Twin Paradox Problem: Do Twins Age Differently?

[stevendaryl]

Dalespam, not trying to challenge your deeper knowledge in relativity, but I do think there is something to discuss here.

Agreed there are inertial frames in SR, but the whole point in SR is that none of them are 'preferred'.

Any kinemtic acceleration will have to be completely mutual between two frames. This means that any and all time dilation will be mutually equal, and no experiment should be able to establish actual measurable velocity time dilation between two bodies moving with a relative velocity w.r.t each other, since the relative velocities are also mutually equal.

This is clearly not the case in experiments, as actual velocity time dilation does provenly exist.

The prediction of SR is that the elapsed time on an ideal clock is given by the formula

$\tau = \int{\sqrt{1-(v/c)^2} dt}$

where $dt$ and $v$ are measured in any inertial coordinate system. That formula clearly predicts that in the situation in which two clocks start at the same spot, depart, and reunite, the clock moves inertially will show the greater elapsed time.

The difference in the ages of the two twins is a prediction of SR; it's the most basic prediction.

 

[arindamsinha]

...the clock [that] moves inertially will show the greater elapsed time

My question was - how in a kinematic framework like SR do we decide which clock moves inertially?

The difference in the ages of the two twins is a prediction of SR; it's the most basic prediction.

Yes, this I understand. Point is, what determines which twin is travelling, given a purely kinematic framework like SR?

 

[Nugatory]

My question was - how in a kinematic framework like SR do we decide which clock moves inertially?

Strap an accelerometer onto the side of the clock. In SR, as long as it always reads zero, we're moving inertially. Or we could plot the path of the clock in space-time, if that path is a straight line the clock is moving inertially.

It's more interesting in GR. There's a moderately entertaining variant of the twin paradox in which the traveling clock is turned around by doing a tight hyperbolic orbit around some massive object, thereby staying in free fall for the entire trip. There's still no ambiguity about what each clock should read at journey's end, but AFAIK the only general way of calculating those times is to do the line integral of $d\tau$ along the paths.

 

[arindamsinha]

Strap an accelerometer onto the side of the clock. In SR, as long as it always reads zero, we're moving inertially. Or we could plot the path of the clock in space-time, if that path is a straight line the clock is moving inertially.

Nugatory, you've lost me.

Trying to guess - if an accelerometer (whatever that is) is strapped to the traveling twin's clock, it will not read zero at certain times - and therefore that twin will be considered not to be in an inertial frame? Is this what you are trying to say?

 

[Nugatory]

Nugatory, you've lost me.

Trying to guess - if an accelerometer (whatever that is) is strapped to the traveling twin's clock, it will not read zero at certain times - and therefore that twin will be considered not to be in an inertial frame? Is this what you are trying to say?

Yes. An accelerometer is a device that measures (non-gravitational) accelerations, and it will read a positive value when the traveller's spaceship takes off, zero while the ship is coasting outbound, a negative value while the ship is reversing directions at turnaround, zero while the ship is coasting inbound, and a positive value when the returning ship slows to rejoin the stay-at-home twin.

 

[arindamsinha]

That was my point Nugatory - when I said that a kinematic solution like SR cannot resolve this issue.

When we say 'the traveller's spaceship takes off', we are referring to a clear preferred frame (presumably Earth). We cannot equally say in reverse 'the Earth-bound twin's planet takes off', recording that phenomenon on an accelerometer strapped to the Earth-bound twin.

Dynamics gets involved here, providing a reason why only one of them is considered accelerating, and that takes it outside of the domain of SR which is based on kinematics alone.

Thanks for helping bring out clearly the point I am trying to make.

 

[Austin0]

Quote by Austin0

You are the one who apparently thinks it is telling you something significant.

I don't know what gave you that impression, but you're completely off track here.

Well i may well be off track but how should I interpret the following if you don't think it is significant and that distant simultaneity has actual temporal meaning??

At any given time, the traveling twin has an instantaneous inertial reference frame. In each of those frames, there is a different notion of "the current age of the stay-at-home twin". Changing frames means changing your notion of what events are simultaneous, which changing your notion of what the "current" age of the distant twin is. That's not an error, it's just a fact.

