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Old 24th July 2018, 11:56 PM   #41
SDG
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Originally Posted by JeanTate View Post
I'm sorry to hear that.

If it's so, then I guess what you consider to be a serious problem with relativity will remain entirely with you.


Have you been able to estimate how many ISF members, who post in threads in this section, are (or have been) teachers? University lecturers? How many do you think have a BSc, an MSc, a PhD? What fields of study do you think they specialized in? How many of those with an MSc or PhD in physics (or related field, such as astrophysics) do you think needed to address one aspect or other of relativity, as part of the work they did on their thesis? How about (advanced) degrees in other fields, such as engineering or mathematics (where an intimate understanding of cycloids is required), how many ISF members do you think there are with such degrees? In terms of their day-to-day work (or former, if retired), how many ISF members do you think need to deal with one aspect of relativity or another? How about cycloids (and related analytic forms)? And so on.

If you do not have at least some appreciation for answers to at least some of those questions, would it be a rational thing to conclude that you, SDG, do not understand your audience?
It appears to me that 0 of the smart people here use cycloids based on replies I am getting. I might be wrong.

If a standing wheel in a train car that moves at a constant straight line velocity starts to be rotationally accelerated then points on the wheel will start to move on a cycloid trajectory in the ground reference frame.

Is everybody OK with this?
I'll wait for feedback on this first statement before I continue so we can progress without confusion.

Can any of the smart people here comment on the first post of the other thread about the conservation of energy of the simple pendulum?
Is the poster right about the equations?
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Old 25th July 2018, 12:06 AM   #42
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Originally Posted by JeanTate View Post
Who cares?

Unless "he" is you, and until you post something here that's concise and conveys the key point unambiguously (without lots of unrelated stuff, for example), why should anyone who posts in this part of ISF care?

For avoidance of doubt, the internet is replete with material that appears to be physics, but contains misunderstandings and errors, of many kinds, at many levels. If an ISF member is interested in discussing something in one of those places, fair enough; if no one is interested, why kick up a fuss?

tl;dr: life is too short to spend on what is almost certainly confused thinking (at best).
I thought it is noble to know the truth. That's the reason I do care.
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Old 25th July 2018, 01:25 AM   #43
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Originally Posted by SDG View Post
It appears to me that 0 of the smart people here use cycloids based on replies I am getting. I might be wrong.
It would help if you actually posted a coherent question to reply to. At the moment, pretty much all you've said is, "It's possible to detect absolute linear motion because cycloids." Without a mathematical analysis, that claim is impossible to evaluate. So, yes, you're wrong; your dissatisfaction with your replies indicates that you haven't in fact framed a coherent question.

Originally Posted by SDG View Post
If a standing wheel in a train car that moves at a constant straight line velocity starts to be rotationally accelerated then points on the wheel will start to move on a cycloid trajectory in the ground reference frame.
No. If, and only if, the wheel is rotating at a constant angular velocity, then any point on the wheel describes a cycloid in the ground reference frame. If it is being rotationally accelerated then its motion will be more complicated.

Originally Posted by SDG View Post
Can any of the smart people here comment on the first post of the other thread about the conservation of energy of the simple pendulum?
Is the poster right about the equations?
The poster is trying to expand out a term which is neglected in the textbook example due to the approximation that the mass on the pendulum occupies a point. Post 4 is therefore a complete explanation of the discrepancy; the first post is trying to analyse a different problem to the textboox example in which the spatial extent of the bob is finite, and in real life this is one of the additional terms which means that the simple pendulum equation is not perfectly accurate.

Dave
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Old 25th July 2018, 01:58 AM   #44
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Originally Posted by The Man View Post
Nope, I put in all the effort it was worth. As I recall the pendulum is freely rotating so in a reference frame rotating with the pudendum it would just be swinging back and forth like a non-rotating pendulum.
Interesting, but NSFW
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Old 25th July 2018, 02:37 AM   #45
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Originally Posted by Dave Rogers View Post
It would help if you actually posted a coherent question to reply to. At the moment, pretty much all you've said is, "It's possible to detect absolute linear motion because cycloids." Without a mathematical analysis, that claim is impossible to evaluate. So, yes, you're wrong; your dissatisfaction with your replies indicates that you haven't in fact framed a coherent question.
To understand the questions requires reading of the other forum.
There are maybe four posts with math in the other forum. Is it OK to repost the stuff here?

Quote:
No. If, and only if, the wheel is rotating at a constant angular velocity, then any point on the wheel describes a cycloid in the ground reference frame. If it is being rotationally accelerated then its motion will be more complicated.
Well, it is going to be more complicated, but not that much. What happens is that after every tiny delta time dt of the acceleration the cycloid becomes more curtate. When the acceleration continues then through ideal cycloid, the cycloid will become prolate. So the motion trajectory is changing cycloid. It is not that complicated.

Quote:
The poster is trying to expand out a term which is neglected in the textbook example due to the approximation that the mass on the pendulum occupies a point. Post 4 is therefore a complete explanation of the discrepancy; the first post is trying to analyse a different problem to the textboox example in which the spatial extent of the bob is finite, and in real life this is one of the additional terms which means that the simple pendulum equation is not perfectly accurate.

