#### Howdy, Stranger!

It looks like you're new here. If you want to get involved, click one of these buttons!

#### Categories

• Nov 18, 2016

Hi Arthur,

Let's imagine a Newton's Cradle involving a single ball crashing into a row of nine balls at what we will call the "near end". If the ball at the far end moves away as a consequence of a deformation wave which begins at the point of impact, the incoming ball at the near end would have rebounded long before the deformation wave reached the far end.

The whole mystery of the Newton's Cradle is that the incoming ball at the near end doesn't rebound, and that since it works best with hard balls, then deformation doesn't come into the analysis. Either (i) something leaves the incoming ball and travels through the row of nine balls, or (ii) something shifts along from the row of nine balls, leaving the far end ball behind, or (iii) a mixture of (i) and (ii), all depending on the actual absolute motions.

If it is claimed that the signal moves through a Newton's Cradle at 3-6 km/sec, then it is acknowledging that the Cradle is in a state of absolute rest, because they are assuming that the signal travelled the full distance of the cradle.

How do they then explain it if they choose to use the frame of reference in which the in coming ball is at rest?

Best Regards
• Nov 18, 2016

Hi David,

Quote: Let's imagine a Newton's Cradle involving a single ball crashing into a row of nine balls at what we will call the "near end". If the ball at the far end moves away as a consequence of a deformation wave which begins at the point of impact, the incoming ball at the near end would have rebounded long before the deformation wave reached the far end.

I believe that this is a wrong statement. I am trying to figure out why you have arrived at such a statement. The reason I believe is that you have a wrong mental picture of the inner workings of the deformation wave. What I mean by that will be clear, I hope, from the rest of this communication.

Before jumping to "a Newton's Cradle involving a single ball crashing into a row of nine balls", let us consider a simpler case of a Newton's Cradle involving a single ball crashing into one ball, the two balls being absolutely identical. Please note: having switched from the complicated case of "1+9" to simple case of "1+1", we did not lose at all the "mystery" you are referring to, for the incoming ball at the near end doesn't rebound in this simple case - just like it doesn't rebound in the case of "1+9". Therefore, to resolve the "mystery", all we need to do is to gain insight into the process that takes place in the simple case of "1+1".

Now, you wouldn't doubt that some deformation process takes place in both balls in the "1+1" cradle, would you? This deformation process begins at the very moment when the balls get in touch with each other, and it ends when the balls bounce from each other (to simplify the analysis, we assume that the collision is perfectly matched, so there are no residual waves left in the balls after the collision). Like any physical process, the deformation process will take some time to complete; let us denote this time Δt (Δt > 0 !). Now let us ask: is Δt a function of the speed of collision V (and by V I mean the relative speed at which the balls approach each other, for I cannot make sense of the notion of absolute velocity)? My answer is no: Δt does not depend on the speed of balls collision. What is your answer?

Let us ask another question: is Δt a function of the diameter D of the balls? My answer is yes: Δt = f(D). What is your answer?

Another question: assuming that the material, of which the balls are made, is always the same what can we say about the nature of the function f(D)? I assert that it is close to linear, i.e. Δt ≈ kD, where k is some constant. How can I prove it? The best proof, of course, would be an experimental evidence. In the absence of such an experiment, the best I can do is to give you my rationale for asserting Δt ≈ kD, and I will give it next.

Let the diameter of the balls be D = 1 cm, and assume that they collide with relative velocity V which is infinitesimally small: V -> 0. Let assume that we have measured the collision time and we got some numerical value for it, Δt. We would like to know now the collision time for two bigger balls with, say, D=5 cm, colliding at infinitesimally small velocity V -> 0. It is a tough call, and we cannot be sure what the collision time in this case would be.

So, let us consider a case which is somewhat similar but amenable to exact analysis. Consider two rows of balls - a blue row and a red row (see picture: https://www.dropbox.com/s/yu1ieyfbc461b0w/Newton'sCradleMystery.pdf ). Each row consists of 5 small balls, each small ball 1cm in diameter. The distance between the small balls in each row is infinitesimally small: s -> 0. Fix the coordinate system with the laboratory. The balls in the blue row are all moving with the same speed V/2 from right to left, while the balls in the red row are all moving with equal velocity -V/2, i.e. the two rows are on the collision course.

What will happen when the blue and red rows collide? We'll have a lot of "internal" localized collisions. How many such "internal" collisions will occur? The precise answer is 1+3+5+7+9 = 25.

A final question: How long will it take for the process of the "blue row / red row collision" to complete? We know that each "internal" collision takes Δt, so a hasty conclusion would be that the process of "blue row / red row collision" will take 25Δt. Why this answer is wrong? Because some of the 25 "internal" collisions happen simultaneously. And what is the correct answer? The correct answer should be obvious to everyone who takes a close look at the picture I have provided: the process of "blue row / red row collision" will take 9Δt.

