How Does the Twin Paradox Challenge Our Understanding of Time Dilation?

[newTonn]

Hai ,I am new to the forum.
I have some basic questions regarding the same old twin paradox.
Twin A stays at Earth and Twin B is the traveller.
from the A,s frame we all agree that Time is dilated for B because it is moving(relative to A) near to the speed of light.
Now switch on to B,s frame.from B,s frame we have to agree that A is moving(relative to B-in opposite direction).So obviously,A,s time also should be dilated as per SR.
Do this dilation cancell each other ,and they will be of same age when they meet again?.
This is a thought experiment only.So i am expecting some logical answers (not mathematical)which can help me to get out of my ignorance.

 

[Tzemach]

No only the twin who is actually moving will age slower, an easy way to imagine this is that the square of your velocity in space plus the square of your velocity in time will equal the square of the speed of light.

I know you didn't want maths but this is very basic.

Time velocity2 + Spatial velocity2 = C2

Think of it a bit like a triangle where the sum of the hypotenuse is equal to the sum of the squares of the other two sides, each twin has a different shaped triangle, they still follow the same equation that rules the universe.

 

[newTonn]

No only the twin who is actually moving will age slower, an easy way to imagine this is that the square of your velocity in space plus the square of your velocity in time will equal the square of the speed of light.

I know you didn't want maths but this is very basic.

Time velocity2 + Spatial velocity2 = C2

Think of it a bit like a triangle where the sum of the hypotenuse is equal to the sum of the squares of the other two sides, each twin has a different shaped triangle, they still follow the same equation that rules the universe.

I am really sorry.what did you mean by actually moving.Do you mean some energy should be involved(like the thrust of rocket) for time to be dilated?.or is it simply (as name of theory suggests-Special relativity),as i understand,only because of relative motion.

 

[lufbrajames]

I am really sorry.what did you mean by actually moving.

I think they will both experience some time dilation, because they are both moving relative to space-time.

when Tzemach said

No only the twin who is actually moving will age slower

he was referring to the one that was moving the fasted relative to space-time.

Im not cocmpletely sure, but that's how i understand it.


- James
- Loughborough Uni

 

[cristo]

The twin paradox is discussed in all good SR texts. There's a non-mathematical explanation on john baez's home page that you might want to check out: http://math.ucr.edu/home/baez/physics/Relativity/SR/TwinParadox/twin_paradox.html

 

[adriansd]

Hey guys!

For an object that is in MOTION, as it approaches the speed of light, it causes time to run slower for the object, compared to a stationary object. If the moving object were a person, they would age slower and their stationary identical twin would age faster.

This can prove difficult, but you need to distinguish between A (from original post) moving RELATIVE to B, and Time Dilation of A relative to B due to the motion of B.

Although B is actually in motion, From B's reference frame, you can say that A is moving relative to B. NewTonn appers to understand this adequately.

Because of B's motion near the speed of light, time runs slower for B relative to A, where A is 'normal time'. This is taking A's refrence frame. For our twins, B ages slower and so twin A is older when twin B returns.

Taking B's reference frame, now it has 'normal' time, and time can be considered to run faster for A. For our twins then, A ages faster, and so A is still older than B when twin B returns... nothing to cancel out.

Gawsh I hope I didn't confuse you more.

Cheers

 

[jambaugh]

The key points to concentrate on when parsing the Twin paradox is that the "when" at which one is comparing their ages changes as you consider the two frames of motion. You compare their ages "at the same time" of course. But now in SR "at the same time" i.e. simultaneity changes as you change velocity frames.

Each twin sees the other as younger because each twin's slices of "right now" through space-time is at an angle relative to the other's.

Doe that help? Let me add that this is not "just a matter of perspective" but a physical effect in that if the traveling twin later comes to a stop relative to Earth he will in fact be younger. Contra-wise if the twin left on Earth later catches up with his brother's speed so that they are again at rest to each other then he will be the younger.

Since space and time are unified now space travel implies time travel in a minor way. As I measure distance and duration the faster you travel through (my) space, the faster you also travel through (my) time (speeds here measured relative to your watch). Or as Stevie Miller put it:
Time keeps on slippin', slippin', slippin', ... into the future!

