# The Naked Scientists Forum

### Author Topic: On ‘Kinetic Energy’ and ‘Nuclear Vibration’.  (Read 300 times)

#### RTCPhysics

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##### On ‘Kinetic Energy’ and ‘Nuclear Vibration’.
« on: 30/08/2016 08:33:37 »
The concept of kinetic energy (KE) is based upon the movement of matter, whether it is a particle or a bullet. But as Lagrange and Einstein pointed out, velocity is relative. So if you are moving at the same speed and direction as the bullet, then you can pick it out of the air, check the make and put it back in the air, allowing it to continue on its way. To you, it hasn’t got any kinetic energy and in the vastness of space, it would be as if you were both in ‘suspended animation’ or even that ‘time was standing still’. But you aren’t and it hasn’t, for if you put yourself in front of the bullet and slow down, then time would have ticked on and the kinetic energy of the bullet would be immediately evident to you.

We know that kinetic energy isn’t just 'stored energy' for it also has direction, so the question that needs to be answered is: “Exactly how is kinetic energy held within the atomic structure of a body of matter, in a manner that gives it both a magnitude and direction?” If you think you know the answer to this question, then you don’t need to read on, unless, of course, you are interested enough in the subject of kinetic energy, to find out what the ‘competitive concept’ being proposed here is.

Turning to the mathematics for a moment, kinetic energy is defined as the mass of a particle or body of matter, multiplied by a ‘velocity term’, which equates to half of the velocity raised to the power of two, ½v². Out of this comes the odd fact, that if the velocity of the particle is equal to or greater than 517,532 mph, then this ‘½v² velocity term’ being applied to the ‘mass’ of the particle, equates to a velocity that is greater than the ‘speed of light’. As nothing is thought to move faster than ‘light speed’, then kinetic energy takes on an unreal quality. So how do you interpret this into the reality of the atomic world?

To progress this question, consider two hydrogen atoms travelling at a velocity, v, in opposite directions, that collide ‘head on’ forming a ‘stationary hydrogen molecule’ with KE=mv².  If the molecule is stationary, then exactly what has happened to the kinetic energy of the two colliding hydrogen atoms? As energy can be neither created nor destroyed, the kinetic energy must survive. If it is to remain in the form of movement and there only seems to be two possibilities. Either the kinetic energy is converted into ‘molecular spin’ rotating at the velocity ‘v’, around an axis set at right angles to the original direction of travel of the two hydrogen atoms or it appears as an ‘internal vibration’ between the two hydrogen atoms that created the hydrogen molecule.

If it doesn’t survive as movement, then there is another possibility and this rests upon the idea that energy can be transformed into mass, as Einstein’s iconic formula E=mc² implies.  However, any increase in the masses of the two protons and electrons that make up the hydrogen molecule, would have to be transformed into their more massive ‘sibling particles’, notably, the ‘neutron’ and the ‘electron muon’ respectively. But hydrogen molecules comprised of a pair of neutron nuclei with two electron muons in orbit around them, have yet to be observed.

So kinetic energy transforming directly into mass is a low probability outcome. Further, as ‘hydrogen molecules’ have not been observed to exist in a ‘spin state’, then the option of kinetic energy re-appearing as internal vibration within a molecule, becomes the more realistic outcome. So exactly how does this occur?

As the two hydrogen atoms collide, their nuclear structure is forced together and distorted. The two nuclei of the atoms are the first to be affected, as the kinetic energy of the hydrogen atom is mainly associated with their mass and the collision causes them to be forced together. This movement of the nuclei also affects the electrons and if the two hydrogen atoms are viewed as having a spherical shape created by their electron orbits, then the collision forces their circular orbits into a more elliptical shape, with the minor axis lined up along the original direction of movement of the two atoms.

This shift of atomic nuclei towards each other, initiates an electrostatic repulsion between them, which repels them apart again in the reverse direction. But this equal and opposite reversal between the two nuclei, causes them to overshoot beyond their centralised locations within their electron orbits. This results in their electron orbits changing shape again, moving through the circular orbit and into an elliptical shape again, but this time with the major axes lined up along their original direction of movement.

This 'nuclei overshoot', however, initiates a reversal of their movement apart, as the combined electrostatic forces of their two elliptically orbiting electrons, act in concert draw them back towards each other again, but once more, the kinetic energy of the nuclei cause them to overshoot towards each other.

As there are no internal forces at work that will dampen down the movement of nuclei and electrons within an atom, this alternating process repeats itself perpetually, creating a ‘repetitive nuclear vibration’ between the two hydrogen atoms within the hydrogen molecule. This internal molecular vibration of the nuclei and electrons along the collision path of the two hydrogen atoms is the means by which the kinetic energy, apparently lost by the two colliding hydrogen atoms, is transferred both in magnitude and direction, into the stationary hydrogen molecule.