Dave
Is there a physics text book that warns reader about this? The simple pendulum equations are an approximation if the 'point' representing bob rotates?
I have not found one yet.

http://mathworld.wolfram.com/CurtateCycloid.html
http://mathworld.wolfram.com/ProlateCycloid.html

Last edited by SDG; 25th July 2018 at 02:44 AM.
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Old 25th July 2018, 02:49 AM   #46
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Speaking of links, here’s some deja-vu to save everybody the effort of retyping:
https://forum.cosmoquest.org/archive.../t-151673.html
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Old 25th July 2018, 03:02 AM   #47
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Originally Posted by SDG View Post
To understand the questions requires reading of the other forum.
There are maybe four posts with math in the other forum. Is it OK to repost the stuff here?
Why not simply formulate the complete argument and post it?

Originally Posted by SDG View Post
Well, it is going to be more complicated, but not that much.
And this is the point at which you go off the rails; you're approximating the motion to a cycloid without calculating the effects of that approximation.

Originally Posted by SDG View Post
Is there a physics text book that warns reader about this? The simple pendulum equations are an approximation if the 'point' representing bob rotates?
I honestly don't know. It would normally be swept up in the general statement that the simple pendulum is an idealised model used to understand the dynamic principles involved and that a real pendulum is more complex but the perturbations may be quite small.

Dave
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Old 25th July 2018, 03:54 AM   #48
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Originally Posted by SDG View Post
If a standing wheel in a train car that moves at a constant straight line velocity starts to be rotationally accelerated then points on the wheel will start to move on a cycloid trajectory in the ground reference frame.
Yes, of course they will. What problem do you have with this?

Quote:
Can any of the smart people here comment on the first post of the other thread about the conservation of energy of the simple pendulum?
Is the poster right about the equations?
Why don't you compare them to a textbook? Why must others do that work for you?

Or, you could state explicitly what you think might be wrong with them.

Hans
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Old 25th July 2018, 06:27 AM   #49
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Originally Posted by Kid Eager View Post
Speaking of links, here’s some deja-vu to save everybody the effort of retyping:
https://forum.cosmoquest.org/archive.../t-151673.html
This link takes you to a formatted version, which I think is easier to read and follow; as a URL:
https://forum.cosmoquest.org/showthr...d-all-the-time

It goes on for six pages, and ends with a moderator closing the thread with the comment: "I think this discussion has become pointless as the OP apparently does not understand physics."

SDG: did you know of this CQ discussion before Kid Eager posted a link to it?

From your reading of the CQ discussion, what key aspects of Jaaanosik's proposal/idea do you think were misunderstood?

How, critically, does the content of the Jaaanosik's proposal/idea, as presented in that CQ discussion, differ from the "challenge" you have posted here in ISF?
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Old 25th July 2018, 06:52 AM   #50
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The claim, it seems to me, is that the time dependence of the kinetic energy of a point mass attached to a rotating wheel is different depending on whether it is measured in (a) a frame of reference in which the axis of rotation of the wheel is stationary or (b) a frame of reference in which the axis of rotation of the wheel is moving at constant velocity. Is that correct?

If so, it's specious. The absolute kinetic energy of a mass in motion at constant velocity is dependent on the frame of reference in which it is measured. Since, in this case, the mass is not in motion at constant velocity, it's hardly surprising that the time dependence of its kinetic energy is different. This simply does not contradict SR.

Dave
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Old 25th July 2018, 08:13 AM   #51
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Originally Posted by Kid Eager View Post
Speaking of links, here’s some deja-vu to save everybody the effort of retyping:
https://forum.cosmoquest.org/archive.../t-151673.html


Looking at the fifth post down in that thread, he says:

Quote:
There are two inertial frames in our experiment: ground and hovercraft.
The rotating wheel is the experiment within these two inertial reference frames.
The acceleration has to be the same in both frames as mentioned in the book.
The essential claim here is that "We have an experiment to detect constant linear motion without any signal from the outside". But "the outside" of what?


It's been a while since I did any relativity, but it seems to me that, if you're comparing accelerations as measured in two different inertial frames, than that constitutes information "from the outside" of at least one of the inertial frames. In this case, the information on the acceleration within the ground frame would have to be communicated to the hovercraft frame, in order to determine the linear motion of the hovercraft wrt the ground, defeating the whole purpose of this exercise.

But maybe I'm just a smart person on a skeptic board.
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Old 26th July 2018, 12:06 AM   #52
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Originally Posted by Dave Rogers View Post
Why not simply formulate the complete argument and post it?
Is Latex available here?
Quote:
And this is the point at which you go off the rails; you're approximating the motion to a cycloid without calculating the effects of that approximation.
https://theelectromagneticnatureofth...i9/wheel02.png
Image location copied from the naked scientists forum. The use of the image is for quotation, not to infringe any copyright.

If we imagine a point in the middle of the ball and a point right bellow the ball but on the wheel. These two points will accelerate differently. The point on the wheel will follow "changing" cycloid.
You are right, the point in the middle of the ball will follow a different trajectory that needs different calculation.

Quote:
I honestly don't know. It would normally be swept up in the general statement that the simple pendulum is an idealised model used to understand the dynamic principles involved and that a real pendulum is more complex but the perturbations may be quite small.