 

[cristo]

Gawsh I hope I didn't confuse you more.

Well, you managed to confuse me, and I'd like to think that I know what I'm talking about!

I like to try not to use phrases such as "speed causes time to run slower" as this is ambiguous, and could be taken to mean that a moving observer's proper time will change, which of course is not true.

The correct way of phrasing such a point is that, if body B is traveling away from the earth, and A is "stationary" on Earth then, in A's reference frame, the time taken for B to complete some outward journey is greater than the time as measured in B's reference frame, due to time dilation.

I would point the OP to the discussion on baez's homepage that I linked to before, as well as standard relativity texts as given on the "relativity on the web" section of the homepage.

 

[adriansd]

I like to try not to use phrases such as "speed causes time to run slower" as this is ambiguous, and could be taken to mean that a moving observer's proper time will change, which of course is not true.

That's what happens when you get a biological scientist attempting to translate the physics in his mind, into words that make sense... with actual high-class physicists watching... I'll keep trying, though.

*shies and hides*

 

[Janus]

Hai ,I am new to the forum.
I have some basic questions regarding the same old twin paradox.
Twin A stays at Earth and Twin B is the traveller.
from the A,s frame we all agree that Time is dilated for B because it is moving(relative to A) near to the speed of light.
Now switch on to B,s frame.from B,s frame we have to agree that A is moving(relative to B-in opposite direction).So obviously,A,s time also should be dilated as per SR.
Do this dilation cancell each other ,and they will be of same age when they meet again?.
This is a thought experiment only.So i am expecting some logical answers (not mathematical)which can help me to get out of my ignorance.

The difference lies in the fact that B has to turn around to return to the Earth.

Here's an analogy:

Take two men, A & B. Have them start walking from the same point at the same pace, but in directions 45 degrees relative to the other. Each man measures progress as distance traveled in the direction that he himself is walking at that time.

How does each man judge the progress of the other? From his point of view, the other man is always behind him. (From A's perpective, B has covered less distance in the direction that A is walking than A has, and vice versa for B.)

This is the equivalent of our two twins seeing time dilated for each other. Each sees the other's progress through time as being less than his own.

Now have man B make a 90 degree turn towards man B so that his path will intersect A's. This is the equivalent of Twin B turning around to return to Earth.

This does not effect the Rate of B's progress as far as A is concerned because B is still walking at a 45 degree angle to A's path after the turn. So when B crosses A's path B will still be behind A.

The same is true for B after the turn, but what about during the turn? Let's say that A, from B'd perpective was 70 meters behind and 100 meters to the right when he makes the 90 degree turn towards A. As he makes the turn, A will swing into a position 100 meters in front of him and 70 meters to the right. So from his position, A has suddenly jumped ahead of him.

For twin B this is the same as saying that during the period when he slows down, and then starts his trip back to Earth, from his perspective time speeds up on the Earth.

Now, man B continues walking in a straight line until his crosses A's path and then turns to start walking in the same direction as A. As he is walking, man A loses some of the head start he got when man B turned, but he never loses it all. B will cross A's path behind A. and when he turns to match A's direction he will stay behind A by a constant distance.

For Twin B, this means that during his return trip, Earth undergoes time dilation and ages more slowly, but never loses all of the time it gained during the turn-around. When twin B reaches Earth, he brakes to a stop and both Twins continue to age at the same rate again, but with B younger than A from the prepective of both Twins.

 

[cristo]

That's what happens when you get a biological scientist attempting to translate the physics in his mind, into words that make sense... with actual high-class physicists watching... I'll keep trying, though.

*shies and hides*

Well, there are lots of high-class physicists here, but I'm not one of them! No, I just wanted to clear up any potential confusion that that phrase tends to cause.

Still, your input is very welcome here!

 

[jtbell]

ThHow does each man judge the progress of the other? From his point of view, the other man is always behind him.

To make this more explicit, each man has to look backwards over his shoulder in order to see the other. This is obviously impossible if both men face in the same direction, but if they face in different directions, as in this example, it follows naturally from the geometry of the situation.

