For Alice to accelerate and then return to go past Bob is almost exactly the same as the Twins Paradox, except Alice has no decelerated when they see each other's clock.PeroK said:We're not talking about the twin paradox!
student34 said:Yes, and that is what I tried to correct for the poster. And I tried telling a different poster that this is basically a Twins paradox scenario.
sdkfz said:The explanation of the twin's paradox is not about one of them "really" having the slowest clock.
I expect the three of us agree about what’s going on: symmetrical time dilation means that both twins correctly find the other clock to be running slow due to time dilation during both legs of the journey; both twins correctly find their own clock be tautologically ticking at one second per second; there are fewer seconds ticked off along the traveler’s path through spacetime than along stay-at-home’s.PeterDonis said:Actually, yes, it is.
Ok, I thought the poster meant "behind", but reading back now, "slower" is what was meant.sdkfz said:At the point where Alice is passing Bob, the elapsed time on her clock may be less than what's on Bobs, but that's got nothing to do with what the relative rate of their clocks are at that time, nor how they perceive the length of each others metre rods.
It doesn't matter what they had for breakfast, either. You are confusing your scenario by having irrelevant stuff happen before the bit that matters.Edit:
I think your underlying issue is you are always searching for some absolute single "truth". One real clock speed, one real length, ...
Your scenario has Bob as a stay-at-home twin and Alice doing the travel. You are trying to make Bob's Universe the "real" one and Alice's not.
That isn't how it works. (Why Alice's clock has less elapsed time and is actually younger when they meet again is for another thread.)
At the point they meet again in your scenario, they are in relative motion. For Bob, Alice's clock is slower, and her rod is shorter. For Alice, Bobs's clock is slower, and his rod is shorter.Carol and Dave could pass each other in space, in a way that neither can tell who accelerated or not to get where they are, let's say they woke up from hibernation. Both can consider themselves as motionless and the other is zooming past. That's fundamental in relativity. If the relative speed between Carol and Dave is 1,000 kph it will have the same "effect" as the same relative speed between Alice and Bob.
Which one? I can't find it.student34 said:I did say that they were at rest initially with each other at 1:00pm in one of my posts.
If you're saying that Alice and Bob were initially at rest, co-located with each other, and then Alice accelerated away from Bob, then she would have to turn around and come back to Bob. None of which I have gathered from your previous posts. You can't keep changing the scenario around and expect it to make sense to other people. You need to start out with a complete, consistent description of what you are talking about.student34 said:This would seem to imply that Alice accelerated first, thus having the slower clock, no?
Ibix said:It's closely analogous to slicing a sausage perpendicular to its length and seeing a circular cross-section and somebody else slicing it diagonally and seeing an elliptical cross section. The sausage is not in a superposition of circular and elliptical - the different people are just making different measurements on it.
I never sausage an analogy before.Ibix said:No. Position and length are well defined. Different frames measure different aspects of a 4d object but both call it length. It's closely analogous to slicing a sausage perpendicular to its length and seeing a circular cross-section and somebody else slicing it diagonally and seeing an elliptical cross section. The sausage is not in a superposition of circular and elliptical - the different people are just making different measurements on it.
Okay, but we are not interested in how old Alice and Bob are. We are interested in Alice's measurement of Bob's metre stick!student34 said:For Alice to accelerate and then return to go past Bob is almost exactly the same as the Twins Paradox, except Alice has no decelerated when they see each other's clock.
And it's not exactly kosher!PAllen said:I never sausage an analogy before.
Post #28PeterDonis said:Which one? I can't find it.
That post still doesn't make it very clear what Alice is doing between 1 pm and 5 pm by Bob's clock. And, as I pointed out in post #76, that doesn't matter anyway if what you're interested in is the length contraction, time dilation, and relativity of simultaneity between Alice and Bob when they pass each other, both moving inertially, at 5 pm by Bob's clock. So all it does is add irrelevant noise to the discussion. As I recommended in post #76, it would be much better just to stick to Alice and Bob passing each other, both moving inertially, in this thread, and talk about other things like "twin paradox" scenarios in a separate thread if you want to do that.student34 said:Post #28
Yes I know what length contraction is. My issue is that the particle at the end of the meter stick seems to exist in two different places, and not just in time but in space.PeroK said:Okay, but we are not interested in how old Alice and Bob are. We are interested in Alice's measurement of Bob's metre stick!