Dave
You might be right, some comments like that would be useful... but I have not found any yet.

Last edited by zooterkin; 27th July 2018 at 02:13 AM. Reason: Rule 5
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Old 26th July 2018, 12:23 AM   #53
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Originally Posted by MRC_Hans View Post
Yes, of course they will. What problem do you have with this?

Why don't you compare them to a textbook? Why must others do that work for you?

Or, you could state explicitly what you think might be wrong with them.

Hans

Hans, I cannot imagine what thinking process goes through your head to conclude that I have a problem with something from this statement:
Quote:
If a standing wheel in a train car that moves at a constant straight line velocity starts to be rotationally accelerated then points on the wheel will start to move on a cycloid trajectory in the ground reference frame.



https://theelectromagneticnatureofth...pendulum01.png
Image location copied from the naked scientists forum. The use of the image is for quotation, not to infringe any copyright.


The equation 5.124 is not correct if the pendulum bob rotates when it falls.
That's the claim from the other thread and it appears to me it is a true statement.

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Old 26th July 2018, 07:36 AM   #54
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I'm afraid the discussion has gone considerably astray. The original assertion has a flaw.

Let's start with a moving platform (traditionally a train) moving with constant velocity V. On the train is a pendulum of mass m and length l, with a bob which is sufficiently small and dense that it can be represented as a point mass. The pendulum swings with an arc which provides a height difference h for the bob. Then, assuming no angular momentum effects, the bob has a standard relationship, maximum v = sqrt (2 g h). Note also that angular effects can be arbitrarily reduced by increasing l. Since, for convenience, we'll want v < V, this works fine. More to the point, angular kinetic energy increases proportionally to linear kinetic energy (you cannot trade one for the other for a fixed geometry), so speaking of linear kinetic energy only causes no problems.

The argument goes like this:

1) At the low point, when the bob is moving forward, KE = 1/2 m (V + v)^2 in the fixed frame.

2) At the low point, when the bob is moving backwards, KE = 1/2 (V - v)^2 in the fixed frame.

3) Since the difference in KE from the bob being at its peak is m g h, the two KEs must be the same.

4) This is only true if v is zero. Alternatively, the peak forward and reverse velocities must be different to conserve energy, and the half-periods must be different. This difference can be used to detect both the magnitude and direction of V, without any observation outside the train. QED.

(The linked thread in the OP went at if from the other end: by using the cycloid as a measure of velocity, it showed that the KEs are different in the different directions. While true enough, this is irrelevant, and the relatively baroque introduction of cycloids distracted folks.)

What the poster did not realize is that a fixed-pivot pendulum already has problems with conservation. The momentum of the two cases is also different (m (V + V) vs m (V - v)), so it's clear that conservation laws don't directly apply. I've not tried to do the calculations, but this must have been dealt with very early on in SR. With no way to rely on conservation of momentum, the apparent problem no longer holds. Or, more precisely, Statement 3 is not true.

Another way to look at it is to observe that the causal model being used is a rotating wheel which rotates at constant velocity. As such it experiences no external forces. This is not true of the the pendulum, which requires constantly changing force to be applied to the pivot.

Last edited by WhatRoughBeast; 26th July 2018 at 07:42 AM.
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Old 26th July 2018, 08:45 AM   #55
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Originally Posted by WhatRoughBeast View Post
I'm afraid the discussion has gone considerably astray. The original assertion has a flaw.

Let's start with a moving platform (traditionally a train) moving with constant velocity V. On the train is a pendulum of mass m and length l, with a bob which is sufficiently small and dense that it can be represented as a point mass. The pendulum swings with an arc which provides a height difference h for the bob. Then, assuming no angular momentum effects, the bob has a standard relationship, maximum v = sqrt (2 g h). Note also that angular effects can be arbitrarily reduced by increasing l. Since, for convenience, we'll want v < V, this works fine. More to the point, angular kinetic energy increases proportionally to linear kinetic energy (you cannot trade one for the other for a fixed geometry), so speaking of linear kinetic energy only causes no problems.

The argument goes like this:

1) At the low point, when the bob is moving forward, KE = 1/2 m (V + v)^2 in the fixed frame.

2) At the low point, when the bob is moving backwards, KE = 1/2 (V - v)^2 in the fixed frame.

3) Since the difference in KE from the bob being at its peak is m g h, the two KEs must be the same.

4) This is only true if v is zero. Alternatively, the peak forward and reverse velocities must be different to conserve energy, and the half-periods must be different. This difference can be used to detect both the magnitude and direction of V, without any observation outside the train. QED.

(The linked thread in the OP went at if from the other end: by using the cycloid as a measure of velocity, it showed that the KEs are different in the different directions. While true enough, this is irrelevant, and the relatively baroque introduction of cycloids distracted folks.)

What the poster did not realize is that a fixed-pivot pendulum already has problems with conservation. The momentum of the two cases is also different (m (V + V) vs m (V - v)), so it's clear that conservation laws don't directly apply. I've not tried to do the calculations, but this must have been dealt with very early on in SR. With no way to rely on conservation of momentum, the apparent problem no longer holds. Or, more precisely, Statement 3 is not true.

