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Non Life Sciences => Physics, Astronomy & Cosmology => Topic started by: Jaaanosik on 29/07/2020 16:27:51

Title: Is angular momentum frame dependent?
Post by: Jaaanosik on 29/07/2020 16:27:51
Hi all,
Is angular momentum frame dependent?
Before going into details how to check it out, I'd like to ask a questions what it would mean for physics,
Jano
Title: Re: Is angular momentum frame dependent?
Post by: Halc on 29/07/2020 16:47:00
Is angular momentum frame dependent?
It seems not. Take an example from relativity.  Earth rotates once every 23:56 hours, but in a frame where it is moving at .8c, it rotates once every 53:33 hours (slower) but has more mass, so the angular momentum seems unchanged.
Title: Re: Is angular momentum frame dependent?
Post by: Jaaanosik on 29/07/2020 17:18:49
Halc,
that's a good point.
What happens to a wheel when the rim of the wheel rotates at 0.7c and the wheel translates at 0.7c.
The wheel axle is 90 degrees to the velocity vector, the wheel is rolling and a trajectory of a point on the wheel rim is a cycloid in the outer reference frame.
The wheel is just rotating in the axle reference frame.
What happens to the angular momentum like here:

(https://i.imgur.com/G8YLStS.png)

From Relativistic Hall Effect paper: https://arxiv.org/pdf/1112.5618.pdf
The angular momentum is symmetrical in the rest/axle frame (a) but it is not symmetrical in the outside frame (b).
What does it mean for physics?
Jano
Title: Re: Is angular momentum frame dependent?
Post by: Halc on 29/07/2020 18:21:05
What happens to a wheel when the rim of the wheel rotates at 0.7c and the wheel translates at 0.7c.
Well, for one thing, you have a wheel that cannot stop spinning without breaking.  A wheel of proper radius 1 spinning at a rim speed of 0.7c has a proper circumference of about ~8.8, as opposed to ~6.3 for a non-spinning wheel of the same radius.  If you slowed the spin of that to a stop, the spokes would be too short and something would break.

Quote
What happens to the angular momentum
It appears to require tensor calculus to answer that, which is a bit above my pay grade, but being tensor calculus, the answer should be frame invariant, so the momentum is the same as the wheel not moving west.
Of interesting note is that the center of gravity of such a wheel migrates away from the axle, or probably better worded as the axle moving away from the center of gravity, since I don't think conservation laws allow the latter to move like that.  So if you tow such a wheel on a long string up to high linear speed, it will migrate to the side a bit as it does so.

The paper does not concentrate on angular momentum calculations much, concentrating instead on the migrations of the geometric centroid and the energy centroid in alternative frames. I think the center of mass would be either the energy centroid or somewhere between the two. Certainly not right at the geometric centroid.
Title: Re: Is angular momentum frame dependent?
Post by: Kryptid on 29/07/2020 19:36:17
I hope this doesn't lead to another discussion about reactionless drives.
Title: Re: Is angular momentum frame dependent?
Post by: Jaaanosik on 29/07/2020 19:47:24
Halc,
the first lines from the abstract:
Quote
We consider the relativistic deformation of quantum waves and mechanical bodies carrying intrinsic angular momentum (AM).
When observed in a moving reference frame, the centroid of the object undergoes an AM-dependent transverse shift...

The angular momentum is the purpose of this paper.

(https://i.imgur.com/zU3fdbf.png)

The paper shows the delta between the angular momentum of the axle frame (a) and the outside frame (b).
The first one in (a) is symmetrical the second one in (b) is not symmetrical, there is a transverse shift.
That's what the paper shows.
Quote
... The antisymmetric AM tensor can be represented by a pair of three-vectors, Lαβ = (H,L), where L = r×p
is the axial vector of the AM, whereas H = pct − (ε/c) r is the polar vector marking the rectilinear trajectory of the particle [13].

What are the implications for physics?
Jano
Title: Re: Is angular momentum frame dependent?
Post by: Jaaanosik on 29/07/2020 19:50:52
I hope this doesn't lead to another discussion about reactionless drives.
Hi Kryptid,
no, just the question, what does it mean for physics?
The angular momentum is frame dependent, as per the paper.
What does it mean for the Noether's theorem?
Jano
Title: Re: Is angular momentum frame dependent?
Post by: Kryptid on 29/07/2020 22:38:25
What does it mean for the Noether's theorem?

Nothing. Noether's theorem still applies.
Title: Re: Is angular momentum frame dependent?
Post by: Malamute Lover on 29/07/2020 22:46:35
I have been through the paper a couple of times and if I am reading it right, total AM (L) is still conserved but the distribution of intrinsic AM (r, 'spin')) and extrinsic AM (p, 'orbital') are frame dependent. But the AM conservation law applies only to L. Even in classical physics r and p are not individually conserved.

I think.

But then it has been a long hot day and I am after all older than at least some of the hills.

And I need  a nap.
Title: Re: Is angular momentum frame dependent?
Post by: Jaaanosik on 29/07/2020 23:06:41
Let us apply an acceleration on the axle like this:

(https://i.imgur.com/6ME2htn.png)

The (a) frame analysis - there is no change of the angular momentum.
The (b) frame analysis - there is a change of the angular momentum because the axle is not in the center of mass.
The wheel is a 'pendulum', the mass is shifted in the +y axis direction,
Jano
Title: Re: Is angular momentum frame dependent?
Post by: Halc on 29/07/2020 23:54:55
Angular momentum seems constant (not dependent on chosen axis of rotation) in an inertial frame where the center of mass is stationary. So Earth, in its own inertial frame has X angular momentum even if the axis is considered somewhere else. This is not true if Earth is moving, so Earth moving linearly at 30 km/sec has less angular momentum around its center of gravity than it does around any other point, like the sun for instance.
That makes angular momentum frame dependent.

Let us apply an acceleration on the axle like this:
You are equivocating acceleration and velocity. That doesn't work. The deformation in the picture is due to linear velocity, not at all due to acceleration.

Quote
The (a) frame analysis - there is no change of the angular momentum.
But the diagram depicts center of mass/deformation, not change of those things.

Quote
The (b) frame analysis - there is a change of the angular momentum because the axle is not in the center of mass.
Does not directly follow. Momentum is not depicted at all in the pictures, but if the momentum vector is applied to the right-side (original) picture, it would be located somewhere other than the axle, but its magnitude and direction (the two elements that matter for a vector) would not necessarily be different.
Title: Re: Is angular momentum frame dependent?
Post by: Malamute Lover on 30/07/2020 00:19:07
Let us apply an acceleration on the axle like this:

(https://i.imgur.com/6ME2htn.png)

The (a) frame analysis - there is no change of the angular momentum.
The (b) frame analysis - there is a change of the angular momentum because the axle is not in the center of mass.
The wheel is a 'pendulum', the mass is shifted in the +y axis direction,
Jano

Quote
This situation is quite similar to the spin Hall effect in various systems, where variations in the intrinsic AM(spin) are compensated at the expense of the centroid shift generating extrinsic AM.
[…]
However, in addition to the shape deformations, a rotating body also acquires mass deformations.  The y >0 and y <0 sides of the wheel have different velocities in the moving frame and their constituent particles acquire different local γ-factors. Owing to this, the y >0 particles become heavier than the y <0 particles.

[page 2 of the article, the same page as the spoked wheel picture]

The spokes on the top are denser, as seen in the picture and heavier. The energy centroid is higher than the geometric centroid. Looks to me like overall angular momentum is conserved, but redistributed.

Title: Re: Is angular momentum frame dependent?
Post by: Jaaanosik on 30/07/2020 04:38:29
Angular momentum seems constant (not dependent on chosen axis of rotation) in an inertial frame where the center of mass is stationary. So Earth, in its own inertial frame has X angular momentum even if the axis is considered somewhere else. This is not true if Earth is moving, so Earth moving linearly at 30 km/sec has less angular momentum around its center of gravity than it does around any other point, like the sun for instance.
That makes angular momentum frame dependent.

Let us apply an acceleration on the axle like this:
You are equivocating acceleration and velocity. That doesn't work. The deformation in the picture is due to linear velocity, not at all due to acceleration.

Quote
The (a) frame analysis - there is no change of the angular momentum.
But the diagram depicts center of mass/deformation, not change of those things.

Quote
The (b) frame analysis - there is a change of the angular momentum because the axle is not in the center of mass.
Does not directly follow. Momentum is not depicted at all in the pictures, but if the momentum vector is applied to the right-side (original) picture, it would be located somewhere other than the axle, but its magnitude and direction (the two elements that matter for a vector) would not necessarily be different.
The +y side has bigger mass than the -y side in the (b) frame.
Let us imagine a rod on the axle with two weights at the end representing the wheel mass.
The W+y weight is at the RE location, on the +y side, the other W-y weight on the -y side, the same distance from the axle.
W+y  has bigger mass than  W-y in the (b) frame. These masses represent the top and the bottom part of wheel, asymmetrical distribution.
W+y  has equal mass to W-y in the (a) frame, symmetrical distribution.
The acceleration is not supposed to change the angular momentum in the (a) frame.
The acceleration is supposed increase the angular momentum in the (b) frame because of the W+y  and W-y mass delta.

I am not equivocating velocity and the acceleration.
I am saying the velocity creates the delta in mass as per the paper.
Now we introduce the acceleration as a new thought experiment.
The delta in mass will increase the angular momentum of the wheel in the (b) frame but it is not suppose to do anything in the (a) frame.
Jano

Title: Re: Is angular momentum frame dependent?
Post by: Jaaanosik on 30/07/2020 04:49:53
Let us apply an acceleration on the axle like this:

(https://i.imgur.com/6ME2htn.png)

The (a) frame analysis - there is no change of the angular momentum.
The (b) frame analysis - there is a change of the angular momentum because the axle is not in the center of mass.
The wheel is a 'pendulum', the mass is shifted in the +y axis direction,
Jano

Quote
This situation is quite similar to the spin Hall effect in various systems, where variations in the intrinsic AM(spin) are compensated at the expense of the centroid shift generating extrinsic AM.
[…]
However, in addition to the shape deformations, a rotating body also acquires mass deformations.  The y >0 and y <0 sides of the wheel have different velocities in the moving frame and their constituent particles acquire different local γ-factors. Owing to this, the y >0 particles become heavier than the y <0 particles.

[page 2 of the article, the same page as the spoked wheel picture]

The spokes on the top are denser, as seen in the picture and heavier. The energy centroid is higher than the geometric centroid. Looks to me like overall angular momentum is conserved, but redistributed.

Can the redistribution happen without new 4-accelerations in the (b) frame?
Can the -y spokes bend without new 4-accelerations in the (b) frame?
How is (a) frame going to see these 4-accelerations?
The (a) frame cannot see different 4-accelerations between +y and -y sides. The (a) frame is symmetrical.
The (b) frame sees these new 4-accelerations between +y and -y sides. The (b) frame is asymmetrical.

What does it mean for physics if two different frame observers do not agree on the accelerations???
Jano

Title: Re: Is angular momentum frame dependent?
Post by: Kryptid on 30/07/2020 04:53:58
What does it mean for physics if two different frame observers do not agree on the accelerations???

Nothing much. That's to be expected for relativity.
Title: Re: Is angular momentum frame dependent?
Post by: Malamute Lover on 30/07/2020 14:42:21
What does it mean for physics if two different frame observers do not agree on the accelerations???

Perfectly normal.

A passenger on a rocket ship experiences constant acceleration. An observer in an inertial frame of reference sees that because of time dilation, the rocket ship acceleration is dropping over time. They disagree about acceleration exactly as expected.


Something to also keep in mind is that the passenger on the rocket ship experiences acceleration while the observer only infers it.
Title: Re: Is angular momentum frame dependent?
Post by: Jaaanosik on 30/07/2020 19:57:09
What does it mean for physics if two different frame observers do not agree on the accelerations???

Perfectly normal.

A passenger on a rocket ship experiences constant acceleration. An observer in an inertial frame of reference sees that because of time dilation, the rocket ship acceleration is dropping over time. They disagree about acceleration exactly as expected.


Something to also keep in mind is that the passenger on the rocket ship experiences acceleration while the observer only infers it.



This requires a good analysis, here is a textbook:

(https://i.imgur.com/QOjHogH.png?1)

(https://i.imgur.com/pIzrV7I.png)

The rotation creates equally spaced world lines in the (a) frame.

(https://i.imgur.com/CUyh6WW.png)

(https://i.imgur.com/kJGT2Tw.png)

Here are the world lines as seen from frame (b), as per the paper:

(https://i.imgur.com/JgJt1L4.png)

The world lines lose the symmetry in (b), they are asymmetric.
This is the problem, two different frames show different analysis.
They do not agree on the world lines. This is not good.
They do not agree on physics,
Jano
Title: Re: Is angular momentum frame dependent?
Post by: Malamute Lover on 30/07/2020 20:08:58
The world lines lose the symmetry in (b), they are asymmetric.
This is the problem, two different frames show different analysis.
They do not agree on the world lines. This is not good.
They do not agree on physics,

Two different reference frames seeing things differently is what relativity is all about. Nonetheless I fall to see that total angular momentum is not conserved.
Title: Re: Is angular momentum frame dependent?
Post by: Jaaanosik on 30/07/2020 20:30:04
The world lines lose the symmetry in (b), they are asymmetric.
This is the problem, two different frames show different analysis.
They do not agree on the world lines. This is not good.
They do not agree on physics,

Two different reference frames seeing things differently is what relativity is all about.

How can this be reconciled with this: "The laws of physics take the same form in all inertial frames of reference."

Quote
Nonetheless I fall to see that total angular momentum is not conserved.
Can you, please, elaborate how you see it?
Jano
Title: Re: Is angular momentum frame dependent?
Post by: Malamute Lover on 30/07/2020 22:25:05
The world lines lose the symmetry in (b), they are asymmetric.
This is the problem, two different frames show different analysis.
They do not agree on the world lines. This is not good.
They do not agree on physics,

Two different reference frames seeing things differently is what relativity is all about.

How can this be reconciled with this: "The laws of physics take the same form in all inertial frames of reference."

The spokes of the wheel are traveling at different speeds as they go around. This is not an inertial frame of reference

.
Quote
Nonetheless I fall to see that total angular momentum is not conserved.
Can you, please, elaborate how you see it?
Jano

As I said earlier.
Quote
This situation is quite similar to the spin Hall effect in various systems, where variations in the intrinsic AM(spin) are compensated at the expense of the centroid shift generating extrinsic AM.
    […]
    However, in addition to the shape deformations, a rotating body also acquires mass deformations.  The y >0 and y <0 sides of the wheel have different velocities in the moving frame and their constituent particles acquire different local γ-factors. Owing to this, the y >0 particles become heavier than the y <0 particles.

    [page 2 of the article, the same page as the spoked wheel picture


The spokes on the top are denser, as seen in the picture and heavier. The energy centroid is higher than the geometric centroid. Looks to me like overall angular momentum is conserved, but redistributed. “

There is a tradeoff between intrinsic and extrinsic AM with the shift in energy centroid generating extrinsic AM compensating for the changes in intrinsic AM, just like the text of the paper says. Where does it say anything about non-conservation of AM? If that had been demonstrated by the paper, it would have been shouted long and loud as something totally new and ultra-important.
Title: Re: Is angular momentum frame dependent?
Post by: Jaaanosik on 30/07/2020 22:27:05
I see the problem this way.
The frame (a) sees the rim of the wheel symmetrically. See the figures 13.14 and 13.15.
The frame (b) sees the rim of the wheel asymmetrically. See the figure 2 of the paper.
Both are the inertial frame observers.
The (a) and (b) observers are not on the rim itself though.
If there is an accelerated observer on the rim of the wheel then this local observer will measure either symmetrical centripetal acceleration as predicted by (a) frame or asymmetrical acceleration where the spacing between 'the rim blocks' changes as predicted by (b) or ... completely something else that neither reference frame predicted.
Jano
Title: Re: Is angular momentum frame dependent?
Post by: Malamute Lover on 30/07/2020 22:31:36
I see the problem this way.
The frame (a) sees the rim of the wheel symmetrically. See the figures 13.14 and 13.15.
The frame (b) sees the rim of the wheel asymmetrically. See the figure 2 of the paper.
Both are the inertial frame observers.
The (a) and (b) observers are not on the rim itself though.
If there is an accelerated observer on the rim of the wheel then this local observer will measure either symmetrical centripetal acceleration as predicted by (a) frame or asymmetrical acceleration where the spacing between 'the rim blocks' changes as predicted by (b) or ... completely something else that neither reference frame predicted.
Jano

As already stated, the outside observer who sees (b) is not looking at an inertial reference frame. The spokes are going faster on top and slower on the bottom relative to overall motion.
Title: Re: Is angular momentum frame dependent?
Post by: Jaaanosik on 30/07/2020 22:31:48
The world lines lose the symmetry in (b), they are asymmetric.
This is the problem, two different frames show different analysis.
They do not agree on the world lines. This is not good.
They do not agree on physics,

Two different reference frames seeing things differently is what relativity is all about.

