12-29-2012, 01:57 PM #1 Member   Join Date: Dec 2012 Location: Boulder, Colorado Posts: 54 The Unaddressed Twin Paradox The twin paradox has probably been discussed more often than any other aspect of special relativity. But the most important issue is almost always ignored: What exactly does the traveling twin conclude about the home twin's current age during the trip? I.e., what exactly IS the traveler's perspective? In particular, IF the traveler can legitimately use the time dilation result during the unaccelerated outbound and inbound legs of the trip (which says that the home twin (she) is ageing more slowly than the traveler (he) is, according to him), then how can the traveler find her older when they are reunited? I have an answer to that question ... in fact, I believe it is the ONLY correct answer. But I'd like to hear other members' thoughts about the question before I give my answer. __________________ Mike Fontenot
 12-31-2012, 07:52 PM #2 Senior Member     Join Date: Jun 2010 Posts: 1,945 First it's important to note that the twin paradox has to do with general relativity, not special relativity. The difference is important, because the so-called paradox is entirely based on the effects of acceleration on the twin who travels. You cannot ignore the acceleration phases, both at the beginning of the trip and at the turn around when the travelling twin begins his return home. If you ignore these phases, then all you have is that under special relativity both twins perceive the other aging more slowly, but they can never get together to compare notes.
10-16-2013, 03:20 PM   #3
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Quote:
 Originally Posted by ChipB First it's important to note that the twin paradox has to do with general relativity, not special relativity. The difference is important, because the so-called paradox is entirely based on the effects of acceleration on the twin who travels. You cannot ignore the acceleration phases, both at the beginning of the trip and at the turn around when the travelling twin begins his return home. If you ignore these phases, then all you have is that under special relativity both twins perceive the other aging more slowly, but they can never get together to compare notes.
I am sorry but you are wrong, the twin paradox has everything to do with SRT. It has nothing to do with acceleration, but with velocity (moving clocks run slow), and the rate of a clock in K when judged from K1, and the rate of a clock in K1 when judged from K (assuming these two frames are in uniform motion). See chapter XII of "Relativity The Special And The General Theory" by A. Einstein, published by Methuen and Co 1920. You will find that chapter XII is in part I which covers special relativity, and general relativity is not even mentioned until part II. The confusion arises when the clocks are brought together to compare readings, and that must involve acceleration. The clocks however, do not need to be brought together, as the Lorentz equation for time tells us what the readings are. So apply the Lorentz equation to find the rate of the clock in K from K1, and you will find it is running slower than K1's clock. Apply the Lorentz equation to find the rate of the clock in K1 from K, and you will find it is running slower than K's clock. There is the paradox - each clock is running slower than the other, which is impossible.

 10-16-2013, 03:51 PM #4 Senior Member     Join Date: Jun 2010 Posts: 1,945 Tom - perhaps we are thinking of different versions of the paradox. The version I was referring to is the one where the astronaut leaves earth and later returns to find that hundreds of years have transpired and his twin is long dead while the astronaut has only aged a few years. It does indeed involve acceleration. See: en.wikipedia.org/wiki/Twin_paradox Last edited by ChipB; 10-17-2013 at 04:42 AM.
10-17-2013, 04:37 AM   #5
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Quote:
 Originally Posted by ChipB Tom - perhaps we are thinking of different versions of the paradox. The version I was referring to is the one where the astronaut leaves earth and later returns to find that hundreds of years have transpired and his twin is long dead while the astronaut has only aged a few years. It does indeed involve acceleration. See: en.wikipedia.org/wiki/Twin_paradox
There is only one clock paradox, but there are many variations to try to "explain" it, and the twin paradox is one such attempt. Using acceleration or gravity simply muddies the waters. So assuming that I have outlined the clock/twin paradox correctly, and you made no comment on it so that assumption is correct, do you agree with the statement that each clock runs slower than the other?

Last edited by tomh4040; 10-17-2013 at 04:39 AM. Reason: inserted "do " before "you agree with..."

