303 wrote:so the twins are the same age then, distant time remains normal, and only the calculated time is off ?
Calculated time becomes measurable when two clocks are brought together again. It is calculated in the sense that when the clocks are distant there times cannot be directly conpared, but can be calculated in any given reference frame.
There is no absolute standard of time. You can compare the two clock times in whatever frame you like, and track calculated time from start to end of the journey to see what is the difference. When the clocks come together again the time difference can be measured and will be the same in all frames.
The Lorentz transformation tells you that B's clock, in A's frame, consistently runs slow relative to A's clock. Thus at the end of the journey B will be younger than A when they meet.
This whole thread is about people (Johan, Happyjack) having a misconceived argument that this can't be true. The two underlying ideas are:
(1) "clocks keep the same time, so must read the same when brought together again. Any LT effect must be "apparent" not "real".
That is misunderstanding time in a relativistic world. Two different
trajectories through spacetime can have different clock times even though they meet. Of course, our intuition tells that that is not possible because there is only one "now", and global time.
(2) The relative velocity and so LT time dilation is symmetrical. Therefore it cannot result in a time difference.
That is the subtlest argument, and in the last few posts we have just seen why it is not true.
The point is that to see which clock will go fastest we can only calculate the time of one (distant) clock relative to the (instantaneous) frame of the other (local). When the clocks are brought together again the calculated time can be read directly. When the clocks are far away calculation is all we have.
So, on the two journeys each clock runs faster than the
calculated time of the other. It is symmetrical. However the B clock has to change frame to come home, and this change of frame alters the A calculated time ahead by double the amount the Lorentz Transformation slowed it down, so that at the end the time difference calculation in B's two rest frames is identical to that in A's single rest frame.
Thus the end result is independent of frame: A's clock has a longer elapsed time than B's clock over the journey.
You need not think of this as clocks running faster or slower. You can think of it as time depending on spacetime path: bent paths have shorter times. Actually this is intuitive. The most highly possible bent path (with highest relative velocity to an inertial reference frame) would be a ray of light travelling out and reflected back. The time elapsed along this ligt-ray path is known to be zero.
Your intuition will make mistakes because you are naturally wanting to
compare times in different frames, and expect the results to be independent of frame. But there is no unique way to compare times of distant clocks in different frames. So the LT can be correct, even though the results it gives are not compatible with a unique global comparison of time.
The LT apprent contradiction in calculated times is possible because two clocks in different frames can only ever meet once, so the apparent discrepancy is resolved by the time shift difference in calculated time of distant objects.
When one clock returns the elapsed time difference can be precisely measured. The LT calculated times can be checked, since when two clocks are together they can be directly compared. There is never inconsistency because in order to return one at least of the clocks must change frame and therefore add an asymmetrical tiume shift component to the calculated elapsed time.
If the two trajectories are symmetrical, with both A and b clocks voyaging out and returning, with the same delta V change at the apex of the voyage, the two clocks would of course end up reading the same.