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Same old twin paradox

by newTonn
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Janus
#19
Jun25-07, 07:50 AM
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Quote Quote by newTonn View Post
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 earths 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 travelling 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 begining 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
#20
Jun25-07, 08:00 AM
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Quote Quote by newTonn View Post
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.
Quote Quote by newTonn
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 [tex]T \sqrt{1 - v^2/c^2}[/tex], where v is his speed relative to the Earth. This equation will work for all inertial observers, though. If the travelling twin has the same speed v relative to the Earth both while he's travelling 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 [tex](t_1 - t_0)[/tex] from the moment of departure [tex]t_0[/tex] to the moment they reunite [tex]t_1[/tex], the traveling twin will have aged by only [tex](t_1 - t_0) \sqrt{1 - v^2/c^2}[/tex]. 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 [tex]\int_{t_0}^{t_1} \sqrt{1 - v(t)^2/c^2} \, dt[/tex]. 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.
Quote Quote by newTonn
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
#21
Jun25-07, 04:52 PM
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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
#22
Jun25-07, 10:52 PM
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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 alltogether 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
#23
Jun26-07, 02:43 AM
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Quote Quote by Janus View Post

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 realy 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
#24
Jun26-07, 03:12 AM
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Quote Quote by pervect View Post
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 travelling 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
#25
Jun26-07, 04:34 AM
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Quote Quote by newTonn View Post
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
#26
Jun26-07, 04:56 AM
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Quote Quote by jtbell View Post
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
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Jun26-07, 05:13 AM
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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
#28
Jun26-07, 05:30 AM
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Quote Quote by jtbell View Post
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
#29
Jun26-07, 05:44 AM
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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 [itex]vL/c^2[/itex]. 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 [itex](0.8c)(4c)/c^2 = 3.2[/itex].
newTonn
#30
Jun26-07, 06:22 AM
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Quote Quote by jtbell View Post
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 [itex]vL/c^2[/itex]. 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 [itex](0.8c)(4c)/c^2 = 3.2[/itex].
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
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Jun26-07, 07:57 AM
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Quote Quote by newTonn
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
#32
Jun27-07, 01:54 AM
P: 11
Quote Quote by newTonn View Post
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
#33
Jun27-07, 02:28 AM
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Quote Quote by ratn_kumbh View Post
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
#34
Jun27-07, 07:37 AM
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Quote Quote by newTonn View Post
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:
[tex]t = \frac{c}{a} sinh \left( \frac{aT}{c} \right) [/tex]

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

They are derived from the standard Relativity equations.
newTonn
#35
Jun27-07, 07:47 AM
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Quote Quote by Janus View Post
They already exist:
[tex]t = \frac{c}{a} sinh \left( \frac{aT}{c} \right) [/tex]

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

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
#36
Jun27-07, 10:17 AM
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Quote Quote by newTonn View Post
Could you please give me an explanation of notations used and if possible,a link to derivations.
The notation is explained on the relativistic rocket page, not sure where you'd find a derivation.
Quote Quote by newTonn
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.


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