# Newton’s Cradle; a Demonstration of Quantum-Geometry Dynamics

Note: to fully appreciate and understand the following discussion, I highly recommend reading the article titled
A Remarkably Simple Proof of the Discreteness of Space . Also, for an explanation of the fundamental mechanism which governs momentum transfer, you may read the article Baseball Physics as Explained by Quantum-Geometry Dynamics.

Everyone is familiar with Newton’s Cradle, a simple mechanism which is thought to demonstrate the law of conservation of momentum. Paradoxically, the physics of Newton struggles to describe and explain the behaviour of this simple system of steel balls. In fact, even the full power of continuous mathematics can only provide an approximation of the behaviour of such a system and even that must be at the cost of undesirable and awkward complications.

The reader may ask why, today, anyone should bother trying to explain such a trivial device when there are so many exciting phenomena to which quantum-geometry dynamics can be applied. Let me ask this question: How can Newton’s cradle be a device demonstrating Newton’s laws of motion when Newton’s laws of motion cannot explain its behaviour? Also, since quantum-geometry dynamics claims to explain the dynamics of motion at all scales (and do so in as a simple and straightforward way), applying it successfully to Newton’s cradle should be easy. Not to mention, an excellent opportunity to test its validity. It also doesn’t hurt that building any version of Newton’s cradle is considerably less expensive than, let’s say, a particle accelerator.

The first thing we must do is formulate the problem in QGD terms. Given a Newton’s cradle system having $n$ steel balls, when lifting then dropping subset consisting of $x$ numbers of balls, the impact with set $y$ number of balls will in motion. The problem will be to predict and explain which balls will be set in motion from the impact as well as determine their momentum and speed.

We have explained earlier articles that QGD defines the momentum of an object $a$ as $\left\| {{{\vec{P}}}_{a}} \right\|=\left\| \sum\limits_{i=1}^{{{m}_{a}}}{{{{\vec{c}}}_{i}}} \right\|$ and the its speed as ${{v}_{a}}=\frac{\left\| {{{\vec{P}}}_{a}} \right\|}{{{m}_{a}}}$ where ${{m}_{a}}$ is its mass in preons(+). Thus momentum and speed are intrinsic properties independent of the frame of reference.

We also know that the preons(-), the elementary particle that dimensionalizes space to form quantum-geometrical space, generating an opposing force to changes in momentum so that it only allows changes in momentum when $\left\| \Delta {{{\vec{P}}}_{a}} \right\|=z{{m}_{a}}$ where $z\in {{N}^{+}}$.

Conservation of momentum (which is a consequence of the conservation of the fundamental momentum of preons(+)) is such that $\displaystyle \sum\limits_{i=1}^{x}{\left\| {{{\vec{P}}}_{{{a}_{i}}}} \right\|}=\sum\limits_{j=1}^{y}{\left\| \Delta {{{\vec{P}}}_{{{a}_{j}}}} \right\|}$ where $\displaystyle \sum\limits_{i=1}^{x}{\left\| {{{\vec{P}}}_{{{a}_{i}}}} \right\|}$ and $\displaystyle \sum\limits_{j=1}^{y}{\left\| \Delta {{{\vec{P}}}_{{{a}_{j}}}} \right\|}$ are respectively the momentum of a group of $x$ balls pulled and released together and the momentum of the group of balls set in motion after impact.

Also since $\left\| \Delta {{{\vec{P}}}_{{{a}_{j}}}} \right\|={z}'{{m}_{a}}$ where ${z}'\in {{N}^{+}}$, we have $\displaystyle \sum\limits_{j=1}^{y}{\left\| \Delta {{{\vec{P}}}_{{{a}_{j}}}} \right\|}=y{z}'{{m}_{a}}$ or $\displaystyle \frac{\sum\limits_{j=1}^{y}{\left\| \Delta {{{\vec{P}}}_{{{a}_{j}}}} \right\|}}{y}={z}'{{m}_{a}}$ and since $\displaystyle \sum\limits_{i=1}^{x}{\left\| {{{\vec{P}}}_{{{a}_{i}}}} \right\|}=\sum\limits_{j=1}^{y}{\left\| \Delta {{{\vec{P}}}_{{{a}_{j}}}} \right\|}$ then $\displaystyle \frac{\sum\limits_{i=1}^{x}{\left\| {{{\vec{P}}}_{{{a}_{i}}}} \right\|}}{y}={z}'{{m}_{a}}$. That is, the momentum transferred must be an integer multiple of the mass of a ball. This all we need to explain and predict the behaviour of a Newton’s cradle system from initial conditions are known. Let’s examine now a few examples showing how the above applies.