Quote by Austin0

Here you are attributing factual reality to the simultaneity as indicated by a hypersurface of clock synchronicity.

I didn't do any such thing. What I said was that the noninertial twin cannot use the time dilation formula to compute the age of the stay-at-home twin, unless he takes into account the jumps due to changes of frames. The spacetime diagram explains why.

As i read this it implies that you think that if he does take into account the jumps indicated by changed planes of simultaneity then he can meaningfully compute the age of the inertial twin.
It also implies that you are attaching temporal meaning to the calculated clock readings at distant locations ,no??
What else do you mean if not that?

my thought on planes of simultaneity in Minkowski diagrams is that they are perfectly meaningful and useful for telling the expected times of relative system clocks at distant locations when dealing with inertial frames.
But as soon as acceleration is included at any point this no longer holds.
This is of course most obvious where the planes intersect but I think it is also raises question of their correspondence to reality everywhere. As this case in point where a sudden non-realistic jump in times is indicated.
So you want to attribute meaning to the areas where they radically diverge but disregard the areas where they converge??
In any case do you really think that equal proper clock readings at distant locations means actual simultaneity?

 

[Rishavutkarsh]

Yes, it is, if you are referring to your comment, "the important thing was that the traveling one was noninertial and the other observer was inertial". It is neither important, relevant, significant, nor is it a requirement for the Twin Paradox.

In the simplest presentation of the Twin Paradox, we talk about the Earth twin as if the Earth had no gravity and no acceleration, which are of course not true and so the Earth and the Earth twin are considered to be inertial for the purpose of discussing the scenario. In this simplest presentation the other twin is non-inertial and so without knowing anything else, we can always say that the traveling twin is the one who ages less because he experienced acceleration whereas the Earth twin did not. This leads many people to falsely jump to the conclusion that it is the acceleration that causes the difference in aging between the twins and they look for explanations (a non-inertial frame or jumping between two frames) that support that incorrect idea. These people tend to reject the simple explanation that I offered in post #42 and quoted in post #52.

But we can complicate the Twin Paradox by having both twins follow exactly the same accelerations. They can both take off in identical spaceships and achieve 90%c. Then one of them immediately turns around and lands back on Earth for the rest of the time while the other twin continues on for a long time before repeating what the first twin did. So now both twins are non-inertial in exactly the same way so we can't jump to the false conclusion that the acceleration is what caused the difference in aging. But we can still analyze this more complicated Twin Paradox using the same inertial Earth reference frame as before. It's the simplest one to use, mainly because it is the one that is used to describe their motions.

In fact, it doesn't matter how the two twins accelerate or what speeds they achieve or what directions they travel in (polygons or circles or some of each) or where they end up together. We can always analyze their individual aging during the entire process from the standpoint of the inertial frame that we use to describe their activities.

Furthermore, with a little more work, we can determine what each twin sees of the other ones clock during the entire scenario, not just at the beginning and the end, and any analysis we do (whatever frame(s) we use) will all provide the same answers. We can make this as complicated as we want. But the more complicated analyses will not provide any more insight or information into what is happening.

so you mean that in fact acceleration has nothing to do with this and it's just used to simplify this. so would you be kind enough to tell us what exactly is seen by the both twins and why? and is acceleration absolute or relative?

 

[Nugatory]

Dynamics gets involved here, providing a reason why only one of them is considered accelerating, and that takes it outside of the domain of SR which is based on kinematics alone.

Thanks for helping bring out clearly the point I am trying to make.

Ah... How does this take it out of the domain of special relativity? SR works just fine for accelerating frames as long as the space-time is flat.

 

[D H]

Strap an accelerometer onto the side of the clock.

Careful there! Our stay at home twin is staying at home, presumably on the surface of the Earth. An accelerometer strapped to the side of an Earth-bound clock will say that the clock is always accelerating upward at 1g.

Since we're talking about special relativity, it would be better to remove gravitation from the picture. Place our stay at home twin in a space station that is orbiting the Sun somewhere beyond Pluto's orbit. The stay at home twin is still subject to the Sun's gravitation, but this is such a tiny, tiny effect that it can be completely ignored. Now you can say "strap an accelerometer onto the side of the clock".

