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Circumventing Newton's third law through Euler Inertial Forces


John2020

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8 minutes ago, Ghideon said:

Please clarify

I apologize, I was referring to the example with nucleus and electron. About the example with the rope the missing part is "until you reach the right or left shoulder" meaning it ascribes an arc equals to π or better let us say just an arc of 1/100 rad (very small)

53 minutes ago, swansont said:

The only place for a centrifugal force could come from would be the nucleus. How would you do that?

An external field would exert a force on the electron opposite the direction of the field. It would similarly exert a force on the proton in the direction of the field. There is nothing centrifugal about that.  

It is just a theoretical exercise even if not practically feasible.

So again, let us assume theoretically for a moment the field strength between the nucleus and the electron may suddenly increase (for unknown reason) to a new value and remains there. The transition occurs while the electron ascribes an arc of 1/100 rad. How this will affect the electron and the system (atom) as a whole?

Edit (1): suddenly increase (for unknown reason) to a new value and remains there.

Edit (2: The transition occur while the electron ascribes an arc of 1/100 rad.

Edited by John2020
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12 minutes ago, John2020 said:

About the example with the rope the missing part is "until you reach the right or left shoulder" meaning it ascribes an arc equals to π or better let us say just an arc of 1/100 rad (very small)

Ok

3 hours ago, John2020 said:

As I said above replace the atom classical model (Bohr model) with a classical mechanics situation. Take a non-rigid rope with a small mass attached. Start to apply a constant angular velocity by spinning it above your head. After some secs and when the small mass is at your left or right shoulder, increase the angular velocity until you reach the. What will happen?

It depends, an almost rigid wire behaves different than a rubber band for instance. It also greatly depends on how much the speed is increased. The non-rigid rope may stretch and the tension in the rope may increase while I try to spin faster. The rope may coil around my hand for some part of a lap. It may take some time (some number of rotations) before the system is stable again. A lot more details are probably needed to make a prediction.

Why such a complicated example? 

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19 minutes ago, Ghideon said:

Why such a complicated example? 

Complicated? I don't think so. I just would like to point out the following behavior (please check it):

1.constant angular velocity -> Tension equals to centripetal force -> Action-Reaction  -> small mass zero radial velocity

2.Transition to larger angular velocity -> A centrifugal force makes its presence (since the rope is non-rigid) affecting the small mass (inertial effect) radial velocity (it was previously zero)

3.Constant angular velocity after transition -> Tension equals to centripetal force -> Action-Reaction -> small mass zero radial velocity

All the above will have as result during the transition time (increase in angular velocity), the small mass will accelerate radially and due to the conservation of momentum the same will be felt by the man rotating the small mass that means it will feel an attraction force towards the displaced mass.

I have experienced the above, haven't you? If not then try it, is very easy.

Edited by John2020
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44 minutes ago, John2020 said:

 So again, let us assume theoretically for a moment the field strength between the nucleus and the electron may suddenly increase (for unknown reason) to a new value and remains there. The transition occurs while the electron ascribes an arc of 1/100 rad. How this will affect the electron and the system (atom) as a whole?

Edit (1): suddenly increase (for unknown reason) to a new value and remains there.

Edit (2: The transition occur while the electron ascribes an arc of 1/100 rad.

The problem with appealing to magic is that you it's a decent bet that some physical law has been violated. And if that happens, all bets are off. Charge conservation is a physical law, and atoms don't behave the way you describe. The Bohr model has a limited usefulness in explaining some things about atomic structure. 

What QM says about this, to the extent it can, is that you would have a new energy level structure, and the electron would make a transition to a lower energy state and give off a photon. The atom is neutral, so there's no net force on it. You do get an induced electric dipole moment in the atom, though.

 

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11 minutes ago, swansont said:

What QM says about this, to the extent it can, is that you would have a new energy level structure, and the electron would make a transition to a lower energy state and give off a photon. The atom is neutral, so there's no net force on it. You do get an induced electric dipole moment in the atom, though.

Let us say instead of increase, we have decrease in field intensity, however the electron has not enough energy to go to the next quantized energy level, thus it will not give off a photon. Wouldn't the electron and the atom behave as the classical example above (rotating a mass with a non-rigid rope)? 

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1 hour ago, John2020 said:

Complicated? I don't think so. I just would like to point out the following behavior (please check it):

1.constant angular velocity -> Tension equals to centripetal force -> Action-Reaction  -> small mass zero radial velocity

2.Transition to larger angular velocity -> A centrifugal force makes its presence (since the rope is non-rigid) affecting the small mass (inertial effect) radial velocity (it was previously zero)

3.Constant angular velocity after transition -> Tension equals to centripetal force -> Action-Reaction -> small mass zero radial velocity

All the above will have as result during the transition time (increase in angular velocity), the small mass will accelerate radially and due to the conservation of momentum the same will be felt by the man rotating the small mass that means it will feel an attraction force towards the displaced mass.

I have experienced the above, haven't you? If not then try it, is very easy.

