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### Author Topic: Conservation of Momentum, at an atomic level  (Read 4075 times)

#### manu3d

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##### Conservation of Momentum, at an atomic level
« on: 17/08/2013 22:43:14 »
Hi everybody!

In classical mechanics, in the context of perfectly elastic objects colliding, the equations regarding the conservation of momentum easy enough to deal with. Problems arise when I start asking myself... sure, but what's happening at an atomic level when two objects collide?

Specifically, I started to wonder: let's say I have a static object consisting of 10 identical atoms connected together in a linear molecule. We'll call it the Rod. Laying around I have a particle accelerator (as you do) and I use it to shoot a single atom identical to the other ten, let's call it the Projectile, toward the Rod. Having (poor) human eyesight but superhuman luck, I manage to align the accelerator and the Rod so that it is essential a 1D collision: the Projectile impacts with only the first atom of the Rod, and its trajectory is aligned with the direction defined by the Rod.

Thinking in terms of classical mechanics it's easy: we are dealing with a mass 1m with speed s heading toward a static mass 10m. With a perfectly elastic collision the outcome is a static Projectile and a Rod moving at 1/10th of the Projectile's speed. Great, simple, logical. But how does the universe "knows" that the Rod has a mass of 10m and therefore it must be accelerated to 0.1s? The Projectile only impacts with the first atom in the Rod and I feel it should transfer to it its whole momentum. Which in turn should be transferred to the next atom and so on until the 10th atom. Given that this last atom is attached to the previous one, the momentum should return backward all the way to the first atom and then to the static projectile which would then restart in the opposite direction with its initial speed. Clearly this is not what happens and my logic is flawed. But why?

How is the speed s of the Projectile redistributed evenly to the atoms of the Rod so that they all decide to move along at 0.1s? The equations of conservation of momentum imply a knowledge of all the masses and velocities in action, to define the entities being dealt with. But there is a point in time in which the knowledge of the impact and the characteristics of the Projectile's motion are known only to the first atom in the Rod. How does that "knowledge" propagates to the other atoms and calculates their collective motion after the impact? I mean, the universe doesn't really know group connected atoms ito objects, does it? Am I entering the realms of quantum mechanics or is this something that can be explained classically?

Thank you for your help!

Manu

[EDITED 18/6/2013: made explicit we are dealing with a 10-atom molecule]
« Last Edit: 18/08/2013 10:49:04 by manu3d »

#### alancalverd

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##### Re: Conservation of Momentum, at an atomic level
« Reply #1 on: 18/08/2013 12:28:32 »
First, the collision need not be perfectly elastic. Momentum is conserved under all interactions.

Then you need to ask how your atoms are connected. Classically, if you model them as rigid billiard balls in contact, you can analyse the collision as between two rigid bodies of mass m and 10m. More realistically, you can consider them as connected by springs, to produce a molecule that can compress and stretch, so some of the kinetic energy will end up as vibration of the molecule.

#### manu3d

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##### Re: Conservation of Momentum, at an atomic level
« Reply #2 on: 19/08/2013 23:24:11 »

Concerning the elasticity of the collision, you are right of course. I just thought a perfectly elastic collision would be the easiest to deal with for the purpose of describing an example setup.

Concerning your turning the focus on how the atoms are connected I feel you are right on the mark. In the past few days I was more or less thinking along the same lines: the bonds between atoms is where everything happens.

But what and how happens? Even if we simplistically imagine the bonds as springs, allowing for a compressible/stretchable molecule as you suggest, I remain fuzzy on how these springs interact to bleed the speed to the 1/10th necessary to keep the equations balanced. I mean, let's break it down a little: 1) The impact occurs, the first atom in the molecule acquires the full speed of the projectile. 2) As the first atom is set in motion by the impact, the bond between the first and second atom is compressed, storing potential energy. 3) Eventually the bond reaches a point when it must release this potential energy by extending and setting the second atom in motion, which in turn compresses the second bond and so on until the 10th. As I imagine it, the peak of potential energy stored in the bonds moves along the molecule in a wave-like pattern, flowing and re-flowing, back and forth endlessly. Is this correct? But even if this picture is correct, it remains the question: how do the bonds decide to collectively store 9/10th of the speed of the impact instead of, say, 1/5th, 0.1379457% or even all of it?

