting for the Doppler effect). OK, this makes sense, since the ship's twin (instantaneously) switched frames from one in which he left Earth 6 yrs ago to one in which he left Earth 27.33 yrs ago, as determined from his (current) frame.

If the ship's twin can "see" the Earth clock "jump time" during the instantaneous turnaround, why wouldn't he "see" his own clock "jump time" during the (instantaneous) turnaround?

Thanks,
Al

 

[matheinste]

Hello Al68

Quote:-

----why wouldn't he "see" his own clock "jump time" during the (instantaneous) turnaround?
----

I haven't thought about the first part of your question. But in answer to your last part, nobody ever sees anything different happen to their own clock. It just carries on as normal.

Matheinste.

 

[granpa]

well he never really 'sees' anything jump. due to the shift in simultaneity his 'calculation' of the Earth twins time will jump.

 

[Al68]

well he never really 'sees' anything jump. due to the shift in simultaneity his 'calculation' of the Earth twins time will jump.

Yeah, he "calculates" that since (in his return frame) he left Earth 27.33 yrs ago, that Earth's clock should read 16.4 yrs. (27.33 * 0.6).

But also, he could watch Earth's clock through a telescope and adjust for the Doppler effect. According to the textbook resolution, it would read 16.4 yrs just after the turnaround(in ship's frame).

Thanks,
Al

 

[Al68]

Hello Al68

Quote:-

----why wouldn't he "see" his own clock "jump time" during the (instantaneous) turnaround?
----

I haven't thought about the first part of your question. But in answer to your last part, nobody ever sees anything different happen to their own clock. It just carries on as normal.

Matheinste.

Would it be any more abnormal to see his own clock "jump" ahead than to see Earth's clock jump ahead?
After all, if it were real (non-instantaneous) acceleration, the ship's twin would calculate that less time passed on Earth than for him during the turnaround, as measured in the ship's frame.

 

[granpa]

But also, he could watch Earth's clock through a telescope and adjust for the Doppler effect.

that is what i was referring to.

 

[Al68]

But also, he could watch Earth's clock through a telescope and adjust for the Doppler effect.

that is what i was referring to.

OK, sure.
In most of the resolutions I've seen, they use the word "see" the Earth's clock "jump" time, or something similar. Of course, the jump isn't actually seen, but the clock (or signals) is actually seen, and, assuming the ship's twin is aware of the Doppler effect, he will know that what he saw was the Earth's clock instantaneously advance ahead during the (instantaneous) turnaround.

Thanks,
Al

 

[MeJennifer]

the clock (or signals) is actually seen

I know that some consider this nitpicking but I think this is essential in understanding relativity. We cannot directly observe the clock. The only thing we can observe are the light signals from the clock, and light signals cause "tricky" effects when they are exchanged between two objects which move with respect to each other or between two objects in a curved spacetime.

The only directly observable effect in relativity are the proper accumulated time differentials between objects that follow different paths in spacetime from one event to another.

 

[MeJennifer]

the clock (or signals) is actually seen

I know that some consider this nitpicking but I think this is essential in understanding relativivty. We cannot directly observe the clock. The only thing we can observe are the light signals from the clock, and light signals cause "tricky" effects when they are exchanged between two objects which move with respect to each other or which have a different gravitational potential.

The only directly observable effect apart from light signals are the proper accumulated time differentials between objects that follow different paths in spacetime from one event to another.

 

[Fredrik]

If the ship's twin can "see" the Earth clock "jump time" during the instantaneous turnaround, why wouldn't he "see" his own clock "jump time" during the (instantaneous) turnaround?

That's a pretty strange question. You're asking why one clock, at one event in space-time, isn't showing two different times in two different frames. How could it? It's just one event.

Consider a car that crashes into another car. That's one event. What you're asking is like asking why the cars didn't miss each other in another frame.

 

[Al68]

That's a pretty strange question. You're asking why one clock, at one event in space-time, isn't showing two different times in two different frames. How could it? It's just one event.

Yes, I know it's a strange question. Absurd, in fact. But we're assuming that the ship can just switch frames without accelerating. And the actual time since he left earth(in his new frame) has changed according to SR.
And the ship's twin "sees" the Earth clock show two different times corresponding to the two different times that the two different ship frames would judge as the time elapsed since the ship left earth. In the ship's new frame, at the time of the turnaround, the ship's twin left Earth 27.33 years ago, and the Earth's clock should read 16.4 yrs, accordingly. He can observe Earth's clock instantaneously "jump time", which is also strange, even though this fact is somewhat hidden by the Doppler effect.

 

[granpa]

Yeah, he "calculates" that since (in his return frame) he left Earth 27.33 yrs ago, that Earth's clock should read 16.4 yrs. (27.33 * 0.6).

why would he 'calculate' that? he has only to look at his clock to see how much time has passed.

 

[granpa]

If the ship's twin can "see" the Earth clock "jump time" during the instantaneous turnaround, why wouldn't he "see" his own clock "jump time" during the (instantaneous) turnaround?

the simple answer is because the shift in simultaneity that occurs during acceleration depends on how far away the object is. up close nothing much changes but farther and farther away it changes more and more.

