Understanding the Twin Paradox: Time and Length Contractions Explained

[granpa]

suppose we start with a long line of stationary, evenly spaced and perfectly synchronized clocks along the x axis. if the stationary twin is at the origin and at t=0 the other twin passes the origin moving at relativistic speed with gamma=2 then from the point of view of the stationary twin the moving twin is length contracted to 1/2 his normal length. but despite this moving twin, due to loss of simultaneity, will see the line of clocks as no longer synchronized and therefore as being contracted to 1/2 its normal length. that's obvious. presumably, due to loss of simultaneity, each individual clock will seem to be running at 1/2 its normal speed even though the time read off as each individual clock passes would be running at twice normal speed. i think that's right.

but what happens when the moving twin suddenly stops? now all the clocks are synchronized and there is no length or time contraction. in particular, from the point of view of the moving twin, what time does the non-moving twins clock show just before and just after the moving twin suddenly stops?

what happens then when the moving twin begins moving back toward the origin? what does he see?

 

[Histspec]

Look at this example - (My value for gamma differs from yours a little bit):

Let us assume three clocks A, B, and C. A and B are at rest in S and C is at rest in S', while the relative velocity between S and S' is 0.5c.

In S:
A and B are synchronous and the distance between them is 1 light-second.
At t=0 clock C starts from A and arrives at t=2 at B. In S, clock C is time-dilated and therefore shows the t'=1,73..s. The same is happening when C travels back from B to A. So t=4s and t'=3,46..s..

In S':
We are no changing the frame and are looking into S'. In that frame, A and B are not synchronous, becaus of relativity of simultaneity B starts 0,577..s before A. Due to Length-contraction the distance between A and B is 0,866..Light seconds.
When A starts at t=0, B is showing t= 0,5s (Not 0,577..s because B is time dilated).

Now we can answer your question:

but what happens when the moving twin suddenly stops? now all the clocks are synchronized and there is no length or time contraction. in particular, from the point of view of the moving twin, what time does the non-moving twins clock show just before and just after the moving twin suddenly stops?

When B is just before C, C shows t'=1.73..s. Both A and B are time-dilated, so A shows t=1.5s and B shows t=2s.
When B stops at C, A jumps forward in time to be synchronous with B, so A and B show t=2s.
When B is just after C, A jumps forward in time again and shows t= 2.5..s.
When A stops at C, C shows t'=3,46..s. Both A and B are time-dilated, so A shows t=4s and B shows t=4.5s. So when A and C are coming togeter, in both frames A shows 4s and C shows 3.46..s - so A is older.

 

[granpa]

thank you. I figured it was something like that.

now I presume that this shift in A's time from B's point of view as B accelerates has something to do with general relativity. (of course, the time on A's clock doesn't really change when B accelerates. it just appears to B to do so). I suggest another thought experiment.

suppose the length between A and B is some absurdly large distance (jillions of light years) and C, having already stopped at B, instead of changing speed instantly, accelerates at 1 g for some length of time (about a year) till it reaches relativistic speed. the time spent accelerating is such a small part of the total transit time that it can be ignored. during this acceleration the time on A's clock continually shifts (from B's point of view). how is this equivalent to gravity?

I'm just trying to understand general relativity here. I don't need a detailed mathematical explanation.

 

[yuiop]

... presumably, due to loss of simultaneity, each individual clock will seem to be running at 1/2 its normal speed even though the time read off as each individual clock passes would be running at twice normal speed. i think that's right.

That would be better stated as {each individual clock will seem to be running at 1/2 its normal speed even though the time read off as each individual clock passes would show an elapsed time that is twice that shown on his own clock.}

When he is moving the line of clocks is no longer synchronised and he sees each clock as having a positive offset right from the outset and the further away each clock is the greater the offset proportional to Do*v/c^2 where Do is the proper distance measured in the rest frame of the line of clocks. Thereafter he see the line clocks advancing at 1/2 the rate of his own clock but the elapsed time on each clock that he passes is twice that shown on his own clock because as far as he is concerned they had a head start.

