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Forgetting Hooke's law for a minute why, from a microscopic perspective (preferably quantum) on up to a macroscopic one, does a spring under tension exert a force?

I was thinking that there might be an analogy between the high and low pressure states of an air wave and the density of the mass distribution of the compressed and stretched states of an iron spring, but I don't know nearly enough about solid state physics to even guess.

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Because springs represent a macroscopic example of an ubiquitous type of potential - the simple harmonic oscillator. –  crasic Jun 6 '11 at 8:40

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up vote 6 down vote accepted

You could make an analogy between the pressure distribution of a sound wave and the mass density distribution of a realistic spring undergoing vibrations, but it wouldn't give you the explanation you're looking for. As a matter of fact, that would be more like explaining a sound wave in terms of springs, rather than what you're trying to do, i.e. explaining a spring in terms of waves.

Although I'm not intimately familiar with the details, basically what goes on at the microscopic level of a spring is that, when the spring is at equilibrium, the atoms are set in some sort of rigid structure. Any given pair of atoms has a potential energy which is a function of the distance between those two atoms, so the entire spring has a potential energy determined by all the distances between every possible pair of atoms:

$$U = \sum_{i,j} U_{ij}(r_{ij})$$

In equilibrium, the spring will take a shape which minimizes this total potential energy.

If you think about it, a metal spring might typically be formed by heating some metal to make it malleable (or even melting it), and then forming it into the desired shape before it cools. The heat allows the atoms to move around relatively freely so that they can reach the equilibrium configuration that minimizes their potential energy, then once the spring cools, they are frozen in place.

Of course, the atoms are not completely frozen in place. As I see that Georg has already written in his answer, the potential energy between two atoms ($U_{ij}(r_{ij})$) has a minimum at their equilibrium distance and goes up on either side. If you add some energy into the system, say by exerting a force on it, you can get the atoms to move closer together or further apart. When you stretch or compress a string, you are really just doing this to all the (pairs of) atoms in the spring simultaneously. The atoms will, of course, "try" to return to their equilibrium position, i.e. they will "try" to minimize their potential energy, and this is what you feel as the restoring force of a spring under tension.

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Expanding on my comment.

A spring, or anything that can be modeled by a string is any interaction that can be described by a potential that is proportional to $x^2$ the displacement. Why is this potential important? Qualitatively, it is the solutions to this potential (i.e. the motion of a particle that sits in this potential) are the solutions of a simple harmonic oscillator i.e. Sine Waves.

In a very hand wavy approach to this. You can think that to the lowest order the universe is host to 2 ubiqutious potentials - the $1/x$ potential that is responsible for our inverse square laws and the $x^2$ potential responsible for oscillations and harmonic motions. A constant potential will exert no force and an $x^1$ potential will exert a force on an object that is constant at all distances - clearly unphysical.

More specifically, a simple equation to model the potential "felt" by two neutral atoms clsoe to each other is given in the Lennard-Jones potential A graph of which is shown below. This type of potential is somewhat ubiquitous (in one form or another) when taking about intermolecular interactions. There are refinements, expansions, better models, etc. But they all have one thing in common a "minimum" or a "cup" in the potential which represents the stable configuration. A particle/atom/crystal/etc. that exists in that region of the potential is "stable". Nudge it in one direction and it will naturally "fall" towards the low point of the potential.

Why is this important? Well, to first approximation the region near the low point of the potential is effectively a $x^2$ potential - a simple harmonic oscillator. Thus any system that is modeled by this potential will respond like a harmonic oscillator in the limit of a really small nudge. I.e. it behaves like a spring. Due to the magic of ignoring all mathematical rigor, effectively any potential that vaguely looks like the L-J potential, in fact any potential that has a local minimum like such, will to first order look like a $x^2$ potential in the small region around the minimum.

enter image description here

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I would appreciate an explanation for the downvote –  crasic Jun 6 '11 at 11:14
    
This is 1st for rechurning, second for nonsense of harmonic oscillator. For elastic behaviour the exact form of the potential is irrelevant. ""like a harmonic oscillator in the limit of a really small nudge. I.e. it behaves like a spring."" Not all elastic materials follow Hookes law. (Which should be forgotten here, remember? –  Georg Jun 6 '11 at 11:40
    
@Georg Provided that the above graph is of the potential between two particles and that it has a continuous 2nd derivative at the minimum, which is the rest point, it follows Hookes law. Now, I don't personally understand the harmonic oscillator part and find the $x^2$ to obviously contradict the graph, but that's not my area. –  AlanSE Jun 7 '11 at 14:13

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