Stupider-er Twins Question

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  • #1
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
 

Answers and Replies

  • #2
matheinste
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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.
 
  • #3
granpa
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well he never really 'sees' anything jump. due to the shift in simultaneity his 'calculation' of the Earth twins time will jump.
 
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  • #4
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
 
  • #5
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.
 
  • #6
granpa
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But also, he could watch Earth's clock through a telescope and adjust for the Doppler effect.

that is what i was referring to.
 
  • #7
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
 
  • #8
MeJennifer
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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.
 
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  • #9
MeJennifer
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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.
 
  • #10
Fredrik
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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.
 
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  • #11
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.
 
  • #12
granpa
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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.
 
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  • #13
granpa
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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.
 
  • #14
Fredrik
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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.
 
  • #15
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
 
  • #16
yuiop
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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.
 
  • #17
Fredrik
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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.
 
  • #18
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
 
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  • #19
Fredrik
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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.
 
  • #20
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
 
  • #21
Fredrik
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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.
 
  • #22
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
 
  • #23
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
 
  • #24
Fredrik
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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.
 
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  • #25
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
 
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  • #26
Fredrik
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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

[tex]t=\frac 1 v x+3.6[/tex]
[tex]t=-\frac 1 v x+20[/tex]

(with v=0.8). So they will meet at (11.8,6.56) in Earth's frame.
 
  • #27
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

[tex]t=\frac 1 v x+3.6[/tex]
[tex]t=-\frac 1 v x+20[/tex]

(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
 
  • #28
Fredrik
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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.
 
  • #29
phyti
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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.
 

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  • #30
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
 
  • #31
Fredrik
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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).
 
  • #32
matheinste
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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.
 
  • #33
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
 
  • #34
matheinste
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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.
 
  • #35
Fredrik
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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.
 

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