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One's beard would be longer! 
A lot of seasoned scientists and academic nice peeps know this.Where you have the case of the twins and one of them buggers off in a space craft at nearly the speed of light and then returns back at some time to find that he (or she) is younger than the twin left at home !What I want to know is .....exactly how long did the space bound twin go away for then ?Say the space-bound twin goes away for five years......comes back but is only two years younger than his brother....was he away for five years or two years ?Is it both ?
Things get interestinger when you have triplets, one stays on the ground, one goes off in a spacecraft to the celestial north, the other off to the celestial south, and both later return. What are their relative ages, and why?
Well, with respect to what is the distance being measured?
In fact, there is no real paradox involved, or at least, it is easily resolved.For the twins to see each other again after all those years, at least one of them has to get back with his brother. This will involve acceleration -i.e. slowing down and then traveling back to the other twin - more speeding up and slowing down. The conditions for special relativity no longer apply as this is no longer an inertial frame (constant velocity with no acceleration or gravitational field), so, if one has been stationary or in an inertial frame and the other has been speeding up and slowing down, GENERAL relativity comes into play for one of them, so they will have had experienced different relativistic effects and one will appear older than the other. Time slows down under conditions of acceleration or a gravitational field.The one who did the return journey will appear, to BOTH of them, to be the younger one.If they both go away and come back, they will look the same age.
Actually (even if it's not obvious at all, even for me) they say the acceleration-deceleration of the starship can last so little time that, even if it's very large in intensity, its effects on the time differences becomes negligible.
Surely their must be a paradox if you look at the problem from the point of view of each twin in turn and only consider SR. Treat one twin (A) as 'stationary' and he will see his returning brother (B) (the one who did all the traveling) as younger. Now treat the other one (B) as 'stationary'. He will see his brother (A) as the one who did the traveling and, so, will see A as younger.
Neither of the twins is 'actually stationary' it is all relative.That is the paradox.The only difference in the two situations, above, is that only A, in the first case and B in the second case are in inertial frames. The other twin is in a non inertial frame and, therefore, can't be subject to simple SR calculations.
It's true, as I said too, that the paradox comes from the fact they are not in inertial ref. frame one respect to the other, but GR is not necessary to solve the paradox. They are two different things.
The only way there would be no paradox would be if there were NO apparent different in the ages of the twins when they got together again.This seems to be our main issue.In your thought experiment, the twins can both see that one has gone faster than the other because one will arrive first (to punch the time clock, put there by another person, at the other end) - one of them will expect to age differently than the other and, as you say, there would be no paradox here.
The paradox arises when they both end up at the same place and the same time (from a third observer's point of view).
After some more thought:QuoteActually (even if it's not obvious at all, even for me) they say the acceleration-deceleration of the starship can last so little time that, even if it's very large in intensity, its effects on the time differences becomes negligible.To accelerate to significant speeds would actually take considerable time and the energy involved to account for the relativistic mass increase (ΔΕ= Δm csquared is relevant) would be non-negligible and there is plenty of scope for the 'time factor' to be distorted a significant amount, too. The Mossbauer effect, for example, relies on very narrow resonances in metal atoms and detects time dilation by using the effect on the absorption of RF waves by tightly bound atoms. It is sufficiently sensitive to detect the difference in gravitational field (an acceleration) on the rate time progresses between two different floors in a laboratory building. So it is quite easy to disturb time a measurable amount in non-inertial frames.
Ok, now consider this: Instead of a 4 l.y. distant star, the travel is to a 40 l.y. star. Accelerations and decelerations of B can last exactly the same as in the previous 4 l.y. travel, but, now, they end up with an age difference of 100years - 60years = 40 years, not 4. So, how can accelerations and decelerations have to do with it?
As far as each twin is concerned, in my very simple model, there is no 'fixed ' earth or distant star. In fact we have no fixed grid in space for our measurements, at all.They are both in a deserted part of deep space. I am not allowing you to have any more in your experiment than this.They could both be going at any speed you like at the start of the experiment. The only things they can see or measure are each other and their ships and their relative motion.Apart from the fact that one has used his engines (which must play a part in resolving the paradox), when they start to move apart, their relative motions are equal and opposite.Now take up my previous argument. Neither twin knows which of them has actually 'moved', only that they have changed position, relative to each other. The actual distances they each observe the other one to have traveled is not really relevant. What is relevant is the fact that, when they finally meet (when their space time vectors coincide again) they will have different lengths of beard!SR tells us that when the traveling one (the one who uses his rockets) returns to the original spot will have aged less because of time dilation and he will appear younger.But, from his point of view, it was the other one who went off at speed and then returned. So the other one will appear younger to him.