Nugatory, you've lost me.

Trying to guess - if an accelerometer (whatever that is) is strapped to the traveling twin's clock, it will not read zero at certain times - and therefore that twin will be considered not to be in an inertial frame? Is this what you are trying to say?

An accelerometer is the ideal device for detecting whether an object is moving inertially. Accelerometers measure proper acceleration. If you have a Wii, or a gamepad, or a smart phone, you have an accelerometer. That's how those systems know you are moving the controller. They are also used to detect vibrations in equipment and in the Earth, and to sense acceleration in aircraft, missiles, and spacecraft .

 

[arindamsinha]

Ah... How does this take it out of the domain of special relativity? SR works just fine for accelerating frames as long as the space-time is flat.

I am referring to the 'reciprocality' of all observations in SR as a kinematic theory. Reciprocality is not conserved if we somehow ascribe a unilateral acceleration to one body but not the other in a two-body situation, unless we are defining one of them to be a preferred frame of observation (which is contrary to SR).

This is why I state it takes it out of the domain of SR. GR does seem to resolve this nicely, bringing mass and gravity (i.e. momentum conservation) into the picture, providing a tie-breaker on 'who is moving'.

This is a good discussion. Let me know your thoughts.

 

[Nugatory]

Careful there! Our stay at home twin is staying at home, presumably on the surface of the Earth. An accelerometer strapped to the side of an Earth-bound clock will say that the clock is always accelerating upward at 1g.

Since we're talking about special relativity, it would be better to remove gravitation from the picture. Place our stay at home twin in a space station that is orbiting the Sun somewhere beyond Pluto's orbit. The stay at home twin is still subject to the Sun's gravitation, but this is such a tiny, tiny effect that it can be completely ignored. Now you can say "strap an accelerometer onto the side of the clock".

By "the clock" I meant the traveling twin's clock, and I was hoping no one would catch the oversimplification with respect to the stay-at-home twin. Well, better luck next time :-)

But seriously, kidding aside, yes you are right. It's remarkable how many presentations of the twin paradox describe stay-at-home as "unaccelerated" yet leave him stuck on the surface of the Earth living his entire life subject to a constant 1g acceleration.

 

[D H]

It's remarkable how many presentations of the twin paradox describe stay-at-home as "unaccelerated" yet leave him stuck on the surface of the Earth living his entire life subject to a constant 1g acceleration.

In those presentation's favor, that 1g acceleration (better: being deep in the Sun's and the Earth's gravity well) is a tiny effect. It's about 20 seconds of time dilation compared to that inertial counterpart in deep space for a forty year round trip.

 

[Rishavutkarsh]

So can anyone tell if acceleration is not absolute what exactly causes the difference if ageing ?

 

[harrylin]

My question was - how in a kinematic framework like SR do we decide which clock moves inertially? [..]

SR is not a kinematic framework. It's similar to classical mechanics in which the derivation of the transformation of motion of an object wrt a platform that is rotating relative to another one is purely kinematics. Dynamics comes in when one identifies for example the one platform as being in uniform straight motion, in which case the laws of mechanics are valid for motion wrt that platform and not for motion wrt the other platform. The same is true for SR.

 

[harrylin]

Strap an accelerometer onto the side of the clock. In SR, as long as it always reads zero, we're moving inertially. [..]

No, that's wrong for exactly the reason that you mention next. See post #104: in the original example, the acceleration that breaks the symmetry can not be detected by a simple accelerometer.

It's more interesting in GR. There's a moderately entertaining variant of the twin paradox in which the traveling clock is turned around by doing a tight hyperbolic orbit around some massive object, thereby staying in free fall for the entire trip. There's still no ambiguity about what each clock should read at journey's end, but AFAIK the only general way of calculating those times is to do the line integral of $d\tau$ along the paths.

Although SR does not account for gravitational time dilation, its effect is probably negligible for that example.

 

[pervect]

So can anyone tell if acceleration is not absolute what exactly causes the difference if ageing ?