It's more complicated than you realise or at least more complicated than you describe.

1) Probably correct in ideal situation 

2) not necessarily, If you just rotate faster then the rope and the ball will lag behind and curl around your hand. If the rope is free to move without your hand getting in the way (rope already almost parallel to the ground for instance) you need to analyse the difference in velocity. Increasing the rotation too fast gets you out of the rhythm and the ball may stop rotating. 

I have of course rotated a ball on a string faster and faster. That can be achieved by pulling the rope slightly harder while at the same time increasing the angular velocity to compensate. Turning to speed up a rotating ball by just rotating faster may or may not work. Compare a contemporary hammer throw (ball on string) vs Scottish hammer throw (ball on a shaft).

 


 

 

 

1 hour ago, John2020 said:

Let us say instead of increase, we have decrease in field intensity, however the electron has not enough energy to go to the next quantized energy level, thus it will not give off a photon. Wouldn't the electron and the atom behave as the classical example above (rotating a mass with a non-rigid rope)? 

If we change the laws of physics then things will change. Other than that, hard to tell. What are you trying to achieve by analysing non-physical situations? Why the focus on things that can't happen, and can't be correctly described within the laws you try to apply?

 

Edited by Ghideon
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26 minutes ago, Ghideon said:

Increasing the rotation too fast gets you out of the rhythm and the ball may stop rotating. 

The idea is that you are in sync with the rotating ball (in rhythm). Or imagine the rope cannot curl (being rigid) but is non-rigid radially. Would you agree with the conclusion I shared above?

Edited by John2020
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16 minutes ago, John2020 said:

The idea is that you are in sync with the rotating ball (in rhythm). Or imagine the rope cannot curl (being rigid) but is non-rigid radially. Would you agree with the conclusion I shared above?

In what frame of reference are you doing the analysis? Why is there a centrifugal force only in step 2? Are you using different frames of reference in steps 1 and 3?
 

 

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11 minutes ago, Ghideon said:

In what frame of reference are you doing the analysis? Why is there a centrifugal force only in step 2? Are you using different frames of reference in steps 1 and 3?

Let us say that both are rotating in outer space, while the ball rotates counterclockwise , the man rotates clockwise.

Edit: changed the rotation direction.

Edited by John2020
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7 minutes ago, John2020 said:

Let us say that both are rotating in outer space, while the ball rotates counterclockwise , the man rotates clockwise.

 

In what frame of reference are you doing the analysis? (You do know what a frame of reference is?) Are we looking at the man from a stationary position beside him or are we rotating along with the man so that the string is attached in origo and the ball appears at rest when angular velocity is constant?

Why is there a centrifugal force only in step 2?

 

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1 minute ago, Ghideon said:

In what frame of reference are you doing the analysis? (You do know what a frame of reference is?) Are we looking at the man from a stationary position beside him or are we rotating along with the man so that the string is attached in origo and the ball appears at rest when angular velocity is constant?

Why is there a centrifugal force only in step 2?

In step 2 there is a centrifugal force because we have to do with a non-rigid rope that implies when the angular velocity increases, a centrifugal force will take place.

We are seeing the rotating man and ball from a stationary position.

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3 minutes ago, John2020 said:

OK and when you rotate with the man in the same direction.

Ok, we are in a rotating frame of reference where fictitious forces may apply. Why is there a centrifugal force only in step 2? 

There is no point in analysing this slightly more complicated example until there is some common baseline regarding the basics.

Edited by Ghideon
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1 minute ago, Ghideon said:

Ok, we are in a rotating frame of reference where fictitious forces may apply. Why is there a centrifugal force only in step 2? 

There is no point in analysing this slightly more complicated example until there is some common baseline regarding the basics.

As I mentioned previously, when the ball rotates with constant angular velocity, you have just the tension and the centripetal force -> action-reaction. On increasing the angular velocity for a short time (step 2), a centrifugal force will additionally rise that pushes the mass radially outwards. Finally, on step 3 the angular velocity is kept constant (after the increase) that will imply again, we have just tension and centripetal force -> action-reaction.

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34 minutes ago, John2020 said:

As I mentioned previously, when the ball rotates with constant angular velocity, you have just the tension and the centripetal force -> action-reaction. On increasing the angular velocity for a short time (step 2), a centrifugal force will additionally rise that pushes the mass radially outwards. Finally, on step 3 the angular velocity is kept constant (after the increase) that will imply again, we have just tension and centripetal force -> action-reaction.

You said that but it does not seem to be correct. In a rotating frame of reference there is always centrifugal force* unless angular velocity is zero.


step 1) centrifugal force and centripetal force are in balance

step 2) 
2.1) the centrifugal force is, for a short time, greater than the centripetal force
2.2) the centripetal force will increase and be greater in magnitude than centrifugal force for a short while

step 3)  centrifugal force and centripetal force are in balance and have a greater magnitude than in step 1.