Another way to look at it is to observe that the causal model being used is a rotating wheel which rotates at constant velocity. As such it experiences no external forces. This is not true of the the pendulum, which requires constantly changing force to be applied to the pivot.
Mmm, ok, then. However, assuming the train does not move at a relativistic speed (trains usually don't), AND assuming it moves in a straight line at constant speed, then from the observational frame of the train, the pendulum will act the same regardless of the speed of the train, including the speed zero.

From a stationary observation frame, the pendulum will seem to speed up and slow down, alternately.

You will get no clue to the speed of the train by watching the pendulum from the train.

Hans
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Old 26th July 2018, 08:56 AM   #56
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This seems like a frame change error (as most SR/GR errors seem to be).

The KE only differs from outside the train (the "fixed frame" observer). In that case, KE is different because of the KE added in one direction from the trains motion. They're the same inside the train (where the KE does NOT included the velocity added by the train).

The difference between the forward and backward KEs (fixed frame) should equal 2 mgh (if I'm reading it correctly): the KE converted during the "backswing" plus the KE converted during the "foreswing". Forward is train KE + pendulum KE, backwards is train KE minus pendulum KE.

Or am I missing something here?
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Old 26th July 2018, 09:09 AM   #57
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Originally Posted by WhatRoughBeast View Post
What the poster did not realize is that a fixed-pivot pendulum already has problems with conservation. The momentum of the two cases is also different (m (V + V) vs m (V - v)), so it's clear that conservation laws don't directly apply. I've not tried to do the calculations, but this must have been dealt with very early on in SR. With no way to rely on conservation of momentum, the apparent problem no longer holds. Or, more precisely, Statement 3 is not true.
Good point, and one that hadn't occurred to me - a simple pendulum is not a closed system; another of the simplifications is that the suspension point is fixed, meaning that in effect it is attached to an infinite mass. So, if a pendulum is swinging on a moving train of finite mass, then one of two conditions must apply:

(a) The forward velocity of the carriages is not constant, but includes a small perturbation such that the sum of the momenta of the pendulum bob and of the remainder of the train remains constant;

or,

(b) There is a periodic force applied to the carriages which is equal and opposite to the horizontal component of the force exerted on the carriages by the pendulum string at either end of its swing.

As an illustration, one could simply hang a pendulum from a frame on a set of low-friction bearings; if the mass of the frame is comparable with that of the pendulum bob, one would observe the frame moving backwards and forwards in opposition to the movement of the pendulum bob. My guess would be that the analysis treats this differently in the stationary and moving reference frames, and that this is the source of the apparent SR violation.

Dave
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Old 26th July 2018, 09:16 AM   #58
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Originally Posted by Dave Rogers View Post
Good point, and one that hadn't occurred to me - a simple pendulum is not a closed system; another of the simplifications is that the suspension point is fixed, meaning that in effect it is attached to an infinite mass. So, if a pendulum is swinging on a moving train of finite mass, then one of two conditions must apply:

(a) The forward velocity of the carriages is not constant, but includes a small perturbation such that the sum of the momenta of the pendulum bob and of the remainder of the train remains constant;

or,

(b) There is a periodic force applied to the carriages which is equal and opposite to the horizontal component of the force exerted on the carriages by the pendulum string at either end of its swing.

As an illustration, one could simply hang a pendulum from a frame on a set of low-friction bearings; if the mass of the frame is comparable with that of the pendulum bob, one would observe the frame moving backwards and forwards in opposition to the movement of the pendulum bob. My guess would be that the analysis treats this differently in the stationary and moving reference frames, and that this is the source of the apparent SR violation.

Dave
Yes, another approximation is the assumption that the mass of the pendulum is infinitesimal compared to that of the train. But even in a scenario where it isn't, the linear speed will not enter the equation from the reference frame of the train. What will happen is that it will be modulated by the movement of the pendulum.

Hans
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Old 26th July 2018, 02:35 PM   #59
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There are a great many other assumptions/approximations, if one is trying to get close to a model of reality. And almost any of us ISF members could add more to any list created.

The extent to which these "matter" (i.e. make a difference to any real observations of a pendulum in a train)? That's for whoever wants to explore the difference between models and reality to decide.

Here are a few I quickly came up with:

- variation in "g": it differs with distance between the Earth's center and whatever one is considering; it also depends on the distribution of mass in the pendulum+train's vicinity

- electrical and magnetic fields: all matter responds to these, to one degree or another, and it's nigh on impossible to create a pendulum in a train that is free of these

- size changes due to changes in temperature: I very much doubt that all parts of any real pendulum+train will reach a constant temperature, and keep it; all real materials will change in size due to changes in their temperature

- "air": even if the system is in a "vacuum", it can never be perfect, and so there'll always be friction
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Old 27th July 2018, 12:29 AM   #60
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Originally Posted by JeanTate View Post
This link takes you to a formatted version, which I think is easier to read and follow; as a URL:
https://forum.cosmoquest.org/showthr...d-all-the-time

It goes on for six pages, and ends with a moderator closing the thread with the comment: "I think this discussion has become pointless as the OP apparently does not understand physics."

SDG: did you know of this CQ discussion before Kid Eager posted a link to it?