In the example shown in the figure on the left, $x=3$which leave $2$ balls on the right. Now, we know that the momentum can only be transferred in integer multiples of ${{m}_{a}}$, and since the three balls carry an integer multiple of ${{m}_{a}}$ (equivalent to an integer number of balls), if it were to be transferred to the remaining two balls, the resulting change of momentum for each ball would be a non-integer. Non-integer changes in momentum are forbidden by structure of quantum-geometrical space.

So the total momentum that can be transferred to the two red balls is $2\left\| {{{\vec{P}}}_{a}} \right\|$ , $1\left\| {{{\vec{P}}}_{a}} \right\|$each, but since the total momentum of the blue balls is $3\left\| {{{\vec{P}}}_{a}} \right\|$that leaves us with $3\left\| {{{\vec{P}}}_{a}} \right\|-2\left\| {{{\vec{P}}}_{a}} \right\|=1\left\| {{{\vec{P}}}_{a}} \right\|$ . Since the remaining momentum cannot be transferred, it will be kept by the blue balls. But here again, it cannot be divided between the three blue balls, which would imply them having momentums that are non-integer multiples of ${{m}_{a}}$. The remaining momentum will be kept by one of the balls, which evidently is the blue ball on the right. The result will be as shown in the figure on the left.

The example above shows a configuration where $x=4$, leaving one ball. Applying what we’ve learned above, we see that four is divisible by one. Then the entire momentum of the blue balls can be transferred to the red ball. Using QGD definitions we gave at the beginning of this article, we find that the speed of red ball is four times that of the speed of the group of blue balls.

Now, let’s make thing a little more complicated by using a cradle where the balls are of different mass.

The cradle above is made using balls having different mass. With the large balls being exactly twice as massive as the smaller balls. In the figure above, $x$ will not represent the number of balls pulled, since we have balls of two different masses, but the equivalent number of small balls. So here $x=3$. Using the same logic as above, the momentum of the blue balls will be transferred in its entirety to one ball.

Now, some readers may ask why can’t the momentum of the blue balls be transferred to three red balls instead since such change in momentums are integer multiples of ${{m}_{a}}$ (here ${{m}_{a}}$ is the mass of a small ball). The answer is that if a ball can transfer its momentum it will. Transfer of momentum is compulsory if possible and in the example above, the successive balls from impact can transfer the momentum and so they do. The last red ball has no other ball to transfer the momentum to, so it must carry it. And its speed will be three times that of the group of blue balls before impact.

Allow me one last example using the cradle above.

In the above figure $x=7$ , but the momentum cannot be transferred completely to the red balls. $6\left\| {{{\vec{P}}}_{a}} \right\|$can be transferred to the six red balls on the left, leaving us with $1\left\| {{{\vec{P}}}_{a}} \right\|$. But $2\left\| {{{\vec{P}}}_{a}} \right\|$is required to move the large red ball and fractional changes in momentum being forbidden, the remaining momentum cannot be transferred to it. As a result, it must be kept by one of the blue balls. The blue ball that will carry the remaining $1\left\| {{{\vec{P}}}_{a}} \right\|$ with be the one on the left. And since it cannot move towards the right, to conserve momentum, it will have to be reflected back to the left (see figure below).

I leave it to the reader to figure out what in this last example when the blue ball on the left and the group of red balls on the right swing back to impact the remain balls. But that shouldn’t be difficult using the notions presented in this article.

Finally, as discussed in the article titled A Remarkably Simple Proof of the Discreteness of Space , the transfer of momentum produces a nearly instant acceleration of the balls that are set in motion from the impact. Such an acceleration, if space were continuous and speed and momentum described classically, would require a nearly infinite force. The balls that are set in motion do not accelerate from rest to the speed after impact by going through all the intermediary speeds between them. The balls, QGD predicts, will move instantly go from rest to the speed that corresponds to the momentum transferred to them. And only after impact will they slow down from the effect of gravity. Testing that prediction should be easy enough using a high speed camera.

What we have shown here is that Newton’s cradle can be used to demonstrate at our scale QGD’s concepts of speed, momentum and conservation of momentum which we have used so far to describe the motion of particles and structures at the fundamental scale. It follows that the laws of motion at all scales are consequences of the fundamental scale interactions. As in the previous article, A Remarkably Simple Proof of the Discreteness of Space, we have shown that physics at our scale is the observable consequences of physics at the fundamental scale.

Click here to see a well-made slow motion video of Newton’s cradle in action. Also, readers may be interested in reading Rocking Newton’s Cradle by S. Hutzler, G. Delaney, D. Weaire and F. McLeod) as an example of how complicated it can be to describe even this simple system using classical physics.