By analogy, our traveling twins are "facing in different directions" in spacetime, including time as a dimension along with the three spatial dimensions. When the traveling twin turns around and begins his journey home, he also "turns" to "face in a new direction" in spacetime. This causes his "perspective" on the stay-at-home to change abruptly (including the age of the stay-at-home in the traveler's reference frame), even though the stay-at-home has done nothing himself, and notices nothing happening to him.

 

[JesseM]

Hai ,I am new to the forum.
I have some basic questions regarding the same old twin paradox.
Twin A stays at Earth and Twin B is the traveller.
from the A,s frame we all agree that Time is dilated for B because it is moving(relative to A) near to the speed of light.
Now switch on to B,s frame.from B,s frame we have to agree that A is moving(relative to B-in opposite direction).So obviously,A,s time also should be dilated as per SR.
Do this dilation cancell each other ,and they will be of same age when they meet again?.
This is a thought experiment only.So i am expecting some logical answers (not mathematical)which can help me to get out of my ignorance.

The basic equations of special relativity such as the time dilation equation only apply in inertial reference frames--frames which are not accelerating, so an observer at rest in such a frame feels no G-forces. In the twin paradox, if the traveller moves away from the Earth for a while and then turns around to return so they can compare ages on Earth, then he must accelerate during the turnaround, so he doesn't have a single inertial frame of his own throughout the journey, and therefore he can't use the ordinary time dilation equation to predict how much the Earth twin will have aged.

 

[JesseM]

Hey guys!

For an object that is in MOTION, as it approaches the speed of light, it causes time to run slower for the object, compared to a stationary object. If the moving object were a person, they would age slower and their stationary identical twin would age faster.

This can prove difficult, but you need to distinguish between A (from original post) moving RELATIVE to B, and Time Dilation of A relative to B due to the motion of B.

Although B is actually in motion, From B's reference frame, you can say that A is moving relative to B. NewTonn appers to understand this adequately.

In relativity there is no such thing as being "actually in motion", and there is no objective truth about which of two inertial observers is "actually" closer to the speed of light. Speed can only be defined relative to a particular frame, and the laws of physics work exactly the same in every inertial frame; so if two observers are moving apart at relativistic speeds, each one can consider themselves to be at rest and the other to be moving close to the speed of light, neither perspective is more correct than the other.

If the velocity that two observers are moving apart is changing, though, there is an absolute truth about which one is accelerating--it'll be the one that experiences G-forces. Unlike velocity, acceleration is absolute in SR.

 

[newTonn]

Taking B's reference frame, now it has 'normal' time, and time can be considered to run faster for A. For our twins then, A ages faster, and so A is still older than B when twin B returns... nothing to cancel out.

Gawsh I hope I didn't confuse you more.

Cheers

Taking B,s reference frame,i couldn't understand,why time is running faster for A.in this frame,A is moving as faster as B(as viewed from A's frame) in space time,so his time should run slower as much as that of B (as viewed from A's frame).

 

[newTonn]

The basic equations of special relativity such as the time dilation equation only apply in inertial reference frames--frames which are not accelerating, so an observer at rest in such a frame feels no G-forces. In the twin paradox, if the traveller moves away from the Earth for a while and then turns around to return so they can compare ages on Earth, then he must accelerate during the turnaround, so he doesn't have a single inertial frame of his own throughout the journey, and therefore he can't use the ordinary time dilation equation to predict how much the Earth twin will have aged.

Time dilation is caused due to the relative motion(between two inertial frames) isn't it?.In this case One of the frame is accelerating.So the equation for time dilation is not apllicale to both frames.or if there is any equation of time dilation between one inertial frame and one accelerating frame,it is applicable to both in the same manner.
So when traveller turns around and accelerate,whatever equation applicable for traveller,is true for the men at rest also.

Finally both will give the same result of same ages for both twins.Am i correct?

 

[newTonn]

The difference lies in the fact that B has to turn around to return to the Earth.