How can this be reconciled with this: "The laws of physics take the same form in all inertial frames of reference."

The spokes of the wheel are traveling at different speeds as they go around. This is not an inertial frame of reference

.
Quote
Nonetheless I fall to see that total angular momentum is not conserved.
Can you, please, elaborate how you see it?
Jano

As I said earlier.
Quote
This situation is quite similar to the spin Hall effect in various systems, where variations in the intrinsic AM(spin) are compensated at the expense of the centroid shift generating extrinsic AM.
    […]
    However, in addition to the shape deformations, a rotating body also acquires mass deformations.  The y >0 and y <0 sides of the wheel have different velocities in the moving frame and their constituent particles acquire different local γ-factors. Owing to this, the y >0 particles become heavier than the y <0 particles.

    [page 2 of the article, the same page as the spoked wheel picture


The spokes on the top are denser, as seen in the picture and heavier. The energy centroid is higher than the geometric centroid. Looks to me like overall angular momentum is conserved, but redistributed. “

There is a tradeoff between intrinsic and extrinsic AM with the shift in energy centroid generating extrinsic AM compensating for the changes in intrinsic AM, just like the text of the paper says. Where does it say anything about non-conservation of AM? If that had been demonstrated by the paper, it would have been shouted long and loud as something totally new and ultra-important.


The question: Is accelerated observer on the wheel rim going to see/observe/measure the deformation that is not predicted by (a)?
Jano
Title: Re: Is angular momentum frame dependent?
Post by: Jaaanosik on 30/07/2020 22:41:27
I see the problem this way.
The frame (a) sees the rim of the wheel symmetrically. See the figures 13.14 and 13.15.
The frame (b) sees the rim of the wheel asymmetrically. See the figure 2 of the paper.
Both are the inertial frame observers.
The (a) and (b) observers are not on the rim itself though.
If there is an accelerated observer on the rim of the wheel then this local observer will measure either symmetrical centripetal acceleration as predicted by (a) frame or asymmetrical acceleration where the spacing between 'the rim blocks' changes as predicted by (b) or ... completely something else that neither reference frame predicted.
Jano

As already stated, the outside observer who sees (b) is not looking at an inertial reference frame. The spokes are going faster on top and slower on the bottom relative to overall motion.
My apologies, I introduced a new scenario when the axle is accelerated.
Having said that, my last couple of posts are about the textbook and the paper.
There is no acceleration of the axle here, just to make it clear.
Both, (a) and (b) are inertial observers looking at the rotating wheel.
The question stands, is the accelerated wheel rim observer going to see/observe/measure the deformation or not.
What prediction/observation wins for the accelerated observer on the wheel rim? Is it (a) or (b)?
Jano
Title: Re: Is angular momentum frame dependent?
Post by: Malamute Lover on 30/07/2020 23:00:14
I see the problem this way.
The frame (a) sees the rim of the wheel symmetrically. See the figures 13.14 and 13.15.
The frame (b) sees the rim of the wheel asymmetrically. See the figure 2 of the paper.
Both are the inertial frame observers.
The (a) and (b) observers are not on the rim itself though.
If there is an accelerated observer on the rim of the wheel then this local observer will measure either symmetrical centripetal acceleration as predicted by (a) frame or asymmetrical acceleration where the spacing between 'the rim blocks' changes as predicted by (b) or ... completely something else that neither reference frame predicted.
Jano

As already stated, the outside observer who sees (b) is not looking at an inertial reference frame. The spokes are going faster on top and slower on the bottom relative to overall motion.
My apologies, I introduced a new scenario when the axle is accelerated.
Having said that, my last couple of posts are about the textbook and the paper.
There is no acceleration of the axle here, just to make it clear.
Both, (a) and (b) are inertial observers looking at the rotating wheel.
The question stands, is the accelerated wheel rim observer going to see/observe/measure the deformation or not.
What prediction/observation wins for the accelerated observer on the wheel rim? Is it (a) or (b)?
Jano

I repeat, the observer of (b) is not looking at an inertial reference frame. There are continuous velocity changes as the wheel goes round (velocity being a vector). That is the whole point of the paper is that the relativistic effects seen in (b) are due to acceleration and not even straight line acceleration. This is why the energy centroid moves. And as I have already quoted from the paper, that movement balances the books on AM conservation.
Title: Re: Is angular momentum frame dependent?
Post by: Jaaanosik on 30/07/2020 23:19:55
...
I repeat, the observer of (b) is not looking at an inertial reference frame. There are continuous velocity changes as the wheel goes round (velocity being a vector). That is the whole point of the paper is that the relativistic effects seen in (b) are due to acceleration and not even straight line acceleration. This is why the energy centroid moves. And as I have already quoted from the paper, that movement balances the books on AM conservation.

The observer (a) is not looking at an inertial reference frame as well.
There is an observed velocity change for (a) observer too.
The (a) sees one centripetal acceleration to the center of the wheel and the rim blocks are equally spaced.
The (b) sees normal and tangential accelerations but their sum should point to where?
Energy centroid, different location?
If an observer is on the rim and a measurement is done, what is the direction of the acceleration?
Is the spacing/contraction change of the rim blocks real for the rim observer or not?
Jano
Title: Re: Is angular momentum frame dependent?
Post by: Halc on 30/07/2020 23:25:22
I see the problem this way.
The frame (a) sees the rim of the wheel symmetrically. See the figures 13.14 and 13.15.
The frame (b) sees the rim of the wheel asymmetrically. See the figure 2 of the paper.
Both are the inertial frame observers.
The (a) and (b) observers are not on the rim itself though.
If there is an accelerated observer on the rim of the wheel then this local observer will measure either symmetrical centripetal acceleration as predicted by (a) frame or asymmetrical acceleration where the spacing between 'the rim blocks' changes as predicted by (b) or ... completely something else that neither reference frame predicted.
The question stands, is the accelerated wheel rim observer going to see/observe/measure the deformation or not.
What prediction/observation wins for the accelerated observer on the wheel rim? Is it (a) or (b)?
Jano
I understand the question. No observer on the rim is going to see a symmentrical wheel. The close parts appear bigger, and the far side appears smaller.  It doesn't need to spin or move for that to be true.

As for the view, it is the same for any observer on the wheel at any point on the wheel.  They're all going to see the exact same thing which is the rest of the wheel going off into the distance to each side and curving up above in the 'up' direction.  There would be no way to tell where on the wheel you are in the picture or if the wheel is moving. This is exactly per principle of relativity which say there is no way, from within a box, to determine inertial motion of the box.

Of course the observer on the rim can tell that the wheel is spinning, which way, and how fast. But those proper facts are the same for any observer on the rim.
Title: Re: Is angular momentum frame dependent?
Post by: Kryptid on 30/07/2020 23:31:17
...
I repeat, the observer of (b) is not looking at an inertial reference frame. There are continuous velocity changes as the wheel goes round (velocity being a vector). That is the whole point of the paper is that the relativistic effects seen in (b) are due to acceleration and not even straight line acceleration. This is why the energy centroid moves. And as I have already quoted from the paper, that movement balances the books on AM conservation.

The observer (a) is not looking at an inertial reference frame as well.

If you are talking about two accelerating frames, those do not necessarily have to agree with each other. Accelerating frames are not inertial.
Title: Re: Is angular momentum frame dependent?
Post by: Malamute Lover on 30/07/2020 23:44:49
...
I repeat, the observer of (b) is not looking at an inertial reference frame. There are continuous velocity changes as the wheel goes round (velocity being a vector). That is the whole point of the paper is that the relativistic effects seen in (b) are due to acceleration and not even straight line acceleration. This is why the energy centroid moves. And as I have already quoted from the paper, that movement balances the books on AM conservation.

The observer (a) is not looking at an inertial reference frame as well.
There is an observed velocity change for (a) observer too.
The (a) sees one centripetal acceleration to the center of the wheel and the rim blocks are equally spaced.
The (b) sees normal and tangential accelerations but their sum should point to where?
Energy centroid, different location?
If an observer is on the rim and a measurement is done, what is the direction of the acceleration?
Is the spacing/contraction change of the rim blocks real for the rim observer or not?
Jano

In (a), that is as seen by an observer who is stationary with respect to the wheel, the AM is constant because the wheel is moving symmetrically. The motion of the wheel does not involve any speed changes, only direction. An observer on the rim will be on a wheel located in an inertial frame of reference and will see no difference in acceleration as the wheel goes around. This is also the case with an observer on the rim in (b). The wheel is still located in an inertial frame of reference to this observer. For an observer on the rim, there is no difference between (a) and (b). The presence or absence of an observer on the rim changes nothing. Neither sees any motion other than the wheel spinning.

In (b), that is as seen by an observer who sees the wheel moving at 0.7 c, the motion of the wheel does involve speed changes, the top goes faster and the bottom goes slower than the axle. The observed relativistic effects are different. The energy centroid that this observer sees has moved upward where the spokes are going faster and have more mass. Yet, as the paper says, the changes in the intrinsic AM are compensated by the changes in the extrinsic AM. Total AM is conserved.
Title: Re: Is angular momentum frame dependent?
Post by: Jaaanosik on 31/07/2020 11:54:13
(https://i.imgur.com/kJGT2Tw.png)

The Fig. 13.15 shows 'the rim blocks'. Is the length contraction real? Are the blocks changing the length for real?
... or it is just a visual effect?
We connect the A0 block and B0 block with a string.
Is the string going to break when the wheel accelerates and the gap grows between A1 block and B1 block?
Jano
Title: Re: Is angular momentum frame dependent?
Post by: Halc on 31/07/2020 12:07:41
The Fig. 13.15 shows 'the rim blocks'. Is the length contraction real?
It's very real. Length contraction is not just a visual effect. It is entirely real. If radius is held constant as depicted, then real gaps must form between the blocks, which will be quite apparent to any observer.
I brought this up in post 3.

Quote
We connect the A0 block and B0 block with a string.
Is the string going to break when the wheel accelerates and the gap grows between A1 block and B1 block?
The string will break, just like in Bell's spaceship 'paradox'.  The only difference here is that they're moving in a circle instead of a straight line, but length contraction is real in both cases.
Title: Re: Is angular momentum frame dependent?
Post by: Malamute Lover on 31/07/2020 14:31:35
The Fig. 13.15 shows 'the rim blocks'. Is the length contraction real?
It's very real. Length contraction is not just a visual effect. It is entirely real. If radius is held constant as depicted, then real gaps must form between the blocks, which will be quite apparent to any observer.
I brought this up in post 3.

Several problems here.

In the Ehrenfest paradox, the circumference contracts, not expands, since the wheel is moving.  From the viewpoint of a non-rotating observer at the center of the circle, the blocks would be seen to shrink along with any gaps between them. This could be seen by comparing the observed length of the blocks compared to their width. An observer riding on the rim would not notice any difference in the size of the blocks or the gaps.

Why are you making the circumference larger? And why are the gaps larger?

The radius is assumed to be constant. This is incorrect. The size of the radius can only be determined with a measuring rod. If we use the rotating stick as the measuring rod, we will see something interesting. It will not appear to be rigid. Light coming the rod from near the rim will take longer to reach the observer in the center and will show the stick at an earlier time in its rotation. The stick will appear to curve and have a length greater than the expected radius.

But if the observer in the center holds a measuring rod out to the circumference until sparks fly when it touches the rim, the measured radius would be as expected for a non-rotating wheel. (Actually, the measurement will be a bit too long as it would take time for the light of the sparks to reach the center. But since the speed of light is known, an adjustment could be made for this.) This result contradicts the radius implied by the observations made of the blocks on the spinning wheel.

Length contraction is not real. It is relative. (Hey, a catchy name for a theory.)

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We connect the A0 block and B0 block with a string.
Is the string going to break when the wheel accelerates and the gap grows between A1 block and B1 block?
The string will break, just like in Bell's spaceship 'paradox'.  The only difference here is that they're moving in a circle instead of a straight line, but length contraction is real in both cases.

Bell’s spaceship ‘paradox’ is not a paradox at all. It is impossible to synchronize separated clocks. The two spaceships cannot start at the same time. There is not such thing as the same time.
Title: Re: Is angular momentum frame dependent?
Post by: Jaaanosik on 31/07/2020 15:15:25
Halc,
that's the answer I expected.

Mamalute Lover,
if you deny Bell's spaceship 'paradox' then you are going against Einstein and his original 1905 paper.
He starts his paper with a definition of simultaneity:
http://hermes.ffn.ub.es/luisnavarro/nuevo_maletin/Einstein_1905_relativity.pdf

If you deny the simultaneity then you have nothing to back up your statements.
You cannot argue this is like this because the relativity says so. You denied the relativity.
Do you have your own math? Do you have your own hypothesis?
Jano

(https://i.imgur.com/jefPfY2.png)
(https://i.imgur.com/nC3fihP.png)
(https://i.imgur.com/iP2pLjQ.png)
Title: Re: Is angular momentum frame dependent?
Post by: Jaaanosik on 31/07/2020 15:31:03
The Fig. 13.15 shows 'the rim blocks'. Is the length contraction real?
It's very real. Length contraction is not just a visual effect. It is entirely real. If radius is held constant as depicted, then real gaps must form between the blocks, which will be quite apparent to any observer.
I brought this up in post 3.

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We connect the A0 block and B0 block with a string.
Is the string going to break when the wheel accelerates and the gap grows between A1 block and B1 block?
The string will break, just like in Bell's spaceship 'paradox'.  The only difference here is that they're moving in a circle instead of a straight line, but length contraction is real in both cases.
Halc,
if you look at the (b) frame analysis:

(https://i.imgur.com/G8YLStS.png)

Is the deformation real then?
The wheel is already accelerated and it has a constant rotation.
If an A1 rim observer connects a string to B1 rim observer at the bottom then is it fair to expect the string to be broken at the top?
Is the (a) axle observer going to see the string broken?
Do we have a multiverse here?
The strings not broken for (a) but broken for (b)?
Jano
Title: Re: Is angular momentum frame dependent?
Post by: Halc on 31/07/2020 17:40:17
Several problems here.

In the Ehrenfest paradox, the circumference contracts, not expands, since the wheel is moving.
I didn't state otherwise, in the quote of mine to which you responded or in post 3. Given a spinning radius of 1, it has a contracted circumference of 6.283 and a rest circumference (proper circumference) of 8.796. That's contracting due to spin.  Either the spokes get shorter (reeled in??) as it spins, or the thing perhaps needs to be manufactured already spinning.

  From the viewpoint of a non-rotating observer at the center of the circle, the blocks would be seen to shrink along with any gaps between them.[/quote]Given fixed length radius, the blocks would have no gaps between them when stationary, and as they shrink, gaps would form, allowing more blocks to be inserted in them if you like. You seem to suggest that the gaps shrink as well, which is wrong.  Picture a bunch of roller coaster cars, touching each other while parked in a circular track. As they speed up, the track doesn't change size at all, but gaps must form between the cars as they contract. That's what the recent diagram depicts.  The ring circumference will contract if there's no track and the cars are bolted together. In that case there never are gaps, but the radius goes down as the ring circumference contracts.

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This could be seen by comparing the observed length of the blocks compared to their width. An observer riding on the rim would not notice any difference in the size of the blocks or the gaps.
He'd very much notice the gaps forming, which were not there at all before. I agree that he'd not notice the ratio of length/width of his own car changing. We're talking about fixed radius here. Fixed spokes or a track or something.

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Why are you making the circumference larger?
I'm not.
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And why are the gaps larger?
Different example. Don't mix them. The text of mine you quoted is about gaps forming with blocks moving at a fixed radius.  Post 3 talks about a solid ring contracting as it rotates, reducing the radius. Nothing gets larger with speed.

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The radius is assumed to be constant.
Only in the example in post 29, where the radius is held constant with detached objects moving around that fixed path at ever increasing speeds.

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The size of the radius can only be determined with a measuring rod.
We have one. You can't measure a stationary track or fixed length spokes moving only perpendicular to their length?  Post 29 shows a sort of track. No spokes. The grey boxes seem to be the track, not spinning with the red stuff.

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If we use the rotating stick as the measuring rod, we will see something interesting. It will not appear to be rigid. Light coming the rod from near the rim will take longer to reach the observer in the center and will show the stick at an earlier time in its rotation. The stick will appear to curve and have a length greater than the expected radius.
Appearances or no, the stick (spoke) is in fact straight in an frame where the axis of rotation is stationary. Light 'appearing' to curve is Coriolis effect that you get in a rotating frame. Light does not move in straight lines in a rotating frame. If you photograph the thing from a distant point on the axis, the spoke-stick will appear straight since light takes about equal time to get from any location of the stick to the point of view.