The shortest distance in space between two points is a straight line. In flat space, there is only one straight line between two points. On a (large) curved surface, there can be several geodesics (the curved space equivalent of a straight line) connecting two points of unequal length connecting the same two points, so only one of the lines is the shortest distance. So a given sraight line between two points isn't necessarily the shortest, but the shortest curve is a straight line.

Similarly, the worldline with the longest proper time between two points in space time is a geodesic - the natural motion of a particle experiencing no forces other than inertial and gravitational forces.. Similar to the situation in curved space, there can be several worldlines of different proper time connecting two points in space time, only one of which is the longest.

In both cases (space and space-time) if you make the region of space-time "small enough", the multiple paths will not exist there will be only one geodesic or

It's just geometry - and the twin "paradox" is rather similar to the triangle inquality in Eulidean geometry. The velocity change is rather like the angle

It's a natural feature of nature that a clock moving under the influence of no external forces moves along a geodesic, which is a path which extermizes proper time. This is the relativistic princple of Hamilton's principle.

 

[stevendaryl]

As i read this it implies that you think that if he does take into account the jumps indicated by changed planes of simultaneity then he can meaningfully compute the age of the inertial twin.

I don't know what you mean by "meaningfully" here, but at any moment, the traveling twin is at rest in some momentary inertial reference frame. According to that frame, the stay-at-home twin has a certain age. So there is a computable function F(t) giving the age of the stay-at-home twin in the frame in which the traveling twin is at rest at time t, according to the traveling twin's clock. The function is dependent on the acceleration profile of the traveling twin, and changes discontinuously at the points where the traveling twin's velocity changes discontinuously.

I didn't make any claims about how "meaningful" this function is--what makes a mathematical function meaningful or not meaningful? But when the twins get back together, the relationship between the age A_earth of the stay-at-home twin and the age A_traveler of the traveling twin will be given by
A_earth = F(A_traveler)

 

[harrylin]

So can anyone tell if acceleration is not absolute what exactly causes the difference if ageing ?

That opinion was held by Einstein and he gave his answer in his paper on the twin paradox, see post #104.

However that solution is unpopular today, as mentioned here:
http://math.ucr.edu/home/baez/physics/Relativity/SR/TwinParadox/twin_gr.html

As mentioned in post #127, the best known answers are "stationary ether" and "block universe"; but the first is, again, unpopular today, and the second remains debated. See for example https://www.physicsforums.com/showthread.php?t=583606. (Note the reasons for locking that thread, at the end!)

Perhaps most physicists use the "shut up and calculate" solution.

 

[stevendaryl]

So can anyone tell if acceleration is not absolute what exactly causes the difference if ageing ?

Here's an analogy that might help, or might not. In Euclidean geometry, you hop in a car in New York City and travel for 1000 miles (according to your car's distance meter). Where (at what point in space) do you end up? It depends on the direction you traveled.

In Minkowsky geometry, you hop in a spaceship and travel for 10 years (according to your spaceship's clock). Where (at what point in spacetime) do you end up? It depends on the velocity you traveled. You're traveling in both space and time. What spatial location you end up at after 10 years of travel, and ALSO what temporal coordinate you end up at after 10 years of travel depend on the path you took through spacetime.

 

[Rishavutkarsh]

The shortest distance in space between two points is a straight line. In flat space, there is only one straight line between two points. On a (large) curved surface, there can be several geodesics (the curved space equivalent of a straight line) connecting two points of unequal length connecting the same two points, so only one of the lines is the shortest distance. So a given sraight line between two points isn't necessarily the shortest, but the shortest curve is a straight line.

Similarly, the worldline with the longest proper time between two points in space time is a geodesic - the natural motion of a particle experiencing no forces other than inertial and gravitational forces.. Similar to the situation in curved space, there can be several worldlines of different proper time connecting two points in space time, only one of which is the longest.

In both cases (space and space-time) if you make the region of space-time "small enough", the multiple paths will not exist there will be only one geodesic or

It's just geometry - and the twin "paradox" is rather similar to the triangle inquality in Eulidean geometry. The velocity change is rather like the angle

It's a natural feature of nature that a clock moving under the influence of no external forces moves along a geodesic, which is a path which extermizes proper time. This is the relativistic princple of Hamilton's principle.