How can there not be a paradox there? if A>B, you can't say B>A.But, if you allow for the effect of acceleration on one of them (which one of them could feel as a 'weight force' on him), to alter the rate of progress of time on his ship. That is a GR effect. As has been said previously, effects on the passage of time have been observed, due both to high speeds ( the muon observations - SR) and due to gravitational fields (the Mossbauer effect - GR).
a non inertial reference frame doesn't necessarily mean that it is accelerating, it can mean that it was accelerated.
What you need to pay attention to is where the measurement of time takes place. In this case it is the earth.
Sophie, it's called "gravitational time-dilation":http://en.wikipedia.org/wiki/Gravitational_time_dilation
It would seem, with a first logical reasoning, that the twin who has used engines "should" see the other twin as younger too. But this is only a "kind-of-logical-intuitive reasoning",
Nowhere is stationary,
QuoteSophie, it's called "gravitational time-dilation":http://en.wikipedia.org/wiki/Gravitational_time_dilationUncle Albert equates gravity and acceleration, doesn't he? They are precisely the same thing; Newton2.
A little acceleration for a long time or a lot for a little time will both have the effect of changing the RATE of the clocks on the traveller's ship for ever, once the engines are turned off.
Minkowski recast Einstein's version of Special Relativity (SR) on a new stage, Minkowski spacetime. The Twin Paradox has a very simple resolution in this framework. The crucial concept is the proper time of a moving body.
As you see, B accelerates around point (L,T) but how much it accelerates doesn't show in the evaluation.
Code: [Select]As you see, B accelerates around point (L,T) but how much it accelerates doesn't show in the evaluation.The fact is that he DID accelerate, and did not stay in an inertial frame, - which is what got him to the speed with which you are doing your calculations. It is only the fact that he accelerated that makes him different from the 'stationary' twin. You cannot argue with that, I am sure.
Without an accelerometer, neither of the twins would, initially, know whether it was he or his brother or both of them who was 'moving away' (trains moving out of stations etc.). That was the start of the original idea of SR.
As far as I can see, your calculations are, of course, fine but the link in my last post evaluates how the acceleration has produced precisely the sort effect that your calculations show. This is not surprising; there are often many ways of showing a result / killing a cat. However - the link takes more into account and doesn't rely on any talk of a fixed reference frame or a journey to any particular star. It just describes my simple two twin situation - out in the middle of nowhere - and determines which of them ages slower.
If you consider the space-time diagram of a twins paradox type experiment, you can see that the twins form a triangle.Acceleration means that the triangle isn't perfect, that the points of the triangle are actually slightly rounded by the acceleration.However, this can be made to be a very small effect, by taking the limit of high accelerations.What the twin paradox says is very similar to the geometrical theorem that the shortest distance between two points is a straight line, or eqivalently, the "triangle inequality" that says that the sum of the length of two sides of a triangle is always longer than the remaining third side.In the case of the twins paradox, though, the statement becomes that the observer following a geodesic path (analogous to the straight line, especially in the flat geometry of SR) has the longest proper time. The time is the longest, but the distance in the Euclidean geometrical analogy is the shortest. This has to do with the difference between Euclidean geometry with it's ++++ metric signature, and the Lorentzian geomery of SR with it's -+++ signature.Acceleration (the curvature at the tips of the triangle) really has very little to do with this geometrical result.That is why people say that the twin paradox isn't about the rounding of the corners due to the acceleration - it's due to the angle between the observers. This angle on a space-time diagram is simply a change of velocity. What is important is not the rate at which the velocity changes (the accleration) , but what is often called "delta-V", the change in velocity.The geometrical analogy of the "twins pardox" would be the "triangle paradox". The triangle paradox would say:If I go directly from A to B, I travel the shortest distance. And if I go from A to C to B, I travel a longer distance. It's "paradoxical" (?) that when I go from A to C to B, that this distance is longer than going straight from A to B. Furthermore, if I look at the distance from A to B, that's shorter than going from A to C to B. Now AB is the shortest distance! Of course people gemerally don't get confused by the "triangle paradox" - it's really not that much more difficult not to get confused by the "twin paradox" either.