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But if the observer in the center holds a measuring rod out to the circumference until sparks fly when it touches the rim, the measured radius would be as expected for a non-rotating wheel. (Actually, the measurement will be a bit too long as it would take time for the light of the sparks to reach the center. But since the speed of light is known, an adjustment could be made for this.)
If the radius is 1, then the light will take time 1 to reach the observer in the center despite it not moving along the spoke. No adjustment is needed.

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This result contradicts the radius implied by the observations made of the blocks on the spinning wheel.
The blocks are not making any measurement of the radius. They measure proper circumference.

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Length contraction is not real. It is relative.
You're denying that gaps form between the blocks? By your posts above, it seems so. Let me know how that works for you. I stand by my statement that contraction is objectively real and is well illustrated by the Ehrenfest scenario. There is no frame in which those gaps do not form.

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Bell’s spaceship ‘paradox’ is not a paradox at all.
With that I'll agree. None of them are paradoxes.

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It is impossible to synchronize separated clocks. The two spaceships cannot start at the same time. There is not such thing as the same time.
This is nonsense.
The ships are initially stationary, and begin identical proper acceleration at the same time relative to the frame in which they are stationary. Are you in denial now that clocks can be synced in a given inertial frame? Einstein gives some nice examples of ways to do exactly that.

You appear to be searching for cop-out excuses to avoid explaining a scenario that you apparently don't understand. The ship at the rear could even start out a little before the other, putting initial slack in the string. As the ships and the string gain speed and contract, the string will eventually break, but it takes a bit longer due to that initial slack.

Is the deformation real then?
The deformation into an ellipse is frame dependent, but that doesn't make it not real. M-L seems to think otherwise.

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The wheel is already accelerated and it has a constant rotation.
Both wheels have identical proper angular velocity. That's not the same as constant rotation. The angular velocity of an object is frame dependent since it can be used as a clock, and time is dilated in a frame in which the object is moving.

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If an A1 rim observer connects a string to B1 rim observer at the bottom then is it fair to expect the string to be broken at the top?
If the two wheels are different wheels in the same frame, then the string breaks same as if I attach a string between a moving car and a parked one.
I don't think you mean that, but I don't know what you mean by 'A1 rim' and 'B1 rim'.  They seem to be references to different objects in relative motion, which probably breaks the string.
If you mean a string from one side of a wheel to the other side of the same wheel, then no, that string will not break for either wheel so long as they keep spinning.  The spokes already serve as such a string.

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Is the (a) axle observer going to see the string broken?
I cannot figure out where you are putting your string. The (a) axle observer cannot see anything different than the (b) axle observer since, per principle of relativity, linear motion cannot be locally detected. The two wheels might be the same wheel, just considered in two different inertial frames.

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Do we have a multiverse here?
????  What brings this up?
Quote
The strings not broken for (a) but broken for (b)?
Your description made it sound like one string between the two wheels, so if one sees it break, the other will see the same string break.
Title: Re: Is angular momentum frame dependent?
Post by: Jaaanosik on 31/07/2020 19:32:21
...
Is the deformation real then?
The deformation into an ellipse is frame dependent, but that doesn't make it not real. M-L seems to think otherwise.

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The wheel is already accelerated and it has a constant rotation.
Both wheels have identical proper angular velocity. That's not the same as constant rotation. The angular velocity of an object is frame dependent since it can be used as a clock, and time is dilated in a frame in which the object is moving.

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If an A1 rim observer connects a string to B1 rim observer at the bottom then is it fair to expect the string to be broken at the top?
If the two wheels are different wheels in the same frame, then the string breaks same as if I attach a string between a moving car and a parked one.
I don't think you mean that, but I don't know what you mean by 'A1 rim' and 'B1 rim'.  They seem to be references to different objects in relative motion, which probably breaks the string.
If you mean a string from one side of a wheel to the other side of the same wheel, then no, that string will not break for either wheel so long as they keep spinning.  The spokes already serve as such a string.

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Is the (a) axle observer going to see the string broken?
I cannot figure out where you are putting your string. The (a) axle observer cannot see anything different than the (b) axle observer since, per principle of relativity, linear motion cannot be locally detected. The two wheels might be the same wheel, just considered in two different inertial frames.

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Do we have a multiverse here?
????  What brings this up?
Quote
The strings not broken for (a) but broken for (b)?
Your description made it sound like one string between the two wheels, so if one sees it break, the other will see the same string break.


Halc,
This is the bottom of the cycloid as seen from (b).
Please, ignore A0 and B0, it should say A1 and B1 because the speed is close to 0 in the (b) frame, the bottom part of the cycloid, therefore there is almost no gap between the A1 and B1 rim blocks from (b) point of view.
The string is attached here at the bottom when there is very tiny gap between the A1 and B1 rim blocks.

(https://i.imgur.com/kIphyF0.png)

The wheel makes a half a turn.

(https://i.imgur.com/JXGaEmA.png)

The gap grows. Is the string going to break?
I agree both observers will see the same result, either it breaks or it does not.
There is no multiverse, two different outcomes for two different observers. :)
Jano

Title: Re: Is angular momentum frame dependent?
Post by: Halc on 31/07/2020 20:50:56
This is the problem in physics.
Over a hundred years after the SR paper and the relativists do not agree on the Lorentz contraction.
Physicists are in complete agreement over these kinds of trivial implications of SR.
If by 'relativists' you mean the sort of people that hang out on forums like this, then yes, there are always those who either don't understand or simply choose to deny the prevailing view for whatever purpose. Over a thousand years since it's been shown that the world is round, and the Earthlings still do not agree on its shape. So what? The people who matter do agree about it. Neither theory has been in contention for a long time.

This is the bottom of the cycloid as seen from (b).
Ah, thankee for the closeup. I didn't even see the subscripts and realize that those were the points you were talking about. Yes, a string from those two points must break as there is no extra string to bridge the gap that forms between those two blocks when the blocks start moving around that fixed-radius path.

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The wheel makes a half a turn.
It probably makes a whole lot of them. There's no need for the angular speed to ramp up almost instantly. The gaps form even if it takes millions of rotations to get up to speed.

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The gap grows. Is the string going to break?
Yes.

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I agree both observers will see the same result, either it breaks or it does not.
So will any observer not on the ring. The string breaking is an objective fact, meaning length contraction is real.
Title: Re: Is angular momentum frame dependent?
Post by: Jaaanosik on 31/07/2020 21:48:28
Halc,
I am not sure we understand each other. I'll try again.
There is no more angular acceleration in (a) frame.
The wheel is already up to speed, accelerated and it has the shape on the right from this figure 13.15.
The gaps are already created in the (a) frame.

(https://i.imgur.com/kJGT2Tw.png)

We also know that the wheel rim blocks in (b) have 0 speed at the bottom of the cycloid.
The question is if it is OK to assume the spacing is like this for the (b) reference frame.

(https://i.imgur.com/0XYmbOR.png)


Almost no space at the bottom of the cycloid and a bigger gap at the top.
That's why it is a half a turn, 180 degrees, from the bottom to the top for already rotating wheel as seen from (b).
Jano
Title: Re: Is angular momentum frame dependent?
Post by: Halc on 31/07/2020 22:24:56
Halc,
I am not sure we understand each other. I'll try again.
There is no more angular acceleration in (a) frame.
The wheel is already up to speed, accelerated and it has the shape on the right from this figure 13.15.
The gaps are already created in the (a) frame.
You're mixing two very different examples, so this doesn't make sense.
(a) and (b) are the colorful picture from reply 2.  There are no blocks or gaps in that example.
In that example, (a) and (b) are the same spinning wheel viewed in two different inertial frames. There is no angular acceleration involved. Being the same wheel both pictures spin at the same proper angular velocity (but not at the same angular velocity).

The grey/red picture from post 29 depicts the blocks, stationary on the left (t0) and spinning on the right (t1) at some fixed radius track.  The gaps form as the angular speed increases. Near light speed, the 'rim' is all gaps and an arbitrarily large number of blocks could fit in. Neither of these two wheels has linear motion.

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We also know that the wheel rim blocks in (b) have 0 speed at the bottom of the cycloid.
(a) and (b) do not depict blocks.

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The question is if it is OK to assume the spacing is like this for the (b) reference frame.
There is no spacing involved in the (a) (b) picture. You're mixing scenarios.  (a) and (b) are the same wheel in different inertial frames. They can't differ in any objective way because they're the same thing.

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That's why it is a half a turn, 180 degrees, from the bottom to the top for already rotating wheel as seen from (b).
I think you're talking about A0 and B0.  The half turn is irrelevant. The gaps form because the wheel on the right is spinning, and both A0 and B0 travel continuously all around the circuit, not just a half turn.
Title: Re: Is angular momentum frame dependent?
Post by: Jaaanosik on 31/07/2020 23:04:10
Right, if the wheel looked like this:

(https://i.imgur.com/7XcoEOu.png)

No need for the outside 'track', spokes keep the rim blocks together.
I hope the same question about the spacing makes sense now,
Jano
Title: Re: Is angular momentum frame dependent?
Post by: Malamute Lover on 31/07/2020 23:19:26
Several problems here.

In the Ehrenfest paradox, the circumference contracts, not expands, since the wheel is moving.
I didn't state otherwise, in the quote of mine to which you responded or in post 3. Given a spinning radius of 1, it has a contracted circumference of 6.283 and a rest circumference (proper circumference) of 8.796. That's contracting due to spin.  Either the spokes get shorter (reeled in??) as it spins, or the thing perhaps needs to be manufactured already spinning.

Of course, the spokes experience Lorentz contraction. They are moving too. Because they are in motion, and the light from the outer portions takes longer to reach the observer in the center, they appear curved. That is, the observer does not need to look along the length to notice Lorentz contraction. Yes, the spokes get shorter as the rim contracts.  Since the rim is moving along with the blocks, it will experience exactly the same Lorentz contraction as the blocks. No increasing gap sizes.

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From the viewpoint of a non-rotating observer at the center of the circle, the blocks would be seen to shrink along with any gaps between them.
Given fixed length radius, the blocks would have no gaps between them when stationary, and as they shrink, gaps would form, allowing more blocks to be inserted in them if you like. You seem to suggest that the gaps shrink as well, which is wrong.  Picture a bunch of roller coaster cars, touching each other while parked in a circular track. As they speed up, the track doesn't change size at all, but gaps must form between the cars as they contract. That's what the recent diagram depicts.  The ring circumference will contract if there's no track and the cars are bolted together. In that case there never are gaps, but the radius goes down as the ring circumference contracts.

But as we have seen, the radius does shrink because the spokes also experience Lorentz contraction and that is visible to the observer in the center.

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This could be seen by comparing the observed length of the blocks compared to their width. An observer riding on the rim would not notice any difference in the size of the blocks or the gaps.
He'd very much notice the gaps forming, which were not there at all before. I agree that he'd not notice the ratio of length/width of his own car changing. We're talking about fixed radius here. Fixed spokes or a track or something.

Sorry, no fixed radius when the wheel is spinning.

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Why are you making the circumference larger?
I'm not.
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And why are the gaps larger?
Different example. Don't mix them. The text of mine you quoted is about gaps forming with blocks moving at a fixed radius.  Post 3 talks about a solid ring contracting as it rotates, reducing the radius. Nothing gets larger with speed.

The increasing gap size points to the blocks shrinking but the rim not shrinking. That is equivalent to the rim expanding relative to the blocks. But the rim will contract as much as the blocks. No gaps.

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The radius is assumed to be constant.
Only in the example in post 29, where the radius is held constant with detached objects moving around that fixed path at ever increasing speeds.

Detached objects following a circular track requires an additional force to keep them on track. That is, there must be something comparable to a rim to exert that inward force. This changes nothing. Whatever is keeping the blocks on track will be subject to the same Lorentz contraction.

If you want something that holds the blocks on track but is not itself moving, you are introducing all sorts of complications. For one thing, from the viewpoint of the blocks, the whatever it is outside will be the thing that is moving and will therefore be Lorentz contracted and have a smaller circumference/radius. For another, the force holding the blocks in place will result in the outside influence thing being set in motion and the blocks slowing down. (Newton #3)

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The size of the radius can only be determined with a measuring rod.
We have one. You can't measure a stationary track or fixed length spokes moving only perpendicular to their length?  Post 29 shows a sort of track. No spokes. The grey boxes seem to be the track, not spinning with the red stuff.

(https://i.imgur.com/kJGT2Tw.png)

The spinning disk (grey) and the red boxes on it will all undergo Lorentz contraction along the direction of motion. This will reduce the circumference and therefore the radius. No broken strings.

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If we use the rotating stick as the measuring rod, we will see something interesting. It will not appear to be rigid. Light coming the rod from near the rim will take longer to reach the observer in the center and will show the stick at an earlier time in its rotation. The stick will appear to curve and have a length greater than the expected radius.
Appearances or no, the stick (spoke) is in fact straight in an frame where the axis of rotation is stationary. Light 'appearing' to curve is Coriolis effect that you get in a rotating frame. Light does not move in straight lines in a rotating frame. If you photograph the thing from a distant point on the axis, the spoke-stick will appear straight since light takes about equal time to get from any location of the stick to the point of view.
If you photograph the whole thing from a distant point. You will see something different from what an observer at the center sees. But it will still exhibit Lorentz contraction since it is moving with respect to that distant observer. Lorentz contraction is relative. It is not real.

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But if the observer in the center holds a measuring rod out to the circumference until sparks fly when it touches the rim, the measured radius would be as expected for a non-rotating wheel. (Actually, the measurement will be a bit too long as it would take time for the light of the sparks to reach the center. But since the speed of light is known, an adjustment could be made for this.)
If the radius is 1, then the light will take time 1 to reach the observer in the center despite it not moving along the spoke. No adjustment is needed.

The observer in the center will not know that contact has been made until a light signal comes back from the rim. In the meantime, the observer is still pushing the rod out beyond the point of contact.  The length of measuring rod that has been pushed out is too much. An adjustment needs to be made to the measured length.

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This result contradicts the radius implied by the observations made of the blocks on the spinning wheel.
The blocks are not making any measurement of the radius. They measure proper circumference.

No, the blocks are measuring Lorentz contracted circumference because they are in motion. No gaps.

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Length contraction is not real. It is relative.
You're denying that gaps form between the blocks? By your posts above, it seems so. Let me know how that works for you. I stand by my statement that contraction is objectively real and is well illustrated by the Ehrenfest scenario. There is no frame in which those gaps do not form.

I am denying that gaps form because the circumference is contracting along with the blocks. There is no frame in which the gaps do form.

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It is impossible to synchronize separated clocks. The two spaceships cannot start at the same time. There is not such thing as the same time.
This is nonsense.
The ships are initially stationary, and begin identical proper acceleration at the same time relative to the frame in which they are stationary. Are you in denial now that clocks can be synced in a given inertial frame? Einstein gives some nice examples of ways to do exactly that.

As I explained in a prior post, the clocks are out of sync when the engines start because each space ship sees a delay in the start of the other space ship’s engine. As a consequence, each space ship sees the other one going slower than itself because it has not been accelerating for as long. No common frame of reference. It is not possible to start the engines at the same time. Distance = time delay.

You appear to be searching for cop-out excuses to avoid explaining a scenario that you apparently don't understand. The ship at the rear could even start out a little before the other, putting initial slack in the string. As the ships and the string gain speed and contract, the string will eventually break, but it takes a bit longer due to that initial slack.

No cop outs going on. I understand the scenarios perfectly. You are the one being inconsistent in the application of Lorentz contraction. It applies to the circumference as much as to the blocks.

As I described in the previous post, even starting at the same prearranged time on previously synchronized clocks, each spaceship thinks the other one is going slower with opposite expectations. Even if you manage to get them agreeing that they are both going at the same speed, the two spaceships and the string would all be in a common inertial reference frame and they would see no Lorentz contraction and no broken string.

As I pointed out in my prior post, an observer in a different inertial reference frame would see the entire complex of ships and strings equally Lorentz contracted and no broken string.

If a string was tied inside a spaceship along the direction of travel, and the spaceship accelerated (slowly) to a high speed, would the string break? It is not material objects that get contracted, it is space itself by appearing to be at different orientations in Minkowski spacetime to different reference frames. This is the point you are missing.

Lorentz contraction is relative, not real.
Title: Re: Is angular momentum frame dependent?
Post by: Malamute Lover on 31/07/2020 23:46:34

The question: Is accelerated observer on the wheel rim going to see/observe/measure the deformation that is not predicted by (a)?
Jano

An observer on the rim is going to see deformations not present in (a) because the different parts between the rim and the center are going at different speeds. But it will not be the same deformations as an observer at the center would see..
Title: Re: Is angular momentum frame dependent?
Post by: Malamute Lover on 31/07/2020 23:51:01
...
Is the deformation real then?
The deformation into an ellipse is frame dependent, but that doesn't make it not real. M-L seems to think otherwise.