I do understand what you are trying to explain but how can we differentiate between the paths took by the twins (tell that which one will age faster) as speed is relative and acceleration has nothing to do with this. we can say that for the moving twin the stationary twin is moving with the same velocity so when they meet ie come at same point in spacetime
how can this be said that the traveling will be younger and he took the shorter path.

What determines whether the path took by anything will be longer or shorter?

 

[stevendaryl]

What determines whether the path took by anything will be longer or shorter?

The proper time for any path through spacetime in SR is given by

$\tau = \int{\sqrt{1-(v/c)^2} dt}$

where $v$ and $t$ are measured in any inertial coordinate system. While $v$ and $t$ are both relative to a coordinate system, the integral has the same value in every coordinate system. It's clear that if there is a coordinate system in which $v=0$ for some path, then that path will have the longest proper time.

This is very closely analogous to path lengths in Euclidean geometry. The length of a path through 2-D Euclidean space is given by

$L = \int{\sqrt{1+m^2} dx}$

where $m$ (the slope of the path, relative to the x-axis) and $x$ are relative to a coordinate system, but the integral has the same value in any coordinate system (well, any system in which $m$ is well-defined; it diverges for vertical paths). It's clear that if there is a coordinate system in which $m=0$ for some path, then that path will have the shortest length.

In SR, acceleration (change of $v$) doesn't come into play in calculating proper time, but it is provable that the unaccelerated path has the longest proper time. In Euclidean geometry, bending (change of $m$) doesn't come into play in calculating path length, but it is provable that the unbent path has the shortest path length.

 

[Rishavutkarsh]

The proper time for any path through spacetime in SR is given by

$\tau = \int{\sqrt{1-(v/c)^2} dt}$

where $v$ and $t$ are measured in any inertial coordinate system. While $v$ and $t$ are both relative to a coordinate system, the integral has the same value in every coordinate system. It's clear that if there is a coordinate system in which $v=0$ for some path, then that path will have the longest proper time.

This is very closely analogous to path lengths in Euclidean geometry. The length of a path through 2-D Euclidean space is given by

$L = \int{\sqrt{1+m^2} dx}$

where $m$ (the slope of the path, relative to the x-axis) and $x$ are relative to a coordinate system, but the integral has the same value in any coordinate system (well, any system in which $m$ is well-defined; it diverges for vertical paths). It's clear that if there is a coordinate system in which $m=0$ for some path, then that path will have the shortest length.

In SR, acceleration (change of $v$) doesn't come into play in calculating proper time, but it is provable that the unaccelerated path has the longest proper time. In Euclidean geometry, bending (change of $m$) doesn't come into play in calculating path length, but it is provable that the unbent path has the shortest path length.

how can you say that the path took by the moving twin is the shorter one?

 

[stevendaryl]

how can you say that the path took by the moving twin is the shorter one?

From the formula for proper time:
$\tau = \int{\sqrt{1-(v/c)^2} dt}$

There is no inertial coordinate system in which the velocity of the traveling twin is always $v=0$, but there is an inertial coordinate system in which the velocity of the stay-at-home twin is always $v=0$.

 

[phyti]

2 clocks:
It comes down to which clock loses the least amount of ticks.
Clocks A and B move at speed a and b respectivlely.
A and B part,and B returns to A.
If the B clock runs slower than A on the outbound segment, it must take an inbound segment at -a>v<a, avoid losing more ticks.
Since there is no speed by which a clock can gain time, the loss is permanent, and the gain on the inbound segment does not compensate for the loss on the other.
Time dilation is a function of speed (v/c). Acceleration only sets the rate of td by establishing a different speed.
The twin case with one returning is the simplest of all cases, and would require a period of acceleration. That makes one path so different from the other that it would seem to be the explanation.


jumping frames:
The time 'gap' results from eliminating the time required for deceleration and acceleration, when simplifying the problem.

 

[Dale]

Agreed there are inertial frames in SR, but the whole point in SR is that none of them are 'preferred'.

Any kinemtic acceleration will have to be completely mutual between two frames.

