A2 is co-located with B1 at a single event. It happens just once. Comparing the time of that event with events at different places depends on your reference frame, because simultaneity is relative. But in any reference frame, the events happens only once. No. In every reference frame, the event of A1 meets B2 happens before the event of A1 meets C2. "When A1 is in transit from B2 to C2" isn't well defined at B1, because simultaneity is relative. How does B1 decide when A1 is in transit from B2 to C2?
And how does B1 know when it is between 4:45 and 5:00? By looking at B1's own clock, right? So, when B1's clock reads betwen 4:45 and 5:00, B1 looks toward A1 to see A2. After B1's clock reads 5:00, B1 looks away from A1 to see A2. Of course, in A2's rest frame, it seems that A1 is not between B2 and C2 at that the time B1's clock is between 4:45 and 5:00, but that doesn't stop B1 from looking in the correct direction, right?
Yeah, I suppose that's right, but then that means that A1 is simultaneously between B2 and C2 and not between B2 and C2. How can it be that A1 can move from B2 to C2 while A2 hasn't yet reached B1 and then move again from B2 to C2 after A2 has passed B1?
Once again, you are confusing frames. Your "paradox" is resolved when you include the well-understood fact that simultaneity is frame-dependent. Which observer sees A1 move from B2 to C2, apart from A1? Which observer sees A2 hasn't reached B1? The problem you're having seems to be tied to this lack of specificity; each observer has their own clock, and time is only fixed for that observer by that (local) clock.
Isn't it true that in relativity all frames of reference are equally true and there no preferred frame of reference? So, it seems there are two possibilities... 1. A1 moves from B2 to C2 before A2 reaches B1 and then again after A2 passes B1. 2. A1 moves from B2 to C2 once, but A2 both hasn't reached B1 and has already passed B1.
Interesting thread. Your 1st sentence - I didn't know that that was the case. If so, interested in the reply.
There is no preferred frame of reference, yes. That's wrong, A1 moves once from B2 to C2, according to A1. No, A2 reaches B1 once, according to A2. So any other observer will see either event occur once. The times these events occur are different for A1 and A2 because there is no preferred frame of reference for (non-comoving) clocks. You could argue that clocks have only one "preferred" frame, which is the local frame, which means observers can only refer to local time. That's what relativity says: the time events occur is relative for non-comoving observers.
No, not really, because there is no reference frame in which A1 is in two places at once. It means that when B1's clock is between 4:45 and 5:00, A1's position at that time depends on what reference frame you use. Try thinking of it by comparing "Now" with "Here" When C2 passes A1 as A1's clock reads 5:00, they both ask "What was Here 15 minutes ago?" A1 says "I was here, B2 was here, and you were somplace else." C2 says "I was here all alone. You were someplace else." Both are correct, but they disagree because "Here" is relative, and there are using different references to establish what "here" means. When A2 passes B1 as B1's clock reads 5:00, they both ask "Where is A1 Now?" A2 says "A1 is passing B2, and A1's clock reads 4:45" B1 says "A1 is passing C2, and A1's clock reads 5:00" Both are correct, but they disagree because "Now" is relative, and there are using different references to establish what "now" means.
You're just restating the paradox. When A2 passes B1, A1 is passing B2 according to A2, meaning that A1 moves from B2 to C2 after A2 passes B1 in the C2-B2-A2 frame of reference. When A2 passes B1, A1 is passing C2 according to the A1-B1 frame of reference, meaning that A1 moved from B2 to C2 before A2 reached B1. So while A1 was moving from B2 to C2, A2 both hadn't reached B1 and had already passed B1. Here's a reposting for anyone fuzzy about the situation. Viewed from the A1-B1 reference frame at 4:45: Code: C2 B2 A2(.8c---->) A1 B1 (4:45) (4:45) Viewed from the A1-B1 reference frame at 5:00: Code: C2 B2 A2(.8c---->) A1 B1 (5:00) (5:00) Viewed from the A2-B2-C2 reference frame when the clock at A1 reads 4:45 and the clock at B1 reads 5:00: Code: C2 B2 A2 A1 B1 (<-----.8c) (4:45) (5:00) Viewed from the A2-B2-C2 reference frame at rest a little later... Code: C2 B2 A2 A1 B1 (<-----.8c) (5:00) (5:15) How can it be that A1 travels from B2 to C2 while A2 has not yet reached B1 and then again A1 travels from B2 to C2 after A2 has already passed B1? Obviously we have a problem, while A1 is moving from B2 to C2, A2 has either not yet reached B1 or A2 has already passed B1, both situations can't simultaneously be true.
Sorry, but the only one who is fuzzy about the situation seems to be you. You recieved some excellent feedback on your scenario and you should go back and read it. Your consistent error is that you are mixing frames.
I've read the answers and it's been a bunch of nonsense. Just one example: So A2 is both next to B1 and not next to B1 when the clock at A1 reads 5:00, but that doesn't mean A2 is in two different places at the same time. Oh, I see, we have a contradictory result but it's ok because the clock at A1 reading 5:00 has no absolute meaning. What a bunch of nonsense. The clock at A1 reading 5:00 does have meaning, it's an event. Relativity of simultaneity is what's causing the problem in the first place! The event of the clock at A1 reading 5:00 is simultaneous with A2 passing B1 in one frame of reference but not in the other frame of reference.. So what is causing the problem or the paradox is then somehow offered as a solution. Or how about this nonsense.. The situation I've posted above leads to a contradictory result between two different frames of reference. If somebody wants to say you can't do that because that's "mixing frames", that's fine, but it doesn't resolve the problem.
Draw a circle and a square on a piece of paper. Place it on a table so the the circle is to the right of the square. Have someone else stand on the opposite side of the table and ask them to what side the circle is to the square. He will say that the circle is to the left of the square. Oh no, we have a paradox! How can the square be both to the right of and to the left of the square?. This is exactly the type of "paradox" you insist on seeing with relativity of simultaneity. With the square and circle, which is to the right of which depends on the observer, and there is no way to see both to the right of each other. It meaningless to say "I see the circle to the right and you see the circle to the left, so the circle is both to the left and right and we have a paradox". In the same way, the measurement of time and space are dependent on the inertial frame of reference, and different frames of reference will measure the time and distance between events differently. And it is just as meaningless to say, "When A1 reads 5:00 in one frame that B1 and A2 are next to each other, but in another frame when A1 reads 5:00 B1 and A2 are not next to each other, so A2 is both next to B1 and not next to B1 at the same time", as was to say that the circle was both to the left and right of the square. Your refusal or inability to grasp this is just that, your your refusal or inability, and it has no bearing on the validity of the concept of the relativity of simultaneity nor of Relativity in general.
Hey, I just noticed this. You are correct in that assessment. You did indeed get contradictory results BETWEEN 2 FRAMES. Hooray, you win the kewpie doll. The different frames give different results which means in one frame 2 events may be simultaneous but in another frame they will NOT be simultaneous. You have discovered that Einstein was right - fine job!
So, basically in the A1-B1 frame of reference A1 moving from B2 to C2 is simultaneous with A2 racing towards B1. In the A2-B2-C2 frame of reference A1 moving from B2 to C2 is simultaneous with A2 racing away from B1. So we just sit back and say "That's relativity of simultaneity! There's no paradox and no problem." Ummm... yeah....sure.....