Acceleration (the curvature at the tips of the triangle) really has very little to do with this geometrical result.
I can see that we are agreeing about all of this apart from the fact that you say the acceleration is irrelevant.
If it weren't for the fact that one was accelerating, there would be no difference in velocity so no effect.
QuoteAcceleration (the curvature at the tips of the triangle) really has very little to do with this geometrical result.The velocity difference gives you the ANGLES in your triangle. which is what gives you something to calculate with. If you had instantaneous speed change then you would have sharp corners, which would look pretty but is not practical. Velocity is merely the integral over time of acceleration - in practice, giving curves, not sharp corners. This detail (curved or sharp) is not really relevant.Your 'delta v' is a factor from outside SR and accounts for the angle.The only paradox is there when one ignores this.The total change in velocity, of course, depends on the acceleration and how long it's accelerating for (the same as impulse (force times time) can be used in conventional dynamics problems. I can't see why this presents you with a difficulty. During the time our twin is changing velocity, it is not in an inertial frame. The reason behind all your calculations is what happens whilst it is in a non- inertial frame.
The only situation in which you can ignore GR is when they are coasting past each other, without any velocity change. If you have acceleration , you can't ignore GR. But why should you want to? 
Sorry, Sophie, but *where exactly* did I use GR to make my calculations?ds2 = (cdt)2 - (dx)2is the Minkowsky metric, valid only in a "Flat" space-time, that is, in SR and NOT in GR.
QuoteSorry, Sophie, but *where exactly* did I use GR to make my calculations?ds2 = (cdt)2 - (dx)2is the Minkowsky metric, valid only in a "Flat" space-time, that is, in SR and NOT in GR.Your geometry is impeccable and it gives the right result from the point of view of the twin who stays behind because HE is in an inertial frame. There is nothing wrong with your construction to show what happens on the straight bits, but you are including changes in velocity at the corners, which, as I keep saying, involves the traveler not being in an inertial frame all the time. This is where the GR bit comes in. It is this effect on him that makes him observe the one who stays behind as being older than him when he returns. It is the traveler who experiences the time-rate changes at the corners.
If you don't allow for this then what is the difference between the observations of the two? There is no reference grid, so why, apart from the acceleration factor, will the experiences not be symmetrical and paradoxical? Go through the same procedure for the travelling twin and obey the rules; he is not in flat space time all the 'time' so he must see things differently; you can't use your Minowski metric in the same way for him. But this is just going round in circles and that, btw, is not an inertial frame, either!
You are sounding quite convincing now. I'm all in favour of using the simplest method for getting an answer and I have some chance of following SR in my brain. BUT (and there's got to be a but) it seems that, using your argument, no one need ever use GR to explain anything. You imply that one can always explain the effects of general relativity by referring your measurements to some inertial frame, elsewhere. This can't be true for all cases, or they wouldn't have bothered to invent GR. Can we, possibly, reconcile our differences with the following scenario?Your traveling twin goes near a huge dead star, which he cannot see, because it is a black dwarf. The gravitational field of this object introduces an acceleration (we, surely, must use GR in this case, mustn't we?), which he measures and which he takes into account when he analyses his observations of his non-traveling brother. He attributes his relativistic effects to acceleration due to his engines and does your calculations , making the appropriate corrections etc. to decide how much younger he will be at the end of the experiment.Can he possibly still get the correct answer about his relative age after this experience?
If he can, then I can't see why GR should ever be necessary. If he can't, then what is the difference between the acceleration due to his engines and the acceleration due to the gravitational field that he enters?
I am not being argumentative for the sake of it now; I am genuinely mystified.
Is the traveller actually capable of making all the corrections without taking GR into account? Unless he makes allowances for his time dilation (due to acceleration) can he, in fact, determine where he is and how fast he's going? I guess you will say "yes".
Have you an example of a similar type of experiment where GR would HAVE to be used, so I can get the picture properly?
I notice that not many other people are getting involved with this.Is there anyone else who can resolve this?