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Let's assume a setup with a static linear molecule with three identical atoms connected by bonds and a single atom, identical to the other three, being shot at the molecule. Let's also assume that everything happens in 1D and can therefore be illustrated by this simple scheme:

(projectile) A -> A-A-A (molecule)

Where the direction of the arrow is the direction of motion of the projectile and the dash signs in the molecule are the bonds between atoms.

[EDIT - must change next paragraph as the description does not fit what would happen: stand-by, no time to do it right now!]

Assuming a low-energy, non-relativistic, perfectly elastic collision, the projectile will stop dead in its track while the molecule will be set in motion at 1/3rd of the projectile's speed, as it has 3 times the projectile's mass (conservation of Momentum). Balancing also the equations for the Kinetic Energy easily reveals that 2/3rds of projectile's kinetic energy must have been locked away in internal vibrations of the molecule.

Now, it's easy for us to calculate these numbers by considering the projectile and the molecule as just two separate systems and conveniently attaching a total mass to each. But in reality the universe doesn't group atoms into molecules, does it? Each bond in the setup doesn't "know" about the existence of the other bond and each atom could be said to be unaware of all but the atoms bonded or colliding with it.

In this context, not "knowing" total masses, how does the universe "calculates" that in this case the molecule must store 2/3rds of the projectile kinetic energy into internal vibrations, no more, no less?

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  • $\begingroup$ ""Assuming a low-energy, non-relativistic, perfectly elastic collision,"" means that the molecule will store nothing at all, but the single atom impactor will recoil carrying away part of the engergy. Before starting questions resting on wrong assumptions, read about elastic collisions. $\endgroup$
    – Georg
    Aug 25, 2013 at 18:28
  • $\begingroup$ Georg, I've read about elastic collision but I might still be making mistakes. Are you saying that my description of what happens is not correct? Are you saying that the projectile will not only stop after impact but also bounce backward? I have in mind some experiments we did in the physics lab during high school (20 years ago!) on an air-rail, and I remember little carriages stopping on their track, not bouncing back. But I could be wrong. $\endgroup$
    – manu3d
    Aug 26, 2013 at 20:05
  • $\begingroup$ Yes of course! The impactor will stop dead only if both masses are equal! If a ball hits the ground, does the earth fly away? $\endgroup$
    – Georg
    Aug 28, 2013 at 10:15
  • $\begingroup$ Ah,thanks for reminding me about the case where that happens. That helps: I will reformulate the question in light of it. Thankfully, the specific example is not the focus of the question. $\endgroup$
    – manu3d
    Aug 30, 2013 at 14:12

1 Answer 1

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The universe doesn't calculate anything. Energy and it's conservation and everything that follows from it is a human-made concept, a physical model. It describes the things that happen in nature quite well (in the right circumstances), but it is only a model. In how far this lends some kind of "reality" to the concept/model I couldn't say. It's more of a philosophical question. Physical laws only describe how nature works, but not why it works the way it does.

Edit:

Concerning the physical content of the collision you described: I think of it in the following way: The projectile hits the first atom of the molecule, let's call it A1, and transfers its momentum to it. Now, this atom moves in the direction along the bond axis directly towards the next atom, A2. While it does so, it has to work against the spring force of the bond between A1 and A2. Thus, kinetic energy of movement is converted to kinetic energy of vibration. A2 will now be accelerated in the direction of the next atom, A3, and it has to work against the spring force of the bond between A2 and A3 and(!) the restoring force brought about by the bond between A1 and A2. Again kinetic energy is converted to vibrational energy. But now you have a system of coupled oscillators. This process will be repeated for every atom in the molecule until every bond vibrates. So, it is not true that

Each bond in the setup doesn't "know" about the existence of the other bond and each atom could be said to be unaware of all but the atoms bonded or colliding with it.

They are coupled through the spring forces of the bonds and thus each atom "feels" the mass of every other atom in the molecule as well as the bond strength of every bond in the molecule.

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  • $\begingroup$ Thanks for your answer Philipp. My suspicion is that there is an answer at a quantum-mechanical level, or perhaps even at a classical level thinking bonds as springs with oscillations partially cancelling each other. But I can see how we are otherwise potentially close to the edge of metaphysics and therefore philosophy. Let's stick to the physical world then, and let's see if anything comes out. $\endgroup$
    – manu3d
    Aug 22, 2013 at 22:50
  • $\begingroup$ @manu3d I added some more physical thoughts to my answer, which I hope are helpful. $\endgroup$
    – Philipp
    Aug 22, 2013 at 23:14
  • $\begingroup$ Dear Philipp, thank you, your later edit is much appreciated. Indeed it is a good qualitative description of how things happen. Clearly the molecule's atoms would like to move at the same speed as the projectile, so much so that if no bonds were involved we'd be dealing with a newton's cradle and the last atom would depart the molecule with the speed of the projectile. What is left unclear is how the individual interactions between single atoms and bonds generate that split 2/3rds (vibration) + 1/3rd (net motion) without "counting" the atoms involved prior to the impact. $\endgroup$
    – manu3d
    Aug 23, 2013 at 12:13
  • $\begingroup$ Concerning your end-of-answer comment, I disagree. Of course all atoms in a molecule "feel" -indirectly- each other, as a movement or impact somewhere in the molecule is likely to be "felt" everywhere else in the molecule. -However-, if you think of a linear molecule 1000 atoms long, unless you invoke some of the strangest (and minimal) quantum effects such as probability waves having non-zero values even at astronomical distances, we can reasonably assert that A1 is highly unlikely to be influenced by A1000 and therefore, for all intents and purposes, doesn't know about it. $\endgroup$
    – manu3d
    Aug 23, 2013 at 12:20
  • $\begingroup$ @manu3d Concerning your first comment: Unfortunately, I can't give you the exact equations on a microscale for this. Maybe one would get to it by solving the equations of motion for a system of coupled harmonic oscillators. This might be a question for physics.SX. Concerning you second comment: I see your point. Maybe I have used the word "feel" injudiciously here, since during the first impact the influence of the other atoms on the movement of A1 will be very small. The "feel" part comes later when the bond starts to swing as then A2 is influenced by A3 and so on. That's how I meant it. $\endgroup$
    – Philipp
    Aug 23, 2013 at 14:32

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