Twin Paradox: Explaining Symmetry & Age Difference

In summary: The time that any observer directly measures (the one that makes him age) is the so-called proper time \tau, defined as\tau=\int \sqrt {1 - \frac{v(t)^2}{c^2}}dt, where v(t) is the velocity and c is the speed of light. An observer who experiences a change of velocity (an acceleration, such as the twin who's leaving the earth) now gets a different value for the proper time that has passed in his frame of reference than one who remains at constant velocity (the twin on earth). Therefore the situation is not symmetric.The anti-symmetry in the
  • #1
teodorakis
88
0
Hi, i want to ask about the basic explanation of twin paradox. In the explanation it says one twin has to accelarate to come back and the symmetry is broken and so one twin is older than other. Could you explain this symmetry and aging relation?
 
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  • #2
it's simply an extension or an illustration to explain time dialation
 
  • #3
teodorakis said:
Hi, i want to ask about the basic explanation of twin paradox. In the explanation it says one twin has to accelarate to come back and the symmetry is broken and so one twin is older than other. Could you explain this symmetry and aging relation?

Well, without any acceleration, the system would be symmetric, and you couldn't distinguish one of the brothers as the moving one. But since the who's leaving the planet has to accelerate in order to leave the Earth and to turn around in order to get back to it (which is also an acceleration), the symmetry is broken. One can calculate how acceleration affects the proper time of the brother and therefore ends up with the age difference.
 
  • #4
Polyrhythmic said:
Well, without any acceleration, the system would be symmetric, and you couldn't distinguish one of the brothers as the moving one. But since the who's leaving the planet has to accelerate in order to leave the Earth and to turn around in order to get back to it (which is also an acceleration), the symmetry is broken. One can calculate how acceleration affects the proper time of the brother and therefore ends up with the age difference.

i mean the symmetry is broken so one of them ages more than other does not satisfy me could you tell me in a more simple way from basics may be?
thanks
 
  • #5
when one travels at a speed close to that of light, the time elapsed and length of the journey which he will measure will be different from that of a stationary observer by a constant and hence his values will be less than that of the observer.
this is not quite explicit but 2moro i give you comprehensive explanation because am about going offline.
 
  • #6
teodorakis said:
i mean the symmetry is broken so one of them ages more than other does not satisfy me could you tell me in a more simple way from basics may be?
thanks

The time that any observer directly measures (the one that makes him age) is the so-called proper time [tex]\tau[/tex], defined as

[tex]\tau=\int \sqrt {1 - \frac{v(t)^2}{c^2}}dt, [/tex]

where v(t) is the velocity and c is the speed of light. An observer who experiences a change of velocity (an acceleration, such as the twin who's leaving the earth) now gets a different value for the proper time that has passed in his frame of reference than one who remains at constant velocity (the twin on earth). Therefore the situation is not symmetric.
 
  • #7
ok i can get the anti-symmetry in the situation and this breaks the paradox but i have a difficulty to understand that how this anti-symmetry cause the difference in the age disagreement. Also i want to ask that the accelerating one sync his clocks differently by jumping into other lattice of clocks and meters(so this is relative to other twin,right?) but other one still stays in same refence frame and sees other one slowing or fastening?, woooow, my head will explode and i even can't seem to ask what i mean.
 
  • #8
The twin paradox can be understood by looking at the proper times of the two twins. You have to use two (arbitrary) curves C1 and C2 in spacetime connecting two spacetime points A and B (twin 1 starts at A with his rocket and returns to meet twin 2 again). The proper times of the two twins are

[tex]\tau_i = \int_{C_i}d\tau = \int_{t_A}^{t_B} dt \sqrt{1-v^2(t)}[/tex]

The first integral measures something like a generalized length in four-dim. spacetime along the curve Ci. This formulation has the advantage that a) it shows the asymmetry that is entirely due to the curves taken by the two twins and b) shows that in principle the definition of the proper time does not depend on any external clocks and meters. The second integral can be evaluated in any coordinate system, i.e. the time t need not be the proper time of either twin.
 
  • #9
Here's a possible explanation:
The apparent intuitive paradox lies in the fact that time must "accelerate" for the twin who is accelerating (i.e. the "inertial" twin must appear to the accelerating twin to age faster as the twin accelerates). Otherwise we derive a contradiction of a form similar to the contradiction encountered when one is trying to find the orientation of a moebius strip.
The question then is this: what is special about an accelerating reference frame that can increase the "proper time" registered between two external events that are moving?
In other words, when calculating the "interval" or "proper time" between two (inertial) events in spacetime in the accelerating frame we need to add some positive terms that depend on the acceleration of the observer.

Intuitively, you can imagine that all "particles" visible from a reference frame get an "allowance" to spend on moving either in time or in space. From the accelerating reference frame, a non-accelerating particle gets to move "for free" while advancing its own clock as well. However, you can imagine that the "free move" can also be used to advance the particle's proper time if certain conditions are met.