From your reading of the CQ discussion, what key aspects of Jaaanosik's proposal/idea do you think were misunderstood?

How, critically, does the content of the Jaaanosik's proposal/idea, as presented in that CQ discussion, differ from the "challenge" you have posted here in ISF?
It appears to me Jaaanosik was on the right track with pointing out variation of the kinetic energy but it was lacking the the conservation of energy and proper analysis of the whole system.
His current proposal seems to be more matured.
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Old 27th July 2018, 02:33 AM   #61
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The current issue from the other forum is an acceleration of the wheel as shown in the post 52.
The first delta time is 1s and \omega goes from 0 to 0.01rad/s in the first dt.
The delta of the kinetic energy of the ball in the train reference frame after 1s will be

1/2 m_ball v^2 + 1/2 I_ball \omega^2

This applies to every ball on the wheel in the train reference frame.
Now let us compare the bottom ball and the top ball velocities in the ground reference frame

https://theelectromagneticnatureofth...01r1v0.1_v.png
Image location copied from the naked scientists forum. The use of the image is for quotation, not to infringe any copyright.

So if delta v in the ground frame is 0.01m/s for the bottom and the top balls then it means the balls should have the same rotational kinetic energy because all balls where accelerated with the same amount of energy. The same amount energy was pushed into them.
The issue is though, the ground reference frame \omega is not equal for the bottom and the top.
So if the top ball T gets less rotational energy it has more energy left for the translational kinetic energy.
The bottom ball B absorbs more energy into the rotational kinetic energy less energy into the translational kinetic energy.
The end result will be different delta v for the bottom and the top balls assuming the same amount of energy was pushed into the balls and the balls absorb different amount of the rotational kinetic energy.
This delta v can be measured from within the train without any signal from "the outside".

https://theelectromagneticnatureofth...01r1v0.1_w.png
Image location copied from the naked scientists forum. The use of the image is for quotation, not to infringe any copyright.


This is how I understand Jaaanosik's latest claim.
Please, can we have a civilized discussion about it?
What do I miss?


Edited by zooterkin:  Edited for rule 5.
See this post for the sites you can hotlink to.

In addition, if you do link to an image on a site that permits hotlinking, please ensure the image is not so large that it disrupts the display of posts on the form. If the image is large, please use the 'imgw' tag.

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Old 27th July 2018, 02:47 AM   #62
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Originally Posted by SDG View Post
So if the top ball T gets less rotational energy it has more energy left for the translational kinetic energy.
The bottom ball B absorbs more energy into the rotational kinetic energy less energy into the translational kinetic energy.
The end result will be different delta v for the bottom and the top balls assuming the same amount of energy was pushed into the balls and the balls absorb different amount of the rotational kinetic energy.
This delta v can be measured from within the train without any signal from "the outside".

[...]

What do I miss?
The most obvious thing is this:

The top and bottom balls are rigidly attached to opposite sides of the wheel. By definition, therefore, their velocities in the same frame of reference as the centre of the wheel are equal and opposite, so in that frame of reference there cannot be any difference between them to measure.

Dave
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Old 27th July 2018, 02:55 AM   #63
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Originally Posted by SDG View Post
The current issue from the other forum is an acceleration of the wheel as shown in the post 52.
The first delta time is 1s and \omega goes from 0 to 0.01rad/s in the first dt.
The delta of the kinetic energy of the ball in the train reference frame after 1s will be

1/2 m_ball v^2 + 1/2 I_ball \omega^2

This applies to every ball on the wheel in the train reference frame.
Now let us compare the bottom ball and the top ball velocities in the ground reference frame

https://theelectromagneticnatureofth...01r1v0.1_v.png
Image location copied from the naked scientists forum. The use of the image is for quotation, not to infringe any copyright.

So if delta v in the ground frame is 0.01m/s for the bottom and the top balls then it means the balls should have the same rotational kinetic energy because all balls where accelerated with the same amount of energy. The same amount energy was pushed into them.
The issue is though, the ground reference frame \omega is not equal for the bottom and the top.
So if the top ball T gets less rotational energy it has more energy left for the translational kinetic energy.
The bottom ball B absorbs more energy into the rotational kinetic energy less energy into the translational kinetic energy.
The end result will be different delta v for the bottom and the top balls assuming the same amount of energy was pushed into the balls and the balls absorb different amount of the rotational kinetic energy.
This delta v can be measured from within the train without any signal from "the outside".

https://theelectromagneticnatureofth...01r1v0.1_w.png
Image location copied from the naked scientists forum. The use of the image is for quotation, not to infringe any copyright.


This is how I understand Jaaanosik's latest claim.
Please, can we have a civilized discussion about it?
What do I miss?
Of course we can. It helps specifying what you are asking about
.
What you miss is that the wheel (or any part of it) does not exchange energy with the train. Or to be precise, it does not interact with the linear movement of the train. (IF the wheel is not in balance, it will, like the pendulum, alternatively add and subtract some kinetic energy, but the sum will be zero. In other words, the velocity of the train will oscillate slightly, but this will be independent of the actual velocity of the train.)

The apparent cycloid movement of a point on the wheel is an illusion that comes from watching it from a stationary point.

