yup, your post is the 4th time I've encountered this explanation. That might make me sound stubborn but really, all the explanations are identical. When we draw a diagram from the earth twin's perspective, that scenario is correct. If we draw from the space twin's perspective, that is also correct...But only if we use Galilean velocity addition. So the space twin receives certain signals at higher velocities. Since they are sending light signals to one another, no velocity addition allowed. So the space twin spends more than half the time getting redshift. If thats wrong, perhaps my error stems from misinterpreting the frame switching.
This is, of course, false. You are starting to sound more and more like martillo, there is a very good explanation in his thread that shows the error in your thinking.
Martillo's a nice guy, he was just very excitable that day. Anyway, all I see are his posts. I would like a good explanation, despite our agreement.
apologies for all the double posts. I was once wondering why only certain posts needed approval and I experimented a bit too many times.
Eram, if we are still talking about the situation described in the OP... Two travelers who start off in opposite directions from a common inertial frame of reference, each of those travelers will measure the two legs of their journey to take the same amount of time. As they travel away from their starting point, they will see the other as redshifted. As they travel back toward the starting point, they will see the other as blue shifted. But since both the outward and return trip takes the same time according to their own clocks, they will see both the redshift and blue shift for the same duration of time. It really does not matter from which frame of reference you examine the problem, each observer will measure each leg of either traveler's trip as taking the same time, in both direction.., as long as they are using their own clocks to measure time. The three observers will not agree on how much time has passed, but they will agree that the trip away, takes the same time as the trip back. Thus all observers will agree that the duration of redshift and blue shift, is equal. They just won't agree on the total elapsed time.
that's interesting. the original scenario is actually like two twin paradoxes put together, so i thought better settle that first.
It would work out the same in either case. The red and blue shifts, say more about relative velocities, than they do about elapsed time. Each observer must measure elapsed time by their own clock, and while they will not agree on the total, they will agree that each leg of the traveler's trip, took the same time. As example only... The traveler may say, I spent half a year going and half a year comming back, while the stationary observer might say, the traveler took one year going and one year comming back.
This isn't quite true. At the moment each twin turns around there will be a period of time which the other twin appears not to be shifted at all. Think about it.
While true, it does not change how long they measure the two legs of the journey to take and thus how long they see the other as red or blue shifted. All it does is add an undetermined time between the red and blue shifts where they, at least appear, to share an inertial frame with the other.., thus no red or blue shift.
I suppose that's true. But in janus' explanation, there is a resultant asymmetry. Why not so in this case?
If you are referring to post #80 Here Janus is saying the same thing I have been saying, when measured by the clock in either frame the time is the same going as returning. In his example Janus, explained the situation including the speed of light delays and simultaneity of relativity... And I stand corrected... He is correct, in pointing out that when one takes into account the time of light delay, the traveling twin will have been on his/her way home for some time before the stay at home twin sees the turn around with a telescope. As a result the stay at home twin will not see the traveling twin redshifted and blue shifted for equal times. I should have caught this myself when RJBerry questioned my post in post #90. Sometimes an attempt to simplify, winds up complicating or even just plain wrong. Anyway, the example Janus provided does describe how things would be seen, when accounting for time of light delays.
I've modified my space-time diagrams to include light signals in order to show the Doppler shift as seen by the observers: First from the Earth twin frame. Please Register or Log in to view the hidden image! The space twins send signals to each other continuously. For simplicity, I just show the ones sent at whole year intervals. At the scale of these diagrams. c is at a 45° to the horizontal. Notice how during the 4 yrs of his trip he receives just 1 years worth of signals from the other twin. He then does his turn around and starts receiving signals at a one to one ratio for three years until year 7 of his trip. At this point he has received 4 years of signals from the other space twin. this is also when he first notes that he other half has turned around himself. From then on, he receives signals at a rate of 4 to 1 and get 4 years worth of signals in the last year of his trip, adding up to a total of 8 yrs worth of signals received during the trip which equals the length of his own trip. Same thing, as seen according to the frame of one of the twin's outbound legs (or the inbound the other, take your pick) Please Register or Log in to view the hidden image! Again, the light signals are at 45° angles (Light speed is invariant). If you check where the light signals from the twins intersect the world lines of the other, you see that they match exactly what we see in the first diagram, each space twin receives exactly 8 yrs of signals from the other. Also note that in both diagrams, the light signals from each space twin reach the Earth twin at a rate of 1 year per 2 for 8 yrs and then 2 per 1 for 2 years, meaning that the Earth twin gets 8 yrs of signals from the other 2 during the 10 year span he himself measures.
Yes, it is a nice color painting. (But that depends on artistic taste.) At school, when I drew a diagram with axis and I have not noticed the units involved, the teacher does not let me continue and I was getting the minimum grade so grave the mistake was considered. So, for me this "diagram" is a lovely colorful graphics.
Do you mean that you don't know what each eaxis is? Time is on the vertical axis. Displacement is on the horizontal axis.