 

[Fredrik]

Yes, I know it's a strange question. Absurd, in fact.

I don't think you do. What happens at an event has absolutely nothing to do with coordinates. Space-time is completely coordinate independent.

But we're assuming that the ship can just switch frames without accelerating.

It has to accelerate, but there's no upper bound on how fast it can accelerate, so we simplify by assuming that the acceleration takes (almost) no time at all. It's obvious from the space-time diagram that this doesn't change anything relevant.

And the actual time since he left earth(in his new frame) has changed according to SR.

Not the "actual time". The thing that changes is the difference between the time coordinates of the event where he left Earth and the event where he turned around. That difference obviously depends on what coordinate system we're using.

And the ship's twin "sees" the Earth clock show two different times corresponding to the two different times that the two different ship frames would judge as the time elapsed since the ship left earth.

No. The clock on Earth will show different times at the two events on Earth that are simultaneous with the turnaround event in the two frames.

He can observe Earth's clock instantaneously "jump time", which is also strange,

Not really. The clock must show different times at different events along its world line. If if didn't, it wouldn't be a clock. So the fact that it does can certainly not be as strange as it would be to have a clock show two different times at one event, which is impossible by definition.

 

[Al68]

OK, let's say that a buoy passes Earth at 0.8c (at rest with ship) at t = 10 yrs Earth time, t' = 6 yrs ship time. How far apart are the ship and buoy before the turnaround, as measured in each frame?

Thanks,
Al

 

[yuiop]

OK, let's say that a buoy passes Earth at 0.8c (at rest with ship) at t = 10 yrs Earth time, t' = 6 yrs ship time. How far apart are the ship and buoy before the turnaround, as measured in each frame?

Thanks,
Al

The Earth observer will say the bouy and ship are 8 light years apart while the ship observer will say the bouy and ship are 13.333 light years apart. Note that at the same time the ship observer will say Earth and ship are 4.2 light years apart.

 

[Fredrik]

OK, let's say that a buoy passes Earth at 0.8c (at rest with ship) at t = 10 yrs Earth time, t' = 6 yrs ship time. How far apart are the ship and buoy before the turnaround, as measured in each frame?

If I assume that you meant that "Earth at t=10 years" is an event on the buoy's world line, then I'm getting the same answers as kev: 8 light-years in Earth's frame and 13.333... light-years in the ship's frame.

Note that "Earth at t=10 years" is not the same as "Earth at t'=6 years". The latter is actually the same as "Earth at t=3.6 years".

Some details: The event on the buoy's world line that's simultaneous in the ship's frame with the turnaround event is located at (t,x)=(-7.777...,-14.222...). (Solve t=1/0.8x+10=0.8x+3.6 for t and x). A Lorentz transformation tells us that the coordinates in the ship's frame of this event are (t',x')=(6,-13.333...). The turnaround event has coordinates (t',x')=(6,0), so the distance is -13.333... light-years.

Kev also said that "at the same time the ship observer will say Earth and ship are 4.2 light years apart". I'm getting a different result. At the turnaround event, the coordinates of Earth in the ship's frame are (t',x')=(6,6*(-0.8))=(6,-4.8), so the distance between Earth and the buoy is -4.8-(-13.333...)=8.5333... light-years.

 

[Al68]

If I assume that you meant that "Earth at t=10 years" is an event on the buoy's world line, then I'm getting the same answers as kev: 8 light-years in Earth's frame and 13.333... light-years in the ship's frame.

Note that "Earth at t=10 years" is not the same as "Earth at t'=6 years". The latter is actually the same as "Earth at t=3.6 years".

Some details: The event on the buoy's world line that's simultaneous in the ship's frame with the turnaround event is located at (t,x)=(-7.777...,-14.222...). (Solve t=1/0.8x+10=0.8x+3.6 for t and x). A Lorentz transformation tells us that the coordinates in the ship's frame of this event are (t',x')=(6,-13.333...). The turnaround event has coordinates (t',x')=(6,0), so the distance is -13.333... light-years.

Kev also said that "at the same time the ship observer will say Earth and ship are 4.2 light years apart". I'm getting a different result. At the turnaround event, the coordinates of Earth in the ship's frame are (t',x')=(6,6*(-0.8))=(6,-4.8), so the distance between Earth and the buoy is -4.8-(-13.333...)=8.5333... light-years.

I think Kev meant 4.8 ly for the distance between the ship and Earth in the ship's frame. I have to assume, though, that since the buoy is at rest with the ship, the ship's twin and an observer on the buoy will consider the turnaround event to be simultaneous with the buoy passing Earth at t'=6 yrs. And, since the ship and buoy are at rest, their distance should be the same at any time in Earth's frame prior to the turnaround (in Earth's frame).
So how far apart are the ship and buoy in Earth's frame prior to the turnaround?