..but what happens when the moving twin suddenly stops? now all the clocks are synchronized and there is no length or time contraction. in particular, from the point of view of the moving twin, what time does the non-moving twins clock show just before and just after the moving twin suddenly stops?

Lets say he is looking through a telescope at the non moving twins clock and it shows 2.00.00 PM just before he stops. 1f he stops within one second then he will see a time of somewhere between 2.00.00 PM and 2.00.01 PM on the non moving twin's clock. In other words, he will notice very little change in elapsed time of the non moving twin's clock. What he will notice, is that just before he stopped the non moving twins clock was advancing at rate of 1/2 a second to every second that passes on his clock and just after he stops the non moving twin's clock has speeded up to the same rate as his own clock. When he has stopped he will notice that the clock that he has stopped next to, shows an elapsed time that is twice that of is own clock. If he allows for light travel times, he will notice that all the clocks in the line of clocks show twice the elapsed time of his own clock.

what happens then when the moving twin begins moving back toward the origin? what does he see?

For the sake of argument let's say the clock that he stopped next to, shows an elapsed time of two hours while his own clock shows a journey time of one hour. If he moved back to the non movig twin so slowly that there was no further time dilation of his own clock, then when he got back his clock would be one hour behind the non moving twins clock when they are standing next to each other once again. However, no matter how slow he moves, there will be some time dilation and the slower he moves the longer that additional time dilation will be over so that in practice his clock will be slightly more than a hour behind the non moving twin's clock.

[EDIT] When I stated that when the moving twin looks at the non moving twins clock through a telescope I should have added that he would have to allow for a classic doppler shift in his calculations.

 

[yuiop]

thank you. I figured it was something like that.

now I presume that this shift in A's time from B's point of view as B accelerates has something to do with general relativity. (of course, the time on A's clock doesn't really change when B accelerates. it just appears to B to do so). I suggest another thought experiment.

Time dilation is not necessarily due to acceleration. I read an analogy in the context of "the clock postulate" that went something like this. Imagine a cyclist on a cold but calm day. When he accelerates from rest to a velocity relative to the air he feels a wind chill factor that cools him down more than when he is at rest. It is not the acceleration that causes the wid chill, but the motion relative to the air. The acceleration simply changes his motion relative to the air. When he stops accelerating and sentles down to a cruising speed he continues to be cooled at greater rate by the wind chill factor even though there is no longer any acceleration. For wind chill factor you could substitute time dilation gamma factor.

 

[granpa]

i know that velocity results in time dilation. that isn't what i was referring to. when B stops then the line of clocks are all synchronous. before that they were out of synch. that is a time shift.

 

[yuiop]

thank you. I figured it was something like that.

now I presume that this shift in A's time from B's point of view as B accelerates has something to do with general relativity. (of course, the time on A's clock doesn't really change when B accelerates. it just appears to B to do so). I suggest another thought experiment.

suppose the length between A and B is some absurdly large distance (jillions of light years) and C, having already stopped at B, instead of changing speed instantly, accelerates at 1 g for some length of time (about a year) till it reaches relativistic speed. the time spent accelerating is such a small part of the total transit time that it can be ignored. during this acceleration the time on A's clock continually shifts (from B's point of view). how is this equivalent to gravity?

I'm just trying to understand general relativity here. I don't need a detailed mathematical explanation.

Ok, let's try and create the equivalent situation in a gravitational context. C is sitting on the surface of the Earth experiencing a constant acceleration of 1g. A is high up and his clock rate is faster than that of C because he experiences less gravitational time dilation. To make the situation completely equivalent C observes A to moving away from him at a progressively greater rate in your kinetic example so we will have to do the same in the gravitational example and have A accelerating upwards. A's upward acceleration will cause his clock to kinetically time dilate and classic doppler shift due to motion going away from C will make his clock appear to be slowing even more as far as C is concerned. As you can see the situation, is is not simple for your example and proportional effects of kinetic time dilation have to be weighed against the gravitational time dilation. Without a detailed mathematical numerical analysis it would be difficult to conclude anything.