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The wheel is already accelerated and it has a constant rotation.
Both wheels have identical proper angular velocity. That's not the same as constant rotation. The angular velocity of an object is frame dependent since it can be used as a clock, and time is dilated in a frame in which the object is moving.

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If an A1 rim observer connects a string to B1 rim observer at the bottom then is it fair to expect the string to be broken at the top?
If the two wheels are different wheels in the same frame, then the string breaks same as if I attach a string between a moving car and a parked one.
I don't think you mean that, but I don't know what you mean by 'A1 rim' and 'B1 rim'.  They seem to be references to different objects in relative motion, which probably breaks the string.
If you mean a string from one side of a wheel to the other side of the same wheel, then no, that string will not break for either wheel so long as they keep spinning.  The spokes already serve as such a string.

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Is the (a) axle observer going to see the string broken?
I cannot figure out where you are putting your string. The (a) axle observer cannot see anything different than the (b) axle observer since, per principle of relativity, linear motion cannot be locally detected. The two wheels might be the same wheel, just considered in two different inertial frames.

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Do we have a multiverse here?
????  What brings this up?
Quote
The strings not broken for (a) but broken for (b)?
Your description made it sound like one string between the two wheels, so if one sees it break, the other will see the same string break.


Halc,
This is the bottom of the cycloid as seen from (b).
Please, ignore A0 and B0, it should say A1 and B1 because the speed is close to 0 in the (b) frame, the bottom part of the cycloid, therefore there is almost no gap between the A1 and B1 rim blocks from (b) point of view.
The string is attached here at the bottom when there is very tiny gap between the A1 and B1 rim blocks.

(https://i.imgur.com/kIphyF0.png)

The wheel makes a half a turn.

(https://i.imgur.com/JXGaEmA.png)

The gap grows. Is the string going to break?
I agree both observers will see the same result, either it breaks or it does not.
There is no multiverse, two different outcomes for two different observers. :)
Jano

There are no gaps because the circumference of the rim has been subjected to Lorentz contraction as well. You are applying contraction only to one part of a moving system when you should be applying it to all of the system.
Title: Re: Is angular momentum frame dependent?
Post by: Malamute Lover on 31/07/2020 23:54:22
I see the problem this way.
The frame (a) sees the rim of the wheel symmetrically. See the figures 13.14 and 13.15.
The frame (b) sees the rim of the wheel asymmetrically. See the figure 2 of the paper.
Both are the inertial frame observers.
The (a) and (b) observers are not on the rim itself though.
If there is an accelerated observer on the rim of the wheel then this local observer will measure either symmetrical centripetal acceleration as predicted by (a) frame or asymmetrical acceleration where the spacing between 'the rim blocks' changes as predicted by (b) or ... completely something else that neither reference frame predicted.
Jano

As already stated, the outside observer who sees (b) is not looking at an inertial reference frame. The spokes are going faster on top and slower on the bottom relative to overall motion.
My apologies, I introduced a new scenario when the axle is accelerated.
Having said that, my last couple of posts are about the textbook and the paper.
There is no acceleration of the axle here, just to make it clear.
Both, (a) and (b) are inertial observers looking at the rotating wheel.
The question stands, is the accelerated wheel rim observer going to see/observe/measure the deformation or not.
What prediction/observation wins for the accelerated observer on the wheel rim? Is it (a) or (b)?
Jano

Neither, the observer on the rim in (b) is yet another frame of reference from the outside observer.  The observed deformations will be different again. Lorentz contraction is relative, not real.
Title: Re: Is angular momentum frame dependent?
Post by: Kryptid on 01/08/2020 00:42:19
Lorentz contraction is relative, not real.

What about it being relative makes it not real?
Title: Re: Is angular momentum frame dependent?
Post by: Malamute Lover on 01/08/2020 03:25:59
Lorentz contraction is relative, not real.

What about it being relative makes it not real?

By relative, I mean that it is observer dependent, which is how Einstein meant it.. Different observers are seeing different things. Which one is real? As I said earlier, what is real is the situation in Minkowski spacetime, Observers in different reference frames see different aspects of that. Nobody sees the 'real; real because it exists in 4 dimensions and we do not see that way.
Title: Re: Is angular momentum frame dependent?
Post by: yor_on on 01/08/2020 12:19:36
This one reminds me a lot of the spinning disk in where different speeds ,depending on where you measure its spin from its center, leads to different contractions, fracturing and breaking it up, according to some interpretations, before you even get to measuring it. Unless made from some very elastic material, probably needing to be 'infinitely' elastic :)

https://en.wikipedia.org/wiki/Ehrenfest_paradox
=

We have apparently black holes spinning close to 'c' though?  Then again, they are themselves a singularity. Then we have the stuff it 'sucks in' (drag with it)  which should follow that spin, I think? You have locally defined a geodesic (uniform motion, aka being locally 'weightless') for each object, but what about the spin here?

And then there is the spaghettification of course, by tidal forces (different gradients of gravity). Which means that even in the absence of any weight, locally defined, gravity still will split you into pieces. the point being that you can't define both a geodesic and a weight. If you define it as those gradients inferring a weight of the object you just left a geodesic. And the same goes for the spin as that also should be experienced as a weight, normally. So the stuff dragged by a black holes spin must then still be a geodesic, if I'm thinking correct, no matter the black holes 'spin' as defined from a far away observer?

https://www.nasa.gov/mission_pages/chandra/news/08-003.html
https://en.wikipedia.org/wiki/Spaghettification
Title: Re: Is angular momentum frame dependent?
Post by: Jaaanosik on 01/08/2020 20:37:11
Hi all,
Is the Lorentz Contraction real or not real.
What a beautiful 'can of warms' we got open here. :)
We all want to find out the truth.
If we come close to finding out the truth we will also find the answer to the question of this thread: Is angular momentum frame dependent?

I suggest we all put aside our convictions what is the correct answer and we develop an argument together; we find the answer together.
There are good points on both sides, to show that the LC is not real and also to show that the LC is real.
First, let us discuss how we can show the LC is not real.
I'll make a statement and I suggest we get an agreement if the statement can lead towards the answer.
If we say yes, then we will analyze and prove the statement.

If the Special Relativity is reciprocal then the Lorentz Contraction is not real.

Please, let us discuss the statement above. Do we agree it is a true statement?
Jano
Title: Re: Is angular momentum frame dependent?
Post by: Kryptid on 01/08/2020 20:45:29
If the Special Relativity is reciprocal then the Lorentz Contraction is not real.

Please, let us discuss the statement above. Do we agree it is a true statement?
Jano

What do you mean by special relativity being "reciprocal"?
Title: Re: Is angular momentum frame dependent?
Post by: Jaaanosik on 01/08/2020 20:51:31
If the Special Relativity is reciprocal then the Lorentz Contraction is not real.

Please, let us discuss the statement above. Do we agree it is a true statement?
Jano

What do you mean by special relativity being "reciprocal"?

Two inertial observers see each other clocks going slower.
Two inertial observers see each other Lorentz Contracted.
Whatever the first inertial observer can say about the second one then the second observer can say the same things about the first one.
There is no preferred reference frame.
Jano
Title: Re: Is angular momentum frame dependent?
Post by: Kryptid on 01/08/2020 20:59:49
Their observations are both equally real.
Title: Re: Is angular momentum frame dependent?
Post by: Jaaanosik on 01/08/2020 21:41:28
Here is a new thread to discuss the LC: https://www.thenakedscientists.com/forum/index.php?topic=80208.0
Let us keep this thread to the Angular Momentum discussion.

Title: Re: Is angular momentum frame dependent?
Post by: CPT ArkAngel on 02/08/2020 01:58:57
You cannot consider the wheel has a single frame but only as a collection of infinitely small points, each its own frame because they all have a different acceleration. The mistake is that they forget the additional delays between the frames. The proof of that is if you consider all point frames individually, angular momentum is conserved. It is simply an invariant under special relativity because the speed of light is constant and the mass increases as the length decreases.  There is zero problem.
Title: Re: Is angular momentum frame dependent?
Post by: Malamute Lover on 02/08/2020 04:18:03
Lorentz contraction is relative, not real.

What about it being relative makes it not real?

How about different observers seeing different things? Who is right?

Title: Re: Is angular momentum frame dependent?
Post by: Kryptid on 02/08/2020 04:31:51
How about different observers seeing different things? Who is right?

They both are right. Someone travelling near the speed of light might experience only a few hours of the passage of time if they traveled to Alpha Centauri, but someone on Earth watching them through a telescope will see that it took them over 4 years to get there. You can't say that one is right and one is wrong in that case either. They are both correct in their own reference frames. It's the same with length contraction.
Title: Re: Is angular momentum frame dependent?
Post by: CPT ArkAngel on 02/08/2020 05:10:06
Malamute, you forgot that the energy of acceleration cannot propagate faster than the speed of light. There is a delay between the front and the back as soon as you apply a force on an object. Thus the front and the back cannot be calculated as if they were in the same frame. SR and GR are classical theories. In classical physics, any object is inside space and time, it is embedded. Space and time are represented by a continuum, it is not divisible in chunk. The only way to represent an object which is not embedded in space is to picture it by infinitely small particles separated by space so the space is still a continuum. In Newtonian physics, you can considered objects has having a fixed length because space and time are not interwoven, but space is still a continuum and object are also embedded in space, it is just that you can make abstraction of the continuum to solve the problem when the object is solid.
Title: Re: Is angular momentum frame dependent?
Post by: yor_on on 02/08/2020 12:25:45
It's a nice idea and question Jaaanosik. You can always define different frames of reference to that wheel or disk . Let us assume that a spinning disk, or wheel, breaks down, fractures, due to spins being at different speeds, as defined from its center. One of tenets of relativity is that you always will find a logic for why it does, no matter where you as a observer is situated. So presuming you are 'at rest' with its spin at some location you should be able to define a locally coherent explanation for why it fractures, just as you should be able to do with f.ex the https://en.wikipedia.org/wiki/Unruh_effect

Actually it's deeper than than just relativity. It's a question of how this universe is ordered. By logic or by magic.
Title: Re: Is angular momentum frame dependent?
Post by: Malamute Lover on 02/08/2020 19:06:39
How about different observers seeing different things? Who is right?

They both are right. Someone travelling near the speed of light might experience only a few hours of the passage of time if they traveled to Alpha Centauri, but someone on Earth watching them through a telescope will see that it took them over 4 years to get there. You can't say that one is right and one is wrong in that case either. They are both correct in their own reference frames. It's the same with length contraction.

Does anyone see the string break?
Title: Re: Is angular momentum frame dependent?
Post by: Malamute Lover on 02/08/2020 19:22:39
Malamute, you forgot that the energy of acceleration cannot propagate faster than the speed of light. There is a delay between the front and the back as soon as you apply a force on an object. Thus the front and the back cannot be calculated as if they were in the same frame. SR and GR are classical theories. In classical physics, any object is inside space and time, it is embedded. Space and time are represented by a continuum, it is not divisible in chunk. The only way to represent an object which is not embedded in space is to picture it by infinitely small particles separated by space so the space is still a continuum. In Newtonian physics, you can considered objects as having a fixed length because space and time are not interwoven, but space is still a continuum and object are also embedded in space, it is just that you can make abstraction of the continuum to solve the problem when the object is solid.

The rod is being pushed and pulled. A rod being pulled will not break, stretch or whatever until it cannot propagate the energy wave any further. If it were not attached to anything it would simply move in the direction it is being pulled. In the present example it is attached at the other end and the energy wave from the pulled end will meet the energy wave from the pushed end in the middle and the waves will cancel out and the entire rod will be moving at the same speed. Continued pulling and pushing energy (ongoing acceleration) will propagate in the same manner. If we are talking about string, there will be slack on the pushed end that will take up the energy from the pulled end.
Title: Re: Is angular momentum frame dependent?
Post by: Jaaanosik on 02/08/2020 20:18:09
How about different observers seeing different things? Who is right?

They both are right. Someone travelling near the speed of light might experience only a few hours of the passage of time if they traveled to Alpha Centauri, but someone on Earth watching them through a telescope will see that it took them over 4 years to get there. You can't say that one is right and one is wrong in that case either. They are both correct in their own reference frames. It's the same with length contraction.
Please, check this post: https://www.thenakedscientists.com/forum/index.php?topic=80208.msg610136#msg610136
The traveler watches a couple of hours on the Earth to take 4 years on the ship?
Correct?
Jano
Title: Re: Is angular momentum frame dependent?
Post by: Kryptid on 02/08/2020 22:51:15
Does anyone see the string break?

I'm not sure, but if one does, then they both do.

The traveler watches a couple of hours on the Earth to take 4 years on the ship?
Correct?

Nope. The ship must accelerate into order to reach such a very high velocity. That is an acceleration that is not experienced by the observer on Earth. So the situation is not symmetrical.
Title: Re: Is angular momentum frame dependent?
Post by: Colin2B on 02/08/2020 23:55:04
Lorentz contraction is relative, not real.

What about it being relative makes it not real?

How about different observers seeing different things? Who is right?
What do you mean by real?

In the example @Kryptid gives further down a traveller can be seen by earth observers to take (say) 4yrs to travel to a distant star - based on the distance they measure. The traveller, however, sees the distance to the star length-contracted and so takes less time to travel that distance. The experience is real for both of them. (Time dilation and length contraction are effectively the same thing).

If you think these effects are not real you will have to rewrite what we understand about electricity and magnetism and also a great deal of chemistry. Good luck with that  ;)
Title: Re: Is angular momentum frame dependent?
Post by: Malamute Lover on 03/08/2020 00:04:23
Lorentz contraction is relative, not real.

What about it being relative makes it not real?

How about different observers seeing different things? Who is right?
What do you mean by real?

In the example @Kryptid gives further down a traveller can be seen by earth observers to take (say) 4yrs to travel to a distant star - based on the distance they measure. The traveller, however, sees the distance to the star length-contracted and so takes less time to travel that distance. The experience is real for both of them. (Time dilation and length contraction are effectively the same thing).

If you think these effects are not real you will have to rewrite what we understand about electricity and magnetism and also a great deal of chemistry. Good luck with that  ;)

What is real is what is going on in 4D Minkowski spacetime. We time bound observers only get to see a slice at a time depending on our reference frames. Multiple observers in different reference frames see different things. None of them see the real thing. Relativistic effects are relative not objective.
Title: Re: Is angular momentum frame dependent?
Post by: Halc on 03/08/2020 00:39:04
The gaps between blocks are said to appear because the radius of the circle is taken to be constant.
There are actually three distinct scenarios.
1) You have a solid ring that spins. It contracts as you spin it.  A shrinking ring (that reduces in radius from any perspective) seems awful real to me.  Using this scenario, I can pass one wedding ring through another identical one by spinning it.  That's real contraction, or it couldn't be done. It isn't observer dependent.
You are sort of describing such a thing below, except with superfluous spoke that will bend because they're too long, so they serve no purpose other than to be deformed by being squashed.

2) Spoke scenario, or the roller coaster track, which is essentially the same scenario.  Here the radius is held constant by the non-contracting straight spoke, or by the stationary track.  There is no solid ring, but a series of detached adjacent blocks.  If there are spoke, you have essentially a row of independent pendulums.  If a track, you have a row of 'bumper cars'.  Spin it up and gaps form between the blocks, and more can be inserted if you like.
Observers in any frame will agree on this, but you seemingly are in denial of it.

3) The actual Ehrenfest scenario where he takes a non-Euclidean 3 dimensional solid (a spinning cylinder) and declares it paradoxical when its non-Euclidean properties are illustrated. If the object is rigid, it shatters as soon as you attempt to change its angular speed.  That shattering is an objective effect that any observer in any frame will witness.  There are other ways to create non-Euclidean objects.  Find a neutron star and build a sort of feeding trough that encircles it at some low altitude, say 10 km wide and thick. Fill that trough will some material that hardens into some rigid object, and remove the trough. In normal space, the outside radius will be 60π km greater than the inside radius, but with this object the difference will be less than that, depending on how close to the star you build it.  Remove the object from the vicinity of the neutron star and it shatters just like the Ehrenfest object. Ehrenfest found a way to create such an object in flat spacetime using length contraction, but it doesn't work if length contraction isn't real. 

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It is my contention that the spokes will maintain their length but will curve due to differing time dilation along the length. The fixed length of the spokes but bent into a curve will result in a reduced radius and the rim holding the blocks will contract as expected. No gaps will appear.
If you content that the curving spokes is what draws the blocks in, that wouldn't work if the blocks didn't actually contract since they would not fit in the smaller radius.  If the radius of a spinning wheel does contract, that is a real effect visible to any observer, so not a relative effect.
This assertion that it is the spokes bending and curving, pulling in the separate blocks is contrary to all accepted physics. It is a fantasy assertion that you cannot support. It seems you are one of those that will hold to your assertions forever rather than show a willingness to learn.