Thanks,
Al

 

[Fredrik]

Wait a minute. Does the buoy pass Earth at t=10 or t'=6? It's impossible to answer without knowing that. If you meant t'=6, kev and I both gave you the answer to the wrong question.

"Earth" represents a vertical line in the space-time diagram. "t=10" represents a horizontal line. "t'=6" represents a line with slope v (like one of the blue lines in my diagram). The Earth line intersects both the t=10 line and the t'=6 line, but not at the same event.

If you meant t=10, then we already gave you the answer. It's 8 light-years. This is very obvious. Earth is always at x=0, by definition of the unprimed coordinates, and the ship turns around at t=10, x=0.8*10=8, so the distance is 8-0=8.

 

[Al68]

Wait a minute. Does the buoy pass Earth at t=10 or t'=6? It's impossible to answer without knowing that. If you meant t'=6, kev and I both gave you the answer to the wrong question.

"Earth" represents a vertical line in the space-time diagram. "t=10" represents a horizontal line. "t'=6" represents a line with slope v (like one of the blue lines in my diagram). The Earth line intersects both the t=10 line and the t'=6 line, but not at the same event.

If you meant t=10, then we already gave you the answer. It's 8 light-years. This is very obvious. Earth is always at x=0, by definition of the unprimed coordinates, and the ship turns around at t=10, x=0.8*10=8, so the distance is 8-0=8.

I meant t'=6. The buoy passes Earth simultaneously (in the outbound ship/buoy frame) with the ship turnaround.

Thanks,
Al

 

[Fredrik]

OK, so the buoy passes Earth at t=3.6 years. At this time in Earth's frame, the ship is 3.6*0.8=2.88 light-years from Earth. Since the ship and the buoy move at the same velocity, the distance (in all frames) between them will be constant. So it's still 2.88 light-years in Earth's frame when the rocket turns around at t=10.

The distance in the ship's frame is 2.88/gamma=4.8 light-years.

 

[Al68]

OK, so the buoy passes Earth at t=3.6 years. At this time in Earth's frame, the ship is 3.6*0.8=2.88 light-years from Earth. Since the ship and the buoy move at the same velocity, the distance (in all frames) between them will be constant. So it's still 2.88 light-years in Earth's frame when the rocket turns around at t=10.

The distance in the ship's frame is 2.88/gamma=4.8 light-years.

OK, after the turnaround, in the ship's new frame, where is the buoy?

Thanks,
Al

 

[Al68]

OK, after the turnaround, in the ship's new frame, where is the buoy?

OK, I'll try an answer. The ship turnaround is at t'=6 yrs in ship/buoy frame, t = 3.6 yrs in Earth frame. Since the buoy is at Earth and can observe Earth's clock locally, the turnaround event occurs at 3.6 yrs in Earth frame, 2.88 ly away from earth. After the turnaround, buoy is still at Earth moving toward ship. Ship arrives at Earth at t' = 12 yrs on ship's clock, t = 7.2 yrs on Earth's clock. Earth twin is younger when they reunite. Earth twin will observe ship's clock to run slower the whole time, but "jump" ahead during the turnaround, since it was not simultaneous (earth's velocity relative to the ship changed). I won't bother drawing a spacetime diagram, just switch the ship and Earth around on the diagrams all over the net.

Oh, yeah, I didn't mention the space station in my "resolution". Oh, well, the textbook resolution doesn't mention a buoy, so that's OK.

And I didn't follow the rule that says we have to consider who "felt" acceleration during the turnaround. I guess that's OK, too, since that rule isn't in Einstein's original SR.

So, did I resolve this version of the "twins paradox", or am I missing something?

Thanks,
Al

 

[Fredrik]

OK, I'll try an answer. The ship turnaround is at t'=6 yrs in ship/buoy frame, t = 3.6 yrs in Earth frame.

You seem to have misunderstood how to do this (and maybe also what an event is). The turnaround is an event that has coordinates (10,8) in Earth's frame. The event (3.6,0) is the event where the buoy passes Earth. These two events are simultaneous in the ship's frame before the turnaround, but it's absolutely not true that...

the turnaround event occurs at 3.6 yrs in Earth frame, 2.88 ly away from earth.

The event (3.6,2.88) in Earth's frame is the event on the ship where the ship's clock shows 2.16 years.

There is only one turnaround event and it's at (10,8) in Earth's frame, at (6,0) in the ship's frame before the turnaround, and at (0,4.8) in the buoy's frame (if the buoy's clock is set to 0 when it passes Earth).

OK, after the turnaround, in the ship's new frame, where is the buoy?

The event on the buoy's world line that's simultaneous in the ship's new frame with the turnaround event is at (11.40,6.24) in Earth's frame. Let's call this event Z. If we take the turnaround event to be the origin of the ship's new coordinates, then event Z has coordinates (0,-1.05) in the ship's new frame.

It will be a good exercise for you to try to obtain these results, but don't start with the calculations. The first thing you should do is draw a space-time diagram (from Earth's point of view) so you can see that these results are reasonable. You should think really hard about how to draw simultaneity lines.