Look carefully at what you wrote. "None of them are preferred", where "them" refers to different inertial frames. If there is any kinematic acceleration between the frames then at least one of the frames is non inertial, so "none of them are preferred" does not even apply. There is no need for anything to be mutual or symmetric between an inertial and a non inertial frame, nor between two non inertial frames.

SR postulates the equivalence of all inertial frames, not the equivalence of all possible frames.

 

[arindamsinha]

... If there is any kinematic acceleration between the frames then at least one of the frames is non inertial, so "none of them are preferred" does not even apply. ...

Excellent point. I think we should agree that SR is not the right framework to explain why there is asymmetric measurable time dilation between two bodies (like GPS and Earth surface clocks). We need to look at GR for this explanation.

PS: In the above, I am referring to the velocity time dilation part only, not the gravitational time dilation.

 

[Rishavutkarsh]

2 clocks:
It comes down to which clock loses the least amount of ticks.
Clocks A and B move at speed a and b respectivlely.
A and B part,and B returns to A.
If the B clock runs slower than A on the outbound segment, it must take an inbound segment at -a>v<a, avoid losing more ticks.
Since there is no speed by which a clock can gain time, the loss is permanent, and the gain on the inbound segment does not compensate for the loss on the other.
Time dilation is a function of speed (v/c). Acceleration only sets the rate of td by establishing a different speed.
The twin case with one returning is the simplest of all cases, and would require a period of acceleration. That makes one path so different from the other that it would seem to be the explanation.


jumping frames:
The time 'gap' results from eliminating the time required for deceleration and acceleration, when simplifying the problem.

as said earlier what would you say when both twins go through same acceleration?
both start at the same time when they both reach 90%C one returns back while other keeps going , how will you describe this?
they both went through same acceleration and retardation but still the moving one is younger {when they meet after 10 years (say)} . this is the point i am confused about.
Any help would be greatly appreciated.

 

[pervect]

I do understand what you are trying to explain but how can we differentiate between the paths took by the twins (tell that which one will age faster) as speed is relative and acceleration has nothing to do with this. we can say that for the moving twin the stationary twin is moving with the same velocity so when they meet ie come at same point in spacetime
how can this be said that the traveling will be younger and he took the shorter path.

What determines whether the path took by anything will be longer or shorter?

Short answer: the geometry determines it.

The spatial analogy is that the shortest path between two points is a straight line. If you imagine a triangle, the sum of the lengths of the two sides of the triangle is always greater than or equal to the hypotenouse, and it's only equal when the "triangle" is degenerate.

The twin paradox is just the space-time version of the triangle inequality.

If you are in the flat space-time of special relativity , there will be one and only one path between any two points in that space-time that is straight line motion. We'll call the two points the "origin" point and the "destination" point. Straight line motion means pretty much the same thing in flat space-time as it does in Newtonian theory as you have true inertial frames.

The two points must be fixed in both space and time, and there must be enough time for light to reach and pass the destinaton point from the first for a material body to be able to do the same. The technical term for such a path is a "timelike separation" between points, to insure that a "timelike path" exists between them. For the rest of this post I'll presuppose that such a condition is satisfied.

Thus if you specify the both the origin point and the destination point, the geometry determines the unique timelike path that is also a geodesic that connects them. This path will be the path of maximum proper time. It will be represented by a body that moves in "natural motion", or "uniform motion". You can also think of this as a body that is at rest in some inertial frame, the one inertial frame that contains both the origin and destination points at the origin point of the inertial frame.

If you are in the curved space-time of general relativity, there may be more than one path if you consider a large enough time interval. But let's consider the case first where the spacing between points is close, and there's only one path.

Then the path that experiences the most proper time will be the unique path of a body in free fall - free fall paths determine geodesic motion - that starts at the origin point and ends at the destination point.

Example: If you consider a point on the surface of the earth, and another point at the same location 1 second later, the path of maximum proper time will be that of an object thrown upwards, such that it begins at the origin point and ends at the destination point.

This path will be unique up until you get a separation in time between the oriigin and destination points of one orbital period of the Earth (which is about 90 minutes, IIRC), at which points you'll have many possible paths you can take. The orbital paths will generally have short proper times and hence not be what you're looking for. It's the path that gets you furthest away from the Earth in the shortest possible time that will be the path of maximum proper time. If you look at the equations , and choose a particular coordinate system, you can talk about this path as a balancing act between gravitational time dilation which is minimzed by getting far away, and velocity time dilation, which is minimzed by not moving quickly. But it's more helpful (though more abstract) to think of the path as just being determined by the space-time geometry.