Now how can we identify a reference frame as being absolutely inertial (why can't we say that the accelerating twin is actually at rest and the other twin is accelerating?) At an intuitive level, we can say that acceleration necessarily occurs when there is some interaction between two particles; if a reference frame is associated with the orientation and location of a particle, then it is inertial if the particle is not interacting with any other particle (although this definition may not deal with deeper issues).
 
  • #10
Polyrhythmic said:
Well, without any acceleration, the system would be symmetric, and you couldn't distinguish one of the brothers as the moving one. But since the who's leaving the planet has to accelerate in order to leave the Earth and to turn around in order to get back to it (which is also an acceleration), the symmetry is broken. One can calculate how acceleration affects the proper time of the brother and therefore ends up with the age difference.

So, imagine I am stood on Earth holding a clock. My friend Jack passes me in his super fast spaceship traveling at say 0.7c He also has a clock and as he passes me his clock is synchronised with mine. He then continues his journey.

A second friend Jill is on a return journey to Earth in her super fast spaceship coming from the opposite direction to Jack.

Jill also has a clock and as Jack and Jill pass each other, Jill’s clock is synchronised to Jack’s.

Jill then continues her journey and as she passes me on earth, I take a reading of her clock.

As there has been no acceleration between the synchronisation of the clocks, does Jill’s clock read the same as mine as she passes me?
 
  • #11
rede96 said:
So, imagine I am stood on Earth holding a clock. My friend Jack passes me in his super fast spaceship traveling at say 0.7c He also has a clock and as he passes me his clock is synchronised with mine. He then continues his journey.

A second friend Jill is on a return journey to Earth in her super fast spaceship coming from the opposite direction to Jack.

Jill also has a clock and as Jack and Jill pass each other, Jill’s clock is synchronised to Jack’s.

Jill then continues her journey and as she passes me on earth, I take a reading of her clock.

As there has been no acceleration between the synchronisation of the clocks, does Jill’s clock read the same as mine as she passes me?

No, you would get a different time, but this time the situation is symmetric. In principle it's not distinguishable which of you was moving.
 
  • #12
Polyrhythmic said:
No, you would get a different time, but this time the situation is symmetric. In principle it's not distinguishable which of you was moving.

So how would I know which system (i.e. my clock time or the traveling clock time) had 'aged' less?
 
  • #13
teodorakis said:
i mean the symmetry is broken so one of them ages more than other does not satisfy me could you tell me in a more simple way from basics may be?
thanks
Relativity is a geometric theory (both special and general rtelativity). Do you understand geometrically how a straight line connecting two points will always be shorter than a bent line connecting the same two points. In SR an accelerated observer has a bent worldline and because of the Minkowski geometry bent worldlines are shorter than straight ones.
 
  • #14
rede96 said:
So how would I know which system (i.e. my clock time or the traveling clock time) had 'aged' less?

That's the point, you wouldn't know. In your system, the other one's clock would experience time dilation and vice versa. If there's no acceleration involved, the system is symmetric.
 
  • #15
The basic explanation is given by DaleSpam in post #13 which is related to the formula in my post #8. The benefit is that it works w/o using synchronization of clocks, w/o exchanging signals, and that it can be generalized to GR with curved spacetime.
 
  • #16
ok could you please just explain it to me without jumping into minkowski space and worldline stuff and some complicated formulas, i mean in terms of what the accelerating twin and inertial twin "see" each other in whole trip?
 
  • #17
teodorakis said:
ok could you please just explain it to me without jumping into minkowski space and worldline stuff and some complicated formulas, i mean in terms of what the accelerating twin and inertial twin "see" each other in whole trip?
I think that's not possible. Using clocks, exchange of light signals etc. will in the very end result in reasoning based on Minkowski space as well.

There are some basic facts: One can draw "straight" lines (geodesics) in Minkowski space which are generalizations of straight lines as you know them from ordinary Euclidean geometry. Along these straight lines there is no acceleration and the length of the world line = the proper time has an extremum. If one twin does not travel along such a straight line the situation is no longer symmetric. He feels acceleration, the length of his world line is different hence his proper time is different, too. This is similar to a straight line and a curved line in Euclidean space where again the two lengths differ (but there you can't interpret length as proper time). So just draw a straight line and a curved line which intersect exactly twice. The proper time for the two twins between the start and the end of the trip is something like the length measured along the two lines.

The only problem with this picture is that in Minkowski 4-space a line that is straight in 3-space but along which the twin feels acceleration is no longer straight but curved in 4-space! But that only means that measuring "length" or "proper time" in Minkowski space is different; you can't do that using a sheet of paper and a ruler. The benefit of this picture is that there is no need to exchange signals between the two twins.
 