 

[Al68]

You seem to have misunderstood how to do this (and maybe also what an event is). The turnaround is an event that has coordinates (10,8) in Earth's frame. The event (3.6,0) is the event where the buoy passes Earth. These two events are simultaneous in the ship's frame before the turnaround, but it's absolutely not true that...
The event (3.6,2.88) in Earth's frame is the event on the ship where the ship's clock shows 2.16 years.

There is only one turnaround event and it's at (10,8) in Earth's frame, at (6,0) in the ship's frame before the turnaround, and at (0,4.8) in the buoy's frame (if the buoy's clock is set to 0 when it passes Earth).The event on the buoy's world line that's simultaneous in the ship's new frame with the turnaround event is at (11.40,6.24) in Earth's frame. Let's call this event Z. If we take the turnaround event to be the origin of the ship's new coordinates, then event Z has coordinates (0,-1.05) in the ship's new frame.

It will be a good exercise for you to try to obtain these results, but don't start with the calculations. The first thing you should do is draw a space-time diagram (from Earth's point of view) so you can see that these results are reasonable. You should think really hard about how to draw simultaneity lines.

Hi Fredrik,

Thanks for your response.

I only presented my "resolution" to show that the math will work my way just by re-defining the event at (6,4.8) in the ship's frame and (3.6,2.88) in Earth's frame, instead of "at the space station". This is defining a different event, which was my point. It's like saying the Earth switched frames instead of the ship. I only did it to show that the math will work either way, not because I don't understand how to do it the way you did. I understand how, I'm just not sure why to do it that way instead of the way in my "resolution". Also in my "resolution", there is no space station. The distance traveled to the turnaround point is defined as the distance between the ship and buoy.

In your results, the twin on the ship will never observe the buoy pass the ship, if I read it right. Will the buoy observe the ship pass it?

Thanks,
Al

 

[Fredrik]

In your results, the twin on the ship will never observe the buoy pass the ship, if I read it right. Will the buoy observe the ship pass it?

It was the position coordinate that was negative, not the time coordinate. It's obvious from a space-time diagram that they will meet. The equations of the two world lines in Earth's frame are

t=\frac 1 v x+3.6
t=-\frac 1 v x+20

(with v=0.8). So they will meet at (11.8,6.56) in Earth's frame.

 

[Al68]

It was the position coordinate that was negative, not the time coordinate. It's obvious from a space-time diagram that they will meet. The equations of the two world lines in Earth's frame are

t=\frac 1 v x+3.6
t=-\frac 1 v x+20

(with v=0.8). So they will meet at (11.8,6.56) in Earth's frame.

OK. I (wrongly) interpreted the negative position coordinate to mean behind the ship.

Well, yes, these results look reasonable, but they don't explain why we must choose Earth's frame to define the distance between the origin and turnaround point. Of course if we do, we must calculate everything the way you did.

Again my question is why do we have to do it that way, not how to do it that way. I don't know how else to word it. Why can't we choose to have a ship leave Earth and turnaround at (6,4.8) in the ship frame and (3.6, 2.88) in Earth's frame instead of the traditional twins paradox? Define the distance in the ship's frame and treat the Earth like it "switched frames" relative to the ship?

Thanks,
Al

 

[Fredrik]

Well, yes, these results look reasonable, but they don't explain why we must choose Earth's frame to define the distance between the origin and turnaround point.

In this problem, we're told to calculate a bunch of stuff related to a rocket that leaves Earth at v=0.8c and reverses its direction after 10 years in Earth's frame. The problem clearly specifies how to draw the world space-time diagram. The turnaround event is at (10,8).

If we instead are told to calculate a bunch of stuff related to a rocket that leaves Earth at v=0.8c and reverses its direction after 6 years in its own frame, that's the same problem. This specification of the problem will lead to the same space-time diagram.

I'm not quite sure what it is you're describing, but it seems to be a completely different problem at best (and possibly contradicting itself). You seem to be assuming not only that it's the Earth that changes velocity, but also that it goes on a much shorter trip than the ship in the original problem.

Again my question is why do we have to do it that way, not how to do it that way. I don't know how else to word it. Why can't we choose to have a ship leave Earth and turnaround at (6,4.8) in the ship frame and (3.6, 2.88) in Earth's frame instead of the traditional twins paradox? Define the distance in the ship's frame and treat the Earth like it "switched frames" relative to the ship?

If you understand what happens to simultaneity lines when an object changes its velocity, you should see that it matters which one of the objects involved changes its velocity. And if you understand time dilation, you should see that the length of the trip matters too.

 

[phyti]

McJennifer;

We cannot directly observe the clock. The only thing we can
observe are the light signals from the clock,

You have just defined 'perception' which is what is understood
when using the term 'observe'. All perception is indirect and
historical (after the fact/event). Give us something new.

Al68;

If the ship's twin can "see" the Earth clock "jump time" during
the instantaneous turnaround, why wouldn't he "see" his own
clock "jump time" during the (instantaneous) turnaround?

