I Because of length contraction, does absolute position exist?

[PeroK]

I don't think this is correct. Only one of them would have the slower clock. It depends which one did the acceleration from when they were at rest with one another. In this case Alice did the acceleration, and so only Alice's clock is slower.

What if they both accelerated equally in order to achieve their relative velocity? Like both went round the Large Hadron Collider in opposite directions?

 

[student34]

What if they both accelerated equally in order to achieve their relative velocity? Like both went round the Large Hadron Collider in opposite directions?

I guess both would have slower clocks relative to clocks on Earth, but Alice's would still slower since she would have had to do more of the acceleration to get to a 50% length contraction of Bob's meter stick.

Are you saying that I am wrong about my correction?

 

[PeroK]

I guess both would have slower clocks relative to clocks on Earth, but Alice's would still slower since she would have had to do more of the acceleration to get to a 50% length contraction of Bob's meter stick.

Are you saying that I am wrong about my correction?

Yes, you couldn't be more wrong, as the saying goes! The Lorentz Transformation is symmetric. There is no "absolutely at rest" frame and no "accelerated and really moving" frame.

That is, in fact, the principle of relativity that dates back to Galileo.

 

[student34]

Yes, you couldn't be more wrong, as the saying goes! The Lorentz Transformation is symmetric. There is no "absolutely at rest" frame and no "accelerated and really moving" frame.

That is, in fact, the principle of relativity that dates back to Galileo.

So do you also not accept the explanation for the Twins paradox?

 

[PeterDonis]

I don't think this is correct. Only one of them would have the slower clock.

This is incorrect. Each one is moving inertially, so each one perceives the other's clock to be running slow.

It depends which one did the acceleration from when they were at rest with one another.

There is no requirement that they ever were at rest relative to one another.

If they started out at rest, and one of them accelerated away, and then turned around and accelerated back, and came to rest again relative to the other, while the other one remained inertial the whole time, then the one who traveled would have less time elapsed on their clock (and both of them would agree about this). But that is not time dilation; that's a "twin paradox" scenario, which is something different.

And, as above, there is no requirement that two people who happen to be moving relative to one another now were ever at rest relative to one another in the past, so not every scenario will even meet the requirements for being a "twin paradox" scenario where there is an invariant (i.e., not frame dependent) answer as to whose clock has more or less elapsed time. In many cases there is simply no invariant answer to that question; all you have is the various frame-dependent answers for each one's frame.

 

[PeterDonis]

So do you also not accept the explanation for the Twins paradox?

Please see my post #55 just now. The scenario we have been discussing in this thread is not the same as the twin paradox scenario.

 

[PeterDonis]

Alice's would still slower since she would have had to do more of the acceleration to get to a 50% length contraction of Bob's meter stick.

Even in "twin paradox" type scenarios, this reasoning is not correct. Length contraction is still frame-dependent no matter what the past history of either observer was.

 

[sdkfz]

So do you also not accept the explanation for the Twins paradox?

You are missing what puts "paradox" in "twin's paradox".

i.e. that for both the other has a slower clock (while in relative motion).

The explanation of the twin's paradox is not about one of them "really" having the slowest clock.

 

[PeterDonis]

The explanation of the twin's paradox is not about one of them "really" having the slowest clock.

Actually, yes, it is. As I said in post #55, the difference in elapsed time on the twins' clocks when they come back together is an invariant, not frame-dependent. So yes, one of them "really" has aged less.

 

[sdkfz]

Actually, yes, it is. As I said in post #55, the difference in elapsed time on the twins' clocks when they come back together is an invariant, not frame-dependent. So yes, one of them "really" has aged less.

Of course one has really aged less, that's the other side of the "paradox". I'm not contradicting that.

 

[student34]

This is incorrect. Each one is moving inertially, so each one perceives the other's clock to be running slow.There is no requirement that they ever were at rest relative to one another.

If they started out at rest, and one of them accelerated away, and then turned around and accelerated back, and came to rest again relative to the other, while the other one remained inertial the whole time, then the one who traveled would have less time elapsed on their clock (and both of them would agree about this). But that is not time dilation; that's a "twin paradox" scenario, which is something different.

And, as above, there is no requirement that two people who happen to be moving relative to one another now were ever at rest relative to one another in the past, so not every scenario will even meet the requirements for being a "twin paradox" scenario where there is an invariant (i.e., not frame dependent) answer as to whose clock has more or less elapsed time. In many cases there is simply no invariant answer to that question; all you have is the various frame-dependent answers for each one's frame.

I did say that they were at rest initially with each other at 1:00pm in one of my posts. This would seem to imply that Alice accelerated first, thus having the slower clock, no?

 

[student34]

You are missing what puts "paradox" in "twin's paradox".

i.e. that for both the other has a slower clock (while in relative motion).

The explanation of the twin's paradox is not about one of them "really" having the slowest clock.

All I am saying is that Alice accelerated relative to Bob, and so Alice would have the slower clock. The other poster said that they would both see each other's clock as slower. I do not agree with that. Am I wrong?

 

[PeroK]

So do you also not accept the explanation for the Twins paradox?

We're not talking about the twin paradox!

 

[sdkfz]

All I am saying is that Alice accelerated relative to Bob, and so Alice would have the slower clock. The other poster said that they would both see each other's clock as slower. I do not agree with that. Am I wrong?

While in relative motion, no matter how they got there, two observers will consider the others' clock to be slower. It's symmetrical. You are wrong.