10-17-2013, 04:51 AM   #6
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Quote:
 Originally Posted by tomh4040 do you agree with the statement that each clock runs slower than the other?
Yes, as stated in my original response in post #2 of this thread 10 months ago:

Quote:
 Originally Posted by chipb under special relativity both twins perceive the other aging more slowly, but they can never get together to compare notes.
The Twin Paradox, however, is much more difficult to understand and explain as it involves aspects of asymmetry of acceleration so that the twins end up at the same place but with vastly different experiences of how much time has passed. They don't both see each other as being younger but rather both recognize that one twin is definitely older than the other. It cannot be explained using Special Relativity alone.

Last edited by ChipB; 10-17-2013 at 04:55 AM.

10-17-2013, 09:29 AM   #7
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Quote:
 Originally Posted by ChipB The Twin Paradox, however, is much more difficult to understand and explain as it involves aspects of asymmetry of acceleration so that the twins end up at the same place but with vastly different experiences of how much time has passed. They don't both see each other as being younger but rather both recognize that one twin is definitely older than the other. It cannot be explained using Special Relativity alone.
Why should the clocks (or twins) be brought together to be compared? The clocks are either synchronous or they are not, whether they are together or not, or whether they are being observed or not. The LTs tell us what we want to know, and you accept the statement "So apply the Lorentz equation to find the rate of the clock in K from K1, and you will find it is running slower than K1's clock. Apply the Lorentz equation to find the rate of the clock in K1 from K, and you will find it is running slower than K's clock..."
You accept that as a fact, apparently without realising the contradiction of having K > K1 and K1 > K.

10-17-2013, 03:37 PM   #8
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Quote:
 Originally Posted by tomh4040 So apply the Lorentz equation to find the rate of the clock in K from K1, and you will find it is running slower than K1's clock.
The Lorentz equations can do more than give you the time dilation result and the length contraction result. They can also tell you about SIMULTANEITY, which is critical for understanding the twin "paradox" ... i.e., they can tell you, at each instant in the observer's life, what the current age of the other twin is (according to the observer). When you do that for the traveling twin (the one who turns around), you find that when he accelerates in the direction toward his twin, he will conclude that the home twin's age rapidly increases. It is that rapid increase during the turnaround that results in the TOTAL ageing of the home twin being greater than the traveler's total ageing, in spite of the fact that the home twin ages more slowly than the traveler (according to the traveler) during each of his constant velocity portions of the trip. The result is that both twins agree about their two ages at the end of the trip, even though they disagree about their respective ages during the rest of the trip.

The information about simultaneity given by the Lorentz equations can be more quickly and easily obtained by using the CADO equation, described here:

and here:

"Accelerated Observers in Special Relativity", PHYSICS ESSAYS, December 1999, p629.
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10-18-2013, 11:21 AM   #9
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Quote:
 Originally Posted by tomh4040 Why should the clocks (or twins) be brought together to be compared? The clocks are either synchronous or they are not, whether they are together or not, or whether they are being observed or not.
Sorry, but SR tells us that in fact what each observer percieves of the other person's clock is dependendt on their relative velocities and dustance between them. So no - the twins do not percieve that theiur clocks are synchronous or not - except when they are next to each other. I was careful in my phrasing earlier - that each twin perceives that the other's clock is running slow.

Quote:
 Originally Posted by tomh4040 The LTs tell us what we want to know, and you accept the statement "So apply the Lorentz equation to find the rate of the clock in K from K1, and you will find it is running slower than K1's clock. Apply the Lorentz equation to find the rate of the clock in K1 from K, and you will find it is running slower than K's clock..." You accept that as a fact, apparently without realising the contradiction of having K > K1 and K1 > K.
Again, I accept that each percieves that the other's clock is running slower than his own. I do not accept your phrasing that K>K1 and K1>K. This article may help you understand the issue better:
http://en.wikipedia.org/wiki/Relativity_of_simultaneity

10-18-2013, 12:49 PM   #10
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Quote:
 Originally Posted by ChipB First it's important to note that the twin paradox has to do with general relativity, not special relativity.
Special relativity is perfectly capable of handling the twin "paradox". General relativity is needed only when actual (physically real) gravitational fields are present. There are no such physical gravitational fields in the twin "paradox" scenario.

Taylor and Wheeler show how to determine the traveler's perspective in the twin "paradox" scenario, using only special relativity, in their example (Example 49) in their "Spacetime Physics" book, pp. 94-95. Brian Greene in his NOVA series on the "Fabric of the Cosmos" gets the same result, again using only special relativity.