The 'time jump' is not something perceived, but a fictitious
change resulting from a change of (pseudo simultaneous)
reference frames.
As mathienste said, the observer detects no change of his clock
(because it isn't moving relative to him).

Fredrik makes a good point of using a space-time diagram to
sort things out.

The included image shows the space twin perceives 2 yrs of
earth events on the outbound leg, and 18 yrs on the inbound leg.
If we allow an instantaneous turnaround, then the ship clock
still shows 6 yrs. and would catch up on the return trip.

You can also eliminate the turnaround by using two ships which
cross at the 8 lyr destination and synchronize their clocks.

 

[Al68]

I'm not quite sure what it is you're describing, but it seems to be a completely different problem at best (and possibly contradicting itself). You seem to be assuming not only that it's the Earth that changes velocity, but also that it goes on a much shorter trip than the ship in the original problem.

Yes and yes.

If you understand what happens to simultaneity lines when an object changes its velocity, you should see that it matters which one of the objects involved changes its velocity. And if you understand time dilation, you should see that the length of the trip matters too.

Yes to both again.

If we consider the Earth to be the object that changes direction, won't we have to draw the ship's worldline as vertical and get completely different results?

Thanks,
Al

 

[Fredrik]

If we consider the Earth to be the object that changes direction, won't we have to draw the ship's worldline as vertical and get completely different results?

Yes. That was my point. (It doesn't have to be vertical because the slope of the line depends on what frame we're using, but it's a straight line in all inertial frames, and it's vertical in the ship's own frame, which would be the most convenient frame to use).

 

[matheinste]

Hello Al68.

Quote:-

---If we consider the Earth to be the object that changes direction, won't we have to draw the ship's worldline as vertical and get completely different results?--

Yes if the Earth changes direction, that is undergoes the acceleration and the ship does not, then the results will be different but then the scenario is different. In that case the ship's worldline will be a vertical straight line in a standard spacetime diagram.

Which body ghanges direction, in the basic stay at home and traveller case is the all important point. There are more complicated examples but it all comes down to the same thing.

In all such scenearios it all comes down to that the clock carried by the one who deviates most from a straight line on the spacetime diagram will show the shortest proper time. Proper time by definition being the time shown by a clock in which it remains at rest. Bear in mind that a clock is always at rest relative to itself moving inertially or not.

Matheinste.

 

[Al68]

Hello Al68.

Quote:-

---If we consider the Earth to be the object that changes direction, won't we have to draw the ship's worldline as vertical and get completely different results?--

Yes if the Earth changes direction, that is undergoes the acceleration and the ship does not, then the results will be different but then the scenario is different. In that case the ship's worldline will be a vertical straight line in a standard spacetime diagram.

Which body ghanges direction, in the basic stay at home and traveller case is the all important point. There are more complicated examples but it all comes down to the same thing.

In all such scenearios it all comes down to that the clock carried by the one who deviates most from a straight line on the spacetime diagram will show the shortest proper time. Proper time by definition being the time shown by a clock in which it remains at rest. Bear in mind that a clock is always at rest relative to itself moving inertially or not.

Matheinste.

That sounds like, although the "twins paradox" scenario is resolved as presented, all we have to do is redefine the scenario, define the turnaround point as a distance measured in the ship's frame, consider that the Earth accelerated relative to the ship, and have the Earth twin age less, since SR only cares about coordinate acceleration and not "proper" acceleration. What am I missing?

Thanks,
Al

 

[matheinste]

Hello Al68.

Quote:-

--consider that the Earth accelerated relative to the ship, and have the Earth twin age less,---

The Earth twin would age less if the Earth accelerated and the ship did not, but in these scenarios it is not usually the Earth that accelerates because this is not a very realisteic option and so would not help to pose the twins "paradox". As normally stated the twins "paradox" is meant to show a real difference in age from a realistic scenerio and present this as a paradox, which we all know can be resolved with SR. If we made the Earth in effect the "traveller" then we use an impractical and unrealistic, though not impossible scenario which lessens the effect of the "paradox" as usually given by making it wholly improbable anyway to the learner of SR at who it is aimed as an example for study.

Quote:--

--SR only cares about coordinate acceleration and not "proper" acceleration.---

The whole point here is that one accelerates and the other does not. I don't know how you define proper acceleration and coordinate acceleration, but what we care about in this scenario is absolute acceleration as detected by an accelerometer. Only one, the ship or the Earth experiences this in our present example. As normally proposed it is the ship which experiences the acceleration. If the Earth experiences it and the ship does not then,as you say, the result is reversed and the Earth twin ages less.

I don't think you are missing anything you may just have the wrong idea about acceleration. Acceleration is absolute and physical and coordinate independent.

Matheinste.

 

[Fredrik]

That sounds like, although the "twins paradox" scenario is resolved as presented, all we have to do is redefine the scenario, define the turnaround point as a distance measured in the ship's frame, consider that the Earth accelerated relative to the ship, and have the Earth twin age less, since SR only cares about coordinate acceleration and not "proper" acceleration. What am I missing?

I don't get why you think you can replace one problem with another that doesn't look anything like the original.