Apologies if this sounds quite confusing: it's just a reflection of the state of my head thinking about it!

Again, thank you Alan!

Manu

#### alancalverd

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##### Re: Conservation of Momentum, at an atomic level
« Reply #3 on: 20/08/2013 00:00:49 »
The classical analogy breaks down at the quantum level, but by the time you have constructed a device capable of aligning 10 atoms in a straight line and firing another one precisely along the axis, you will probably have recruited a theoretical physicist with a blackboard and more time than I have to explain it and calculate the outcome!

However the billiard ball and spring model can get you a long way towards a description of molecular vibrations. In effect, the compression wave travels along the line at the "speed of sound" in the molecule, and the wave will indeed be reflected in a macroscopic assembly. So you end up with a resonating loaded spring whose total momentum is conserved but whose ends wobble back and forth at a frequency determined by the masses and spring rates. This has important implications in engineering - you can't transmit a pulse along a pushrod any faster than the speed of sound in the metal, which is why overhead camshafts replaced pushrods in high-performance piston engines: at high speeds, the pushrods resonate and bend instead of actuating the valves at the correct times.

#### evan_au

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##### Re: Conservation of Momentum, at an atomic level
« Reply #4 on: 20/08/2013 11:32:41 »
I think a large part of the mechanical interaction will occur when the pulse of movement (after having bounced off the "free" end of the rod) propagates back to the last atom of the rod, which is still in contact with the ball. This pulse of motion will cause the rod to push the ball away from the end of the rod.

The relative speeds of the rod and ball will be determined by this wave.

Some of the pulse of movement will propagate back into the rod, but much will be carried away by the ball.

#### manu3d

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##### Re: Conservation of Momentum, at an atomic level
« Reply #5 on: 20/08/2013 12:56:06 »
Dear Alan and Evan, thank you for your replies!

By the sound of it I'm not going to get an idea of why the bonds end up storing exactly 9/10th of motion of the projectile without getting into the specifics of quantum mechanics. Any lead on what to look for in this context? I.e. specific terms to search on google, tutorials or papers on the matter? There's plenty on quantum mechanics but I cannot find specific information on what happens during short events such as molecules bouncing on each other or joining together.

On the other hand, this thread made me think a little about the numbers of my simple example. The momentum of projectile and rod must be conserved from before to after the impact. Assuming Mp = 1 and Vp = 1 (p = projectile) and assuming that the impact is perfectly elastic, the projectile stops dead on its track and as Mr = 10, Vr = Vp / 10 = 0.1 (r = rod), that is the Rod is set in motion at 1/10th the speed of the projectile. -However-, if we look at the balancing of the kinetic energy something is missing. The projectile -before- the impact has a kinetic energy of 1/2 * Mp * Vp^2 = 0.5 * 1* 1^2 = 0.5. The Rod -after- the impact has a kinetic energy of 1/2 * Mr * Vr^2 = 0.5 * 10 * 0.1^2 = 0.05 or 1/10th of the projectile. In short, 9/10th of the kinetic energy is "missing". I guess this is proof what you Alan were saying about kinetic energy going into internal vibrations?

Thanks again for your help!

#### evan_au

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##### Re: Conservation of Momentum, at an atomic level
« Reply #6 on: 21/08/2013 11:02:59 »
I don't think you need to go to quantum mechanics to solve this problem. The behavior of conservation of momentum during collisions is apparent in macro-scale objects, so it does not relay on quantum effects.

It should be possible to simulate the behavior during collisions by dividing the objects into finite-sized elements which are much larger than an atom or molecule - or even a bacterium. These elements each have a size, mass and elasticity, but can ignore quantum effects. These elements are effectively connected by springs, as suggested by Alan.

Solutions to such problems are easily solved by matrix methods, if you have access to a supercomputer. But even today's PC is as powerful as a previous generation of supercomputers, so this should be quite feasible with some considerable programming effort.

See: http://en.wikipedia.org/wiki/Finite_element_method

#### yor_on

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##### Re: Conservation of Momentum, at an atomic level
« Reply #7 on: 21/08/2013 21:13:12 »
Yep, internal vibrations is the way to go as I gather it.