 

[granpa]

classic doppler shift due to motion going away from C will make his clock appear to be slowing even more as far as C is concerned.

this is factored out when talking about time dilation (and loss of simultaneity) as is light travel time.

 

[Janus]

suppose the length between A and B is some absurdly large distance (jillions of light years) and C, having already stopped at B, instead of changing speed instantly, accelerates at 1 g for some length of time (about a year) till it reaches relativistic speed. the time spent accelerating is such a small part of the total transit time that it can be ignored. during this acceleration the time on A's clock continually shifts (from B's point of view). how is this equivalent to gravity?

I'm just trying to understand general relativity here. I don't need a detailed mathematical explanation.

The thing to keep in mind is that gravitational time dilation is due to a difference in potential rather than difference in local g force. Thus a clock runs faster at the top of a mountain than at sea level not because gravity is weaker, but because it is higher in Earth's field. If we were to replace Earth's gravity field with an uniform one (one that didn't fall off with increased distance from the center of the Earth), the Clock on the mountain would still run faster, even though it experiences the same gravitational force as one at sea level. In fact, it the difference in clock rates will actually be greater, for the very reason that Earth's gravity in the first example falls off with distance, which leads to a smaller difference in potential than the uniform field does.

When B accelerates at one g towards A, this is the equivalent of a uniform gravity field of 1 g strength extending from B to A, from B's point of view. Since A is vastly higher than B in this "field" from B's point of view, A is at a vastly greater potential and runs vastly faster than B. The greater the distance from A to B, the faster A runs according to B.

 

[dude222]

When the moving twin accelerates, he sees a linear spatial gradient in the speed of light, and all things under a pseudo-force equal to the force per unit mass he experiences times the energy of the particle under examination. The linear spatial gradient in the speed of light is equal to the force per unit mass he experiences. Note that I am using a slightly different definition of force, since the velocities in the momentum formulas are normalized by light speed. Since the pseudo-force is proportional to energy, it is like gravity. I have posted a calculation on my hi5 account, under the name johnwilliams22. I hope this helps.

 

[matheinste]

Hello dude222

Quote

------When the moving twin accelerates, he sees a linear spatial Gradient in the speed of light, and all things under a pseudo-force equal to the force per unit mass he experiences times the energy of the particle under examination. The linear spatial gradient in the speed of light is equal to the force per unit mass he experiences. Note that I am using a slightly different definition of force, since the velocities in the Momentum formulas are normalized by light speed. Since the pseudo-force is proportional to energy, it is like gravity. I have posted a calculation on my hi5 account, under the name johnwilliams22. I hope this helps.-------

What is that in layman's terms.

Matheinste.

 

[granpa]

Janus:
wikipedia backs you up. you are 100% right.

 

[granpa]

http://www.sysmatrix.net/~kavs/kjs/addend4.html

The standard version of the twin paradox has an astronaut twin – call her Stella – accelerating away from her Earthbound twin sister, Terra. Stella rockets through the cosmos at relativistic speeds, then later turns around and flies home, eventually landing back on Earth.

But in order to eliminate pesky accelerations, we'll say that the astronaut twin, Stella, was already in motion as she passes over Earthbound twin Terra. At the pass-by, their clocks each start at zero. Then, instead of later turning around, Stella eventually passes another astronaut, Alf, moving equally fast in the opposite direction, who adopts Stella's clock reading and continues back toward Earth. When Alf arrives at Earth, he passes over homebound Terra without slowing, at which time their clock readings are compared. Thus there are no accelerations and the scenario can be examined with the simplest (SR) arithmetic, like what's been shown in diagrams thus far.

 

[matheinste]

Hello again.

I thought that the twins (non)paradox required them to be in the same frame as each other at the start and end of the traveling twins journey.

Matheinste.

 

[granpa]

tehy have to be at the same place at the same time.

 

[matheinste]

Hello again

In your scenario above the traveller never returns.