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First some comments about time dilation. The pilot of a rocket ship that has accelerated to 0.99 c will experience a time dilation factor of about 7. The clock on the ship will be running 7 times slower than before the acceleration.
No frame reference, so that statement is ambiguous.  It will be running 7 times slower relative to the frame in which the ship is moving at .99c.  Not saying you're wrong, just sloppy.  It's running at normal rate relative to the ship of course.

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That is, watching landmarks go by, he sees himself as traveling at almost 7 c.
If there is a grid with mileposts he can watch, then yes. You can call that his proper velocity if you want, which is what you get by simply multiplying his proper acceleration by his proper time under that acceleration. There's no limit to proper velocity since it accumulates additively, not relativistically.

Question for the rest of the world: What are metric mileposts or mile-markers?  Just kilometer markers?  I live close to Canada and still don't hear the term used ever. What do they call (in English) the little signs on the side of the road?

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(Either that or he thinks the distances have shrunk.) But after several years of sightseeing, he decelerates and gets back home again, and finds that his twin brother has aged much more than him. He realizes that acceleration really did slow his clock and that he was not traveling as fast as he thought. He was really traveling much slower.
What an absolute way of putting it, but yes, in the brother's frame, he was travelling at nearly c. In his own frame, he wasn't moving at all, only the length-contracted markers were. He knows very well that those moving markers do not mark distances properly and can not move faster than light, as evidenced by the fact that he can see the one's coming at him.  If they were moving at nearly 7c, they'd not be visible at all until they had passed, just like you can't hear a supersonic jet coming at you. So he knows those markers are moving at about .99c and are nowhere near a km apart. To assert otherwise is to assert light moving faster than c, something I'm starting to see a lot of here.

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Unlike time dilation due to different relative inertial speeds, time dilation resulting from acceleration is real and has real consequences. This is the resolution of the so-called Twin Paradox.
So I have four observers Alice, Bob, Charlie and Denise.  Alice was always on Earth.  Charlie came to Earth with Denise and Charlie accelerated to Earth frame to marry Alice.  Bob accelearted to match speeds with Denise to marry her.  Denise and Alice have thus never accelerated, and the other two have.
So according to your statement above, time dilation of the clock on Earth is real according to Charlie because he accelerated to Earth frame, but it is not real for either Alice or Denise, neither of whom have ever accelerated.  Meanwhile dilation of the clock on the ship with Bob and Denise is real to Bob (having accelerated to that frame), but not to Denise right there with him.
Sounds very inconsistent that the same clock is really dilated and also not.

Also, length contraction in all 3 scenarios at the top of this post is very real by your definition since there is a very real consequence (observed by anybody) in all three situation.

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In the case of the spinning wheel, different portions of each spoke are traveling at different speeds. The time dilation factor increases as one travels up the spoke from the center.
Agree. Any clock nearer the center will run objectively faster than one further out.  ISS clocks run faster than sea-level clocks for this reason, but GR must be invoked to compute exactly how much.

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A clock carried up from the axle to a certain point and back down again will have recorded less time than a clock that stayed at the center. The further up the clock is taken, the greater the discrepancy when brought back.
Also how long it spent out there. You'd have to integrate the curve.
If this is not a reasoning why length contraction doesn't actually occur, then you're getting way off topic.

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When the wheel was spun up, the different parts experienced different accelerations, and/or different centripetal accelerations as the wheel turns.
In scenario 3, and also the spoked wheel in post 2, nothing was spun up. There was never any acceleration. In scenario's 1 & 2, there is angular acceleration involved.

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This is acceleration-based time dilation and it is therefore real. At each level, the clock is slower than at the center and the real speed is less than an observer at that level thinks it is.
Oh, so if I build a clock on the rim of an already-spinning wheel, it will stay in sync with the clock at the axle? If not, what do you mean by this distinction between acceleration-based real time dilation and not-real?
Suppose Bob passes Earth at .866c without acceleration, syncing his clock to Earth when he's in its presence. Both say zero.  After a year, he smashes the clock works, freezing it at its reading of 1 year. He turns around, builds a new clock, sets it to zero and smashes the workings again, freezing it at 1 year as well. He then decelerates.  Now he has two clocks that have never accelerated (at least not while working) that cumulatively account for 2 years, but 4 years have gone by on Earth.
I'm sorry, but your stories are getting sillier and sillier, seemingly in a desperate attempt to not admit you've made a mistake. Einstein never mentioned any distinction between real and fake time dilation, or real and fake length contraction.  Measurements using light would be different than empirically measured if length contraction wasn't real.  All you have to do is time light as it goes from one end to the other and back in a moving train.  Measure the time with stationary synced clocks (OK, you have tried to counter this by flat out denial of the possibility of syncing stationary clocks). It will take way too long if the train is not really contracted.  A history of acceleration of the train plays no role in the equations involved.


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Since the angular displacement would vary at each level, the spoke would curve.
Time dilation has no effect on the curvature of the spokes.  They're effectively strings. Time dilation does have an effect on the angular velocity of the wheel.  An observer at the axle would measure a smaller angular velocity than one at the rim, which is why for relativistic wheels, we specify linear rim speed, not radians per second. The spokes are straight unless there is angular stress on them such as the wheel being accelerated by torque being applied at the hub, but no such torque exists in our examples. All measurement are done in steady state.  Yes, if I put enough change in the rate of torque on a real bicycle wheel, the change would need to propagate up the spokes to the rim at the speed of sound, causing a momentary wave to travel up the spoke. That would bend it a bit just like a sideways yank on a garden hose causes a bend to travel up the hose.  A freely spinning wheel has straight spokes so long as there is any tension on them.
Title: Re: Is angular momentum frame dependent?
Post by: Jaaanosik on 03/08/2020 03:49:17
Lorentz contraction is relative, not real.

What about it being relative makes it not real?

How about different observers seeing different things? Who is right?
What do you mean by real?

In the example @Kryptid gives further down a traveller can be seen by earth observers to take (say) 4yrs to travel to a distant star - based on the distance they measure. The traveller, however, sees the distance to the star length-contracted and so takes less time to travel that distance. The experience is real for both of them. (Time dilation and length contraction are effectively the same thing).

If you think these effects are not real you will have to rewrite what we understand about electricity and magnetism and also a great deal of chemistry. Good luck with that  ;)

The example presented by Kryptid is in question, therefore your argument based on Kryptid's comment is in question as well,
Jano
Title: Re: Is angular momentum frame dependent?
Post by: Jaaanosik on 03/08/2020 03:56:18
...

Nope. The ship must accelerate into order to reach such a very high velocity. That is an acceleration that is not experienced by the observer on Earth. So the situation is not symmetrical.

There are examples where physicists play the tag scenario without the acceleration.
Meaning a spaceship flies by with the required velocity.

(https://i.imgur.com/6MCSVif.png)

The acceleration is not important, but the relative velocity is.
Please, see how the red full line is an 'average' of the acceleration world line.
Please, can we take this discussion to the reciprocal thread?
Jano
Title: Re: Is angular momentum frame dependent?
Post by: Malamute Lover on 04/08/2020 00:50:40
The gaps between blocks are said to appear because the radius of the circle is taken to be constant.
There are actually three distinct scenarios.
1) You have a solid ring that spins. It contracts as you spin it.  A shrinking ring (that reduces in radius from any perspective) seems awful real to me.  Using this scenario, I can pass one wedding ring through another identical one by spinning it.  That's real contraction, or it couldn't be done. It isn't observer dependent.
You are sort of describing such a thing below, except with superfluous spoke that will bend because they're too long, so they serve no purpose other than to be deformed by being squashed.

The geometry of the spinning ring is non-Euclidean. It was not only accelerated up to speed, it is under continuous acceleration as it spins. SR is not a good guide here. The acceleration requires the use of GR. Because the spacetime is curved, the distance from the spinning ring to the non-spinning ring is greater. It is following a curve and is longer than the straight-line Euclidean distance. An observer on the spinning ring will think the non-spinning ring is expanding so no problem with fitting. Which is which? As with GR problems in general, who is undergoing acceleration? But the question can only be settled by bringing the two rings into a common inertial reference frame and comparing clocks, just like with the Twins. It cannot be settled by comparing observations.

2) Spoke scenario, or the roller coaster track, which is essentially the same scenario.  Here the radius is held constant by the non-contracting straight spoke, or by the stationary track.  There is no solid ring, but a series of detached adjacent blocks.  If there are spoke, you have essentially a row of independent pendulums.  If a track, you have a row of 'bumper cars'.  Spin it up and gaps form between the blocks, and more can be inserted if you like.
Observers in any frame will agree on this, but you seemingly are in denial of it.

The spokes bend. They cannot stay straight because they have different clock rates along the length. They are not getting squashed. They would act the same even if there were no rim or blocks. The bending of the spokes and the contraction of the blocks match because they both involve the same Lorentz factor. Are you denying time dilation?

As with the two spinning rings above, the geometry has become non-Euclidean because of the accelerations (startup and ongoing). (I really did not want to involve GR because I made an Unbreakable Vow to not ever index another tensor.)

3) The actual Ehrenfest scenario where he takes a non-Euclidean 3 dimensional solid (a spinning cylinder) and declares it paradoxical when its non-Euclidean properties are illustrated. If the object is rigid, it shatters as soon as you attempt to change its angular speed.  That shattering is an objective effect that any observer in any frame will witness.  There are other ways to create non-Euclidean objects.  Find a neutron star and build a sort of feeding trough that encircles it at some low altitude, say 10 km wide and thick. Fill that trough will some material that hardens into some rigid object, and remove the trough. In normal space, the outside radius will be 60π km greater than the inside radius, but with this object the difference will be less than that, depending on how close to the star you build it.  Remove the object from the vicinity of the neutron star and it shatters just like the Ehrenfest object. Ehrenfest found a way to create such an object in flat spacetime using length contraction, but it doesn't work if length contraction isn't real. 

Ehrenfest wrote his paradox before General Relativity was developed and did not know about curved spacetime. Ehrenfest assumed 3D Euclidean geometry, not non-Euclidean as you stated. Ehrenfest assumed that the radius could not change, which it certainly can in non-Euclidean geometry when one of the dimensions is time. He therefore assumed contraction was real, which is what led to his claim of a paradox.

A clock on the surface of the cylinder will run slower than a clock inside the cylinder. To an observer on the surface (and therefore to the mechanical properties of the cylinder) the same number of revolutions a minute are taking place even though the cylinder Lorentz contracts and its rotational speed increases to conserve angular momentum. An observer inside with a faster clock will be surprised to not see the cylinder shatter just as he is surprised when his Twin comes back much younger than him.

Contraction is relative, not objective.

It was consideration of Ehrenfest that gave einstein a clue on how to proceed to deal with non-inertial frames in a non-Euclidean geometry.

Spacetime is not going to be flat in the vicinity of a neutron star. (Assume a non-spinning neutron star to avoid certain non-linearities.) Any measurement you try to make on the object will be distorted in the same degree as the object. To an observer on the object, it is a proper circle. It appears distorted to an outside observer because of the different shape of spacetime. When it is brought up into a different shaped spacetime it will be deformed and become damaged.

How will it be deformed? An interesting fact about gravity wells is that as you go lower into the well, a circle equidistant from the center will not lose as much circumference as is loses radius. This is the opposite of the rotating circle model. In that model, centripetal acceleration will increase as one moves outward. The 4D spacetime resembles a curved bottom bowl with curvature increasing as one moves outward from the center.

(https://assets.katomcdn.com/q_auto,f_auto,w_300/products/095/095-1195132/095-wc11951.jpg)

When it comes to gravity, which according to GR is the same as acceleration (why things fall down faster and faster), the acceleration increases as you move inward. A gravity well looks like this.

(https://demos.smu.ca/images/stories/Pics/how_to/gravity_well/gridsun.png)

As the radius decreases (following the curve) the circumference does not shrink as much as expected by Euclid. More material can be packed into the circumference than in a flatter spacetime.

A circle constructed around the circumference of a gravity well and brought up (along a polar axis) will have too much material in the circumference.  That much material does not fit into the now smaller circumference relative to the radius. It will be crushed around the circumference. Where does the energy come from to crush it? From the force that lifted it out of the gravity well.

(As the circle is brought up along the polar axis it will be subject to differing gravity gradients along its parts. Let us assume that the material has good tensile strength but poor crushing resistance.)

As I said, a gravity well and a rotating circle have opposite spacetime curving. The equivalent of lifting a ring out of a gravity well would be in the rotating circle scenario would be pushing a ring down toward the axle. This is independent of any Lorentz contraction. The ring around the neutron star was not rotating. A ring of stationary solid material being pushed down toward the axle would also be crushed, because it is being forced into a flatter spacetime with less space than it currently occupies.

In both scenarios, the crushing would be experienced by the ring itself and would be objective. But in neither case does it have anything to do with Lorentz contraction, which derives from relative motion.

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It is my contention that the spokes will maintain their length but will curve due to differing time dilation along the length. The fixed length of the spokes but bent into a curve will result in a reduced radius and the rim holding the blocks will contract as expected. No gaps will appear.
If you content that the curving spokes is what draws the blocks in, that wouldn't work if the blocks didn't actually contract since they would not fit in the smaller radius.  If the radius of a spinning wheel does contract, that is a real effect visible to any observer, so not a relative effect.
This assertion that it is the spokes bending and curving, pulling in the separate blocks is contrary to all accepted physics. It is a fantasy assertion that you cannot support. It seems you are one of those that will hold to your assertions forever rather than show a willingness to learn.

What I contend is that the blocks bend exactly as expected from their speed. They are not ‘drawn in’. There is nothing to hold them up since the spokes are not straight. The rim of the spinning wheel will contract exactly as if there were never any spokes. I explained exactly how the curving happens. It is due to different clock rates at different points on the spoke and the fact that the spoke is moving in a circle. Do you deny that the clocks will have different rates at different points on the spoke?  Again it is non-Euclidean geometry going on because acceleration is involved.

It seems that you are the one unwilling to learn.

Whether and how much contraction an observer will see depends on the observer. An observer on the spinning wheel sees nothing different because contraction is only visible from another reference frame. Observers on different locations on a spoke and going at different speeds would see contraction but disagree on how much contraction. And an observer in the center would disagree with all of them. Relative, not objective.

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First some comments about time dilation. The pilot of a rocket ship that has accelerated to 0.99 c will experience a time dilation factor of about 7. The clock on the ship will be running 7 times slower than before the acceleration.

No frame reference, so that statement is ambiguous.  It will be running 7 times slower relative to the frame in which the ship is moving at .99c.  Not saying you're wrong, just sloppy.  It's running at normal rate relative to the ship of course.

I did provide a reference frame. The frame in which the ship is moving at .99 c is the one before the acceleration, which I explicitly mentioned right there: “accelerated to 0.99 c”. No sloppiness. We are not going ad hom., are we?

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That is, watching landmarks go by, he sees himself as traveling at almost 7 c.
If there is a grid with mileposts he can watch, then yes. You can call that his proper velocity if you want, which is what you get by simply multiplying his proper acceleration by his proper time under that acceleration. There's no limit to proper velocity since it accumulates additively, not relativistically.

You have to bring two observers into the same reference frame to judge which one is right, that is, which one underwent acceleration. That will be the one with the slower clock.

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(Either that or he thinks the distances have shrunk.) But after several years of sightseeing, he decelerates and gets back home again, and finds that his twin brother has aged much more than him. He realizes that acceleration really did slow his clock and that he was not traveling as fast as he thought. He was really traveling much slower.
What an absolute way of putting it, but yes, in the brother's frame, he was travelling at nearly c. In his own frame, he wasn't moving at all, only the length-contracted markers were. He knows very well that those moving markers do not mark distances properly and can not move faster than light, as evidenced by the fact that he can see the one's coming at him.  If they were moving at nearly 7c, they'd not be visible at all until they had passed, just like you can't hear a supersonic jet coming at you. So he knows those markers are moving at about .99c and are nowhere near a km apart. To assert otherwise is to assert light moving faster than c, something I'm starting to see a lot of here.

His observation is that he is traveling faster than c. Known landmarks are whizzing by at 7 c. Why can’t he conclude that Einstein was wrong? Again, the only way to tell the difference is to bring him back into a frame of reference that did not experience acceleration (which he did) and compare clocks. Or calendars as the case may be.

Title: Re: Is angular momentum frame dependent?
Post by: Malamute Lover on 04/08/2020 00:51:33

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Unlike time dilation due to different relative inertial speeds, time dilation resulting from acceleration is real and has real consequences. This is the resolution of the so-called Twin Paradox.
So I have four observers Alice, Bob, Charlie and Denise.  Alice was always on Earth.  Charlie came to Earth with Denise and Charlie accelerated to Earth frame to marry Alice.  Bob accelearted to match speeds with Denise to marry her.  Denise and Alice have thus never accelerated, and the other two have.
So according to your statement above, time dilation of the clock on Earth is real according to Charlie because he accelerated to Earth frame, but it is not real for either Alice or Denise, neither of whom have ever accelerated.  Meanwhile dilation of the clock on the ship with Bob and Denise is real to Bob (having accelerated to that frame), but not to Denise right there with him.
Sounds very inconsistent that the same clock is really dilated and also not.