So you mean , acceleration is the cause for less ageing of B.
Then i think either we have to re-define time dilation(including equations),or we have to admit that time dilation is not real and only a perspective.
it is as in the case of ball dropped vertically from a moving train.
For the man inside train will see the ball falling vertically.
the man standing at platform will see the ball falling in a parabola.
A man in sun will see the ball falling in a different path,with respect to the Earth's motion around sun.
A man at centre of milky way will see the ball falling in a different path,incorporating both the movement of Earth around sun and sun around galactic centre

and so on...until the real and true path.Everybody is seeing the same event from their own perspectives.The event is one and same but the position of observer's in space are different.if we consider light is transmitted instantaneously,you will find that,the time and interval of event is same for all observers.The confusion comes when we introduce the time for light to reach the observer.
Let us take the case of the duration of event.
let the ball took 1 sec to reach the floor of train(as per nearest observer).
consider an observer far away,and is moving at a speed of 150,000km / sec.
let the image of dropping of ball from hand reach him after several years.The image of ball hitting the floor should reach him after 1 second.but at this time he is not in that position in the space(he is traveling at a speed of 150000 km/per hour).So the image has to travel again 150,000km to reach him.but when image reaches there (in half seconds).he is 75000 km ahead of the image. and so on...and finally when image reaches him it will take more than the actual duration of event.
if we analys this properly, we can find out that the observer inside the train see a vertical path of ball because of his relative position in space with respect to ball.
similarly the far away observer(Moving at 0.5c) ,see a more duration for the event because of his relative position in the space when he sees the beginning of event and the end of event.
in other words,there is an absolute path for an event and there is an absolute time for event.ONLY THE PERSPECTIVE MATTERS.

 

[pervect]

newTonn, have you read any of the FAQ's about this topic? For instance "The sci.physics.faq about the twin paradox"?

This really has been covered before, and you don't seem to be grasping the idea that it's the relativity of simultaneity that is necessary and sufficient to resolve the "paradox".

 

[Janus]

So you mean , acceleration is the cause for less ageing of B.

Only from B's perspective, From A's perspective it was only B's relative motion that led to the age difference.

Then i think either we have to re-define time dilation(including equations),or we have to admit that time dilation is not real and only a perspective.

The equations are perfectly fine as they are, Both A and B have to use them during the those periods when their relative motion is constant in order to get the correct age difference at the end. Besides, they have been verified by experiment.
And if you want to say that time dilation is not "real", then you have to say that the age difference at the end of the trip is not "real" and that the results of said experiments are not "real".

it is as in the case of ball dropped vertically from a moving train.
For the man inside train will see the ball falling vertically.
the man standing at platform will see the ball falling in a parabola.
A man in sun will see the ball falling in a different path,with respect to the Earth's motion around sun.
A man at centre of milky way will see the ball falling in a different path,incorporating both the movement of Earth around sun and sun around galactic centre

and so on...until the real and true path.

There is no "real and True path", you can only measure the path relative to some reference system, and no refernce system is "truer" than another.

Everybody is seeing the same event from their own perspectives.The event is one and same but the position of observer's in space are different.if we consider light is transmitted instantaneously,you will find that,the time and interval of event is same for all observers.The confusion comes when we introduce the time for light to reach the observer.
Let us take the case of the duration of event.
let the ball took 1 sec to reach the floor of train(as per nearest observer).
consider an observer far away,and is moving at a speed of 150,000km / sec.
let the image of dropping of ball from hand reach him after several years.The image of ball hitting the floor should reach him after 1 second.but at this time he is not in that position in the space(he is traveling at a speed of 150000 km/per hour).So the image has to travel again 150,000km to reach him.but when image reaches there (in half seconds).he is 75000 km ahead of the image. and so on...and finally when image reaches him it will take more than the actual duration of event.
if we analys this properly, we can find out that the observer inside the train see a vertical path of ball because of his relative position in space with respect to ball.
similarly the far away observer(Moving at 0.5c) ,see a more duration for the event because of his relative position in the space when he sees the beginning of event and the end of event.
in other words,there is an absolute path for an event and there is an absolute time for event.ONLY THE PERSPECTIVE MATTERS.