In any event, it's the geometry that determines the path of longest proper time in space-time, just as it's geometry that defines the shortest path in space.

And if you take some path that's less than optimum, in space you travel a longer distance, in space-time (assuming a purely timelike path) you travel a shorter time.

There's a little more to be said about the case where you have multiple paths in space (or in space-time), but the basics are that globally you have to try all possible geodesic paths and test them to find which is the longest / shortest. The path you're looking for will be a geodesic, but any particular geodesic may or may not be the one that you want.

 

[Jeronimus]

From the formula for proper time:
$\tau = \int{\sqrt{1-(v/c)^2} dt}$

There is no inertial coordinate system in which the velocity of the traveling twin is always $v=0$, but there is an inertial coordinate system in which the velocity of the stay-at-home twin is always $v=0$.

The answer is more complicated than this.

There is no inertial reference frame for the first and second phase either, in which the velocity of the traveling twin is always v=0.
There is an inertial reference frame however with an observer at rest who considers both rockets to be flying away of him at the same vrel seen from his point of view, after the instantaneous acceleration, being in phase2.

Any of the two twins could decide to turn around with the same results. Whoever turns will be the twin who aged less. (So there seems to be some symmetry BETWEEN the two twins up until phase 3, but it will not be easy to explain why)

The symmetry breaks in the turn around phase 3. Not at the initial acceleration phase 1, nor at phase 2 when they move at vrel relative to each other, nor at phase4 when the turn around twin accelerates instantaneous back into the inertial reference frame the stay at home twin is at rest with.
In fact, in phase 4 (when the twins are next to each other at vrel), either of the twins could decide to accelerate instantaneous into the inertial reference frame his twin brother is at rest with. It would not matter. The twin who accelerated at a distance (turn around) is the one who aged less.

I tried to warp my mind around this, but in the end, it gets too complicated. One of the issues is that we are talking about a symmetry between the two twins, not between two frames. The symmetry seems to be broken with any object which is not local to the acceleration. (and yet, it might not be broken, considering the next sentence)
Also, while the accelerating twin might claim that everything else is accelerating, we are not talking just accelerating along the x-axis anymore, but also accelerating along the t axis.
Let alone if we assume a real, non-instantaneous acceleration, then we can talk only about getting arbitrary close to clocks being synced after the acceleration in phase1 & 4 still. We can however choose such a high acceleration, that the effect becomes negligible small.edit: When we talk about symmetry, we usually mean along the x axis. They move at vrel relative to each other. That is along the x axis. But what if they move at vrel to each other along the t axis? This is where i get stuck for the moment.

 

[stevendaryl]

The answer is more complicated than this.

There is no inertial reference frame for the first and second phase either, in which the velocity of the traveling twin is always v=0.
There is an inertial reference frame however with an observer at rest who considers both rockets to be flying away of him at the same vrel seen from his point of view, after the instantaneous acceleration, being in phase2.

Yes, but the integral for proper time is an invariant; it has the same value in every inertial coordinate system. So if there exists one coordinate system in which one twin has $v=0$ throughout, then we can use that coordinate system to compute the integral.

 

[ghwellsjr]

As long as we ignore gravity or pretend that it does not exist, you can define and analyze any scenario in a single Inertial Reference Frame (IRF). This is the easiest way to solve any Special Relativity problem. All you have to do is keep track of the tick rate of each clock based on its speed in that IRF. It doesn't matter if a clock is at rest in that IRF or if it accelerates, it doesn't matter.

Stevendaryl's oft-repeated equation for calculating the Proper Time on any clock may look intimidating but it is very simple as long as we invoke instant accelerations whenever we want a clock to change its speed. Then we can chop up the activity of any clock into intervals when it is moving at a constant speed and simply keep track of how long and how fast each clock is moving in our single IRF.