A fundamental flaw in the argument is arguing that you can observe something from the moving reference frame, but needing a stationary reference frame to show it.

Hans
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Old 27th July 2018, 02:59 AM   #64
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Originally Posted by Dave Rogers View Post
The most obvious thing is this:

The top and bottom balls are rigidly attached to opposite sides of the wheel. By definition, therefore, their velocities in the same frame of reference as the centre of the wheel are equal and opposite, so in that frame of reference there cannot be any difference between them to measure.

Dave
We could easily assume the "wheel" consisted of balls on strings, but it would make no difference: The components of the wheel do not exchange energy with the linear movement of the train.

The presumed observation would be correct if the train was accelerating, but then also trivial.

Hans
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Old 27th July 2018, 04:54 AM   #65
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Originally Posted by Dave Rogers View Post
The most obvious thing is this:

The top and bottom balls are rigidly attached to opposite sides of the wheel. By definition, therefore, their velocities in the same frame of reference as the centre of the wheel are equal and opposite, so in that frame of reference there cannot be any difference between them to measure.

Dave
Oops, trying to describe what is in the other forum and I forgot to mention that he mentioned the balls are on elastic rods. The rods are flexible. The balls have a limited freedom of motion.
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Old 27th July 2018, 05:10 AM   #66
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Originally Posted by SDG View Post
Oops, trying to describe what is in the other forum and I forgot to mention that he mentioned the balls are on elastic rods. The rods are flexible. The balls have a limited freedom of motion.
Then you have to factor in the elastic energy caused by flexure of the elastic rods. But I still don't see how there can be any difference between what's observed in a moving or a stationary frame of reference, because whatever turning moment is applied to the wheel is defined within the moving frame of reference. I think the actual result may be that, if the turning moment is defined in the stationary frame of reference, then there is a different result in the moving frame of reference; but that requires that part of the experiment takes place in the stationary frame, negating the supposition that it takes place entirely in the moving frame.

And I agree that the balance between rotational and translation energy for the balls is different when measured in the ground reference frame and when measured in the train reference frame; but again that doesn't violate SR, because there is no measurement entirely within the moving reference frame that can detect this difference.

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Old 27th July 2018, 05:20 AM   #67
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Originally Posted by MRC_Hans View Post
Of course we can. It helps specifying what you are asking about
.
What you miss is that the wheel (or any part of it) does not exchange energy with the train. Or to be precise, it does not interact with the linear movement of the train. (IF the wheel is not in balance, it will, like the pendulum, alternatively add and subtract some kinetic energy, but the sum will be zero. In other words, the velocity of the train will oscillate slightly, but this will be independent of the actual velocity of the train.)

The apparent cycloid movement of a point on the wheel is an illusion that comes from watching it from a stationary point.

A fundamental flaw in the argument is arguing that you can observe something from the moving reference frame, but needing a stationary reference frame to show it.

Hans
Hans,
Let's assume the wheel with the balls is well balanced. There are many balls equally spaced around the edge of the wheel though we discuss only B, L, T balls at the moment.
I agree with you there will not be any energy exchange with the train. There will not be a barycenter between the wheel and train car, no oscillation between the wheel and train car.

I would not call a reference frame analysis an illusion. This is where I disagree with your analysis.
The ball L is being accelerated in the straight line, linear acceleration in the ground reference frame at time 1s from the above example, therefore the ball is not going to absorb any rotational kinetic energy only translation kinetic energy. That's how I see the claim.
If there is the same amount of energy and nothing goes the rotational kinetic energy then v translational velocity will be bigger.

If you think about what I wrote carefully we might be zooming in on the issue.

SDG
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Old 27th July 2018, 06:39 AM   #68
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Originally Posted by Dave Rogers View Post
Then you have to factor in the elastic energy caused by flexure of the elastic rods. But I still don't see how there can be any difference between what's observed in a moving or a stationary frame of reference, because whatever turning moment is applied to the wheel is defined within the moving frame of reference. I think the actual result may be that, if the turning moment is defined in the stationary frame of reference, then there is a different result in the moving frame of reference; but that requires that part of the experiment takes place in the stationary frame, negating the supposition that it takes place entirely in the moving frame.

And I agree that the balance between rotational and translation energy for the balls is different when measured in the ground reference frame and when measured in the train reference frame; but again that doesn't violate SR, because there is no measurement entirely within the moving reference frame that can detect this difference.

Dave
Dave,
Let's break it down to multiple scenarios.
1. Straight line linear acceleration - the wheel is stationary, locked and the train car accelerates 0.01m/s^2, the train car will move 0.01m in 1s. The point on the wheel, beneath the ball will move 0.01m in 1s in the ground reference frame. The same point will not move at all in the moving train car reference frame. Let's say the elastic rod holding the ball will allow to move the center of the ball back and the ball will move less, 0.007m in one second.

2. Curvilinear acceleration, on the circle - the train car is stationary and the wheel will accelerate 0.01rad/s^2, the wheel will turn 0.01rad in 1s. The point on the wheel, beneath the ball will move 0.01m in 1s on the circle trajectory in both reference frames.