#### JP

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##### Re: Conservation of Momentum, at an atomic level
« Reply #8 on: 22/08/2013 05:06:58 »
One answer is basically that most forces look like springs when you can only displace particles a little bit, so the suggestions about springs was correct.  If you want to Google for this look for "harmonic oscillator approximation for atomic bonds" or something like that.  This is the approximation that treats bonds like springs.  In the rigorous quantum mechanical description a lot of non-intuitive things will happen, since the particles don't behave like classical particles and the bonds don't behave like springs, but in many cases, after you work through all the math, you find that the answer is very close to the simple spring model.

#### yor_on

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##### Re: Conservation of Momentum, at an atomic level
« Reply #9 on: 22/08/2013 23:21:32 »
http://en.wikipedia.org/wiki/Damping

I don't know JP? As I get it a lot of energy disappear in internal vibrational modes?
Am I wrong?

#### JP

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##### Re: Conservation of Momentum, at an atomic level
« Reply #10 on: 23/08/2013 00:30:38 »
Sure, but that's conservation of energy, not momentum.  Usually we can't experimentally keep track of the energy in those modes (they correspond primarily to energy lost as heat) so we can't use conservation of energy to solve the problem.  Conservation of momentum can be used even in those cases.

#### evan_au

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##### Re: Conservation of Momentum, at an atomic level
« Reply #11 on: 23/08/2013 10:45:03 »
The energy that rebounds into the rod after the ball has left will bounce around inside the rod until it eventually dissipates as heat.

#### yor_on

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##### Re: Conservation of Momentum, at an atomic level
« Reply #12 on: 23/08/2013 12:53:53 »
Thinking of it that toy comes to mind, you know, the one with small metal balls hanging from strings, you setting the utmost in motion, letting it hit the near-most, starting a chain reaction. A sweet example both of momentum (transfer) as well as of dampening to my mind.

#### manu3d

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##### Re: Conservation of Momentum, at an atomic level
« Reply #13 on: 24/08/2013 00:01:47 »
evan_au, yor_on, JP, thank you all for your latest contributions to this thread.

evan_au, you mention:
Quote
I don't think you need to go to quantum mechanics to solve this problem. The behavior of conservation of momentum during collisions is apparent in macro-scale objects, so it does not relay on quantum effects.

You are right of course that conservation of momentum can be dealt with at a macroscopic scale, but you might remember that my original line of enquiry is about how that law emerges in a situation (the universe) in which (as far as I can tell) entities such as molecules are just human constructs. The law emerges at the level of atoms and bonds and that is the level that I am failing to grasp. I.e., if I wanted to write a simplified simulator calculating the motion of each atom given the setup I described, the only way to do it at this stage would be to sum, upon impact, the masses of the atoms involved in the two systems, the rod and the projectile. Alternatively I could think to the bonds as springs, as it has been suggested, but then, how would one set the characteristics of the springs so that the resulting vibrations would store the right amount of kinetic energy and leave the rod as a whole with the right net speed? (this might sound confusiong: that's because -I am- confused).

JP, thank you for your suggestion regarding the harmonic oscillator. Even the simplest case (a single atom projectile impacting a molecule made of just two atoms) mystifies me. I can see the projectile impacting the first atom, the first atom being given the full momentum of the projectile, the bond compressing to then transfer -part- of that momentum to the next atom while reflecting back some of it to the first atom. End result: two atoms now vibrating in a way to store (at any given moment!) half of the projectile's kinetic energy AND travelling together at half the speed of the projectile. I can imagine the atoms vibrating with a 90 degree phase difference, so that their combined, vibrational kinetic energy always adds up to the same value, but I don't quite understand how that value emerges.

yor_on, I don't easily see the connection between the issue we are discussing and damping. In a theoretical, "perfect" setup like I've described there wouldn't be any damping. The internal oscillations would continue ad infinitum. =?

#### JP

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##### Re: Conservation of Momentum, at an atomic level
« Reply #14 on: 26/08/2013 18:52:00 »
The internal vibrations maintain conservation of momentum because changes in momentum are due to forces acting on particles, but two particles connected by a bond (or spring) will exert equal and opposite forces on each other.  So if one feels a force pulling it to the left and giving it momentum in that direction, the other is getting an equal and opposite pull to the right, giving it equal and opposite momentum.  Even though there are vibrations of the atoms within their structure, these vibrations are caused by interactions with each other, which means that they'll cancel out.  The only momentum which doesn't cancel out is that which came from outside the structure--the ball that came in and collided with the system initially.