That concludes my heuristic argument for the functional relation Δt ≈ kD. If you accept it, the "mystery" of Newton's Cradle is resolved.

My best.
• Nov 18, 2016

Hi Arthur,

You are assuming that the transfer of kinetic energy between two colliding objects is due to the deformation on contact and the subsequent vibration wave set up within the structure of the objects. But in the Newton's Cradle, we deliberately use hard balls that don't deform.

Let's consider the two ball scenario with soft balls. Ball 2 is at rest. Ball 1 crashes into it. The two balls compress (deform) at the point of contact. When ball 2 recoils, it will push ball 1 back again, and the same will be so if ball 1 crashes into a row of nine balls. Hence if deformation is the cause of the kinetic energy transfer, then I would always expect a recoil. My whole argument is that kinetic energy is not transferred by deformation elasticity, but rather by another internal process.

That's why I used the example of a row of weights on a horizontal frictionless surface, all joined together by springs. If we push at one end, the compression wave will commence to move through the row, but the far end will already be moving before the compression wave reaches it.

Hence there are two modes of activity at play. The springs detract from the amount of kinetic energy that is transferred by virtue of absorbing that energy into internal vibrational energy.

And that's how I see it in the case of all collisions. In an elastic collision, there is no deformation and hence no internal energy waves, and that's why we use hard balls in a Newton's Cradle.

Best Regards
David
• Nov 18, 2016

Hi David,

Quote: "But in the Newton's Cradle, we deliberately use hard balls that don't deform."

We cannot use hard balls in the Newton's Cradle for the simple reason that there are no hard balls in Nature, period. We can assume whatever we want - like absolutely rigid balls, for example - and, admittedly, that can be a very useful idealization for certain purposes, but hardly it can be useful for the purpose you are employing it. Quite the contrary - it can lead you astray badly and, in fact, I believe it already did.

Quote: "Let's consider the two ball scenario with soft balls. Ball 2 is at rest. Ball 1 crashes into it. The two balls compress (deform) at the point of contact. When ball 2 recoils, it will push ball 1 back again, and the same will be so if ball 1 crashes into a row of nine balls."

How so? Have you run a test with a Newton's Cradle, comprised of rubber balls pumped with air? Or, perhaps, you are aware of the results of such an experiment conducted by someone else.

Quote: "Hence if deformation is the cause of the kinetic energy transfer, then I would always expect a recoil."

Before saying "hence", you better have some experimental evidence to back up the assertion that preceded that "hence". If you have one, I would be very surprised ... and, of course, I would be happy to see and analyze it.

Quote: "My whole argument is that kinetic energy is not transferred by deformation elasticity, but rather by another internal process."

Yes, I know. Likewise, my whole argument is that kinetic energy is transferred by deformation elasticity.

My best.
• Nov 18, 2016

Hi Arthur,

OK, so the issue then comes down to whether in a collision,

(1) There is only one mode of activity, or
(2) There are two modes of activity.

You believe that there is only one mode of activity and that it takes the form of a deformation wave which transfers kinetic energy.

Let's look at a Newton's Cradle made of pneumatic balls. The incoming ball will be called ball number 1. It collides with ball 2 which is the first in a row of nine balls numbered 2 to 10.

If this pneumatic Newton's Cradle is to behave identically to the metal Newton's Cradle, then ball number 2 will not move when ball number 1 collides into it.

When ball number 1 collides into ball number 2, both of them will compress and deform at the contact point. A recoil will then occur. I can't see how, bearing in mind that ball number 2 does not budge from its position, that ball number 1 would not then rebound. I further can't see how the deformation in ball 2, which is at the side where the contact takes place with ball 1, would somehow transmit through the row of balls such as to flick ball 10 away.

Now consider a long metal pole. We push it at one end. Does the whole pole move at once? If you think that a compression wave through the pole determines when each part of the pole begins to move, then consider a long horizontal spring on a frictionless surface. We push it at one end. Does the far end move before the compression pulse reaches the far end? I think you'll find that it does. The far end does not wait until the compression wave reaches it. Something else goes ahead of it at a faster speed.

That something else was what I was alluding to in my earlier 2008 paper "The Aether in Rigid Body Collisions". Then I believed that it was a pure aether compression pulse that may even have had infinite speed, but now I think it's a fine-grained rotational wave though the row of balls which has finite speed of enormous magnitude.

Best Regards
David
• Nov 19, 2016

Hi David,

Quote: "OK, so the issue then comes down to whether in a collision,

(1) There is only one mode of activity, or
(2) There are two modes of activity.

You believe that there is only one mode of activity and that it takes the form of a deformation wave which transfers kinetic energy."