Let's assume, for the sake of argument, that Alice is 33 years and 115 days old. And Bob is 25 years and 62 days old.
Now, back to the problem. Let me try to summarise where we are:
1) Bob's metre stick is ##1m## long in its rest frame.
2) In Alice's rest frame the metre stick is traveling at a large fraction of the speed of light. That makes it difficult for Alice to measure - the metre stick isvmoving almost as far as the light it's emitting.
3) Alice needs a reliable definition of length for a moving object.
3a) One definition of length would be: whatever Bob measures. There is some logic in this. Alice simply says that it's too difficult to measure the length of something moving at near light speed and doesn't try to measure it. We would leave the length of a moving object as undefined.
This is, in fact, called the proper length of the metre stick: its length in a frame where the stick remains at rest.
3b) Can Alice do better? Can Alice give some meaning to the concept of length for a moving object?
One idea is to make simultaneous measurements of both ends of the stick. I've given some examples in this thread of what goes wrong if you allow the position of one end of the metre stick to be measured at some time and the other end at a later time. The measurements must be made at the same time.
This is impossible because Alice cannot be in two places at the same time. (Neither can Bob - but as the stick is at rest relative to Bob, it doesn't matter whether he makes simultaneous measurements or not.) And, if the stick was not moving very fast, then Alice would use light signals and neglect the delay in the speed of light. But, the stick is moving too fast for that.
So, Alice has a genuine problem here. The solution is that she needs a friend, who can stand a certain distance away from her, at rest relative to each other and with synchronised clocks. They then measure the time that each end of the stick passes them and compare notes. This is Carol, I guess.
First, Carol stands ##1m## away from Alice (in their rest frame). The front of the metre stick passes Carol at some time ##t_1## (on Carol's watch), the rear of the metre stick passes Carol at ##t_2## (on Carol's watch); the front of the metre stick passes Alice at ##t_3## (on Alice's watch) and the rear at ##t_4## on Alice's watch.
To get the required simultaneous measurements, we need ##t_2 = t_3##. I.e. we need the front of the metre stick to be passing Alice at the same time as the rear is passing Carol.
To get to the point, what they find is that if they stand ##1m## apart, then ##t_2 < t_3##. I.e. the end of the metre stick has passed Carol before the front has reached Alice. I.e. they find that, using this definition of the length of a moving object, the stick is less than ##1m##.
They must repeat the experiment until they get ##t_2 = t_3##. And, if the speed of the rod is ##0.866c##, they find that they must stand ##0.5m## apart to get ##t_2 = t_3##.
This is length contraction.
All I want to talk about is where, objectively, the particle is at the end of the meter stick when Alice and Bob are both at the same location as she passes by.PeterDonis said:That post still doesn't make it very clear what Alice is doing between 1 pm and 5 pm by Bob's clock. And, as I pointed out in post #76, that doesn't matter anyway if what you're interested in is the length contraction, time dilation, and relativity of simultaneity between Alice and Bob when they pass each other, both moving inertially, at 5 pm by Bob's clock. So all it does is add irrelevant noise to the discussion. As I recommended in post #76, it would be much better just to stick to Alice and Bob passing each other, both moving inertially, in this thread, and talk about other things like "twin paradox" scenarios in a separate thread if you want to do that.
In other words, assuming absolute time and simultaneity. This is the relativity of simultaneity biting you again. It's definitely the hardest concept in SR. You just replaced "absolutely" with "objectively".student34 said:All I want to talk about is where, objectively, the particle is at the end of the meter stick when Alice and Bob are both at the same location as she passes by.
What standard of synchronization will you use to judge the position of the meter stick end at the same time when Alice and Bob meet?student34 said:All I want to talk about is where, objectively, the particle is at the end of the meter stick when Alice and Bob are both at the same location as she passes by.
There is no such thing, as you have already been told multiple times in this thread.student34 said:All I want to talk about is where, objectively, the particle is at the end of the meter stick when Alice and Bob are both at the same location as she passes by.