The original problem specifices three events and three frames. If you change any of that, it's a different problem.

 

[Al68]

Hello Al68.

Quote:-

--consider that the Earth accelerated relative to the ship, and have the Earth twin age less,---

The Earth twin would age less if the Earth accelerated and the ship did not, but in these scenarios it is not usually the Earth that accelerates because this is not a very realisteic option and so would not help to pose the twins "paradox". As normally stated the twins "paradox" is meant to show a real difference in age from a realistic scenerio and present this as a paradox, which we all know can be resolved with SR. If we made the Earth in effect the "traveller" then we use an impractical and unrealistic, though not impossible scenario which lessens the effect of the "paradox" as usually given by making it wholly improbable anyway to the learner of SR at who it is aimed as an example for study.

Quote:--

--SR only cares about coordinate acceleration and not "proper" acceleration.---

The whole point here is that one accelerates and the other does not. I don't know how you define proper acceleration and coordinate acceleration, but what we care about in this scenario is absolute acceleration as detected by an accelerometer. Only one, the ship or the Earth experiences this in our present example. As normally proposed it is the ship which experiences the acceleration. If the Earth experiences it and the ship does not then,as you say, the result is reversed and the Earth twin ages less.

I don't think you are missing anything you may just have the wrong idea about acceleration. Acceleration is absolute and physical and coordinate independent.

Matheinste.

By coordinate acceleration I mean change in relative velocity. By proper acceleration I mean as measured by an accelerometer, like you're talking about. But the biggest thing I see that "causes" the ship's twin to age less in the twins paradox is the simple fact that he didn't travel as far relative to Earth as the Earth twin did relative to the ship, each as measured in his own frame. Simple common sense tells me that at 0.8c, a shorter trip equals less elapsed time (t=d/v in any frame). The resolution's conclusion just follows this stipulation. It doesn't resolve the big picture "clock paradox" for scenarios which may be different. Some of the resolutions say that acceleration is the key to the problem, but they claim this as an axiom without showing why this is true. After all, the Earth does accelerate (change velocity) relative to the ship. By definition coordinate acceleration equals change in relative velocity per unit time.

Just as a side note, most on this board are probably aware that Einstein believed the "clock paradox" was unresolvable in SR. He was fully aware of the (now) common resolutions and rejected them. Any thoughts on why he believed this?

Thanks,
Al

 

[Al68]

I don't get why you think you can replace one problem with another that doesn't look anything like the original.

The original problem specifices three events and three frames. If you change any of that, it's a different problem.

Well, because I'm interested in the big picture "clock paradox", not just a single scenario whose conclusion only applies to narrowly defined conditions.

And I'm curious why Einstein thought it was unresolvable in SR, even after being fully aware of the now common resolutions. Since the reason probably isn't that Einstein "didn't understand SR", "didn't understand how simultaneity works", etc.

I'm sure that when Einstein presented the "clock paradox" and said that the ship's twin should be able to claim that the Earth twin aged less, he obviously didn't mean by using the same specified three events and three frames.

Thanks,
Al

 

[matheinste]

Hello Al68.

Quote:-

---Some of the resolutions say that acceleration is the key to the problem, but they claim this as an axiom without showing why this is true----

To start with this point. To get back to the smaller picture let us talk about the role of acceleration in the standard twin paradox. This has been explained very well in many threads in this forum. I may be able to rephrase a few things but better people than me seem not have been able to make it clear to you so I don’t hold out much hope but I will try.

Take a ship and the Earth at rest relative to each other. Now in all that follows the words ship and Earth wherever they appear in the text can be interchanged. Because they are not in relative motion with regards each other let us for our purposes consider them at rest. In a standard spacetime diagram we can represent their common worldline as a vertical line, corresponding to the time axis.

Let the ship moves off in any direction, for our convenience we will show this displacement as being along the horizontal axis in our spacetime diagram. It does not matter whether to the left or to the right. The worldline of the ship will therefore be a straight line let us say to the right, and its angle upwards to the horizontal will be proportional to its velocity which may be anything less than c.
After some time the ship halts and heads back to the left at some speed making again an tilted upward straight line this time to the right to the left until it reaches the Earth again. We will assume that the initial acceleration, the deceleration at the turnaround point, the acceleration back towards Earth and the final deceleration at Earth to be very high for a very short period, this of course being only a thought experiment. We do this to make the time spent in the acceleration phases as short as possible, in theory infinitesimally small. It has in fact no effect on the outcome.

In spacetime the greater the spacetime distance traversed, and on a spacetime diagram the longer the line(s) representing the path taken, the SHORTER the proper time experienced by anything traveling along this path. So the earth, traveling in a straight line, travels the shortest distance and therefore its clocks record the LONGEST POSSIBLE time. This is contrary to common sense but is accounted for by the time dimension being involved in the calculation. I would show the mathematics but I am not yet proficient in latex. Remember spactime distance is not the same thing as spatial distance. The actual speeds and distances are immaterial, the principle is simply that the longest spacetime diagram path has the shortest proper time, that is shows less ageing. Proper time is of course what the ship and Earth experience themselves.