Not quite. Strictly speaking, I believe that there is an infinite number of modes of activity. Elastic deformation wave is just the most ponderable and the least rapid one of all modes. Electromagnetic action is the next mode of activity - more subtle and much more rapid. The next one is the gravitational mode - still more subtle and way more rapid than the speed of light. For the next one we don't even have a name, etc. etc. ad infinitum.

But the key point is this: the last ball at the other end of the Newton's Cradle cannot and will not move AS A WHOLE until the grossest of all the action modes (i.e. that of the elastic deformation wave) arrives. Parts of the ball will feel - one way or the other - all modes of action, but that is irrelevant when we speak of the gross event of ball bouncing.

Regarding the pneumatic Newton's Cradle. If the balls are pumped to only 2-3 atm, then there might be some minor recoil, indeed. But as the air pressure in the balls goes up the recoil will be less and less pronounced, and at some point we cannot distinguish pneumatic cradle from the steel ball cradle.

The following is not an experiment - it is just a simulation - but it will give you an idea of what I am talking about:

My best.
• Nov 19, 2016

Hi Arthur,

Beginning with the end of your letter below, I watched the video, and yes, that was exactly my point. The stiffer the springs, the less will be the recoil of the incoming ball, and the less will be the relevance of the deformation wave.

In an ideal Newton's Cradle where the balls are totally hard, there will be no deformation wave at all, but since it works perfectly, then the kinetic energy must have been transported by a means other than that of a deformation wave.

And so I have to conclude that the transfer of kinetic energy is a very fast process, much faster than the speed of a deformation wave. And in addition to that, we don't even know which direction it is coming from, because we don't know the absolute speeds relative to the luminiferous medium. The deformation wave on the other hand simply moves in both directions from the point of impact, but I don't believe that it carries with it kinetic energy which is imparted elsewhere. That is done by a more subtle kind of wave.

Best Regards
David
• Nov 19, 2016

Hi David,

Quote: "The stiffer the springs, the less will be the recoil of the incoming ball, and the less will be the relevance of the deformation wave."

In my opinion, this is a blatant non sequitur. I am pretty sure that the stiffer the springs, the less will be the recoil of the incoming ball, but the relevance of the deformation wave will be absolutely the same, no matter what the spring stiffness is.

Quote: "In an ideal Newton's Cradle where the balls are totally hard, there will be no deformation wave at all, but since it works perfectly, then the kinetic energy must have been transported by a means other than that of a deformation wave."

Pardon my stiffness, but this is another non sequitur. In an ideal Newton's Cradle where the balls are totally hard, the speed of the deformation wave is simply approaching infinity, that's all.

Quote: "The deformation wave on the other hand simply moves in both directions from the point of impact, but I don't believe that it carries with it kinetic energy which is imparted elsewhere. That is done by a more subtle kind of wave."

If you believe that there is no deformation wave in the Newton's Cradle that carries with it kinetic energy then you should be willing to play the "Russian" Cradle I have invented just for you.

Imagine steel ball cradle of 1000 balls; each ball is of the size of your head, say, D=10 in. Ball 1 is coming in at the speed V=100 m/s and hits the other 999 balls. You put down \$10,000, and I put down \$10,000. Then, before the ball 1 strikes, you put your head between the balls 500 and 501. If you survive, you take home \$20,000; if you don't - I take the \$20,000 and pay for your funeral. That's my "Russian" Cradle. Are you willing to play it, David?

My best.
• Nov 19, 2016

Hi Arthur,

I accept that the deformation wave pulse does carry energy. It carries an element of energy which oscillates between potential and kinetic energy on a molecular scale. But this energy is detracted from the kinetic energy on the large scale that travels through the system, and it corresponds to the energy that is eventually dissipated as heat.

The stiffer the material, the lesser will be the displacement, and hence the lesser the energy that is lost to heat. (from the formula - energy = 1/2kx^2. As k increases, x decreases and hence the energy decreases)

When you mentioned your second mode of activity, i.e. the electromagnetic mode, you were on the right tracks. In my opinion, the kinetic energy is transported by a wave that is similar in form to an electromagnetic wave, but rather than operating through the luminiferous medium, it is operating through the electron orbitals of the atoms in the metal. Hence, rather than it being a linear displacement wave, it is a fine-grained angular acceleration wave through the metal. In other words, it doesn't involve any deformation.

On your Russian cradle, the human head in the middle will deform when the adjacent metal ball receives its kinetic energy and then pushes against the human head with the same high speed impulse as the incoming ball.

Best Regards
David
• Nov 19, 2016

Hi David,

Quote: "On your Russian cradle, the human head in the middle will deform when the adjacent metal ball receives its kinetic energy and then pushes against the human head with the same high speed impulse as the incoming ball."

