ananthu said:
My confusion has only increased now.
If each looks young to the other, then it means no body has actually aged.
Let us assume A moves away from B and and according to time dilation formula, the clock in the frame of A should appear slower to B and similarly the clock in the frame of B should appear slower to A by the same amount. In this case A and B are moving away from each other. Now take the case of return journey of A. Here both A and B will be approaching each other. Here only I want clarification. When they apporach each other should not the opposite happen ie. the clock the reference frame of A who is actually returning should appear to go fast with respect to that of B and similarly the clock of B should appear faster to A?
So when they meet the time differences that occurred during the forward and return journeys should cancel each other effect and only same time should have elapsed for both of them.
Then it is absurd to say tha when A returns to Earth several years have passed on Earth and B looks older to A and so on.
Can anybody explain this in simple terms?
I'll try. You need to use just one frame of reference to analyze a particular scenario. Consider two clocks traveling with respect to each other, let's say they are far apart and coming towards each other and eventually meet and then continue on their way getting farther apart again. You can examine this from the point of view of one of the clocks in which case the other clock will actually be running slower. It will also appear to be running slower but that's because it is actually running slower. Remember, the first clock is stationary in the frame of reference that we are considering and the second clock runs slower because it is traveling at a speed from the point of view of the frame of reference. You also need to be aware that the faster the clock travels, the slower it runs. There's no confusion about that, is there? Or you can examine it from the point of view of the other clock in which case the first clock is running slower. That by itself is understandable, too, correct? Or you can examine it from the point of view that is always halfway between them where the two clocks will now be traveling at the same speed and the two clocks will be running at the same speed. Again, no problem, correct?
In all these cases, as long as the clocks never change the speed at which they travel, they each run at a constant rate. So your idea that the clocks run at one speed while approaching each other and at a different speed while retreating from each other is not correct. All that matters is the actual speed in a particular reference frame.
Now if you want to select a frame of reference to examine the entire twin "paradox", the easiest one to select is the frame where one of the clocks, "A", remains stationary. The other clock, "B", travels away from the stationary one, "A", and so it runs slower. Eventually, "B" turns around and comes back, let's assume at the same traveling speed as before and so the clock continues to run slow at the same rate as before. Eventually, when "B" gets back to "A", it has an earlier time on it. Perfectly understandable, correct?
Now if you want to torture yourself, you can select some other frame of reference such as the one where the first clock "B" travels away from "A" and eventually comes back. But in this frame, "B" will be stationary and the other one, "A" will be traveling in the other direction. But near halfway into the scenario, "B" is going to start moving toward "A, but it obviously has to go faster than "A" in order to catch up with it and so it will be running even slower than "A". When "B" finally catches up to "A" it will have an earlier time on it because it was running even slower than "A". Now this "explanation" is not complete for many reasons but it gives the general idea. I would have to torture myself and you to provide all the gory details. I'm merely trying to illustrate how it is possible to get the same answer as to which twin is younger in the end, even if we start out in the reference frame of the other twin.
There is another frame of reference we could use and that is the one corresponding to "B's" return trip. In this case, "A" will be traveling at one constant speed the entire time while "B" will be traveling at a higher speed. About halfway into the scenario, "B" stops and eventually "A" approaches until they meet. Again, since "B" was traveling at a higher speed to begin with, it will experience a greater slowing down the "A". This "explanation" is also not complete but is just to give a general idea.
Remember, it's the speed of each clock in the reference frame that determines how fast it runs, not the relative speed between the other clock.
And remember, as long as you analyze the complete scenario in just one reference frame, you will get the same final result as you would in any other reference frame, even ones that are not associated with any objects in the scenario.
