Because of length contraction, does absolute position exist?

[Ibix]

But can't we talk about what the distance between the two particles is in Alice's frame of reference?

Sure, but she needs a meter rule too. Both will see the other's meter rule contracted and therefore cannot agree that the front end of the two meter rules will be at the same place at the same time as the back ends are at the same place. Thus they must measure different distances between the particles.

 

[student34]

Sure, but she needs a meter rule too. Both will see the other's meter rule contracted and therefore cannot agree that the front end of the two meter rules will be at the same place at the same time as the back ends are at the same place. Thus they must measure different distances between the particles.

But I am only interested in where the particle is at the meter stick in front of Bob. This will help me understand how there is no superposition of that particle. It is exactly this scenario that is causing me to think that that particle must be in two different locations in space.

 

[Ibix]

It is exactly this scenario that is causing me to think that that particle must be in two different locations in space.

It is (edit: at least in some frames), just not at the same time. Thats what jbriggs444's diagram is showing you. Do you understand it? It's really the easiest way to resolve all relativistic weirdness.

 

[student34]

It is, just not at the same time. Thats what jbriggs444's diagram is showing you. Do you understand it? It's really the easiest way to resolve all relativistic weirdness.

Yeah, I think that diagram will help me. Is one of the frames in motion?

 

[jbriggs444]

Yeah, I think that diagram will help me. Is one of the frames in motion?

Yes. In order to get two different simultaneity conventions, the frames need to be in relative motion.

 

[PeterDonis]

I am only interested in where the particle is at the meter stick in front of Bob.

Then why do you care what Alice measures? You have already said that both particles are at rest relative to Bob, one at each end of the meter stick. That is a perfectly consistent scenario and there is no "superposition" whatever: each particle has a well-defined unique position relative to Bob and his meter stick.

 

[student34]

Yes. In order to get two different simultaneity conventions, the frames need to be in relative motion.

In the example given in the diagram, what is happening with the frames? Why is the meter stick longer in Frame B?

 

[PeterDonis]

It is exactly this scenario that is causing me to think that that particle must be in two different locations in space.

Why would you think that? Each particle is at one end of Bob's meter stick. And that is just as true for Alice as it is for Bob. The particles are moving relative to Alice and Alice's meter stick will measure a different distance between them, but Alice's meter stick will still measure each particle to be co-located with one end of Bob's meter stick. In Alice's frame each particle just moves in sync with its end of Bob's meter stick, instead of being at rest at its end of Bob's meter stick as it is in Bob's frame. The fact of each particle being co-located with one end of Bob's meter stick is an invariant, and is true in all frames.

 

[student34]

Then why do you care what Alice measures?

Because, for Alice, the particle seems to be in a different location than for Bob. When Alice is at x with Bob, the particle is only 50 cm from Alice. But as others have responded, we are also using Alice's clock, which is at a different time than Bob's.

But my response to that is that Alice's time is just some time t earlier than Bob's clock. And at that time the particles were still 100cm apart for Bob. The particles are never 50cm apart other than for Alice's frame. What else can I think other than some kind of superposition? I just don't understand how this is wrong.

 

[PeterDonis]

When Alice is at x with Bob, the particle is only 50 cm from Alice.

"When Alice is at x with Bob" is different for Alice than it is for Bob. Alice and Bob have different notions of simultaneity because of their relative motion.

we are also using Alice's clock, which is at a different time than Bob's.

It's not just that it's "at a different time". It is running at a different rate. To Alice, Bob's clock is running slow; to Bob, Alice's clock is running slow.

In other words, there are actually three frame-dependent effects going on here: length contraction, time dilation, and relativity of simultaneity. And it is impossible to look at just one of these, or even just two, and get a correct picture of what is going on. You have to look at all three in combination. That is what the spacetime diagrams that have been drawn in this thread are doing.

The particles are never 50cm apart other than for Alice's frame.

And they are never 100cm apart other than in Bob's frame. And in any other frame, they would be a different distance apart, and would be moving at a different speed. And the time dilation and simultaneity would also be different.

 

[PeterDonis]

When you measure the length of the stick in frame A, you are measuring from event a1 to event a2 such that a1 and a2 are simultaneous according to frame A.

When you measure the length of the stick in frame B, you are measuring from event b1 to event b2 such that b1 and b2 are simultaneous according to frame B.

It would probably be better to reverse the frame nomenclature here. The natural way to interpret this diagram is that what you are calling frame A is Bob's frame (since that's the frame in which both worldlines, representing the two ends of Bob's meter stick, are at rest) and what you are calling frame B is Alice's frame.

 

[PeterDonis]

Why is the meter stick longer in Frame B? Is Frame A traveling fast?

Per my post #46 just now, as @jbriggs444 has drawn the diagram, "Frame A" is really what you are calling Bob's frame, and the "Frame A snapshot" is Bob's meter stick at an instant of Bob's time. "Frame B" in the diagram is really what you are calling Alice's frame, and the "Frame B snapshot" is Bob's meter stick at an instant of Alice's time. Because the geometry of the diagram is actually hyperbolic rather than Euclidean, the length of the "Frame B snapshot" is actually shorter than the length of the "Frame A snapshot", corresponding to the fact that Alice sees Bob's meter stick length contracted. And the fact that the "Frame B snapshot" is at an angle to the "Frame A snapshot" corresponds to relativity of simultaneity.

 

[PeroK]

Because, for Alice, the particle seems to be in a different location than for Bob. When Alice is at x with Bob, the particle is only 50 cm from Alice. But as others have responded, we are also using Alice's clock, which is at a different time than Bob's.

But my response to that is that Alice's time is just some time t earlier than Bob's clock. And at that time the particles were still 100cm apart for Bob. The particles are never 50cm apart other than for Alice's frame. What else can I think other than some kind of superposition? I just don't understand how this is wrong.

If Alice observes one end of Bob's metre stick at her feet at time $t = 0$ and the other end at her feet at a time a fraction of a second later. She might claim that the metre stick has zero length. Why is that wrong?

Observed position of one end of metre stick: $x_1 = 0$.

Observed position of other end of metre stick: $x_2 = 0$.

Length of rod $L = x_1 - x_2 = 0$.

Why is that wrong?

 

[student34]

Yes. In order to get two different simultaneity conventions, the frames need to be in relative motion.

A meter stick is not a line segment the exists only as a snapshot in time. It has extent in time and is more like a parallelogram extending indefinitely into the past and into the future. When you measure its length, you are taking a snapshot at an instant in time according to some chosen frame.
View attachment 294841
When you measure the length of the stick in frame A, you are measuring from event a1 to event a2 such that a1 and a2 are simultaneous according to frame A.

When you measure the length of the stick in frame B, you are measuring from event b1 to event b2 such that b1 and b2 are simultaneous according to frame B.

I am sorry everyone; this is embarrassing. I just realized that this diagram is a description of my thought experiment. I understand now what is going on. Thanks a lot for this diagram.

 

[student34]

It's not just that it's "at a different time". It is running at a different rate. To Alice, Bob's clock is running slow; to Bob, Alice's clock is running slow.

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.

 

[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.