Anyhow to make the path deviate from the vertical axis of the spacetime diagram requires a change of velocity, that is an acceleration. Every time the spacetime path changes direction, as it must in this scenario, an acceleration is involved. It is the change in path that causes the difference in proper times and for this change in path an acceleration is needed. The acceleration does not cause the clock differences but is needed to alter the spacetime path.

The point of turnaround of the ship can be anywhere but in the scenario we have chosen it is somewhere to the right on the spacetime diagram to the right of the vertical worldline of the earth.

There have been authors who have claimed that GR is necessary to resolve the paradox but it is generally accepted that GR is not needed. As to Einstein, what can I say, I don’t know the facts.

This has all been posted before and anyway it is the best I can do and only hope it is accurate. If anybody spots any mistakes would they please point them out and we can put them right rather than let Al think we are in disagreement over the point.

Matheinste.

 

[Al68]

Hello Al68.

Quote:-

---Some of the resolutions say that acceleration is the key to the problem, but they claim this as an axiom without showing why this is true----

To start with this point. To get back to the smaller picture let us talk about the role of acceleration in the standard twin paradox. This has been explained very well in many threads in this forum. I may be able to rephrase a few things but better people than me seem not have been able to make it clear to you so I don’t hold out much hope but I will try.
..
Let the ship moves off in any direction, for our convenience we will show this displacement as being along the horizontal axis in our spacetime diagram. It does not matter whether to the left or to the right. The worldline of the ship will therefore be a straight line let us say to the right, and its angle upwards to the horizontal will be proportional to its velocity which may be anything less than c.

Thanks for your reply. Your post is very clear, and everything you said is clear to me. It's what everyone is omitting that I'm asking about. Why must we attribute the deviation from the initial common worldline to the observer who underwent proper acceleration? Why can't we just as rightly (according to SR) attribute this deviation to the other observer and draw the Earth's worldline to the right when it's velocity relative to the ship changes (coordinate acceleration)?

Why can't we just randomly choose which observer to draw vertically on the spacetime diagram? I know that sounds like a silly question, but that very question is part of the reason Einstein pursued GR.

Maybe the answer is so intuitively obvious that nobody considers it necessary to mention, but when I ignore my intuition and use only Einstein's 1905 SR paper, I can't find the answer. (And neither could he).

I understand what happens after we decide who's worldline to draw vertically, that's just simple math. But it's that decision that seems, for lack of a better word, Newtonian.

I mentioned Mach's principle in an earlier post, but nothing came of it. I realize that when the ship fires it's thrusters, Earth's velocity changes relative only to the ship, while the ship's velocity changes relative to earth, and relative to every other single body in the universe. This is obviously not considered important by anyone, since nobody brought it up. Is it relevant?

There have been authors who have claimed that GR is necessary to resolve the paradox but it is generally accepted that GR is not needed. As to Einstein, what can I say, I don’t know the facts.

Well, he was one of those authors, although most consider his GR resolution to be flawed.

Thanks,
Al

 

[Mentz114]

Al68,

Why can't we just randomly choose which observer to draw vertically on the spacetime diagram? I know that sounds like a silly question, but that very question is part of the reason Einstein pursued GR.

Because one of them has a curved worldline which cannot be transformed into a straight vertical line by Lorentz transformation.

It sort of follows that to handle force-free acceleration, you need to curve the axes.

M

 

[Al68]

Al68,

Why can't we just randomly choose which observer to draw vertically on the spacetime diagram? I know that sounds like a silly question, but that very question is part of the reason Einstein pursued GR.

Because one of them has a curved worldline which cannot be transformed into a straight vertical line by Lorentz transformation.

I'm asking about the decision that led to us to consider that one of them has a curved worldline, not how we treat the problem afterward.

I don't know how else to put it.

Thanks,
Al

 

[matheinste]

Hello again Al68.

Quote:-

----Why must we attribute the deviation from the initial common worldline to the observer who underwent proper acceleration? Why can't we just as rightly (according to SR) attribute this deviation to the other observer and draw the Earth's worldline to the right when it's velocity relative to the ship changes (coordinate acceleration)?

Why can't we just randomly choose which observer to draw vertically on the spacetime diagram? I know that sounds like a silly question, but that very question is part of the reason Einstein pursued GR.----

A vertical worldline in a spacetime diagram represents a body in inertial motion, for our purposes this is the same as being at rest. If a body accelerates it undergoes a spatial displacement with changing time which would have to be represented by a sloping line to the left or right in the spacetime diagram.

I do not know for sure, this being a quick answer, but I suppose with some corresponding alterations to the slope of the spatial displacement axis, that is horizontal axis, and a corresponding alteration to the other objects worldline, some sort of modified spacetime diagram may be possible in which an accelerated observer may be given a vertical worldline. I would be interested in any answers to this which may be forthcoming. But why complicate things. The outcome remains the same as far as time differentials are concerned.