The effect that you mention is not what Relativity is based on but is instead called the Doppler effect. It is factored out when dealing with Relativistic effects. Time Dilation is what is left over after you account for light signal delay.

If you try to hold on to the concept of absolute time, you will never grasp Relativity, as Relativity abandons that concept.

 

[JesseM]

Time dilation is caused due to the relative motion(between two inertial frames) isn't it?

The time dilation equation can be used for an inertial observer to calculate the elapsed time on any clock, accelerating or inertial (if the clock is accelerating you have to integrate the time dilation equation--see below). And if two observers are moving apart inertially, time dilation works the same way for both of them--each observer measures the other observer's clocks to be running slower than their own, according to their own set of rulers and clocks. If this seems impossible, check out my thread An illustration of relativity with rulers and clocks, where I show two systems of rulers and clocks moving alongside one another at relativistic speeds, and show how each one measures the other ruler to be shrunk and the other ruler's clocks to be slowed down and out of sync.

So when traveller turns around and accelerate,whatever equation applicable for traveller,is true for the men at rest also.

The twin who accelerates cannot use the same equation--specifically, he can't assume that if his own clock has elapsed a time of T since leaving Earth, the Earth-twin's clock will have elapsed a time of T \sqrt{1 - v^2/c^2}, where v is his speed relative to the Earth. This equation will work for all inertial observers, though. If the traveling twin has the same speed v relative to the Earth both while he's traveling away from Earth and while he's returning to it (and the acceleration is treated as instantaneous), then if the Earth twin ages by an amount (t_1 - t_0) from the moment of departure t_0 to the moment they reunite t_1, the traveling twin will have aged by only (t_1 - t_0) \sqrt{1 - v^2/c^2}. If we don't want to treat the acceleration as instantaneous, and instead take into account the possibility that the traveling twin's speed v(t) relative to the Earth is changing throughout the journey, then the amount the traveling twin ages between departing and returning will be given by the integral \int_{t_0}^{t_1} \sqrt{1 - v(t)^2/c^2} \, dt. But again, these equations only work for calculating the amount of aging when the velocity and times are defined in terms of an inertial frame like the Earth-twin's rest frame.

Finally both will give the same result of same ages for both twins.Am i correct?

If they both use correct equations they'll both predict the same answer for their ages when they reunite at Earth, but the accelerating twin cannot use the standard SR equations like the time dilation equation to calculate the answer.

 

[alvaros]

very basic

how do you measure time ?
how do you measure distance ?
how can you see the clock of your twin ?
the satelites of GPS can measure the Earth's frequency ?
if these questions are very stupid, simply ignore.
if anybody answer, please, in plain english.

 

[pervect]

Time can be measured with a clock.
Distance can be measured with a ruler - a very long one.

You can see the clock of your twin with a telescope, but of course you will see a delayed image. Focusing on what you actually see yields the "doppler shift" explanation of the twin paradox, which I rather like because it avoids abstractions.

Understanding the issue of how one determines what time the twin's clock reads "NOW" (as opposed to what I see through my telescope). is very important to a full understanding of the paradox, however, though one can attempt to avoid the issue by avoiding the notion of simultaneity altogether and focus instead on what one can see.

The confusion arises because the notion of "NOW" is different for every observer in relativity. This is known as the relativity of simultaneity, and is one of the classic stumbling blocks that students have in understanding relativity.

 

[newTonn]

If you try to hold on to the concept of absolute time, you will never grasp Relativity, as Relativity abandons that concept.

I will try to explain why am sticking to the concept of absolute time.
let A & B synchronise their clock.(A is at rest and B is travelling-forget about any acceleration-they synchronise clock when they pass each other).
A's reference frame.
After 5 seconds passed in A's clock ,say the time in B's clock is 3seconds,due to time dilation-(forget about calculations,we are analysing the situation logicaly).
B's reference frame.
After 3 seconds passed in B's clock,What will be the time passed in A's clock.
if the time dilation equation is applicable to B's frame(remember there is no acceleration involved-so it is applicable),it should be of course less than 3 seconds(some figure).

