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Where does the spring force come from?

  1. Mar 10, 2006 #1
    We know that atoms are held together by atomic bonds and when one atom is moved away from the one it's bonded to, there is a restorative force that 'pulls' it back. But how can we explain the nature of this force in terms of the electrostatic/electromagnetic forces between charged atomic particles (proton & electron)? In a covalent bond an electron is shared between two atoms (i.e. its wavefunction is spread over a wider area), but how does that explain the restorative force? If I move one atom away from another, I'm moving the positively charged nucleus away from this shared electron, so the electrostatic force between the nucleus and the electron should only get smaller, not larger (as hooke's law predicts).
     
  2. jcsd
  3. Mar 12, 2006 #2
    I always thought it was a property of the structure of the material that it absorbs energy(kinetic) of a force directed at it and depending on the material it will release that energy back in an effort to return to its original structure.
     
  4. Mar 12, 2006 #3
    Initially em and qm forces are ballanced. When we stretch a material in a certain direction the distances between -ve and + ions become slightly bigger than the equilibrium case. This extra distance (space) will become filled with a static field. (Perhaps also a magnetic one which for simplicity i'll ignore).
    (E- field)^2 X space is energy. We will have to supply this energy.
    There's a limit to this process when on bigger distances (say a couple of Angstroms) fields from different ions from the same direction are starting to merge. At that point we are starting to form 2 new surfaces.

    Where's the spell checker?
     
  5. Mar 12, 2006 #4
    I realise i’ve overlooked your main argument.

    When we separate the 2 plates of a (charged disconnected) capacitor, initially the attractive force is not getting smaller. It only starts to decrease when fringe effects need to be taken in account.

    In my view, one of the reasons that the total restorative force is getting bigger is that at first upon stretching not all +ve and –ve ions in one cross section will be effected. The number of restoring force dipoles will go up when stretching increases.
     
  6. Mar 12, 2006 #5

    Ich

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    No, it´s more like in your first post: Initially, the forces are counterbalanced. When you one atom away, the counterbalancing qm force necessarily decreases faster than the em force. As long as you don´t pull too much, this gives a linearily increasing net force. When you pull too much, the decrease in em force becomes significant an the spring breaks.
     
  7. Mar 13, 2006 #6
    OK Ich good point.
    What about the above mentioned wavefunction? When we store more energy in this system will the PE and KE of the restoring electrons increase and therefore will the spread become more localised?
     
  8. Mar 13, 2006 #7

    Ich

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    This link explains the variation of bond lenght with energy (=Thermal Expansion Coefficient).
     
  9. Mar 13, 2006 #8

    vanesch

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    This is a purely quantum-mechanical effect, which cannot be understood in classical terms. What you have to do, is to solve the quantum-mechanical problem of the electron eigenfunctions for a given separation of the nucleae, say, distance R. That means, for a given R, you place two nucleae at this distance, and you calculate the possible wavefunctions of the (relevant) electron system (in other words, the binding electrons). There will be a "lowest energy" state (the ground state), with energy E0, and there will be excited states, with energy E1, E2... (the eigenstates of the Hamiltonian for this system). But remember that we placed the nucleae at distance R ; if we had chosen a distance R', we'd have found different values E0', E1', E2'... for the energy eigenvalues. So in general, the energy eigenvalue is a function of R: E0(R), E1(R), E2(R)...

    Now, you can plot these values as a function of R (the typical potential well curves), and if E0(R) shows a minimum for a certain value R = R0, then this will be the bond length. The step from E0(R0) to E(R-> infinity) is the bond strength (well, almost! I'm putting something under the carpet here). The pit of the potential curve around E0(R0) looks like a parabola in this case (it is a smooth minimum), hence this looks a lot like the potential curve of a spring.
    The other curves E1(R), E2(R) ... do not need to show a minimum: they are called non-binding states ; but they could: we then have an excited binding orbital. If E0(R) also doesn't show a minimum, then there is no possibility for a chemical bond.

    cheers,
    Patrick.
     
  10. Mar 13, 2006 #9
    I can just about follow that. So qm includes both the attractive em force and the repulsive force due to wave mechanics? You see i’m still trying to hang on to some kind of visual picture.

    Patrick cheers to you too, especially on friday.
     
  11. Mar 13, 2006 #10

    vanesch

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    It is tricky to try to picture quantum-mechanical issues in too classical a way. The simple response is that it turns out that the ground state of the quantum-mechanical problem of, say, 2 protons and 2 electrons, is a bound state (a H2 molecule). What I tried to sketch was a semi-classical picture, where the electrons are treated purely quantum-mechanically, but the protons are treated classically, assuming that the protons react so slowly that the electron system settles in a stationary state (say, the ground state) almost immediately before they move appreciably. This is possible because of the big mass difference between protons and electrons, and the technique is actually called the Born-Oppenheimer approximation.
    As such, in the "classical" proton system, there appears a "potential curve" (which is the parametrized eigenvalue E0(R)), which you can take as a genuine potential curve for the classical proton system. Who says "potential" then says "forces", and these forces are difficult to explain. Call them "quantum forces" if you want. They are the result of the electromagnetic interaction in the given quantum mechanical setting, and are different from the classical electromagnetic interaction you would find between 4 classical point charges of course. Does this mean that there is a "new force" in nature ? No, it is an artefact from our approximation scheme. It's a bit like the centrifugal force in a rotating reference frame. If you insist on working in a rotating reference frame, you have a centrifugal force, but it is not an "interaction" between particles. In the same way, if you insist on a semiclassical picture, "quantum forces" appear to act on the protons.

    It is even funnier: we can now quantize the proton system with this "classical potential" E0(R). We will then solve essentially a harmonic oscillator (with some deformations), and the solution to that quantum problem will give you the vibrational states of the molecule.
    The entire system of electron states and proton vibrational states is then a good approximation to the entire quantum-mechanical problem we should in fact have considered from the start (that's the other half of the Born-Oppenheimer approximation, and I think it was Feynman who wrote down a proof that this technique works indeed).
     
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