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MikeFontenot

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.

ChipB

PHF Helper
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.

tomh4040

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.

ChipB

PHF Helper
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

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tomh4040

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?

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ChipB

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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:

chipb said:
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.

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tomh4040

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.

MikeFontenot

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.

ChipB

PHF Helper
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.

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

MikeFontenot

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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