Time dilation can be seen to be real when a clock that has undergone acceleration and brought back into a common inertial frame with a clock that has not accelerated. Try restating your scenario with some common inertial frame clock comparisons.

Also resolve the contradiction that “Charlie came to Earth with Denise” and Denise has never accelerated. I will presume you mean that Alice and Denise have always been on Earth.

If you attempted to restate your argument with clocks, you would find the flaw in it.

You say that Charlie accelerated to Earth frame. That would mean that Earth was already in an accelerated frame compared to Charlie. Earth clocks were already ticking slower than the clock in the frame of Charlie before they accelerated. By accelerating to match Earth frame, his clock will slow. Maybe you meant that Charlie decelerated and his clock speeded up to match the clocks on Earth.

Will the clock used by Charlie match Earth time? Irrelevant because nowhere does it say they were synchronized in the first place. If Charlie (likewise Bob):

(1) had been on Earth in the first place and synchronized his clock with an Earth clock and then
(2) accelerated away and cruised for a while then
(3) decelerated, reversed direction and accelerated again and cruised for a while and then
(4) come back to Earth and decelerated to the Earth frame,

then after all that comparing the clocks would be meaningful. And it would be found that the Earth clocks were ahead of the Charlie clock. His clock had been slowed by the acceleration and had been running slower while he cruised. The decelerations only brought his clock back to Earth speed not below it.

If somehow, Charlie had stayed still and Earth had done all the acceleration and whatnot, then the Earth clocks would be behind the Charlie clock. Acceleration is not relative. It is real. It is felt. If he saw Earth moving away, he would know who really accelerated by whether or not he felt the acceleration. And the ‘feeling’ could be by a high precision accelerometer so there is no doubt about whether it happened.

Now what about Lorentz contraction? Will Charlie subjectively feel or see any contraction? Say he has a pulse of light bouncing back and forth along the length of the ship and the clock ticks every time the light pulse completes a cycle. Since the length of the ship is contracted, it should take less time for the light pulse to go back and forth, right? And that means the clock should be running faster, right?

But Relativity tells us that clocks run slower by the same mechanism that lengths contract. Since the clock Charlie uses is running slower and by the same Lorentz factor, the Lorentz contraction driven clock speed up is exactly balanced by the Lorentz time stretch. It takes longer for the light pulse to make its round trip according to the Charlie clock. Charlie sees his clock running at the same rate as always and the length of the ship the same as always and the speed of light measured the same as always.

Lorentz contraction is relative, not objective.

Also, length contraction in all 3 scenarios at the top of this post is very real by your definition since there is a very real consequence (observed by anybody) in all three situation.

An observer on the rotating circle has a slowed clock and is unaware of the contraction that has taken place. Despite the increased RPM resulting from bringing the circle smaller (conserved AM remember), a revolution takes as long as it always did to this observer because his clock is slowed. The observer is unaware of any contraction.

Different situations, different observations. Relative, not objective.

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In the case of the spinning wheel, different portions of each spoke are traveling at different s
peeds. The time dilation factor increases as one travels up the spoke from the center.
Agree. Any clock nearer the center will run objectively faster than one further out.  ISS clocks run faster than sea-level clocks for this reason, but GR must be invoked to compute exactly how much.

ISS clocks also run slower because they are at orbital speed. Two factors going on. This is very important in setting the clocks in the GPS satellites. (BTW the real problem with GPS satellites is having them know exactly where they are at each moment, not an easy thing considering the Earth’s gravitational field varies from point to point and the Earth is rotating underneath the satellites.)

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A clock carried up from the axle to a certain point and back down again will have recorded less time than a clock that stayed at the center. The further up the clock is taken, the greater the discrepancy when brought back.
Also how long it spent out there. You'd have to integrate the curve.
If this is not a reasoning why length contraction doesn't actually occur, then you're getting way off topic.

The clock rate at each level is what matters, not the increasing time difference. It is not length contraction. It is bending of the spokes because of the differential in rates. This is exactly on topic but it shows that the spokes do not reach out as far as you want them to, so you want to dismiss it.

You do not have to integrate anything. The new circumference of the ring is easily calculated from the amount of contraction due to its speed. (To be clear that is tangent speed, not RPM) The new extension from the axle of the bent spoke is the radius related to that new circumference. They both are both related to the same Lorentz factor.

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When the wheel was spun up, the different parts experienced different accelerations, and/or different centripetal accelerations as the wheel turns.
In scenario 3, and also the spoked wheel in post 2, nothing was spun up. There was never any acceleration. In scenario's 1 & 2, there is angular acceleration involved.

All the components in all scenarios are following circular paths. The vector is changing. That IS acceleration.

In addition:

1) In the case of the wedding rings, to demonstrate that the two rings were originally the same size (without which the demonstration is meaningless), they must start off at rest with each other. One then must be accelerated by being spun.

2) In the case of the spokes, to (allegedly) create gaps, you must spin the wheel, which you explicitly refer to. Acceleration.

3) In the Ehrenfest scenario, the flaw is that Ehrenfest based everything on an assumption of 3D Euclidean space. As I said earlier, General Relativity, which did not exist yet, uses 4D curved spacetime when acceleration is involved as it definitely is in a rotating framework.

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This is acceleration-based time dilation and it is therefore real. At each level, the clock is slower than at the center and the real speed is less than an observer at that level thinks it is.
Oh, so if I build a clock on the rim of an already-spinning wheel, it will stay in sync with the clock at the axle? If not, what do you mean by this distinction between acceleration-based real time dilation and not-real?
Suppose Bob passes Earth at .866c without acceleration, syncing his clock to Earth when he's in its presence. Both say zero.  After a year, he smashes the clock works, freezing it at its reading of 1 year. He turns around, builds a new clock, sets it to zero and smashes the workings again, freezing it at 1 year as well. He then decelerates.  Now he has two clocks that have never accelerated (at least not while working) that cumulatively account for 2 years, but 4 years have gone by on Earth.

“Suppose Bob passes Earth at .866c without acceleration”

Wrong at the start. The is no way that Bob and Earth can have relative speeds of .866 c without acceleration being involved somewhere. If you think it can, come up with a scenario that would explain how it happened other than the bare assumption.

It is clear that Bob has accelerated to .866 c, either personally or by coming from some environment in motion with respect to Earth, or some combination of the two. As a result, his clock is running at half the speed of a clock on Earth, as would be seen if an elapsed time comparison could be made in a common inertial frame.

“Now he has two clocks that have never accelerated”

The first clock was in a reference frame at .866 c as viewed from Earth. The fact that the clock itself never accelerated is irrelevant, the reference frame it is in did.

The second clock was also in a reference frame at .866 c as viewed from Earth. In order to turn around back to Earth, Bob decelerated and then accelerated back up to .866 c as viewed from Earth. Accelerated reference frame again.

Imagine that Bob built a clock after decelerating but before accelerating in the other direction again. That is, the same inertial frame as Earth. Once the ship finished accelerating again and was just cruising, would this accelerated clock tick at the same rate as the new clock Bob built right after the end of acceleration? If Bob set the clock that he built at the turnaround point and the newly constructed clock to zero at the same moment, would they continue to read the same time or not?

If you say that they would not tick at the same rate, then answer this.  If Bob closed his eyes and counted One Hippopotamus, Two Hippopotamus and so on up to Ten, which clock would he agree with about the seconds elapsed. Keep in mind that Bob was accelerated.

It seems that your understanding of Relativity Theory could use some work.


I'm sorry, but your stories are getting sillier and sillier, seemingly in a desperate attempt to not admit you've made a mistake. Einstein never mentioned any distinction between real and fake time dilation, or real and fake length contraction.  Measurements using light would be different than empirically measured if length contraction wasn't real.  All you have to do is time light as it goes from one end to the other and back in a moving train.  Measure the time with stationary synced clocks (OK, you have tried to counter this by flat out denial of the possibility of syncing stationary clocks). It will take way too long if the train is not really contracted.  A history of acceleration of the train plays no role in the equations involved.

All of my statements are derived from straightforward physics textbooks. There is nothing silly about them. I have addressed your arguments in detail, including all of your counterarguments and I see no reason to change my mind about anything. But I do see we are in ad hom territory.

Einstein never mentioned either fake or real time dilation. He talked about, what was that word again? Oh yes, Relativity. Specifically, “On the Relativity of Lengths and Times”, the second numbered section in his 1905 paper On the Electrodynamics Of Moving Bodies (https://www.fourmilab.ch/etexts/einstein/specrel/specrel.pdf)

Measurements of time dilation, length contraction, energy content are all relative to the reference frames of the observer and the observed. Different observers in different frames will disagree on what they see, including Lorentz contraction.  (The energy content part appeared in a different paper.)

Moving on to your thought experiment:

First some questions. What is the train moving relative too? Is there an observer in that reference frame? Without reference toa history of acceleration, how can a degree of motion be assigned?

It is not only length that appears modified to an external observer in a different reference frame. It is time as well. I have already presented this argument earlier but I will repeat it here.

he has a pulse of light bouncing back and forth along the length of the train and the clock ticks every time the light pulse completes a cycle. (One clock solves the sync problem.) Since the length of the train is contracted, it should take less time for the light pulse to go back and forth, right? And that means the clock should be running faster, right?

But Relativity tells us that clocks run slower by the same mechanism that lengths contract. Since the clock on the train is running slower and by the same Lorentz factor, the Lorentz contraction driven clock speed up is exactly balanced by the Lorentz time stretch. It takes longer for the light pulse to make its round trip according to the clock on the train. An observer on the train sees the clock running at the same rate as always and the length of the train the same as always and the speed of light measured the same as always.

However the outside observer sees the length of the train contracted and the clock on the train slower.

Lorentz contraction is relative, not objective.

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Since the angular displacement would vary at each level, the spoke would curve.
Time dilation has no effect on the curvature of the spokes.  They're effectively strings. Time dilation does have an effect on the angular velocity of the wheel.  An observer at the axle would measure a smaller angular velocity than one at the rim, which is why for relativistic wheels, we specify linear rim speed, not radians per second. The spokes are straight unless there is angular stress on them such as the wheel being accelerated by torque being applied at the hub, but no such torque exists in our examples. All measurement are done in steady state.  Yes, if I put enough change in the rate of torque on a real bicycle wheel, the change would need to propagate up the spokes to the rim at the speed of sound, causing a momentary wave to travel up the spoke. That would bend it a bit just like a sideways yank on a garden hose causes a bend to travel up the hose.  A freely spinning wheel has straight spokes so long as there is any tension on them.

Time dilation causes the curvature of the spokes. Or rather the curvature of spacetime on which the spokes lie, this curvature caused by acceleration, both original and centripetal. You cannot have spinning without original acceleration or else you have no basis for talking about degree of motion. Unless you include the original acceleration, the wheel could be motionless.

I tried to explain it in Euclidean terms as the spokes offset according to the clock rates, but I did not realize at that time that you do not believe in time dilation. (Einstein is not just rolling over in his grave. He is doing 0.99 c in a circle around his major axis.) I really did not want to get into GR spacetime without cause. But now there is cause. The spokes reside in acceleration caused curved spacetime. They do not reach as far out as they would in Euclidean space. The rotating wheel also resides in curved spacetime due to acceleration and has a circumference in that spacetime that matches the tips of the curved spokes. This is necessarily the case since the same Lorentz factor controls both.

Remember that curved bowl?

(https://assets.katomcdn.com/q_auto,f_auto,w_300/products/095/095-1195132/095-wc11951.jpg)

The circumference does not grow as fast as the radius as you go out from the center. Or to say that the other way around, the circumference shrinks faster than the radius as you move into flatter spacetime at the center. This is what happens in a gravity well also except that the flatter spacetime is outward instead of inward. Which is why that circle built deep in the gravity well of the neutron star gets squeezed around its circumference as it is lifted out of the gravity well, not enough circumference anymore in flatter spacetime.

Admittedly these are not easy concepts unless you are really used to the subject. But to understand relativity at all, you have to first accept the idea of time dilation.
Title: Re: Is angular momentum frame dependent?
Post by: Jaaanosik on 04/08/2020 13:14:37
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Are you denying time dilation?
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Contraction is relative, not objective.
...

Is the time dilation relative, not objective as well?
Title: Re: Is angular momentum frame dependent?
Post by: Jaaanosik on 04/08/2020 13:40:35
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Time dilation can be seen to be real when a clock that has undergone acceleration and brought back into a common inertial frame with a clock that has not accelerated. Try restating your scenario with some common inertial frame clock comparisons.
...



(https://i.imgur.com/6MCSVif.png)

The acceleration can be taken out for the time dilation analysis.
You can see the SR reciprocal thread for the discussion.
Have you ever heard about the clock postulate?
http://math.ucr.edu/home/baez/physics/Relativity/SR/clock.html
Gamma factor does not include an acceleration,
Jano
Title: Re: Is angular momentum frame dependent?
Post by: Colin2B on 04/08/2020 14:39:21
The example presented by Kryptid is in question, therefore your argument based on Kryptid's comment is in question as well,
It wasn’t an argument being presented, just clarification for Malamute Lover

What is real is what is going on in 4D Minkowski spacetime. We time bound observers only get to see a slice at a time depending on our reference frames.
This all we ever see, even for those things we consider real.

Multiple observers in different reference frames see different things. None of them see the real thing. Relativistic effects are relative not objective.
Being relative does not exclude being real/objective
I’ll pick this up in the other thread when I have time, so to speak  ;)
https://www.thenakedscientists.com/forum/index.php?topic=80208.msg610352#msg610352
Title: Re: Is angular momentum frame dependent?
Post by: Malamute Lover on 04/08/2020 19:50:58
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Time dilation can be seen to be real when a clock that has undergone acceleration and brought back into a common inertial frame with a clock that has not accelerated. Try restating your scenario with some common inertial frame clock comparisons.
...



(https://i.imgur.com/6MCSVif.png)

The acceleration can be taken out for the time dilation analysis.
You can see the SR reciprocal thread for the discussion.
Have you ever heard about the clock postulate?
http://math.ucr.edu/home/baez/physics/Relativity/SR/clock.html
Gamma factor does not include an acceleration,
Jano

The clock postulate goes into GR territory. But then so does acceleration in general. But if acceleration is identical for two clocks originally synchronized in a common inertial frame, if there is an acceleration related effect, it will be the same for both clocks.
Title: Re: Is angular momentum frame dependent?
Post by: Malamute Lover on 04/08/2020 19:55:20
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Are you denying time dilation?
...
Contraction is relative, not objective.
...

Is the time dilation relative, not objective as well?

Not if it is different for different observers and non-existent in the reference frame in which the time dilation is observed by others. You seem to be assuming a Newtonian Absolute Space and Time. They do not exist.
Title: Re: Is angular momentum frame dependent?
Post by: Halc on 04/08/2020 22:07:14
There are actually three distinct scenarios.
1) You have a solid ring that spins. It contracts as you spin it.  A shrinking ring (that reduces in radius from any perspective) seems awful real to me.  Using this scenario, I can pass one wedding ring through another identical one by spinning it.  That's real contraction, or it couldn't be done. It isn't observer dependent.
You are sort of describing such a thing below, except with superfluous spoke that will bend because they're too long, so they serve no purpose other than to be deformed by being squashed.

The geometry of the spinning ring is non-Euclidean. It was not only accelerated up to speed, it is under continuous acceleration as it spins. SR is not a good guide here.
The ring is treated as having negligible thickness, in which case it is Euclidean and can be spun up to speed. If not, it is a case of the concrete think poured around the neutron star, making it scenario 3. It shatters if you try to spin it. If already spinning, a thick ring is non-euclidean and does not undergo any angular acceleration.
SR describes all three scenarios perfectly, but not the neutron star scenario since that involves gravity.

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The acceleration requires the use of GR. Because the spacetime is curved, the distance from the spinning ring to the non-spinning ring is greater.
It doesn’t matter whether GR or SR is used since spacetime is completely flat in the example, and SR handles acceleration just fine. Spinning a ring doesn’t bend spacetime. It just bends the ring.

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An observer on the spinning ring will think the non-spinning ring is expanding so no problem with fitting.
He will think no such thing any more than anybody thinks the traveling twin made everybody on Earth age faster. He knows very well that his ring is the one spinning and shrinking since rotation is absolute, not relative. Everybody knows it, on the ring or not. That’s what makes it a real consequence.

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Which is which? As with GR problems in general, who is undergoing acceleration?
Acceleration is also absolute. There’s no question what undergoes it. Surely you know at least this much.