As for Mach’s principle, I think there would be no difference I the posing of or resolution of the paradox if nothing else existed in the universe. Acceleration would still be absolute.

I wrote this while before Mentz114 posted his last reply. He says all that needs to be said on that point and i suspect that what he has said about the Lorentz transformation answers my question about a modified spacetime diagram. Not possible ??

Mateinste.

 

[matheinste]

Hello Al68

In answer to the curved worldline—

A spacetime diagram plots distance or displacement against time. For constant velocity this plot or graph is a straight line. For accelerated motion it is not.

Matheinste.

 

[Fredrik]

Well, because I'm interested in the big picture "clock paradox", not just a single scenario whose conclusion only applies to narrowly defined conditions.

It's instructive to consider a scenario defined by three time-like straight lines chosen at random (except that no two of them are parallel), where we imagine physical observers traveling on those world lines and comparing their clocks at the events where two lines meet. This scenario contains everything that's relevant from the standard twin "paradox". (Note that there's no acceleration).

And I'm curious why Einstein thought it was unresolvable in SR, even after being fully aware of the now common resolutions. Since the reason probably isn't that Einstein "didn't understand SR", "didn't understand how simultaneity works", etc.

A person who understands SR would never believe that this is unresolvable in SR, and I do believe that Einstein understood SR.

I'm sure that when Einstein presented the "clock paradox" and said that the ship's twin should be able to claim that the Earth twin aged less, he obviously didn't mean by using the same specified three events and three frames.

I don't know how he presented it (or even that he did), so I can only assume that if he said anything like that, he must have meant that a very naive application of the time dilation formulas which completely ignores any other effects due to relativity of simultaneity leads to the result that the Earth twin aged less (and also that the Earth twin aged more).

 

[Fredrik]

Why must we attribute the deviation from the initial common worldline to the observer who underwent proper acceleration? Why can't we just as rightly (according to SR) attribute this deviation to the other observer and draw the Earth's worldline to the right when it's velocity relative to the ship changes (coordinate acceleration)?

Why can't we just randomly choose which observer to draw vertically on the spacetime diagram? I know that sounds like a silly question, but that very question is part of the reason Einstein pursued GR.

I think you're missing an important thing here. We're not trying to find the best possible theory to handle this scenario. We're just trying to find out what special relativity says about it. And it's clear from any formulation of special relativity that straight lines have a very special significance.

Maybe the answer is so intuitively obvious that nobody considers it necessary to mention,

It's not obvious that there are no better theories, but we're not looking for a better theory. The twin paradox is about finding the mistake in an incorrect application of the rules of SR.

I understand what happens after we decide who's worldline to draw vertically, that's just simple math. But it's that decision that seems, for lack of a better word, Newtonian.

"Minkowskian", or "special relativistic" would be pretty good ways to say it.

I mentioned Mach's principle in an earlier post, but nothing came of it. I realize that when the ship fires it's thrusters, Earth's velocity changes relative only to the ship, while the ship's velocity changes relative to earth, and relative to every other single body in the universe. This is obviously not considered important by anyone, since nobody brought it up. Is it relevant?

Mach's principle may be important to someone who knows SR and is trying to find a theory that includes gravity. Such a researcher might decide early on to only consider theories that satisfy some version of Mach's principle and throw away all other theories without further consideration. This is a lot like deciding to only consider theories where coordinate transformations between inertial frames with a common origin preserve the light-cone at the origin, when trying to find SR. (Only not as powerful, because light-cone preservation is almost the entire theory of SR).

 

[Mentz114]

Al68,

I'm asking about the decision that led to us to consider that one of them has a curved worldline, not how we treat the problem afterward.

I don't know how else to put it.

I thought I'd answered that. We didn't choose which one to treat inertially, and which one non-inertially.The twin who travels non-inertially nominates themselves. Acceleration is absolute.

M

[edit] This is what Matheinste and Fredrik are saying also, I think.

 

[Al68]

Hello Al68

In answer to the curved worldline—

A spacetime diagram plots distance or displacement against time. For constant velocity this plot or graph is a straight line. For accelerated motion it is not.

Matheinste.

Why not?

 

[Al68]

A person who understands SR would never believe that this is unresolvable in SR, and I do believe that Einstein understood SR.

Einstein did consider it unresolvable in SR. And said so, and tried to resolve it in GR.

 

[matheinste]

Herllo Al68.

The plot or graph of constant velocity against time is a straight line. For accelerated motion it is not.

You ask why not!

This is basic mathematics and physics of motion. If you do not know this you really should learn it as it is at a very basic level and if you do not understand this you have no chance of understanding anything in physics involving motion.

Matheinste.

 

[Al68]

Al68,


I thought I'd answered that. We didn't choose which one to treat inertially, and which one non-inertially.The twin who travels non-inertially nominates themselves. Acceleration is absolute.

M

[edit] This is what Matheinste and Fredrik are saying also, I think.

Sure proper acceleration is absolute, but coordinate acceleration is not.

And I know we didn't choose which twin travels non-inertially, but we did choose to treat the inertial frame differently.