It IS true that any special relativity problem can be re-formulated as an analogous general relativity problem having fictitious gravitational fields, by using the equivalence principle. When you do that, you get the same answer as is given by special relativity (and the special relativity approach doesn't have to resort to the use of artificial, fictitious gravitational fields). The only value in using the equivalence principle to convert a special relativity problem into an analogous general relativity problem is as a verification of general relativity ... it has no "value-added" in solving the special relativity problem itself, and the GR approach is more complex, cumbersome, and artificial.
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10-24-2013, 03:09 PM   #11
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Quote:
 Originally Posted by ChipB Sorry, but SR tells us that in fact what each observer percieves of the other person's clock is dependendt on their relative velocities and dustance between them. So no - the twins do not percieve that theiur clocks are synchronous or not - except when they are next to each other. I was careful in my phrasing earlier - that each twin perceives that the other's clock is running slow.
It seems that there is some confusion about what "perceive" means, here is the definition :-
>>/pərˈsiv/ Show Spelled [per-seev] Show IPA verb (used with object), per·ceived, per·ceiv·ing.
1. to become aware of, know, or identify by means of the senses: I perceived an object looming through the mist.
2. to recognize, discern, envision, or understand: I perceive a note of sarcasm in your voice. This is a nice idea but I perceive difficulties in putting it into practice.<<
You are bantering semantics here instead of answering the question. The twins do not have to be next to each other to perceive each other's clock. Each twin becomes aware of the other's clock reading, so A becomes aware of B's clock reading less than his own, and B becomes aware of A's clock reading less than his own. To put that in mathemetical/logical notation:
A < B and B < A
MikeFontenot, It would seem that there are no thoughts about the question "then how can the traveler find her older when they are reunited?" so please give us your answer.

10-25-2013, 06:12 AM   #12
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Quote:
 Originally Posted by tomh4040 MikeFontenot, It would seem that there are no thoughts about the question "then how can the traveler find her older when they are reunited?" so please give us your answer.
I already did ... post number 8.
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 10-25-2013, 07:04 AM #13 Super Moderator     Join Date: Apr 2008 Location: On the dance floor, baby! Posts: 1,931 Even if we get the "traveling" observer away from the Earth's gravitational field somehow we still have an acceleration at the far end of the traveler's turn around point. GR has to be applied to the motion as there is an acceleration of some sort, any sort...that's the whole point about the equivalence principle after all. However it can be explained without needing a direct reference to GR. The issue is after the turn around point...the traveling observer is not simply retracing the path...it's a completely new coordinate system. However I'm not going to even try to explain it as I suck at explaining the "paradoxes." MikeFontenont's link seems to be a good source (though I haven't looked at it in detail.) I like Schutz's (long version) explanation in "A First Course in General Relativity," pg 28 - 30. -Dan __________________ I'm a "something shiny" kind of...Oh look! Something shiny! See the forum rules here.
10-26-2013, 12:50 PM   #14
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Quote:
 Originally Posted by MikeFontenot Special relativity is perfectly capable of handling the twin "paradox". General relativity is needed only when actual (physically real) gravitational fields are present. There are no such physical gravitational fields in the twin "paradox" scenario

I read the CADO attachment but did not realise it was yours. It is, however, not applicable in this scenario :-
I set off in my spaceship to Alpha Centauri with the rocket accelerating at 1G. At the half way point it turns round and decelerates at 1G. It comes to a very brief stop at AC without cutting the motor, so immediately accelerates back towards Earth at 1G, again turning and decelerating at the half way point, coming to a halt at Earth. During the whole of the journey, except for the two brief turnarounds, I was experiencing 1G.
My twin stayed at home on earth, and he was also experiencing 1G, except for the brief periods when he went skydiving or jumping for a volleyball etc.
According to Einstein, the effects of gravity and (reaction motor) acceleration are indistinguishable, so my twin and I experienced the same 1G which affected both ourselves and our clocks equally. So which twin aged less and why that one?
To minimise extraneous effects, let's put the earth and the rocket in intergalactic space, between the Milky Way and Andromeda. Now there are no reference points. I can only see the earth, and my twin can only see my rocket. Fom my point of view, the earth has receded and returned. From my twin's point of view, my rocket has receded and returned. I "perceive" his clock running slower than mine, and he "perceives" my clock running slower than his. Who has aged less, me or my twin, and why that one?