The big questions about the scenario 2.
What distance will the center of the ball move on the circle? 0.007m, less than 0.007m or more than 0.007m? Assuming that there is the same amount of energy applied, the same force for the same period of time applied on the bottom of the elastic rod in both scenarios.

SDG
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Old 27th July 2018, 06:50 AM   #69
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Originally Posted by SDG View Post
The big questions about the scenario 2.
What distance will the center of the ball move on the circle? 0.007m, less than 0.007m or more than 0.007m? Assuming that there is the same amount of energy applied, the same force for the same period of time applied on the bottom of the elastic rod in both scenarios.
And there's the error.

Energy is force times distance - the product of the force applied and the distance moved by the object it is applied to. If the same force is applied for the same time in both frames, it doesn't move through the same distance, because the distance covered in the ground frame of reference is the sum of the distance covered in the train frame of reference and the distance moved by the train in that time interval. Your assumption that the same force applied for the same time produces the same change in energy is in fact wrong.

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Old 27th July 2018, 08:42 AM   #70
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Originally Posted by Dave Rogers View Post
And there's the error.



Energy is force times distance - the product of the force applied and the distance moved by the object it is applied to. If the same force is applied for the same time in both frames, it doesn't move through the same distance, because the distance covered in the ground frame of reference is the sum of the distance covered in the train frame of reference and the distance moved by the train in that time interval. Your assumption that the same force applied for the same time produces the same change in energy is in fact wrong.



Dave

If I understood you correctly, you are saying that F=ma is wrong?
The acceleration "a" is not the same in all reference frames?






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Old 27th July 2018, 08:46 AM   #71
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Originally Posted by SDG View Post
If I understood you correctly, you are saying that F=ma is wrong?
The acceleration "a" is not the same in all reference frames?
No, you didn't understand me correctly; I didn't say anything of the sort, or in fact anything remotely like that. I said that the force moves through different distances in the two frames of reference, and therefore does different amounts of work - or, if you like, imparts different amounts of energy to the ball - in the two different frames.

Dave
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Old 27th July 2018, 09:31 AM   #72
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Originally Posted by Dave Rogers View Post
No, you didn't understand me correctly; I didn't say anything of the sort, or in fact anything remotely like that. I said that the force moves through different distances in the two frames of reference, and therefore does different amounts of work - or, if you like, imparts different amounts of energy to the ball - in the two different frames.



Dave


Well if you read my post again, you will see the two scenarios can be analyzed within the ground reference frame. One frame, not many.
It is a comparison of linear and curvilinear accelerations within one frame.

This is about the conservation of energy.




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Old 27th July 2018, 12:50 PM   #73
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Originally Posted by SDG View Post
Hans,
Let's assume the wheel with the balls is well balanced. There are many balls equally spaced around the edge of the wheel though we discuss only B, L, T balls at the moment.
I agree with you there will not be any energy exchange with the train. There will not be a barycenter between the wheel and train car, no oscillation between the wheel and train car.
Fine.

Quote:
I would not call a reference frame analysis an illusion. This is where I disagree with your analysis.
From a scientific POV, it is. It is not a true motion. If you were to analyze the forces involved, you would find they only pertain to the linear motion of the train and the circular motion if the wheel.

Quote:
The ball L is being accelerated in the straight line, linear acceleration in the ground reference frame at time 1s from the above example, therefore the ball is not going to absorb any rotational kinetic energy only translation kinetic energy. That's how I see the claim.
No, it is not. The train is moving at constant speed. The ball has two constant motions:

1) The linear speed of the train.
2) The orbit, where it is constantly accelerated towards the hub of the wheel (centripetal force), and hence a circular movement.

Quote:
If there is the same amount of energy and nothing goes the rotational kinetic energy then v translational velocity will be bigger.
No. No energy is transferred or converted, but the kinetic energy from the rotation of the wheel is vectored in a circle.

Hans
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Old 27th July 2018, 12:53 PM   #74
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Originally Posted by SDG View Post
Dave,
Let's break it down to multiple scenarios.
1. Straight line linear acceleration - the wheel is stationary, locked and the train car accelerates 0.01m/s^2, the train car will move 0.01m in 1s. The point on the wheel, beneath the ball will move 0.01m in 1s in the ground reference frame. The same point will not move at all in the moving train car reference frame. Let's say the elastic rod holding the ball will allow to move the center of the ball back and the ball will move less, 0.007m in one second.
No. You are introducing a new vector: Acceleration of the train. Acceleration of the train will, of course, affect the parts of the wheel and can be observed within the reference frame of the train. This is elementary.

Hans
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Old 28th July 2018, 02:36 AM   #75
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Originally Posted by SDG View Post
Well if you read my post again, you will see the two scenarios can be analyzed within the ground reference frame. One frame, not many.
It is a comparison of linear and curvilinear accelerations within one frame.
If your analysis within one reference frame yields two answers, then one of them is wrong.

Originally Posted by SDG View Post
This is about the conservation of energy.
I'll try and say it again. The work done by a force is the vector product of the force and the displacement through which it moves. The latter is frame dependent. The work done is therefore frame dependent. You are assuming that a force acting on a body for a fixed amount of time produces a frame-independent change in kinetic energy. This is simply wrong. As long as you're making that mistake you'll get erroneous results, which will give the false impression that the absolute motion of the frame of reference can be detected; in fact, what you are detecting is an arithmetical error arising from the fact that you're calculating the energy change in the wrong frame of reference.