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But the question can only be settled by bringing the two rings into a common inertial reference frame and comparing clocks, just like with the Twins. It cannot be settled by comparing observations.
The two rings are in a common inertial frame, and a clock on one runs objectively slower that the other, just as is observed with the ISS.Observation of a clock is completely unnecessary since the one ring passing through the other is objective.  You can measure the two rings with a relatively stationary ruler and observe (from any frame) that the one ring is unchanged and the spinning on is contracted. The clocks are more evidence, but not necessary to observer the real consequence.
As predicted, you’re just refusing to accept hard evidence. It seems you’re not even denying the contraction now, suggesting instead that the stationary ring might have instead expanded due to the proximity of this spinning ring. No theory suggests any such thing.


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Quote from: Halc
2) Spoke scenario, or the roller coaster track, which is essentially the same scenario.  Here the radius is held constant by the non-contracting straight spoke, or by the stationary track.  There is no solid ring, but a series of detached adjacent blocks.  If there are spoke, you have essentially a row of independent pendulums.  If a track, you have a row of 'bumper cars'.  Spin it up and gaps form between the blocks, and more can be inserted if you like.
Observers in any frame will agree on this, but you seemingly are in denial of it.

The spokes bend.
They do not. You have no way to back this fantasy. Yes, time dilation and width contraction varies along its length, but neither has any reason to curve the string, which would have zero effect on that dilation.

I notice you did not address the bumper-car track, which also maintains fixed radius, and apparently has the property of you not being able to think of some way to deny the gaps that form between the cars.

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Quote from: Halc
3) The actual Ehrenfest scenario where he takes a non-Euclidean 3 dimensional solid (a spinning cylinder) and declares it paradoxical when its non-Euclidean properties are illustrated. If the object is rigid, it shatters as soon as you attempt to change its angular speed.  That shattering is an objective effect that any observer in any frame will witness.
Ehrenfest wrote his paradox before General Relativity was developed and did not know about curved spacetime.
There is no curved spacetime in the scenario. Spacetime is completely flat, lacking a source of gravity in the description. He found it paradoxical that a solid could exist in Euclidean Minkowski spacetime that exhibited non-Euclidean properties, but of course SR predicts it.

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Ehrenfest assumed 3D Euclidean geometry, not non-Euclidean as you stated.
If he assumed the object was Euclidean, then he was mistaken.  Spacetime is in that instance, but not the object.  No rotating object can be.

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He therefore assumed contraction was real, which is what led to his claim of a paradox.
Contraction being real is what is demonstrated, because it resolves the paradox. The radius doesn’t change because the spin never does. The object was never stationary, and he does not suggest that it ever was.

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A clock on the surface of the cylinder will run slower than a clock inside the cylinder.
Indeed. An objective consequence admitted by the guy who denies it.  Hmm…
How is this real consequence explained if time dilation isn’t real? It isn’t relative since the observer on the edge also sees the clock in the middle run faster.
“Neither dilation nor contraction are real, except when I have to admit otherwise”.  Great stance.

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To an observer on the surface (and therefore to the mechanical properties of the cylinder) the same number of revolutions a minute are taking place
Now you’ve contradicted yourself again.  We have a stationary marker by which a rotation can be measured.  Both observers agree on what one rotation is, but if their clocks are not running at the same pace, they necessarily measure a different time for one revolution.

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An observer inside with a faster clock will be surprised to not see the cylinder shatter just as he is surprised when his Twin comes back much younger than him.
The twin apparently doesn’t know his physics then, because if he did, there would be no surprise.  The cylinder doesn’t shatter because it was always spinning.

At this point you go into a bend about gravity and GR, which seems a diversion from the more simple SR topic that you need to master first.  One comment though:

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It will be crushed around the circumference. Where does the energy come from to crush it?
It doesn’t take energy to crush something. It seemingly takes force, which means a strong but brittle object can be crushed by expenditure of arbitrarily small energy. The less brittle it is, the more that energy goes into strain and not into failure, so it takes more. I’m assuming insanely brittle and strong materials for our objects else they’d not be able to withstand the centripetal stresses being put on them. Anything else would just fly apart.

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Whether and how much contraction an observer will see depends on the observer.
The cases I enumerate above are observed by anybody. They were chosen for that purpose. The measuring rod between ships is also a real consequence.

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An observer on the spinning wheel sees nothing different because contraction is only visible from another reference frame.
In all three cases, the observer on the wheel very much sees differences, which are pointed out in the cases above. You seem to agree that one ring fits through the other, something to which all observers agree. You don’t seem to have any fake physics that lets you deny the bumper-car-track thing, nor do you seem to deny the non-Euclidean dimensions of the ‘cylinder’.  OK, the  non-Euclidean dimensions are frame dependent.  A stationary observer will measure normal dimensions, but the inability of the object to change its angular speed is an objective observation.


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Quote from: Halc
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First some comments about time dilation. The pilot of a rocket ship that has accelerated to 0.99 c will experience a time dilation factor of about 7. The clock on the ship will be running 7 times slower than before the acceleration.
No frame reference, so that statement is ambiguous.  It will be running 7 times slower relative to the frame in which the ship is moving at .99c.  Not saying you're wrong, just sloppy.  It's running at normal rate relative to the ship of course.
I did provide a reference frame. The frame in which the ship is moving at .99 c is the one before the acceleration, which I explicitly mentioned right there: “accelerated to 0.99 c”. No sloppiness. We are not going ad hom., are we?
It’s not an ad-hom.  It’s sloppy because you’re describing ‘what the pilot experiences’ and a pilot always experiences being stopped.  You’re referencing the pilot frame and also the original frame, which makes it confusing. That’s sloppy.
Secondly, the bolded statement is wrong since no observer can experience time dilation. I can look at the GPS clocks and objectively notice I’m running slower than them, but I still don’t experience that dilation.

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You have to bring two observers into the same reference frame to judge which one is right, that is, which one underwent acceleration.
Acceleration is absolute (at least in Minkowski spacetime). An accelerometer works inside a box. All observers will agree if something has accelerated. That’s twice you’ve made this mistake.

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His observation is that he is traveling faster than c. Known landmarks are whizzing by at 7 c. Why can’t he conclude that Einstein was wrong?
He is free to propose a different theory, but none has been found so far. So are you a relativity denier then? It seems to be your goal here. You resist it at every step of the way.
Such deniers are dime a dozen on sites like this, but then don't go telling me that your stories conform to an established theory and mine don't. I've pointed out several self contradictions with your assertions.

Einstein didn’t just suggest that light speed yielded the same value in any frame. That because quite apparent by all the attempts to measure the difference as was predicted by the prevailing view of the time. So he can’t conclude Einstein was wrong, he’d have to conclude that all the decades of light speed measurement were wrong. Einstein didn’t perform any of those measurements.
Title: Re: Is angular momentum frame dependent?
Post by: Malamute Lover on 06/08/2020 00:52:24
There are actually three distinct scenarios.
1) You have a solid ring that spins. It contracts as you spin it.  A shrinking ring (that reduces in radius from any perspective) seems awful real to me.  Using this scenario, I can pass one wedding ring through another identical one by spinning it.  That's real contraction, or it couldn't be done. It isn't observer dependent.
You are sort of describing such a thing below, except with superfluous spoke that will bend because they're too long, so they serve no purpose other than to be deformed by being squashed.

The geometry of the spinning ring is non-Euclidean. It was not only accelerated up to speed, it is under continuous acceleration as it spins. SR is not a good guide here.
The ring is treated as having negligible thickness, in which case it is Euclidean and can be spun up to speed. If not, it is a case of the concrete think poured around the neutron star, making it scenario 3. It shatters if you try to spin it. If already spinning, a thick ring is non-euclidean and does not undergo any angular acceleration.
SR describes all three scenarios perfectly, but not the neutron star scenario since that involves gravity.

This is incorrect. The thickness is irrelevant. The ratio of the circumference to the radius, which follow a curved spacetime, is less than 2π. The curved radius is no longer in Euclidean geometry.  The circumference is shorter than it was before rotation caused the spacetime curvature.

Circular acceleration (spin up or centripetal) is like gravitational acceleration but the direction of increasing curvature is reversed. To an observer on the axle, the circle has become smaller and faster. But to the time dilated observer on the circle, a rotation takes as long as ever so the speed did not change.

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The acceleration requires the use of GR. Because the spacetime is curved, the distance from the spinning ring to the non-spinning ring is greater.
It doesn’t matter whether GR or SR is used since spacetime is completely flat in the example, and SR handles acceleration just fine. Spinning a ring doesn’t bend spacetime. It just bends the ring.

You are incorrect again. Acceleration bends spacetime. The spacetime is definitely non-Euclidean.

SR does not handle acceleration at all. It is restricted entirely to inertial reference frames.

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An observer on the spinning ring will think the non-spinning ring is expanding so no problem with fitting.
He will think no such thing any more than anybody thinks the traveling twin made everybody on Earth age faster. He knows very well that his ring is the one spinning and shrinking since rotation is absolute, not relative. Everybody knows it, on the ring or not. That’s what makes it a real consequence.

(This is now the two wedding rings, not the spinning wheel.)

Acceleration is absolute in that it is experienced by the accelerating observer. The magnitude of the acceleration is not absolute in that time dilation disassociates proper acceleration from that measured by an inertial frame observer. The traveling twin thinks he is continuing to accelerate at the same rate – which he can feel - and therefore exceeds c as evidenced by the speed of the landmarks going by. Einstein must be wrong, he thinks.  An inertial observer never sees the twin exceed c, inferring that the acceleration is declining.

While rotation is real because it induces acceleration, the magnitude of the rotation is not absolute, again because of time dilation.

An observer on the periphery of the spinning ring will not experience any shrinkage of his ring.  His ring is as big as ever and he can prove that by walking around it and it takes just as many steps as ever, being unaware of any contraction in himself. The other ring must be expanding, he thinks. Who is right? Bring both rings into the same inertial reference frame and compare clocks. That will tell who experienced acceleration.

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Which is which? As with GR problems in general, who is undergoing acceleration?
Acceleration is also absolute. There’s no question what undergoes it. Surely you know at least this much.

Who is undergoing acceleration is the way one tells which is which. I have said this several times very clearly so there was no need for you to misinterpret it.  Yes, I know that much and a lot more.

The fact of acceleration is absolute, being felt by the accelerated body. But the magnitude of acceleration is not absolute because of time dilation. This is why the pilot of the spaceship and the inertial frame observer disagree on how fast the ship is going.

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But the question can only be settled by bringing the two rings into a common inertial reference frame and comparing clocks, just like with the Twins. It cannot be settled by comparing observations.
The two rings are in a common inertial frame, and a clock on one runs objectively slower that the other, just as is observed with the ISS.Observation of a clock is completely unnecessary since the one ring passing through the other is objective.  You can measure the two rings with a relatively stationary ruler and observe (from any frame) that the one ring is unchanged and the spinning on is contracted. The clocks are more evidence, but not necessary to observer the real consequence.
As predicted, you’re just refusing to accept hard evidence. It seems you’re not even denying the contraction now, suggesting instead that the stationary ring might have instead expanded due to the proximity of this spinning ring. No theory suggests any such thing.

Until you compare clocks, you will not convince the observer on the spinning ring that he was wrong and that his ring contracted, which he did not see, and that the other ring did not expand, which he did see.

Quote from: Halc
2) Spoke scenario, or the roller coaster track, which is essentially the same scenario.  Here the radius is held constant by the non-contracting straight spoke, or by the stationary track.  There is no solid ring, but a series of detached adjacent blocks.  If there are spoke, you have essentially a row of independent pendulums.  If a track, you have a row of 'bumper cars'.  Spin it up and gaps form between the blocks, and more can be inserted if you like.
Observers in any frame will agree on this, but you seemingly are in denial of it.

The roller coaster thing and the spoke thing are not the same. In the spoke thing, the blocks are being held against centripetal force by the rim which is moving with them.  In the roller coaster thing, the cars are being held against centripetal force by an outside track which is not rotating with them.  (If it is rotating with them, then this is exactly the spoke scenario, in which the wheel contracts when the geometry becomes non-Euclidean and the spokes curve.)

As the cars accelerate, they push the track in the opposite direction. (Newton’s Third Law.) That is the acceleration will not be as much as expected, half of the energy going into the track. (No problem. They have powerful engines and lots of fuel.) If the mass of the track is the same as that of the cars, the track will rotate at the same speed and it will contract in exactly the right proportion to match the contracting circle of the cars. No gaps.

If the track is more massive than the cars, it will be going slower than the cars going and would not contract as much as if it were equal in mass. But remember that the cars are on a contracting circumference because their speed is bending spacetime. What is holding them in a circular orbit rather than flying out in a straight line and crashing into the track?

As it turns out, the track is going to contract with the cars. How? Lense-Thirring frame dragging. (The names are Austrian and are pronounced LENsuh TEERing).

Rotating masses drag spacetime with them. That is part of GR. It has been demonstrated in polar orbit satellites where gyroscopes precess toward the Earth’s axis. The degree of dragging is related to the radius, the angular momentum of the body and the radius at which the dragging is measured. The speed of the track and the path of the cars is initially the same. But they are spinning in opposite directions. As the cars continue to accelerate, the track will also accelerate but not as quickly because it is more massive.

As the cars get up to relativistic speed, their mass energy (as determined by an inertial frame outside observer) increases as per the Lorentz factor. Where is the energy coming from? Time dilation. The engines are running at the same power level but from the outside the cars are not accelerating as quickly as the drivers think. The energy difference between driver perceived acceleration and external observer perceived acceleration is going into mass-energy.

As with all characteristic affected by the Lorentz factor, the mass-energy grows quickly as lightspeed is approached. The mass-energy of the cars is growing faster than the mass-energy of the slower accelerating more massive track. The L-T frame dragging strength is growing faster for the cars than for the track. The cars are dragging the spacetime frame of the track toward them. Again since it is all related to the Lorentz factor, everything matches and comes out equal. The track and the expected car path shrink (go non-Euclidean) at the same rate. No gaps.

Your naïve, incomplete and partially incorrect knowledge of Special Relativity does no good in situations involving acceleration where General Relativity must be applied.


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The spokes bend.
They do not. You have no way to back this fantasy. Yes, time dilation and width contraction varies along its length, but neither has any reason to curve the string, which would have zero effect on that dilation.

Not a fantasy but a consequence of GR, which you need to deal properly with acceleration. But you are in denial of GR. I find it curious that you had no problem with the spinning wedding ring contracting but you deny that [i[your[/i] circle can contract. Do you really think that the wedding ring could contract in a purely Euclidean space?

You are in denial that acceleration requires GR and that GR uses non-Euclidean geometry to deal with it. If you do not know the rules, do not try to play the game.

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Quote from: Halc
3) The actual Ehrenfest scenario where he takes a non-Euclidean 3 dimensional solid (a spinning cylinder) and declares it paradoxical when its non-Euclidean properties are illustrated. If the object is rigid, it shatters as soon as you attempt to change its angular speed.  That shattering is an objective effect that any observer in any frame will witness.
Ehrenfest wrote his paradox before General Relativity was developed and did not know about curved spacetime.
There is no curved spacetime in the scenario. Spacetime is completely flat, lacking a source of gravity in the description. He found it paradoxical that a solid could exist in Euclidean Minkowski spacetime that exhibited non-Euclidean properties, but of course SR predicts it.

Naïve Minkowski spacetime relates to SR, not GR. It is not non-Euclidean, having been developed years before GR. Minkowski spacetime is 4D but all dimensions are straight. General Relativity is much hairier involving semi-Riemannian manifolds with varying metrics.

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Ehrenfest assumed 3D Euclidean geometry, not non-Euclidean as you stated.
If he assumed the object was Euclidean, then he was mistaken.  Spacetime is in that instance, but not the object.  No rotating object can be.

The rotating object exists in non-Euclidean spacetime. Of course, it is in a non-Euclidean form. What are you trying to say here?

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He therefore assumed contraction was real, which is what led to his claim of a paradox.
Contraction being real is what is demonstrated, because it resolves the paradox. The radius doesn’t change because the spin never does. The object was never stationary, and he does not suggest that it ever was.

If the spin was always there, then the spacetime was always non-Euclidean. Centripetal acceleration, remember.

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A clock on the surface of the cylinder will run slower than a clock inside the cylinder.
Indeed. An objective consequence admitted by the guy who denies it.  Hmm…
How is this real consequence explained if time dilation isn’t real? It isn’t relative since the observer on the edge also sees the clock in the middle run faster.
“Neither dilation nor contraction are real, except when I have to admit otherwise”.  Great stance.

Acceleration is real because it cannot be denied by those who experience it. Acceleration puts different observers in different reference frames. Time dilation is observer dependent.  Different observers will disagree on the rate of a single clock. Those in possession of the clock deny that there is any time dilation, whereas they cannot deny acceleration. But they can argue about the degree of acceleration. Time dilation is relative not objective. Bringing two clocks into a common inertial frame will allow comparison, with the slower clock the one that experienced more acceleration. 