I take it as a refusal to play the game.

After sending out the last letter, it occurred to me that offering you the Russian Cradle game - which, unlike the Russian Roulette, leaves the player with zero chance of survival - was, perhaps, too brutal. So, as an apology, I am offering you a modified version of it now.

Everything remains the same in the rules of the game except one thing: we now glue together the balls from 2 to 500 at the points of contact of neighboring balls, i.e. we glue 2 to 3, 3 to 4, 4 to 5, ... , 499 to 500.

Would you play the game now?

My best.
• Nov 19, 2016

Hi Arthur,

It would still be dangerous. You would still get your head squeezed. When the kinetic energy wave would reach the ball adjacent to your head, it would be restrained from moving alone into your head at high speed as in the first case scenario, and this restraint would be due to the fact that it would have to pull the row of 499 balls behind it. But the reduced speed of impact would be compensated by a proportionate increase in mass, and hence the momentum of the impact would be the same. It would just result in a slower crushing.

The gluing would also add another factor into the mix because pulling against the glue would cause energy to be lost into heat.

Best Regards
David
• Nov 19, 2016

Hi David,

Quote: "It would still be dangerous. You would still get your head squeezed."

O yes, it would be very painful, indeed. So the head will be squeezed badly in both scenarios, i.e. it will be deformed. How come that the head gets deformed but the ball in its place would suffer no deformation at all, i.e. the ball would feel no pain - even if it had nerves?

Don't you see a contradiction here?

My best.
• Nov 19, 2016

Hi Arthur,

The ball is made of a harder material.

In both cases there will be some kinetic energy imparted and some deformation caused. In the case of the human head, the deformation will be substantial. In the case of the metal ball, the deformation will be negligible.

When the deformation is negligible, the wave speed of the deformation will be much greater and may even catch up with the speed of the kinetic energy wave, but they are still two separate modes of activity. I don't know what that speed would be in each case, but did you say a few mails back that it may be in the order of 3-6 km/sec.? That is fast, and it would seem likely to me. It's got to be finite, very fast, but not as fast as light. But we don't know what direction its going in unless we know the absolute speeds of the balls relative to the luminiferous medium.

Best Regards
David
• Nov 19, 2016

Hi David,

Quote: "In my opinion, the kinetic energy is transported by a wave that is similar in form to an electromagnetic wave, but rather than operating through the luminiferous medium, it is operating through the electron orbitals of the atoms in the metal. Hence, rather than it being a linear displacement wave, it is a fine-grained angular acceleration wave through the metal. In other words, it doesn't involve any deformation."

From what you said so far, I gather:

(1) There are two separate waves in the Newton's Cradle - deformation wave and kinetic energy wave;
(2) Kinetic energy wave propagates very fast but not instantaneously (speed, perhaps, in the order of speed of light) while deformation wave propagates (for real steel balls) at much slower rate (perhaps, a few km/s);
(3) The original kinetic energy of the incoming ball splits into the energy of deformation of the ball material, on one hand, and kinetic energy of increased motion (spin and/or orbital) of electrons in the metal, on the other hand;
(4) The share of energy that propagates in the "kinetic mode" is significantly higher than that which propagates in the "deformation mode".

Would you agree that this is a correct summary of your views?

My best.
• Nov 19, 2016

Hi Arthur,

That's a pretty good summary of my position. I am willing however to reduce the speed of the kinetic waves if needs be. And on your point number (4), the proportion of the two effects would depend on the elasticity of the material. The more rigid the material, the more efficient will be the kinetic energy transfer process, providing of course that shattering doesn't occur.

So far I have deliberately avoided discussing electromagnetism and I intend to keep it that way for a while longer. But as a point of information, I should add that all my original researches were in electromagnetism and that this stuff about Newton's Cradles, Galilean relativity, and the inertial forces only came later in my researches as an after thought when I realized the mathematical parallels between the EM forces and the inertial forces.

So a lot of my ideas in these inertial processes are influenced by similar patterns in the luminiferous medium in relation to EM.

For example, for reasons not yet divulged in our discussions so far, I have concluded that EM waves are a propagation of angular displacement in a sea of molecular vortices based on Maxwell's model. It took me quite a few years to realize that EM waves are not about linear displacement.

Only more recently when examining the Newton's Cradle and the fact that it worked best with balls that don't deform, did I merge the angular displacement wave idea with other researches I had done into the origins of the inertial forces. I had been modelling all atoms and molecules, no matter how complex, on spinning entities with a net angular momentum. I started that simplification mainly when I was trying to figure out what is going on in a pivoted gyroscope when the Coriolis force is causing it to defy gravity. The complications in atomic orbitals are for the chemists to worry about.

Best Regards
David