I am glad to see you recognize your error, even if you then try to play it off as if it isn't an error on your part....
"You have to compare clocks in different frames to establish what the change is, if any, because it depends on the relative motion of the two frames.
But, as I said before, if you're not comparing with clocks in other frames, i.e. you're only dealing with events in your comoving frame, it's just irrelevant."
But this is exactly what you are trying to do. Compare the moving twins clock in the here and now, with the frame he WAS in before he began his motion. You can not calculate into the past without invariably comparing the two frames in which clocks ticked at a different rate than in the frame you currently occupy. The twin's clock in the stationary frame, of which you were once part of, does not tick the same rate as the clock in your current frame. It ticks faster.....
We are not just dealing with events in our co-moving frame in the here and now. We are comparing this frame in the here and now, to another frame in the past in which clocks ran at a different rate than they do now due to changes in velocity.....
You can not compare the moving twins frame to the stationary twins frame without adjusting for time dilation.... And since you are not calculating for the rate of time now, but in the past when you were a part of that other frame..... you are left with no choice but to compare two different frames in which time runs at a different rate and adjustments MUST be made......
The twin CAN NOT use his current clock rate to calculate the rate in the other frame before he began acceleration. HIS PAST..... Likewise you can not use your current clock rate to calculate into YOUR PAST..... The clock rates are not the same.....
This is your error. You are invariably comparing two different frames, that run at different rates of time, but trying to use only your current rate of time when it in all actuality speeds up the further back in time you go. This makes your calculation of time passed based upon decay rates in error.
If only you comprehended the
twin paradox…………...
The paradox is about measurement not calculation, the only concern the twins have is their clocks are synchronized before the journey.
The description of the paradox is straightforward.
(1) Twins synchronize their clocks.
(2) One twin makes a round trip.
(3) Twins compare the time elapsed for the journey on their clocks.
The twins find their clocks are no longer synchronized as the travelling twin’s clock has run slower.
This has absolutely nothing to do with adjustments, calculations, or worrying about past events as the clock comparison is made in the same location in space time, in other words in the present.
There are quite a few methods of solving the paradox; one method is using Minkowski spacetime diagrams.
Minkowski spacetime diagrams are used for flat inertial frames and are therefore an approximation for the travelling twin’s accelerated frame of reference.
The approximation is valid as most of the travel time is in an inertial frame; it only the brief period of time when the twin changes direction to return to the stationary twin where there is an accelerated frame.
A Minkowski spacetime diagram is a two dimensional representation of spacetime and is composed of a horizontal x axis for distance and a vertical time axis w = ct.
The world line is the pathway of a particle or photon in spacetime and is the displacement in both space and time.
For a photon travelling in spacetime the displacements in distance and time are equal and the worldline of a photon is a straight line and forms a 45 degree angle with the time and distance axes.
Since a particle can never travel at or exceed the speed of light, the angle of the tangent of a world line of a particle is less than 45 degrees with the time axis.
When the Lorentz transformations are applied to a Minkowski spacetime diagram, the axes are no longer orthogonal, and the scale distance and time scales change which correspond to length contraction and time dilation.
The degree of distortion depends on the velocity of the frame.
We can use spacetime diagrams to illustrate the twin paradox and why it is not a paradox……………
The twins are called Dave and Bob where Bob is the travelling twin.
Bob travels away from Dave at a speed of 0.8c for 3 years after which he changes direction and travels back towards Bob at a speed of 0.8c for another 3 years before reaching him.
The Minkowski spacetime diagram for the twins is illustrated.
Since Dave is the stationary twin he is only moving in time hence his world line is along the time axis.
Bob is travelling in both space and time but less than c and is composed of an outgoing world line for 3 years and an ingoing world line of 3 years before meeting up with Dave.
The dots along Dave’s and Bob’s worldline correspond to yearly intervals.
Note that Bob’s yearly intervals are stretched out due to time dilation relative to Dave as Bob is moving relative to Dave.
The dashed lines and arrows are signals sent by Dave and Bob to each other.
Since the signals travel at c they are parallel to the Light lines.
Both Bob and Dave send signals to each other at yearly intervals.
As the space diagram clearly shows Dave has sent out 10 signals before Bob returned but only received 6 signals from Bob (including one on his return).
Clearly Bob’s clock was running slower than Dave’s and the paradox of each twin ending up younger than the other is broken.
The symmetry is also broken by the use of the Doppler factor:
√(c-v)/(c+v).
When Bob moves away from Dave the Doppler factor is √(c-0.8)/(c+0.8) = 1/3.
When Bob moves towards Dave the Doppler factor is √(c+0.8)/(c-0.8) = 3.
When travelling away Bob receives only one signal from Dave but nine signals for the return journey.
Dave on the other hand receives 3 signals from Bob in the first nine years and the remaining three in one year.
All of this is accomplished without your convoluted and ridiculously wrong approach.