Matheinste.

 

[granpa]

so?? btw, the total elapsed time is exactly the same.

 

[matheinste]

Hello again.

For who.

Matheinste

 

[granpa]

stella + alf

 

[matheinste]

Hello again.

I think i'll pass on this discusion and leave it to those who understand these things.

Goodbye .

Matheinste.

 

[Garth]

Janus:
wikipedia backs you up. you are 100% right.

Although Janus is always 100% right, that does not necessarily follow as a result of being backed up by wikipedia!

Garth

 

[Garth]

But in order to eliminate pesky accelerations, we'll say that the astronaut twin, Stella, was already in motion as she passes over Earthbound twin Terra. At the pass-by, their clocks each start at zero. Then, instead of later turning around, Stella eventually passes another astronaut, Alf, moving equally fast in the opposite direction, who adopts Stella's clock reading and continues back toward Earth. When Alf arrives at Earth, he passes over homebound Terra without slowing, at which time their clock readings are compared. Thus there are no accelerations and the scenario can be examined with the simplest (SR) arithmetic, like what's been shown in diagrams thus far.

Stella and Alf have different hyper-surfaces of simultaneity, this is where the paradox arises.

Alf cannot simply adopt Stella's clock reading as his own.

Garth

 

[granpa]

he doesnt. he just notes how much time elapsed for stella.

time elapsed for stella on outward journey + time elapsed for alf on inward journey=total time elapsed for moving twin=acceleration is not necessary to explain the paradox.

 

[granpa]

scenario:

a long line of one million stationary perfectly synchronized atomic clocks placed one light second apart from the origin along the x axis. at t=0 a rocket passes the origin at gamma=10. at t=0 all clocks, including the one on board the rocket, which are identical, read 0. the rocket uses a strobe to read the clocks as they pass its window. the rocket reaches the last clock at exactly the same moment as another rocket going the opposite direction at gamma=10. at that point the first rocket transmits a message to the second rocket telling them how much time had elapsed since he passed the origin.

an observer at the origin looking through a telescope at the clocks would see them out of synch due to the time it takes for the light to reach him. but after compensating for that he can easily calculate that they are synchronized.

when the rocket passes the origin at t=0, an observer on board the rocket, looking through a telescope at the clocks, would see exactly the same thing but because light seems to him to be moving at c, relative to himself even though he is moving, would calculate that the clocks are out of synch.

we know that he perceives each of the clocks to be ticking at 1/10 its normal rate. if v is his velocity and d is the distance to the clock as measured by a stationary observer then d/v is the time it will take him to reach it which is also the time the clock will read when he does reach it. subtracting from that 1/10th of the time he perceives having passed during the journey one gets the time that he perceives that the clock read when he was at the origin.

 

[matheinste]

Hello again.

One last thought.

If the traveling twin passes the stay at home twin on his outward journey and is the same age he cannot have been born at the same time. If they were born at the same their travel histeory i.e spacetime intervals covered, are differenttime they cannot be the same age. Time dilation between moving frames or something along those lines.

Either way they are not twins in the normal sense of twins.

Matheinste.

 

[neopolitan]

Hello again.

One last thought.

If the traveling twin passes the stay at home twin on his outward journey and is the same age he cannot have been born at the same time. If they were born at the same their travel histeory i.e spacetime intervals covered, are differenttime they cannot be the same age. Time dilation between moving frames or something along those lines.

Either way they are not twins in the normal sense of twins.

Matheinste.

They could be cousins twice removed. But "The Cousins' Paradox" doesn't quite have the same ring.

