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-β
2)
First we analyze Twin A who travels at 0.4c for 100% of the time:
τ
A = 100%√(1-0.4
2) = 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:
τ
B1 = 80%√(1-0.1
2) = 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:
τ
B2 = 20%√(1-0.99999
2) = 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:
τ
B = τ
B1 + τ
B2 = 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.