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Old 28th July 2018, 06:21 AM   #76
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Originally Posted by Hellbound View Post
This seems like a frame change error (as most SR/GR errors seem to be).

The KE only differs from outside the train (the "fixed frame" observer). In that case, KE is different because of the KE added in one direction from the trains motion. They're the same inside the train (where the KE does NOT included the velocity added by the train).

The difference between the forward and backward KEs (fixed frame) should equal 2 mgh (if I'm reading it correctly): the KE converted during the "backswing" plus the KE converted during the "foreswing". Forward is train KE + pendulum KE, backwards is train KE minus pendulum KE.

Or am I missing something here?
When the pendulum bob is viewed from the external (fixed/stationary) frame of reference, the difference between its kinetic energy when it is moving faster than the train (during the foreswing) and when it is moving slower than the train (during the backswing) can easily be much greater than 2mgh because it is also a function of the train speed.

To help understand this, consider that the kinetic energy required to change the speed of a 1 kg mass by 1 m/s from 1 m/s to 2 m/s is 0.5*(1kg)*(2m/s)^2 - 0.5*(1kg)*(1m/s)^2 or only 1.5 Joules and yet the energy required to change its speed by 1 m/s from 5 m/s to 6 m/s is 0.5*(1kg)*(6m/s)^2 - 0.5*(1kg)*(5m/s)^2 or 5.5 Joules, which is 4 Joules more.

The falling/swinging/rising bob within the moving train is thus also affecting an energy transfer either from and to or to and from the train itself.
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Old 29th July 2018, 07:29 AM   #77
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Originally Posted by Hevneren View Post
Interesting, but NSFW
Dang nab it and being at work now I can't find what that typo means.
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Old 29th July 2018, 08:17 AM   #78
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Originally Posted by SDG View Post
If I understood you correctly, you are saying that F=ma is wrong?
The acceleration "a" is not the same in all reference frames?






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Well let's break down the equation in terms of units. What you get is Force (Newton) = Mass (Kilogram) * Acceleration (Meter Second-2). Moving the acceleration term over to the Force side you get Force (Newton) Acceleration-1 (Meter-1 Second2)= Mass (Kilogram).

So as noted by Dave Rogers not only does the distance (Meter) the force is applied depend on the reference frame (even just in Galilean relativity) the relation here (between force and distance) is force over distance (Newton Meter-1) not force times distance as in energy or work. However, since the speed is not relativistic the time the forces are applied are basically the same for both frames and Force times Time (Newton Second) is momentum. That's the basis of the F=ma equation, momentum. The distance the force is applied over the time the force is applied give the average velocity while the force is applied. Combine that with momentum (resulting from that force) by dividing momentum by that average velocity and we get Momentum (Newton Second) Average velocity-1 (Meter-1 Second) = Mass (Kilogram).

As it is the distance traveled while the force is applied that varies in this case it is the Average Velocity that is frame dependent. Again easily determinable as in one frame there is no consistent lateral component of motion. As well as in the purely rotational frame one would consider the average rotational velocity (Radian Second-1) or acceleration as opposed to the linear equivalence (Meter Second-1). Though conversion is a simple matter with a fixed radius.
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Old 30th July 2018, 03:29 AM   #79
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Originally Posted by MRC_Hans View Post
Fine.



From a scientific POV, it is. It is not a true motion. If you were to analyze the forces involved, you would find they only pertain to the linear motion of the train and the circular motion if the wheel.
I have no idea what it means 'not a true motion'. Is there a false motion? What is that?
The acceleration is supposed to be the same for any reference frame.

Quote:
No, it is not. The train is moving at constant speed. The ball has two constant motions:

1) The linear speed of the train.
2) The orbit, where it is constantly accelerated towards the hub of the wheel (centripetal force), and hence a circular movement.



No. No energy is transferred or converted, but the kinetic energy from the rotation of the wheel is vectored in a circle.

Hans
Ball L is accelerated to the center of the will but still it is only straight line acceleration, \omega is 0 in the inflection point in the ground reference frame.
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Old 30th July 2018, 03:59 AM   #80
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Originally Posted by MRC_Hans View Post
No. You are introducing a new vector: Acceleration of the train. Acceleration of the train will, of course, affect the parts of the wheel and can be observed within the reference frame of the train. This is elementary.

Hans
The reason why I introduced the train acceleration is to point out the difference between straight line acceleration and curvilinear acceleration from energy point of view in one reference frame.
I'll try again. Let us simplify the scenarios with considering rigid rods beneath the balls.
1. We attach the ball to the train car floor. We accelerate the train in a straight line to 1m/s velocity then we stop acceleration and we will maintain constant straight line motion.
What energy is in the ball?
1/2m_ball v^2

2. We attach the wheel to a concrete block on the ground. The ball is on the edge r=1m. We accelerate the ball to 1m/s tangential velocity. It means \omega = 1rad/s then we will stop acceleration we will keep constant rotation.
What energy is in the ball?
1/2m_ball v^2 + 1/2I_ball \omega^2

Do we have an agreement?
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