Twins are used as a sort of shorthand (perhaps not intentionally) meaning "two of the same". So the twins could just be two copies of the same person, the traveller takes out some sort of special insurance, by jumping in the Duplomatic 2000 and making a back-up copy of herself. Then she gets fired off into space reaching cruising speed after a very short period of time and sails off to some destination, then turns around, reaches the same cruising speed in the opposite direction (again very quickly) and returns to see how the copy is going. The copy is older, and pissed. They get in a row and the older, stay-at-home copy loses it and goes on a rampage, leading to a confusing car chase with people shooting at each other for no readily apparent reason. In the end, you find out that in fact both were copies and the real original had gone off in another spaceship which went much faster than the first and hence the original is still young and beautiful when she gets back and marries the policeman who solves the case by accurately shooting the bad copy (twice, because you thought she was dead, but she wasn't really the first time). Just before the credits begin to roll, you learn that the policeman is actually a grandson of the copy he just "retired" (he was raised by an Angolan sheep herder, so he never knew).

Oh, hang on, that has nothing at all to do with the original twins' scenario.

Feel free to pick holes in my alternative scenario.

cheers,

neopolitan

(Sorry about that granpa, it was just that the small-mindedness got to me for a moment there )

 

[matheinste]

Hello neopolitan.

Point taken

Matheinste.

 

[Garth]

Stella and Alf have different hyper-surfaces of simultaneity, this is where the paradox arises.

Alf cannot simply adopt Stella's clock reading as his own.

he doesnt. he just notes how much time elapsed for stella.

time elapsed for Stella on outward journey + time elapsed for alf on inward journey=total time elapsed for moving twin=acceleration is not necessary to explain the paradox.

There is no paradox in non-inertial observers experiencing less lapsed time than an inertial one, the paradox would be if two observers, Terra and Stella, each believe that they are in the inertial frame and therefore both conclude that they ought to experience the greater time lapse.

Stella + Alf cannot conceivably consider their combined experience to be inertial, there was a violent acceleration as your clock frame of reference went from Stella to Alf.

Garth

 

[yuiop]

... btw, the total elapsed time is exactly the same.

... For who(?)

stella + alf

ElapsedTime(Stella+Alf) < ElapsedTime(Terra)

For example let Stella's velocity (Vs) = 0.8c and Alf's velocity (Va) = -0.8c

Stella and Terra set their clocks to zero as they pass each other. Stella continues to a station that is one light year away in Terra's frame which takes 1/0.8=1.25 years according to Terra and 1.25*0.6 = 0.75 years by Stella's clock, Alf who is passing the station as Stella arrrives sets his clock to 0.75 years and when when he passes Earth, Alf's clock reads 1.5 years while Terra's clock reads 2.5years.

Of course this does not conclusively prove any observer is ageing slower than the others and to do that a fourth observer (Fred) would have to transfer from Stella's ship to Alf's ship as they passed the station and that would involve violent acceleration for Fred as Garth has pointed out.

 

[granpa]

as i said before:
time elapsed for Stella on outward journey + time elapsed for alf on inward journey=total time elapsed for moving twin=acceleration is not necessary to explain the paradox.

 

[yuiop]

as i said before:
time elapsed for Stella on outward journey + time elapsed for alf on inward journey=total time elapsed for moving twin=acceleration is not necessary to explain the paradox.

All that is necessary is the path through spacetime. The twin that has taken the shortest path through spacetime when they meet again will age the most.

 

[granpa]

All that is necessary is the path through spacetime. The twin that has taken the shortest path through spacetime when they meet again will age the most.

shortest path through spacetime=shortest path through time?

if all objects always move through spacetime at c then all paths of all objects through spacetime are the same length.

 

[yuiop]

shortest path through spacetime=shortest path through time?

if all objects always move through spacetime at c then all paths of all objects through spacetime are the same length.

shortest path through spacetime = longest proper time

this is sort of opposite to the normal intuition of the shortest distance between two points in 3-space being a straight line.

the 4-velocity may be interpreted as c for all objects but they can take different paths between events and the proper time recorded by the object taking the longest path will be the shortest time interval.

 

[yuiop]

THe attached image shows the paths taken through spacetime according to Terra, Stella and Alf. The red path taken by Terra is always the shortest according to any inertial observer so all observers agree that Terra ages the most.

