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https://en.wikipedia.org/wiki/Twin_paradox

this is an article from wikipedia about the twin paradox. an excerpt from "specific example"

enter image description here

How would I calculate the earth years from the traveller's perspective? The calculations are not stated in the article-- if I assume that moving clocks go slower, then from the traveller's perspective he has aged 6 years, shouldn't the earth years be $\frac{6}{\gamma}$ years, which would then be different from the 10 years as calculated in the "earth perspective" part? what am I missing?

Qmechanic
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Lee Laindingold
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3 Answers3

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Your mistake is to ask about the "traveler's perspective". The traveler has one perspective on the outbound flight and a different perspective on the inbound flight. In both perspectives, earth clocks run slow. But the perspectives differ very much on how long ago the earth clocks were set to zero, and hence on what the earth clocks say now.

The outbound traveler, having arrived at his destination, says "right now, the earth clock reads 3.6 years and is running slow" --- which is why it advanced only 3.6 years during my 6 year journey."

Immediately after beginning his return journey, he says "right now, the earth clock reads 6.4 years and is running slow --- so that it will advance only another 3.6 years during my 6 year return journey, and will read 10 years when I get back".

WillO
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You are overlooking the relativity of simultaneity.

On both legs of the journey, time on Earth is dilated (ie appears to be running more slowly) compared with the time in both the outward and return frames of the traveller. On that basis you might suppose the travelling twin would find the stay-at-home twin to be younger, and hence there to be a paradox. However, you have overlooked the fact that at the turn around point, the plane of constant time for the traveller switches from sloping upwards in time away from the Earth to sloping upwards in time towards the Earth. The effect is that the time 'now' on the Earth from the travellers perspective leaps forward at the turn around point, and it is that which causes the elapsed time on Earth to be greater.

Marco Ocram
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  • I find these sorts of comments read a bit oddly! Overlooking implies understanding but not remembering. IMO the OP has not "forgotten" the concept of simultaneity, they have simply never heard about or learned it! When you say someone has overlooked something you are really implying carelessness, when the real issue is lack of learning in the first place (the cause of ~99% of questions here). Perhaps the only honest answer we can give is "you really don't want to know, but study this written (no YouTube!) course on Special Relativity and get back to us with specifics" ;) – m4r35n357 Dec 08 '21 at 10:44
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    Mea culpa. I thought it would be more diplomatic to assume the OP knew all about it! – Marco Ocram Dec 08 '21 at 13:10
  • Not having a go at any one person, much less you, just thinking aloud! Shame that so much of this site is reduced to fire-fighting to answer lazy people . . . who probably won't understand the answers they get, or even read them! – m4r35n357 Dec 08 '21 at 13:43
  • In fact, I do know what relativity of simultaneity is- I find the use of the word "overlooking" quite suitable. As for your comments on "lazy" people (which is probably referring to me here, I guess), reading an article (whether in a textbook, or on a website) on relativity of simultaneity does not guarantee absolute understanding of the whole concept. I am currently in the process of learning special relativity, and the answers I get allow me to better grasp of relativity of simultaneity--this is what the whole point of stackexchange is for, isn't it?for clarification of concepts? – Lee Laindingold Dec 09 '21 at 09:27
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    I must confess, Lee, that I sympathise with anyone new to SR, as it is very counter-intuitive and most of the explanations of it in textbooks etc are dreadful. – Marco Ocram Dec 09 '21 at 10:26
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You've overlooked the significant change in position displacement across the multi-year time displacement.

Let $\Delta t$ be time between events in the observer frame

Let $\Delta t'$ be time between events in observed frame

Let $\Delta s$ be change in displacement between frames across $\Delta t$ as measured by the observer

Let $\Delta s'$ be change in displacement between frames across $\Delta t'$ as measured by the observed frame

Let $v$ be relative velocity between the frames

Then given constant relative velocity:

$\Delta t' = \gamma (\Delta t - v \Delta s/c^2)$

Or, if the observed frame values are known and we want to predict the observer frame,

$\Delta t = \gamma (\Delta t' + v \Delta s'/c^2)$

derivation

g s
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  • Could you explain a bit more in detail? I know these are Lorentz transformations - what do they have to do with the questabove – Lee Laindingold Dec 08 '21 at 04:13
  • Given earth years since launch $\Delta t'$, distance to earth since launch measured in the Earth frame $\Delta s'$, space ship years are $\Delta t$. Given space ship years since launch $\Delta t$, distance to earth measured in the space ship frame $\Delta s$, earth years are $\Delta t'$. – g s Dec 08 '21 at 04:18
  • I’m afraid I can’t really understand that- could you give more detailed explanation instead of merely defining the terms in the Lorentz transformation? – Lee Laindingold Dec 08 '21 at 04:22
  • Maybe I've misunderstood the question. If you take the (false) premise that we can do trillion-gravity acceleration without any consequences, then we start and end in the same place and you can just define $t' = \gamma t$, in which case they've defined $\gamma = \alpha ^{-1} = 1/0.6$ and $6*1/0.6 = 10$ – g s Dec 08 '21 at 05:01