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  • #18
teodorakis said:
ok could you please just explain it to me without jumping into minkowski space and worldline stuff and some complicated formulas, i mean in terms of what the accelerating twin and inertial twin "see" each other in whole trip?
I think you will be better served to exert yourself and learn the "minkowski space and world line stuff". You can't reasonably expect to learn relativity without learning something new. The Minkowski geometrical framework takes all of the seemingly disconnected pieces and formulas and unites them into one coherent and intuitive whole. It is worth your time if you are interested in relativity.
 
  • #19
teodorakis said:
ok i can get the anti-symmetry in the situation and this breaks the paradox but i have a difficulty to understand that how this anti-symmetry cause the difference in the age disagreement. Also i want to ask that the accelerating one sync his clocks differently by jumping into other lattice of clocks and meters(so this is relative to other twin,right?) but other one still stays in same refence frame and sees other one slowing or fastening?, woooow, my head will explode and i even can't seem to ask what i mean.

It's the opposite of what you think: the break in symmetry causes the difference in age agreement.

With full symmetry, each determines the other to age slower due to the incompatible assumptions about who is "in rest". Thus they disagree about who is older at a certain point in time and none can prove to be right as their distant measurements depend on their incompatible assumptions of signal transit times - in short, "mutual time dilation".

If one of the two next turns around and meets up with the first, a direct age comparison is possible and the two agree that the one who changed his velocity aged more - just as the stay-at-home had been measuring all the time.

You can perhaps better understand it by realising that this is predicted with the use of any inertial frame. For example taking the frame in which the traveler is in rest on his first leg, the stay-at-home will appear to age less at first but during the return leg the traveler will appear to age much slower, with the same absolute comparison of ages when they meet.

Harald
 
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  • #20
In the twin paradox acceleration contributes two things.

  • It provide a reference to "know" which twin is in motion relative to the other
  • and it provides a means for the two twins to be reunited, to compare their clocks.

The acceleration itself contributes a "relatively" insignificant part of the overall time dilation and age difference at the conclusion of the hypothetical.

The original paradox compared the the symmetry of observations made from two inertial frames of reference in relative motion to one another with the asymmetry of the end state of clocks in both frames of reference at the end of a journey, in which only one was in was in motion, as compared to a common beginning inertial frame of reference.

Again acceleration is added to the hypothetical to bring it into close approximation with experience. As a matter of common experience the twins cannot begin from a common frame of reference, experience a change in relative velocity and return to a common frame of reference without one (or both) twin(s) experiencing acceleration.

Again, it is not the acceleration that "causes" the difference in age. It is the acceleration of one twin that helps to determine which twin will experience time dilation resulting from velocity.
 
  • #21
Along these lines...
here's a quote from my favorite paper on the Twin Paradox/Clock Effect :

"First, is it reasonable to say that
it is during his acceleration at C
that twin 2 suddenly loses time or age, and that
this loss at C causes the final age difference at B?
The answer is "no."
It is just as unreasonable to blame the acceleration at C
for the total age difference of the twins
as it would be,
in the case of a triangle ACB in the ordinary Euclidean plane,
to say that the larger length of the path ACB,
as compared to the straight path AB,
is caused by a sudden gain of length at the corner C."

The Clock Paradox in Relativity Theory
Alfred Schild,
The American Mathematical Monthly, Vol. 66, No. 1 (Jan., 1959) (pp. 1-18)
http://www.jstor.org/pss/2309916

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\qbezier(0,0)(0,0)(0,100)\put(0,0){A}
\qbezier(0,0)(0,0)(30,50)\put(30,50){C}
\qbezier(30,50)(30,50)(0,100)\put(0,100){B}
\end{picture}
\[

[/tex]
 
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1. What is the Twin Paradox?

The Twin Paradox is a thought experiment in which one twin travels at high speeds through space while the other stays on Earth. When the traveling twin returns, they find that they have aged less than their twin on Earth, leading to a discrepancy in their ages.

2. How does the Twin Paradox demonstrate symmetry?

The Twin Paradox demonstrates symmetry because both twins experience time dilation (a difference in the passage of time) from their own perspective. The twin on Earth sees the traveling twin age slower, while the traveling twin sees the twin on Earth age slower. This shows that there is no preferred frame of reference in special relativity.

3. Why does the traveling twin age less?

The traveling twin ages less because they are moving at high speeds, which causes time to pass slower for them due to the effects of special relativity. This is known as time dilation.

4. Is the Twin Paradox a real phenomenon?

No, the Twin Paradox is a thought experiment used to illustrate the principles of special relativity. In reality, it is not possible for one twin to travel at such high speeds while the other stays on Earth.

5. Can the Twin Paradox be resolved?

Yes, the Twin Paradox can be resolved by taking into account the effects of acceleration and deceleration on the traveling twin. When these effects are considered, it can be shown that the traveling twin will always have aged less than the twin on Earth, regardless of the direction or duration of their travel.

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