How does the duration of acceleration affect the Twin Paradox?

In summary, this conversation discusses the twin paradox and how it relates to general relativity. It is explained that the acceleration of Twin A's ship breaks the symmetry of the situation and causes time dilation, which can be calculated using the time dilation formula. It is also mentioned that in simple cases, the effects of relative speed and gravitational time dilation can be separated. However, in this scenario, gravity is responsible for both effects contributing to the twin paradox. The concept of geodesic paths and proper time is also discussed in relation to general relativity. Additionally, a variant of the twin paradox is mentioned where there is no acceleration involved and the asymmetry between the two paths is evident.
  • #36
Erase from your memory the false statement than SR cannot deal with acceleration.
If so, then STR can't solve twin paradox? bcoz all the articles on internet say that it is acceleration which causes the the traveling twin to age less than the one who stayed on earth.
 
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  • #37
R Power said:
If so, then STR can't solve twin paradox? bcoz all the articles on internet say that it is acceleration which causes the the traveling twin to age less than the one who stayed on earth.

SR can solve it, did solve it, right from when it was first presented by Einstein. Ignoring gravity (or topologically non-trivial universes), acceleration is necessary to have true twin 'paradox', but it is false to say the acceleration causes the age difference, or that the age difference is localized in any way to a period of acceleration.
 
  • #38
SR can solve it, did solve it, right from when it was first presented by Einstein. Ignoring gravity (or topologically non-trivial universes), acceleration is necessary to have true twin 'paradox', but it is false to say the acceleration causes the age difference, or that the age difference is localized in any way to a period of acceleration.
Then what causes the differential aging? How does STR solves the paradox. This may be very elementary question but I don't have solid understanding of relativity. Can you explain how STR solves paradox or can you give me good references where I can get my answer?
 
  • #40
ghwellsjr said:
[..] I'm trying to at least associate the (dilated) times on the clocks with their speeds relative to a single frame of reference, not to each other.

That is also how Einstein (1905) analysed the difference in clock readings and how Langevin (1911) discussed the difference in ages. I find that the clearest way of presenting it.
 
  • #41
R Power said:
Then what causes the differential aging? How does STR solves the paradox. This may be very elementary question but I don't have solid understanding of relativity. Can you explain how STR solves paradox or can you give me good references where I can get my answer?

- Done in post #22; it's not really considered paradoxical in SR.
Note also that Einstein already gave a correct but one-sided analysis in his 1905 paper:
section 4 of http://www.fourmilab.ch/etexts/einstein/specrel/www/
 
  • #42
harrylin said:
That is also how Einstein (1905) analysed the difference in clock readings and how Langevin (1911) discussed the difference in ages. I find that the clearest way of presenting it.

And this I have no problem with at all. This is what I've been saying: pick a frame, accumulate dilated time (=proper time) along different paths. Compare. Result same no matter what frame you use for the analysis. Where I have a problem is attempts to state some other simple rule to predict or explain this - all such that I've ever seen I consider simply wrong.
 
  • #43
R Power said:
Then what causes the differential aging? How does STR solves the paradox. This may be very elementary question but I don't have solid understanding of relativity. Can you explain how STR solves paradox or can you give me good references where I can get my answer?
Since you asked, I will show you how STR solves the twin paradox. We'll use the scenario that PAllen presented in post #24:
PAllen said:
Consider that, while neither twin is ever inertial (due to continuous changes in direction), twin A is always moving at speed .4c in this chosen inertial frame. Suppose twin B is moving .1c for 80% of the coordinate time between separate and meet up, and at .99999c for 20% of the coordinate time.
And I'm going to use the process I described in post #16:
ghwellsjr said:
But this thread and all the discussion up to this point has been about the Twin Paradox where they start out together, separate, and come back together and I'm saying that if you agree to ignore gravity, then you can analyze the scenario in any single inertial Frame of Reference and the "time spent at a higher speed" is defined uniquely in that FoR and the "higher speed" is defined uniquely in that FoR and we're talking about the coordinate time of each body in that FoR and we apply Einstein's time dilation formula to convert coordinate time into proper time for each body and then we see how much proper time has accumulated for each body as it travels at different speeds according to the FoR for whatever coordinate time intervals from the time they separated until the time they reunite and we get the amount that each one aged and subtract them and we have the differential aging and no time disappeared or needs to be accounted for.
I know that's a mouthful but it's really very simple to analyze using Einstein's formula to get the proper time interval, τ, (tau, the time interval on a clock) as a function of its speed, β, (beta, the speed as a fraction of the speed of light), and the coordinate time interval, t, as specified in the Frame of Reference:

τ = t√(1-β2)

First we analyze Twin A who travels at 0.4c for 100% of the time:

τA = 100%√(1-0.42) = 100%√(1-0.16) = 100%√(0.84) = 100%(0.9165) = 91.65%

Now we analyze the first part of Twin B's trip at 0.1c for 80% of the time:

τB1 = 80%√(1-0.12) = 80%√(1-0.01) = 80%√(0.99) = 80%(0.995) = 79.6%

And the last part of Twin B's trip at 0.99999c for 20% of the time:

τB2 = 20%√(1-0.999992) = 20%√(1-0.99998) = 20%√(0.00002) = 20%(0.00447) = 0.09%

Finally we add the two parts of Twin B's trip to get the total time:

τB = τB1 + τB2 = 79.6% + 0.09% = 79.69%

So we see that Twin B with 79.69% of the coordinate time of the scenario ages less than Twin A with 91.65% of the coordinate time.
 
  • #44
PAllen said:
This underscores my belief that all attempts at a simple rule (other than the formula itself) for which twin ages less run into trouble, even in SR.
Consider the previous post as attempt 6), the only one I presented.

I made the statement "time spent at a higher speed" specifically regarding a variant of the Twin Paradox to show that acceleration is not the cause of one twin aging less. In that variant, both twins experience exactly the same acceleration but one twin spends very little time at the higher speed and most of the time at the slower (rest) speed while the other one spends practically all of time at the higher speed.

Here's exactly what I said and I continue to stand by it:
ghwellsjr said:
I'm glad you mentioned both bodies experiencing acceleration because we can have another variant of the Twin Paradox in which both of them accelerate exactly the same except that one returns home immediately while the other one continues far away from home before matching the acceleration of his twin and returning home a lot later. This clearly shows that it's not the acceleration that causes the differential aging but rather time spent at the relatively higher speed that causes the differential aging.
 

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