My thanks to B Burchell for suggesting the last paragraph.

 11-07-2013, 10:50 AM #15 Senior Member     Join Date: Apr 2008 Location: Bedford, England Posts: 668 This paradox does seem to make it appear that speed is relative, but acceleration is not. One difference that I can see is that both the speed and acceleration are actually vectors. For the travelling twin the acceleration and speed vectors are aligned, For the stay at home twin they are not aligned. I have no idea if this is pertinent to the paradox, but it seems plausible. I had a quick glance at the links offered by MikeFontenot and ChipB, but they looked a bit like hard work...
11-07-2013, 11:07 AM   #16
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Quote:
 Originally Posted by MBW This paradox does seem to make it appear that speed is relative, but acceleration is not.
Well said.

Quote:
 Originally Posted by MBW One difference that I can see is that both the speed and acceleration are actually vectors.
Gah! Velocity is a vector. Speed is a scalar.

Quote:
 Originally Posted by MBW I had a quick glance at the links offered by MikeFontenot and ChipB, but they looked a bit like hard work...
Maybe this link will be a bit easier on you.

-Dan
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11-08-2013, 05:17 AM   #17
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More Speed less Velocity

Quote:
 Velocity is a vector. Speed is a scalar.
Oops, I know this and have even picked other people up on it,
I should have been more careful with my terminology.

I tend to jump into the forum and add a quick post in my lunch break, or before supper, etc.
It does meen some things aren't considered as thoughtfully as they should be.
In the future I will try not to rush in quite so carelessly...

Thanks for the alternative link, I will peruse it while finishing my lunch-time sticky bun.