But we know,as from A's frame,if really the time was dilated in B's clock,the time in A's clock should be 5 seconds.

so could anybody explain me ,after passing 3 seconds in B's clock,what will be the time in A's clock.
if it is 5 seconds . why?
if it is less than 3 seconds(dilated ,as from B's frame) .why?

 

[newTonn]

newTonn, have you read any of the FAQ's about this topic? For instance "The sci.physics.faq about the twin paradox"?

This really has been covered before, and you don't seem to be grasping the idea that it's the relativity of simultaneity that is necessary and sufficient to resolve the "paradox".

In all the explanation and illustrations,please note that the man at rest also is moving.Ok i agree this but the things are getting confused there.
To avoid this,we have to put A in a place and B traveling in a closed loop.This is because we are concerned about the relative speed only.
if we can make the loop a circle,we can eleminate one more factor from the problem.The acceleration now restricted to direction only.Speed can be kept constant.
Now if we give a solution to the problem,we can find ,at the end of the loop,both the clocks are showing same time.(i am a lazy man.-it will take months for me to make an illustrated example with figures.).Say after every 22.5 degree of the loop ,let B flashes light to A,and A notes the time in his clock and vice versa.

 

[jtbell]

so could anybody explain me ,after passing 3 seconds in B's clock,what will be the time in A's clock.

It depends on the reference frame. In the reference frame in which B is stationary and A is moving, 1.8 seconds elapse on A's clock while 3 seconds elapse on B's clock. In the reference frame in which B is moving and A is stationary, 5 seconds elapse on A's clock while 3 seconds elapse on B's clock.

 

[newTonn]

It depends on the reference frame. In the reference frame in which B is stationary and A is moving, 1.8 seconds elapse on A's clock while 3 seconds elapse on B's clock. In the reference frame in which B is moving and A is stationary, 5 seconds elapse on A's clock while 3 seconds elapse on B's clock.

So it is only a matter of perspective,No real time dilation isn't it.?it has nothing to do with ageing.isn't it?

 

[jtbell]

To make clear where the resolution of the "paradox" lies, suppose that A has an assistant (A2), and that both of them carry clocks. They are at rest with respect to each other, separated by the distance that B travels in 5 seconds. They synchronize their clocks so that both read zero (in A and A2's reference frame) when B passes A.

As I noted in my previous posting: in A's and A2's reference frame, 5 seconds elapse on A's and A2's clocks while 3 seconds elapse on B's clock; whereas in B's reference frame, 1.8 seconds elapse on A's and A2's clocks while 3 seconds elapse on B's clock.

Nevertheless, if A and B set their clocks so that both read zero when B passes A, then everybody will agree that when B passes A2, A2's clock must read 5 seconds and B's clock must read 3 seconds!

Why doesn't A2's clock read 1.8 seconds when B passes him?

It's because in B's reference frame, A's and A2's clocks are not synchronized. This is because of the relativity of simultaneity that pervect mentioned. In B's reference frame, when B passes A, A's clock reads zero, A2's clock reads 3.2 seconds, and B's clock reads zero; and when B passes A2, A's clock reads 1.8 seconds, A2's clock reads 5 seconds, and B's clock reads 3 seconds.

 

[newTonn]

It's because in B's reference frame, A's and A2's clocks are not synchronized. This is because of the relativity of simultaneity that pervect mentioned. In B's reference frame, when B passes A, A's clock reads zero, A2's clock reads 3.2 seconds, and B's clock reads zero; and when B passes A2, A's clock reads 1.8 seconds, A2's clock reads 5 seconds, and B's clock reads 3 seconds.

Could you please explain me how you arrived (from B's reference)the reading of A2's clock as 3.2 seconds-logic is enough.
Is there any logic or just to match the final reading of 5 seconds after elapsing 1.8 seconds.
or is there any time forwarding mentioned in SR?

 

[jtbell]

If two clocks are at rest with respect to each other and separated by a distance L in one reference frame, then in another reference frame moving (along that same line) with speed v, those two clocks are out of synchronization by the amount vL/c^2. The clock whose position is "ahead" (in terms of the motion of the two clocks in the second frame) runs "behind" in time. This can be derived from the Lorentz transformation equations, the fundamental source from which the length-contraction and time-dilation equations are also derived.