In the more general case, in which two hydrogen atoms with different velocities, collide at an angle rather than head on, then the resulting ‘nuclear vibration’ still occurs, but with a greater or smaller magnitude, dependent upon the velocities of the two colliding atoms. The hydrogen molecule that is created, emerges from the collision with a residual velocity, travelling in a direction that is determined by the original angle of contact between the two hydrogen atoms.

If we consider the kinetic energy of molecules within a liquid, rather than hydrogen gas, then the collisions between the molecules results in the same process of nuclear vibration, transferring kinetic energy between the liquid molecules. It’s a process that is called, ‘Brownian motion’ after its discoverer.

But perhaps the most interesting impact of the transfer of kinetic energy into nuclear vibration is within a solid, such as the bullet referred to in the opening paragraph. For as the molecular vibration surges backwards and forwards across the bullet in its direction of travel, its impact upon an object placed in its path is like a ‘jackhammer’. So if you are out in space travelling alongside that bullet, bear this in mind!

#### GoC

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##### Re: On ‘Kinetic Energy’ and ‘Nuclear Vibration’.
« Reply #1 on: 02/09/2016 22:39:06 »

The following users thanked this post: RTCPhysics

#### alancalverd

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##### Re: On ‘Kinetic Energy’ and ‘Nuclear Vibration’.
« Reply #2 on: 05/09/2016 14:57:53 »
There's quite a bit of nonsense in the OP. To begin with, v is a vector and v^2 is not a velocity or even a speed but the square of the modulus of a vector, i.e. a scalar which can take any value you like without offending the fudamental principles of relativity. If you start with a massive particle and accelerate it towards v = c, its mass increases, as confirmed by experiment, according to the relativistic equations.

And beware of confusing atoms, molecules and nuclei. Nuclear collision is extremely unlikely as nuclei are very, very small compared with the atoms they inhabit, and carry a lot of charge. If it were not so, we could have all the fusion energy we need by simply bashing two bits of coal together!

If you want a conundrum, consider that a photon has kinetic energy but no mass. However Planck and Einstein sorted all that out about 100 years ago, so I wouldn't advise anyone to waste too much time contemplating it - the answer is in the book and the phenomenon of pair production confirms their predictions (with a very significant yield) every day.

#### GoC

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##### Re: On ‘Kinetic Energy’ and ‘Nuclear Vibration’.
« Reply #3 on: 05/09/2016 19:07:55 »
Can you define charge as you are using it? Can you define a proton particle of no mass. Can you explain the creation of mass without mass? Observations are not their own explanations as many scientists suggest. More contemplation is needed not less. RTC contemplates observations not just allowing observations to be their own explanations.

#### alancalverd

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##### Re: On ‘Kinetic Energy’ and ‘Nuclear Vibration’.
« Reply #4 on: 06/09/2016 23:38:26 »
Can you define charge as you are using it?
same definition as everyone else. I won't cut and paste Wikipedia here, but the first pragraph is adequate.
Quote
Can you define a proton particle of no mass.
obviously not - a proton is a massive particle, by definition.
Quote
Can you explain the creation of mass without mass?
mass and energy are equivalent. No explanation needed, and the predicted equivalence is confirmed by measurement. You can derive the equivalence from a simple consideration of radiation pressure: there are plenty of adequate websites but I can't generate decent differential equation symbols on this forum.

#### RTCPhysics

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##### Re: On ‘Kinetic Energy’ and ‘Nuclear Vibration’.
« Reply #5 on: 21/09/2016 15:42:34 »
RTC reply to Post by: alancalverd
« on: 05/09/2016 14:57:53 »

"There's quite a bit of nonsense in the OP. To begin with, v is a vector and v^2 is not a velocity or even a speed but the square of the modulus of a vector, i.e. a scalar which can take any value you like without offending the fudamental principles of relativity. If you start with a massive particle and accelerate it towards v = c, its mass increases, as confirmed by experiment, according to the relativistic equations."

I hoped it would be recognised, that I wrote this paragraph with my tongue in my cheek! Obviously a mistake.

"And beware of confusing atoms, molecules and nuclei. Nuclear collision is extremely unlikely as nuclei are very, very small compared with the atoms they inhabit, and carry a lot of charge. If it were not so, we could have all the fusion energy we need by simply bashing two bits of coal together!"

Nuclei vibration within a molecule is not enough to cause the two nuclei to fuse. The best physical analogy I can offer is a tuning fork, where the prongs vibrate synchronously, but are kept well apart.

"If you want a conundrum, consider that a photon has kinetic energy but no mass. However Planck and Einstein sorted all that out about 100 years ago, so I wouldn't advise anyone to waste too much time contemplating it - the answer is in the book and the phenomenon of pair production confirms their predictions (with a very significant yield) every day."

I tend to take the view that the knowledge available to Einstein and Plank a 100 years ago, hardly matches that which we have today. And new knowledge can lead to new ways of viewing known physical phenomenon, such as pair production, although I have none to offer.

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##### Re: On ‘Kinetic Energy’ and ‘Nuclear Vibration’.
« Reply #5 on: 21/09/2016 15:42:34 »