So in the simple Twin Paradox where one twin remains at rest in the IRF and the other one travels the entire time between departing and reunion at a constant speed (but with instantaneous accelerations to start, change directions and stop) then we use the equation to see that the tick rate on the traveling clock is slower than the coordinate time of the IRF.

In the scenario of this thread since the traveling twin moves at 90%c, the time dilation factor is 0.435890 so the traveling twin's clock runs slower than the coordinate time (which is the same as the other twin's clock rate) by that amount. If the traveling twin was gone for 10 years according to the IRF, then his clock will accumulate 4.35890 years during the trip while the other twin's clock accumulated 10 years.

If you want to propose a more complicated scenario, for example, both twins take off at the same speed (90%c) but one of them gets back after 5 years (IRF) while the other one gets back after 10 years (but otherwise they both experience the same acceleration, just at different times), then the time on the first clock will accumulate 5 times 0.435890 or 2.17945 years during the first 5 years of IRF time plus 5 more years for the second 5 years of IRF for a total of 7.17945 accumulated time on the first clock compared to 4.35890 years for the second twin's clock.

You can do a similarly simple process for any complicated scenario with instant accelerations. If you want to have accelerations that are gradual then you will have to do actual integrations but otherwise the concepts are identical.

Now, if you want, you can transform any scenario defined in one IRF into any other IRF moving with respect to the first one to see how you get the same answer even though all the speeds and intervals may be different. And if you really want to make things complicated, you can engage in frame jumping or non-inertial reference frames but it won't change any result, they just use different speeds and intervals to do the calculations and arrive at the same final result. I don't know why anyone would want to torture them self like that but if you're one of those people, be my guest, just don't claim that any other way than the first and easiest IRF provides any more insight or information into what is actually happening.

 

[ghwellsjr]

By the way, in my previous post I emphasized that the final result comes out the same in all IRF's and other analyses because that is usually the only result that people care about when discussing the Twin Paradox but I want to make it clear that all intermediate results come out the same also. For example, in any of these scenarios, we can consider each tick of every clock to be a different event and calculate the Proper Time on each clock for each of these events and then transform to any other IRF and calculate the Proper Time again and they will all come out the same for each event. In other words, every event that determines the Proper Time on any clock is invariant (in any IRF or any other kind of analysis).

In particular, we can calculate the Proper Time on the traveling twin's clock when he turns around half way through his trip and all IRF's will give the same result. Furthermore, the Doppler analysis that I promoted in post #5 which does not depend on any IRF or other frame(s) also determines the time on the traveler's clock to be the same as any IRF analysis and we can also show that the stationary twin will see that time on the traveler's clock when he finally sees his twin turn around.

What no analysis can do is determine unambiguously what time is on the stationary twin's clock when the traveling twin turns around because that depends on the selected IRF (or other frame) but they will all agree on the time on the stationary twin's clock when he observes the traveling twin turn around.

 

[phyti]

as said earlier what would you say when both twins go through same acceleration?
both start at the same time when they both reach 90%C one returns back while other keeps going , how will you describe this?
they both went through same acceleration and retardation but still the moving one is younger {when they meet after 10 years (say)} . this is the point i am confused about.
Any help would be greatly appreciated.

In the left pic A's speed profile with an acceleration at (1) is the reverse order of B's speed profile with a deceleration at (2). The profiles are symmetric to the center of the line connecting the end points (0)& (3).
The path lengths are equal, their clocks read the same at (0) and at (3). The acceleration/deceleration did not determine the accumulated time.
The right pic has the same path for A, but a stretched path for B. Since the B path is longer, the B clock accumulates less time than the A clock. The acceleration/deceleration did not determine the accumulated time.

Examples of the 'twin' scenario without
acceleration/deceleration, exchanging info while passing, have been used to show the effect of a longer path on a clock. Einstein said in moving the B clock away from the A clock along a random path, then returning to the A clock, the B clock would show less time. He didn't state how it was moved because it's irrelevant. The point was the motion and its effect on processes (clock).

In the simple twin case with A leaving B, then returning, the change in direction is the most obvious difference, and thus is assumed to play apart in the solution.
That's the danger of making conclusions based on special cases.

 

[phyti]

Here's another approach.
View attachment clock rejoin.doc