 11-10-2013, 12:33 PM #18 Junior Member   Join Date: Oct 2013 Location: Leeds, Yorkshire, England Posts: 24 Here is a more detailed version of the thought experiment outlined in my post of 26th October (which so far has received no comments). As before, we take Earth and relocate it far into intergalactic space. The reason for doing this is firstly so that we don’t need to consider the gravity of surrounding stellar bodies, and secondly to remove the motion of the Earth around the Sun and Milky Way from consideration. Next we prevent the Earth from rotating. Likewise we do this to avoid having to consider the SR/GR effects of the rotation speed and the small amount of centrifugal force it provides. A rocket sits on the Earth’s surface with a large supply of fuel. Inside it is a room with living facilities and enough food and oxygen to support an occupant for many months. It also contains an accurate atomic clock. Beside the launch pad is an identically-fitted room. It contains a similar clock that has been synchronised with the one aboard the rocket. There is also a third clock on the opposite side of the Earth that is synchronised with the other two. Two identical twins agree to take part in the experiment. Each will spend the next several months either in the rocket or the Earth room, but neither will know which. Prior to launch, they are both given a sedative and put to sleep. Each twin is then randomly assigned to be moved into either the rocket or the stationary room. The rocket lifts off. At first, very slowly so as not to apply much acceleration, then as it moves further from Earth and gravity decreases, the rocket increases its acceleration to ‘fill in’ what is missing from Earth’s gravity. This acceleration will be steadily increased so that the gravity measured by an on-board accelerometer and felt at all times will be exactly equal to 1G. Shortly after launch, when the acceleration has steadied to 1G, the twins wake up. Neither of them knows which room they are in. The rooms are identical in layout and both experience what appears to be gravity. If they drop something it will accelerate toward the floor at 9.81m/s2, i.e. at 1G. According to the Principle of Equivalence (also called the strong equivalence principle), as proposed by Einstein and frequently described by falling elevators and rising rockets, the situation inside the two rooms is essentially identical. That is, there is no experiment you could devise that would allow either of the twins to determine which room they are in. The fact that the clocks are moving away from each other means they must have a relative velocity, otherwise they would remain a fixed distance apart. Therefore according to SR, time dilation should be occurring and the faster-moving clock should be running more slowly. But since the relative speed between the Earth and rocket is at all times exactly equal from both viewpoints, there appears to be no way of determining which is ‘faster’. As for GR, since the acceleration/gravity situation of both rooms is exactly equal at all times (other than the brief lift-off), according to the Equivalence Principle it would appear we are also unable to favour one clock over the other. So we are left with a conundrum: either we find a way of favouring one clock over the other or we agree that no time difference accumulates between them. After travelling for 10 months, and using a simple classical mechanics calculation, we could determine that the rocket is moving at 87% the speed of light (relative to Earth, which is now relocated outside our galaxy). At this speed we get a Lorentz factor of 2. This might mean that either the rocket or Earth clock is running half the speed of the other. These numbers however are not so important because we mainly care about which clock is ahead of the other, and not by how much (although we are also interested in that!). So let’s just pick 10 months as an arbitrary duration and assume a Lorentz factor of 2 at that point. This factor will be sufficient to override minor clock-drift errors, measurement errors, and brief periods where the acceleration of the rocket is not 1G, such as the launch and rotation (as described later. After travelling for 10 months (according to the local clock) the occupant aboard the rocket will take a sedative. The same will occur at the Earth-room (according to their clock). Both twins will then sleep for a while. The rocket engine will then be stopped, allowing the craft to drift freely in space with no acceleration while it is rotated 180 degrees to face the opposite direction – now pointing at Earth. The engine will then be started again, applying an acceleration force of exactly 1G. Both twins will then wake up. When the rocket-twin awakes, he notices no difference. Just as before, he experiences what feels like a gravitational force of 1G toward the floor. The Earth-twin experiences the same. The rocket is facing the opposite direction and is now decelerating, but by all accounts everything according to the Equivalence Principle is the same. There is still no experiment either twin could perform to determine which is experiencing gravity. Therefore it would seem that according to GR, both clocks should still be running at the same rate. And since the relative velocity is still identical – that aspect never changes – the clocks’ situation is still symmetrical according to SR. An objection here might be that there is a difference because the clocks are experiencing ‘gravity’ in opposite directions. And therefore the clock on the rocket will now be faster (or slower) than the one on the Earth. For those who raise this objection, refer to that third clock placed on the other side of the Earth. It is still in-synch with the first Earth clock and now experiencing gravity in exactly the same direction as the rocket. The deceleration process continues for the same time as the original acceleration process (10 months), at which point the rocket comes to rest relative to Earth. However the engine doesn’t stop. Instead it continues to apply exactly the same amount of force. Deceleration becomes acceleration and the occupant notices nothing unusual. The acceleration continues for the next 10 months (according to the local clock) until the rocket reaches (presumably) the original rotation point. At this point (according to their own clocks), both twins are put to sleep, the rocket is rotated 180 degrees, and then starts to decelerate while pointed away from Earth. Both twins awake and notice nothing unusual in their ‘gravity’ situation. The rocket continues its deceleration in a perfect reverse of its original departure, steadily coming to a stop relative to Earth, and all the while carefully adjusting its acceleration to give an on-board experience of 1G. Just prior to landing, both twins are put to sleep and then awoken after landing. The rocket has now landed beside the replica Earth room. Question: According to the combined rules of SR and GR, and allowing for minor clock-drift errors and the brief periods of launch, landing, and rotation, when the clocks are compared side-by-side, which of them will have recorded more time? And why not the other way around? Also, which of the twins will be older?
 11-10-2013, 12:53 PM #19 Super Moderator     Join Date: Apr 2008 Location: On the dance floor, baby! Posts: 1,931 tomh4040: tldr. I thought that your original post on this was fairly clear. Please check my latest post. There is a link there that should answer your questions. If not, then come back and we'll try to answer your questions about it. -Dan __________________ I'm a "something shiny" kind of...Oh look! Something shiny! See the forum rules here.
11-10-2013, 02:32 PM   #20
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Quote:
 Originally Posted by topsquark tomh4040: tldr. I thought that your original post on this was fairly clear. Please check my latest post. There is a link there that should answer your questions. If not, then come back and we'll try to answer your questions about it. -Dan
That link is to a different scenario. This is a favourite tactic of relativists: change the question to a different one, and answer that. All this talk of changing inertial frames is a smokescreen. Both occupants/clocks are experiencing 1G for the vast majority of the time, so according to GR the clocks stay in sync to a very high degree of accuracy. Both occupants/clocks have the same relative motion, so according to SR the time dilation is equal: ie from the reference frame holding clock1, clock1 > clock2 and from the reference frame holding clock2, clock2 > clock1.