Time dilation and length contraction are relative. Two spaceships not in a common inertial frame will both see the other’s clock as running slow and their own clock as running properly. Who is right? Neither one. Time dilation is relative. Bring them into a common inertial frame and the two clocks will run at the same rate. But one will be behind the other because it got accelerated more. Agreement on who is right cannot not be done in separate reference frames and not by directly comparing clock rates.  Relative, not objective.

A very fast spaceship passes a stationary one, relative to some specific inertial reference frame. Both will see the other as contracted and having a slow clock. If contraction is real then why don’t the observers on the two ships see the other observers get crushed with blood and guts shooting out their sides? And why would that happen? Because another ship passed at a different speed?

Sorry, but you are still wrong. Lorentz contraction is relative, not objective.

Title: Re: Is angular momentum frame dependent?
Post by: Malamute Lover on 06/08/2020 00:59:22
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To an observer on the surface (and therefore to the mechanical properties of the cylinder) the same number of revolutions a minute are taking place

Now you’ve contradicted yourself again.  We have a stationary marker by which a rotation can be measured.  Both observers agree on what one rotation is, but if their clocks are not running at the same pace, they necessarily measure a different time for one revolution.

Nope. The reduced circumference circle is spinning faster because tangential speed is still the same but the increased time dilation makes an observer on the circumference think a revolution takes as long as if the circumference were still longer.

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An observer inside with a faster clock will be surprised to not see the cylinder shatter just as he is surprised when his Twin comes back much younger than him.
The twin apparently doesn’t know his physics then, because if he did, there would be no surprise.  The cylinder doesn’t shatter because it was always spinning.

Based on his first person experience, the traveling twin is convinced Einstein is wrong and speeds greater than c are possible. Not until he gets back into a common inertial frame with his now much older twin is he given evidence that Einstein was right after all and that observations on different reference frames are relative.

At this point you go into a bend about gravity and GR, which seems a diversion from the more simple SR topic that you need to master first. 

I know SR cold which is why I know that when acceleration is in the picture you need to bring in GR and non-Euclidean spacetime. The circumference is smaller and the spokes do not reach as far out because the spacetime they are on is curved.

One comment though:

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It will be crushed around the circumference. Where does the energy come from to crush it?
It doesn’t take energy to crush something. It seemingly takes force, which means a strong but brittle object can be crushed by expenditure of arbitrarily small energy. The less brittle it is, the more that energy goes into strain and not into failure, so it takes more. I’m assuming insanely brittle and strong materials for our objects else they’d not be able to withstand the centripetal stresses being put on them. Anything else would just fly apart.

(We are now talking about the material circle constructed around a neutron star being raised out of the gravity well in a differently shaped spacetime.)

There must be some force involved or there would be no crushing at all. Where does the energy come from? It comes from the same source as the energy used to raise it out of the gravity well. It would resist being forced into a differently shaped spacetime. Note that there is no spinning involved here so this is a different problem from the spinning wheel situations.

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Whether and how much contraction an observer will see depends on the observer.
The cases I enumerate above are observed by anybody. They were chosen for that purpose. The measuring rod between ships is also a real consequence.

You cannot directly see curved spacetime. Our 3D Euclidean oriented brains do not work that way. You can only see the consequences. I have addressed all of the cases you enumerated and shown how your assumptions are wrong because you do not take into account GR and curved spacetime, which must be considered because acceleration is involved.

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An observer on the spinning wheel sees nothing different because contraction is only visible from another reference frame.
In all three cases, the observer on the wheel very much sees differences, which are pointed out in the cases above. You seem to agree that one ring fits through the other, something to which all observers agree. You don’t seem to have any fake physics that lets you deny the bumper-car-track thing, nor do you seem to deny the non-Euclidean dimensions of the ‘cylinder’.  OK, the  non-Euclidean dimensions are frame dependent.  A stationary observer will measure normal dimensions, but the inability of the object to change its angular speed is an objective observation.

The observer on the wheel does not see length contraction which can only be seen from a different reference frame. The observer on the wheel is time dilated relative to another observer and thinks the wheel is the same circumference because it takes the same subjective time to make a revolution relative to a fixed marker. But from a different reference frame nearer to the center, the circumference, being length contracted, is smaller than would be expected from the length of the spokes, a clear indication that spacetime is curved. As with the orbit of Mercury deep in the Sun’s gravity well, the curvature cannot be seem directly but can be inferred from the precession of the elliptical orbit.

An object certainly can change its angular speed by changing the radius. When a spinning skater pulls her arms in, angular momentum is conserved which requires and increase in angular velocity.

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Quote from: Halc
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First some comments about time dilation. The pilot of a rocket ship that has accelerated to 0.99 c will experience a time dilation factor of about 7. The clock on the ship will be running 7 times slower than before the acceleration.
No frame reference, so that statement is ambiguous.  It will be running 7 times slower relative to the frame in which the ship is moving at .99c.  Not saying you're wrong, just sloppy.  It's running at normal rate relative to the ship of course.
I did provide a reference frame. The frame in which the ship is moving at .99 c is the one before the acceleration, which I explicitly mentioned right there: “accelerated to 0.99 c”. No sloppiness. We are not going ad hom., are we?
It’s not an ad-hom.  It’s sloppy because you’re describing ‘what the pilot experiences’ and a pilot always experiences being stopped.  You’re referencing the pilot frame and also the original frame, which makes it confusing. That’s sloppy.

You first accused me of not mentioning the original frame, which I did implicitly by mentioning acceleration from it. Now you are berating me for mentioning both frames.  You are just looking for something, anything, to nitpick about even if you contradict yourself. That sure sounds like ad hom.

And “a pilot always experiences being stopped”? No, the pilot experiences being accelerated to 0.99 c from his original reference frame. If he decelerates and finds himself in a matching inertial frame that is actually accelerated from the starting frame, is he stopped? Do you have any understanding of the subject at all or was this another ad hom attempt that misfired?

Secondly, the bolded statement is wrong since no observer can experience time dilation. I can look at the GPS clocks and objectively notice I’m running slower than them, but I still don’t experience that dilation.

If you were able to see the GPS clock, it would be going at the same rate as your own clock because it was intentionally set to run slower.

To account for the increased rate of the GPS clock due to lower gravitational level at the 20,000 km orbital altitude (45 μ s a day fast) and the time dilation at 14,000 km/s orbital speed  (7 μ s a day slow), the clock on a GPS satellite is set to run 38 μ s a day slow to match mean sea level clocks. If there were another unadjusted clock on the satellite, then it would be seen to be 38 μ s a day fast.  Note however that even on the adjusted clock running at the same rate as yours, there will be a difference in time readings of .067 to .089 seconds depending which of the 24 active satellites you were comparing against. The time difference is how GPS works.

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You have to bring two observers into the same reference frame to judge which one is right, that is, which one underwent acceleration.
Acceleration is absolute (at least in Minkowski spacetime). An accelerometer works inside a box. All observers will agree if something has accelerated. That’s twice you’ve made this mistake.

The mistake is on your side.

Because the accelerometer on the spaceship is time dilated relative to the unaccelerated observer, it’s history will show a steady acceleration all the way up to 7 c just like the pilot felt, while an outside observer will see the spaceship’s acceleration decrease as it approaches 0.99 c. The fact of acceleration is absolute. But because acceleration slows the clock rate as seen by an outside observer but not by anyone on the spaceship, the rate of acceleration is relative.

To prove whose subjective experience is in error requires bringing both clocks into a common inertial frame and comparing them. In this case, the pilot will realize that his clock ran slow even if it did not seem that way to him. He will then accept that Einstein was right after all.

This is basic stuff. Why are you having a problem with it?

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His observation is that he is traveling faster than c. Known landmarks are whizzing by at 7 c. Why can’t he conclude that Einstein was wrong?
He is free to propose a different theory, but none has been found so far. So are you a relativity denier then? It seems to be your goal here. You resist it at every step of the way.
Such deniers are dime a dozen on sites like this, but then don't go telling me that your stories conform to an established theory and mine don't. I've pointed out several self contradictions with your assertions.

Einstein didn’t just suggest that light speed yielded the same value in any frame. That because quite apparent by all the attempts to measure the difference as was predicted by the prevailing view of the time. So he can’t conclude Einstein was wrong, he’d have to conclude that all the decades of light speed measurement were wrong. Einstein didn’t perform any of those measurements.

Acceleration is in the domain of General Relativity. A constant speed of light only leads to Special Relativity.  Because the pilot only studied SR, not GR (and how many have studied GR?) he did not know that acceleration was the explanation of the twin paradox. From his SR based viewpoint, it was perfectly valid for him to say that his twin was traveling away from him and should have been time dilated. In SR, the traveling twin will in fact see the other twin’s clock run slower and consider that proof of the other twin being the one in motion.

To prove whose subjective experience is the more credible, it is necessary to bring both clocks into a common inertial frame to determine who experienced less time.  That is the one who was most time dilated.  I say ‘most’ time dilated because the stay at home twin might have flown X-15 rocket planes a lot and the clock he carried with him at all times is going to be behind the clock on his wall at home. But even that clock is going to have less elapsed time than the one in high orbit around the earth.  Keep in mind that all these clocks will run at the same rate when brought into a common inertial frame.

The pilot disbelieving his own subjective experience because he knew Relativity Theory is just silly. Anyway he only knew the easy part. Which you seem to be having trouble with.

BTW the reason for the attempts to determine if light speed would always be measured at the same value regardless of the state of motion was that Maxwell’s Field Theory of Electromagnetism said that it would and the speed that fell out of the equations was equal to the already measured speed of light. This is why Michaelson and Morley conducted their experiments.  Michaelson continued to believe in the luminiferous ether despite the experiments, and also despite Maxwell’s explanation of why the ether was not necessary.  Einstein was not believed by everyone for some time. There are relativity deniers today, to use your phrase. But there are many more relativity misunderstanders, which includes you, I am afraid.
Title: Re: Is angular momentum frame dependent?
Post by: Halc on 13/08/2020 01:37:02
Mod note:
I have moved all the Bell's stuff into a new topic in New Theories:

Split: Bell's paradox: Does the string break?
https://www.thenakedscientists.com/forum/index.php?topic=80334.0

This thread will track the angular momentum question.  There were definitely ideas being pushed that have no business in the main physics section.

My apologies for the brute scalpel work. Whole posts were moved, and some of the dialog moved definitely concerned this thread, and there are still traces of the Bell's discussion in this thread.
Title: Re: Is angular momentum frame dependent?
Post by: Jaaanosik on 22/08/2020 18:55:37
The paper explains the centroids:

(https://i.imgur.com/G8YLStS.png)

Is the angular momentum calculated in regards to geometric or energy centroid?
Jano
Title: Re: Is angular momentum frame dependent?
Post by: Halc on 23/08/2020 00:05:04
Is the angular momentum calculated in regards to geometric or energy centroid?
Center of mass is what counts, which is the geometric center.

Per wiki:
"angular momentum L is proportional to moment of inertia I and angular speed ω  measured in radians per second.[3]

L=Iω
Unlike mass, which depends only on amount of matter, moment of inertia is also dependent on the position of the axis of rotation and the shape of the matter"
Title: Re: Is angular momentum frame dependent?
Post by: Jaaanosik on 26/08/2020 15:20:19
Is the angular momentum calculated in regards to geometric or energy centroid?
Center of mass is what counts, which is the geometric center.

Per wiki:
"angular momentum L is proportional to moment of inertia I and angular speed ω  measured in radians per second.[3]

L=Iω
Unlike mass, which depends only on amount of matter, moment of inertia is also dependent on the position of the axis of rotation and the shape of the matter"
I guess then the angular momentum appears to be frame dependent because the geometric mass is frame dependent.
The geometric mass is at the center of the wheel for the axle frame and it is shifted for the outside frame.
The outside frame has non-relativistic frames around itself that your calculation should be correct but with shifted center of mass, right?
Title: Re: Is angular momentum frame dependent?
Post by: Halc on 26/08/2020 16:01:20
I guess then the angular momentum appears to be frame dependent because the geometric mass is frame dependent.
That doesn't follow, and you shouldn't guess.
Do the math. The momentum of both pictures is a vector pointing at the PoV, and the one picture has more mass and less angular velocity than the other, so it is unclear from that if angular momentum is frame dependent.

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The geometric mass is at the center of the wheel for the axle frame and it is shifted for the outside frame.
That alone has no effect on momentum which is a vector, and vectors are composed of direction and magnitude, but not location.

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The outside frame has non-relativistic frames around itself that your calculation should be correct but with shifted center of mass, right?
I have done no calculation. I actually don't know the answer, and will not guess.\

My suspicion is that the angular momentum of the wheel is less in a frame in which it is moving.  It takes force to linearly accelerate a spinning wheel, and that force would be applied to the axle, not the frame-dependent center of mass.  As the FDCoM moves upward, the force acting on the axle applies a frame-dependent torque in opposition to the spin of the wheel, which would seem to reduce its frame dependent angular momentum. But maybe I'm wrong. I've not run the numbers.
Title: Re: Is angular momentum frame dependent?
Post by: Jaaanosik on 26/08/2020 17:42:23
...
That alone has no effect on momentum which is a vector, and vectors are composed of direction and magnitude, but not location.
...

That's an interesting comment.
The vector of the same direction and magnitude but not the location.
If the vector is not in the same location within the reference frame then the acceleration effect will be different based on the location. Correct?
What I mean is that the outside frame sees the geometric center shifted.
If the angular momentum vector is shifted to the geometric center then what happens when the axle, now off center, gets accelerated?
Title: Re: Is angular momentum frame dependent?
Post by: Halc on 26/08/2020 21:44:54
[momentum] is a vector ... composed of direction and magnitude, but not location.
That's an interesting comment.
The vector of the same direction and magnitude but not the location.
If the vector is not in the same location within the reference frame then the acceleration effect will be different based on the location. Correct?
What I mean is that the outside frame sees the geometric center shifted.
If the angular momentum vector is shifted to the geometric center then what happens when the axle, now off center, gets accelerated?
If you read my comment, I said the momentum doesn't have a location, so the nonexistent location cannot be shifted anywhere. Any yet I see at least four references to the location of the momentum above (bolded), which makes no sense.

It not having a location means that I can apply a certain torque anywhere on the object and it will have an identical effect on its angular momentum regardless of where the torque is applied.

Similarly, I can apply linear force anywhere on the object and it will have an identical effect on its linear momentum regardless of where the force is applied. That's what it means for momentum not to have a location.
Title: Re: Is angular momentum frame dependent?
Post by: Jaaanosik on 28/08/2020 17:19:54
[momentum] is a vector ... composed of direction and magnitude, but not location.
That's an interesting comment.
The vector of the same direction and magnitude but not the location.
If the vector is not in the same location within the reference frame then the acceleration effect will be different based on the location. Correct?
What I mean is that the outside frame sees the geometric center shifted.
If the angular momentum vector is shifted to the geometric center then what happens when the axle, now off center, gets accelerated?
If you read my comment, I said the momentum doesn't have a location, so the nonexistent location cannot be shifted anywhere. Any yet I see at least four references to the location of the momentum above (bolded), which makes no sense.

It not having a location means that I can apply a certain torque anywhere on the object and it will have an identical effect on its angular momentum regardless of where the torque is applied.

Similarly, I can apply linear force anywhere on the object and it will have an identical effect on its linear momentum regardless of where the force is applied. That's what it means for momentum not to have a location.
That's not the case.
A vector cannot be 'hanging' with no location/position to start.
The linear and angular momentum vectors start in the center of mass.
If a rigid body is pushed by a force vector and this force vector is not going through the center of mass then the end result is a change of linear and angular momentum of that rigid body.
If the rigid body is pushed by the force vector through the center of mass then the end result is only a change of the linear momentum.
Title: Re: Is angular momentum frame dependent?
Post by: Jaaanosik on 16/09/2020 21:19:08
Going back to the relativistic Hall effect paper.


(https://i.imgur.com/G8YLStS.png)

(https://i.imgur.com/Y6ibIMx.png)

The first diagram analysis, right side (b), is done for the first outside frame (1OF).
The second diagram analysis, right side (b), is done for the second outside frame (2OF) that is moving at 2v in the first outside frame.
The relativistic addition is done between the frames so the velocity is not >c.
The rest frame (a) is in the middle between 1OF and 2OF.
Let us consider the velocity 0.866c.
The rest frame moves to the left at v=0.866c in 1OF.
The rest frame moves to the right at v=0.866c in 2OF.
The text book says the centroids are frame dependent.
The (b) analysis is correct for both frames.
The relativity appears to be symmetrical.
The axle acceleration does not predict the same result!
The acceleration in 1OF decreases the angular momentum, but the acceleration increases the angular momentum in 2OF.
This appears to be a contradiction, meaning only one can be correct.