In your example, the time-dilation factor corresponds to a speed of 0.8c, so A and A2 must be separated by 4c in their reference frame. Therefore the amount by which their clocks are out of synchronization in B's frame is (0.8c)(4c)/c^2 = 3.2.

 

[newTonn]

If two clocks are at rest with respect to each other and separated by a distance L in one reference frame, then in another reference frame moving (along that same line) with speed v, those two clocks are out of synchronization by the amount vL/c^2. The clock whose position is "ahead" (in terms of the motion of the two clocks in the second frame) runs "behind" in time. This can be derived from the Lorentz transformation equations, the fundamental source from which the length-contraction and time-dilation equations are also derived.

In your example, the time-dilation factor corresponds to a speed of 0.8c, so A and A2 must be separated by 4c in their reference frame. Therefore the amount by which their clocks are out of synchronization in B's frame is (0.8c)(4c)/c^2 = 3.2.

it seems the clock which is ahead is ahead in time in contradiction with your explanation.so A2's clock should read -3.2 seconds.
could you please give me some links to learn more about this out of synchronisation.

 

[jtbell]

it seems the clock which is ahead is ahead in time in contradiction with your explanation.

Draw a diagram, with A at the left side and A2 at the right. If B moves to the right in A's and A2's frame, then A and A2 move to the left in B's frame, with A "ahead" of A2.

A2's clock reads a later time than A's clock does, in B's frame, so A2's clock is "ahead."

 

[ratn_kumbh]

Time dilation is caused due to the relative motion(between two inertial frames) isn't it?.In this case One of the frame is accelerating.So the equation for time dilation is not apllicale to both frames.or if there is any equation of time dilation between one inertial frame and one accelerating frame,it is applicable to both in the same manner.
So when traveller turns around and accelerate,whatever equation applicable for traveller,is true for the men at rest also.

Finally both will give the same result of same ages for both twins.Am i correct?

Observation for such case should be done from an inertial frame. Since B returns (thus accelerats & it is only felt by B). its frame reference can't be selected. So u can choose A's reference Or any other inertial frame. U'll find that twin who returns, is younger.

 

[newTonn]

Observation for such case should be done from an inertial frame. Since B returns (thus accelerats & it is only felt by B). its frame reference can't be selected. So u can choose A's reference Or any other inertial frame. U'll find that twin who returns, is younger.

Time dilation equation is applicable from one frame to another frame which is moving with a uniform velocity with respect to the first frame.So it is not possible from any frame (at rest or uniform motion),to apply time dilation equation for calculation of time in an accelerating frame(here B).
So somebody has to find an equation for that.The equation contains v(not a),which is the relative velocity(not acceleration)

 

[Janus]

Time dilation equation is applicable from one frame to another frame which is moving with a uniform velocity with respect to the first frame.So it is not possible from any frame (at rest or uniform motion),to apply time dilation equation for calculation of time in an accelerating frame(here B).
So somebody has to find an equation for that.The equation contains v(not a),which is the relative velocity(not acceleration)

They already exist:
t = \frac{c}{a} sinh \left( \frac{aT}{c} \right)

T = \frac{c}{a} sinh^{-1} \left( \frac{at}{c} \right)

They are derived from the standard Relativity equations.

 

[newTonn]

They already exist:
t = \frac{c}{a} sinh \left( \frac{aT}{c} \right)

T = \frac{c}{a} sinh^{-1} \left( \frac{at}{c} \right)

They are derived from the standard Relativity equations.

Could you please give me an explanation of notations used and if possible,a link to derivations.
And please explain me how somebody knows,who is accelerating?

 

[JesseM]

Could you please give me an explanation of notations used and if possible,a link to derivations.

The notation is explained on the http://math.ucr.edu/home/baez/physics/Relativity/SR/rocket.html page, not sure where you'd find a derivation.

And please explain me how somebody knows,who is accelerating?

In flat spacetime (no sources of gravity), the accelerating individual will feel G-forces in the direction of acceleration, the inertial individual will feel weightless.