Even though this complicated internal vibrational motion doesn't violate conservation of momentum, it DOES cause kinetic energy conservation to fail.  As the bonds stretch and bend, they convert kinetic to potential energy, so at any instant in time after the collision, we don't know how much kinetic energy the whole system has.

#### evan_au

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##### Re: Conservation of Momentum, at an atomic level
« Reply #15 on: 26/08/2013 22:29:54 »
To imagine how an atomic bond can be represented as two weights, joined by a spring:
• Most of the mass is in the two nuclei
• The electron cloud around each nucleus has lowest energy when the nucleus is in the center of each electron cloud
• Because the atoms have a chemical bond, the electron cloud has lower energy than the same atoms separated in space. It takes more energy to pull them apart
• However, if the two electron clouds (or their nuclei) get too close together, they repel each other. It takes more energy to push them together
• The lowest energy occurs at the nominal bond length
• There is a theorem in mathematics that if you look at a small range of a curve, it can be approximated a straight line of a certain slope. This slope in the force vs distance curve between the nuclei can be regarded as a spring constant
• Two masses coupled by a spring acts as a harmonic oscillator
• Coherent pressure waves can propagate through a solid made of these atomic bonds. However, they can be scattered by changes in the density or chemical structure of the material (ie different masses of nuclei or springs of different stiffness joining them). If the range of extension is large, the linear approximation is not so accurate and wave propagation is not so effective. All these will turn a coherent pressure wave into random vibrations.
• Random vibrations are felt as heat.
This does not work for very large changes in separation:
• If you crush them too close together, you could get nuclear fusion
• If you pull then too far apart (eg with an electric arc), you end up with a plasma

See: http://en.wikipedia.org/wiki/Covalent_bond (there are other types of bonds too, such as in metals and ionic solids...)

#### yor_on

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##### Re: Conservation of Momentum, at an atomic level
« Reply #16 on: 28/08/2013 13:48:43 »
Assuming that you have a kinetic energy transfered through the balls, disappearing, where did it go? And why? I think of it as the energy disappearing in transformations internally, as into heat etc. You could assume that it has to be taken up by molecules surrounding those balls, but if we put the contraption inside a vacuum chamber then? No molecules surrounding it any longer. And the way it acts must be oscillating internally, taking itself out one way or another. (If I think of it as 'matter waves' acting it also seems possible to assume that they either can quench, or amplify, each other?)

As for something (internal oscillations) would continue ad infinitum. Hmm :) What comes to mind is a perpetuum mobile. Although 'energy' is presumed to be conserved it also transforms, I think that is a rule too. And in the sense of the energy disappearing from the balls, what else can it be?

#### JP

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##### Re: Conservation of Momentum, at an atomic level
« Reply #17 on: 28/08/2013 15:52:50 »
Yor_on, the basic model of balls connected by springs being hit by an outside force (another ball) doesn't have energy losses.  The springs stretch which causes the balls to oscillate/vibrate, and this means that at any instant, some energy is kinetic (in the motion of the balls) and some is potential (in the stretch of the springs).  In most practical problems, we can only measure the bulk properties of matter, so we can't keep track of the motion of individual balls and the springs.  We group this into "internal energy" of the object.

Conservation of energy always holds, so you always have:
Change in Energy = Energy Flow into or out of System.

In this case, we have Internal Energy = Kinetic Energy of all Balls+ Potential Energy in all Springs

If we could calculate this, and we assumed all the balls in the large object were initially at rest with unstretched springs, we'd get:

Kinetic Energy of Sole Ball, Initially = Kinetic Energy of Sole Ball After Collision + Kinetic Energy of all Balls After Collision + Potential Energy in all Springs After Collision.

The problem is that in regular matter, we can't measure the stretches of bonds or motions of individual particles, so this equation doesn't help us (we don't know two of the items on the right-hand side).
« Last Edit: 28/08/2013 15:57:23 by JP »

#### The Naked Scientists Forum

##### Re: Conservation of Momentum, at an atomic level
« Reply #17 on: 28/08/2013 15:52:50 »