 

[granpa]

ok. but what about the usual version of the twin paradox. don't both twins consider the other to be the one that is moving? what do their paths through spacetime look like?

just to be clear, path through spacetime is not the same as 'interval', right?

 

[matheinste]

Hello granpa.

Quote!

----ok. but what about the usual version of the twin paradox. don't both twins consider the other to be the one that is moving?-----

Yes but one is undergoes acceleration, the traveling one

Matheinste.

 

[granpa]

ok. i didnt understand the image the first time but i get it now. so all you are saying is that the one that travels in a straight line without accelerating always travels the shortest route and is always least time dilated. i agree.

but it is a fact that no individual undergoes acceleration during this thought experiment. every individual is already time dilated at the start and remains at that time dilation throughout. the full solution can be calculated without ever referring to acceleration at all. all you need to know is the velocities of the rockets and the distances involved.

when people refer to path through spacetime i always think of Minkowski space. that's quite different.

 

[yuiop]

ok. but what about the usual version of the twin paradox. don't both twins consider the other to be the one that is moving? what do their paths through spacetime look like?

just to be clear, path through spacetime is not the same as 'interval', right?

In the usual version of the twin's paradox, where one twin goes on a long fast journey and returns to the the static twin, their paths will look like the red and green paths in the left most diagram of the picture I posted showing Terra's point of view. It is not possible to depict the point of view according to the traveling twin in a single space time drawing because of the change of direction and it that change of direction that breaks the symmetry and the paradox. The traveling might consider himself to stationary for part of the journey but after he changed direction he would have felt acceleration and would have to conclude that he can not be stationary for all of the "journey".

The symmetry is also broken if each sends signals at yearly intervals on their respective birthdays. When the traveling twin changes direction he sees an immediate increase in the frequency of birthday signals coming from his stationary sibling ,while the stationary twin only sees an increase in birthday frequency of his traveling sibling later on in the experiment. When they count up the number of birthday siganls they each received during the entire experiment it will agree with their difference in ages. As soon as one twin changes direction the symmetry is broken. Before the change in direction, no one can say with certainty which twin is ageing faster. I hope that sort of makes sense :P

 

[yuiop]

ok. i didnt understand the image the first time but i get it now. so all you are saying is that the one that travels in a straight line without accelerating always travels the shortest route and is always least time dilated. i agree.

but it is a fact that no individual undergoes acceleration during this thought experiment. every individual is already time dilated at the start and remains at that time dilation throughout. the full solution can be calculated without ever referring to acceleration at all. all you need to know is the velocities of the rockets and the distances involved.

when people refer to path through spacetime i always think of Minkowski space. that's quite different.

Although I did not put labels on the axes, the paths I drew were plotted on a graph with distance on the hororizontal aaxis and time on the vertical axis which is exactly how Minkowski spacetime diagrams are plotted. There is no difference.

 

[granpa]

i meant that the interval is calculated differently. as i am sure you know.

 

[yuiop]

... but it is a fact that no individual undergoes acceleration during this thought experiment. every individual is already time dilated at the start and remains at that time dilation throughout. the full solution can be calculated without ever referring to acceleration at all. all you need to know is the velocities of the rockets and the distances involved.

In your slightly modified version with 3 observers and no acceleration it is impossible to prove whether Terra, Stella or Alf aged the least.

 

[granpa]

In your slightly modified version with 3 observers and no acceleration it is impossible to prove whether Terra, Stella or Alf aged the least.

you may be right, but it is a fact that it produces exactly the same result as the usual version. the total transit time of stella + alf = the total transit time of the moving twin.

 

[Garth]

you may be right, but it is a fact that it produces exactly the same result as the usual version. the total transit time of stella + alf = the total transit time of the moving twin.

So there is no paradox.

Two observers meet at one event and meet again at a later event.

The non-inertial observer experiences the smaller time elapse.

As I pointed out, Stella+Alf is a non-inertial frame of reference.

Garth

 

[granpa]

i know there is no paradox and the supposed paradox can be explained quite easily without reference to acceleration. all you need to know is the velocity of the rockets and the distances involved.