Personally I think you ought to leave Twins' Paradox stuff out of this thread. Get one thing right at a time.

 

[jbriggs444]

All I am saying is that Alice accelerated relative to Bob, and so Alice would have the slower clock. The other poster said that they would both see each other's clock as slower. I do not agree with that. Am I wrong?

Yes, you are wrong.

Alice accelerated. Her pre-acceleration notions of relative clock rates and of clock synchronization have become irrelevant. In the inertial frame in which Alice is now at rest, her clocks are running normally at one second per second and someone has gone out and synchronized them according to Alice's new standard of rest.

If you want Alice to be using de-synchronized clocks, you need to say so.

When Alice goes to figure out how fast Bob's clocks are ticking, she compares readings on a single Bob clock as it passes by first one Alice clock and then a second Alice clock.

Bob does his comparison the same way. Comparing one Alice clock against two Bob clocks.

 

[student34]

While in relative motion, no matter how they got there, two observers will consider the others' clock to be slower. It's symmetrical. You are wrong.

Personally I think you ought to leave Twins' Paradox stuff out of this thread. Get one thing right at a time.

So in my thought experiment with Bob and Alice, would they both see each other's clock as the same when she goes past him?

 

[sdkfz]

So in my thought experiment with Bob and Alice, would they both see each other's clock as the same when she goes past him?

I can't parse that, sorry.

 

[student34]

I can't parse that, sorry.

What would Bob see on her clock as she goes past him, behind, same or ahead of his?

 

[sdkfz]

What would Bob see on her clock as she goes past him, behind, same or ahead of his?

It's all going to be about initial conditions - how and when they synchronised etc. I found your scenario too muddled to say whose clock would be ahead/behind.
... but that's largely irrelevant to the fact that each will consider the others' clock to be slower than their own.

Edit: reading your scenario again, I would expect Alice's clock to be behind Bob's; but the reason is getting into Twins' Paradox stuff that I still think you ought to have a separate thread about at a later time.

 

[student34]

It's all going to be about initial conditions - how and when they synchronised etc. I found your scenario too muddled to say whose clock would be ahead/behind.
... but that's largely irrelevant to the fact that each will consider the others' clock to be slower than their own.

Edit: reading your scenario again, I would expect Alice's clock to be behind Bob's; but the reason is getting into Twins' Paradox stuff that I still think you ought to have a separate thread about at a later time.

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.

 

[student34]

We're not talking about the twin paradox!

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.

 

[sdkfz]

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.

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.

 

[anuttarasammyak]

Say class members of a space pilot school, Alice, Bob, Carol, Dave, Ellen, ... , go their ways and meet again. If Ellen stays in an IFR during leaving and meeting again she is the oldest. SR is applicable to her observation that the spaceship of their classmates were under the effect of time dilation. The others cannot claim it because they transferred between different momentary IFRs and their SR time dilation observation pieces can not be summed up properly.

 

[Nugatory]

The explanation of the twin's paradox is not about one of them "really" having the slowest clock.

Actually, yes, it is.

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.

Whether we choose to describe this situation as “one of them ‘really’ having the slowest clock” is a matter of personal taste…. But I have to say that I prefer sdkfz’s wording here. It makes it easier to avoid/address a very common beginner’s misunderstanding.

 

[student34]

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.

Ok, I thought the poster meant "behind", but reading back now, "slower" is what was meant.

 

[PeterDonis]

I did say that they were at rest initially with each other at 1:00pm in one of my posts.

Which one? I can't find it.

This would seem to imply that Alice accelerated first, thus having the slower clock, no?

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.

That said, if at the time we are talking about, Alice is flying past Bob inertially, which is what has seemed to be the case up to now in this discussion, none of Alice's past history matters as far as the current length contraction, time dilation, and relativity of simultaneity between Alice and Bob is concerned. It's simply irrelevant. It might be relevant to assessing the total time elapsed on Alice's clock vs. Bob's clock since some past event at which they were together, but it's not relevant at all to assessing what the length contraction, time dilation, and relativity of simultaneity are between Alice and Bob now. And the latter is what this thread is supposed to be about.

I would recommend sticking to the one thing you started with in this thread and starting a separate thread if you want to discuss other things.

 

[ergospherical]

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.

Pork-themed analogies appear to be popular with physicists. Sir Arthur Eddington once said:

"world-wide instants are not natural cleavage planes of time... they are imaginary partitions which we find it convenient to adopt... A pig may be most familiar to us in the form of rashers, but the unstratified pig is a simpler object to the biologist who wishes to understand how the animal functions."

 

[PAllen]

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.

I never sausage an analogy before.

 

[sdkfz]

Normally, physics starts with an ergospherical cow.

 

[PeroK]

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.

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.

 

[PeroK]

I never sausage an analogy before.

And it's not exactly kosher!

 

[student34]

Which one? I can't find it.

Post #28

 

[PeterDonis]

Post #28

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]

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.

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.

 

[student34]

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.

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.

 

[PeroK]

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.

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".

Two observers in relative motion do not agree about a universal "now". Technically, therefore, "where the particle is when $t = 0$ in Alice's reference frame" and "where the particle is when $t'= 0$ in Bob's reference frame" are two different spacetime events.

 

[jbriggs444]

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?

There is no absolute standard of simultaneity. That is why the question you ask is ambiguous.

 

[PeterDonis]

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.

 

[PeterDonis]

The OP question has been answered more than enough times at this point. Thread closed.