 

NOTE: again this only applies when analysing the motion from the rotating frame of reference, rotating with the man along the same axis. From a stationary observer beside the man there are no centrifugal forces, they are fictitious. This is important. Failing to realise the importance of this leads to wrong conclusions.

 

*) To be more precise; Inertial centrifugal force. Reactive centrifugal force is not needed to include at this time.

 

Edited by Ghideon
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7 minutes ago, Ghideon said:

NOTE: again this only applies when analysing the motion from the rotating frame of reference, rotating with the man along the same axis. From a stationary observer beside the man there are no centrifugal forces, they are fictitious. This is important. Failing to realise the importance of this leads to wrong conclusions.

So, what do you predict to happen after all these? The small mass will move for a short time radially and at the same moment what is expected for the man if being in outer space?

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10 minutes ago, John2020 said:

So, what do you predict to happen after all these?

Nothing new*, rotation will continue unhindered as long as there is no further change introduced.

11 minutes ago, John2020 said:

The small mass will move for a short time radially and at the same moment what is expected for the man if being in outer space?

In outer space the analysis need to be different. I was assuming (maybe wrongly) that his was taking place on Earth:

3 hours ago, John2020 said:

I have experienced the above, haven't you? If not then try it, is very easy.

It is not easy to try in outer space. 

But simply put, angular momentum is conserved.   

 

 

*)The man getting tired is one possibility in reality

 

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2 minutes ago, Ghideon said:

*)The man getting tired is one possibility in reality

Correct.

5 minutes ago, Ghideon said:

In outer space the analysis need to be different. I was assuming (maybe wrongly) that his was taking place on Earth:

Next time take the hammer you mentioned and try those steps I proposed and tell what you experience.

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40 minutes ago, John2020 said:

The small mass will move for a short time radially and at the same moment what is expected for the man if being in outer space?

The center of rotation, when man is trying to swing the string + ball around in space, will be somewhere along the string, depening on the mass ratio man vs ball. When ball moves radially for a short while the man moves in the other direction for a short while. On earth we neglect these effects since earth is massive, we consider the earth to be stationary and the man able to  hold his position. 

For a massive ball, such as in hammer throw, the effect on the mans position can't be neglected on earth or in space.

Slightly off topic comparison: The effect on earth is not as easily neglected if we would look at the system earth + moon 

11 minutes ago, John2020 said:

Next time take the hammer you mentioned and try those steps I proposed and tell what you experience.

I have told you what is happening in mainstream physics. You are moving the goal posts by switching frame of reference, location to zero g and now hammer instead of low mass ball.

Please get to the point.

 

 

 

Edited by Ghideon
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2 minutes ago, John2020 said:

Your analysis is simply wrong and contrary to observation.

Then point out the specifics. Trying to describe your fictive scenarios using words opens for mistakes:

4 hours ago, John2020 said:

It is just a theoretical exercise even if not practically feasible.

and

28 minutes ago, John2020 said:

tell what you experience.

Is not the best way to do physics. 

It would be easier to do a mathematical analysis using a proper model. 

Edited by Ghideon
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4 minutes ago, Ghideon said:

Then point out the specifics. Trying to describe your fictive scenarios using words opens for mistakes:

Let others answer this if they like. In the meantime when you have a motor suspended by a thread and start it and let us say starts and stop aftet 1/4 of rotor complete cycle what would the motor do?

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6 hours ago, John2020 said:

Let us say instead of increase, we have decrease in field intensity, however the electron has not enough energy to go to the next quantized energy level, thus it will not give off a photon. Wouldn't the electron and the atom behave as the classical example above (rotating a mass with a non-rigid rope)? 

Dropping to a lower energy doesn’t require energy. Energy is released.

You need an example that’s physically allowed 

 

7 hours ago, John2020 said:

Complicated? I don't think so. I just would like to point out the following behavior (please check it):

1.constant angular velocity -> Tension equals to centripetal force -> Action-Reaction  -> small mass zero radial velocity

2.Transition to larger angular velocity -> A centrifugal force makes its presence (since the rope is non-rigid) affecting the small mass (inertial effect) radial velocity (it was previously zero)

To get a larger angular velocity with no change in tangential velocity, you have to decrease r.  ac = r*w^2 

i.e. pull in. That means you need to increase the centripetal force.

Why do you think there is a centrifugal force?

 

 

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5 hours ago, swansont said:

Dropping to a lower energy doesn’t require energy. Energy is released.

You need an example that’s physically allowed

Just assume for a moment it is possible without releasing the energy over a photon. The nucleus and the electron absorb this energy by converting to radial momentum.

5 hours ago, swansont said:

To get a larger angular velocity with no change in tangential velocity, you have to decrease r.  ac = r*w^2 

i.e. pull in. That means you need to increase the centripetal force.

Why do you think there is a centrifugal force?

 

I mentioned above we have a rope that is non rigid radially that means the tangential velocity may also change. Thus, at that moment as Ghideon also noted, the Centrifugal will be larger than the centripetal resulting in small mass displacement radially.

Edited by John2020
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