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Hooke's Law on a microscopic level

  1. Jan 27, 2014 #1
    Hooke's law states that the force required to stretch/compress a spring is proportional to the distance stretched. Meanwhile, electromagnetic interactions between particles obey an inverse-square law with respect to distance. So, if as a spring is stretched, it's composite particles get farther apart from each other, why does the force required to stretch it increase?

    I know that Hooke's law is only an approximation, but it works quite well. What goes on at the microscopic level which keeps the increased distance between particles from reducing the attractive force between them? If there is a quantum mechanical answer which reveals something special about chemical bonds, I can accept that I am too ignorant of that field to understand the answer as of yet.
  2. jcsd
  3. Jan 27, 2014 #2


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    The deal is that in small distances, the hook's law appears.
    If for examply you have two particles interacting in a distance [itex]a[/itex], and you make a small displacement [itex]r[/itex] you'll get for the potential:
    [itex] V(r+a)= V(a)+ r ( \frac{∂V}{∂r} )_{r=a} + \frac{r^{2}}{2} (\frac{∂^{2}V}{∂r^{2}})_{r=a}+O(r^{3})[/itex]
    Now if at your initial distance everything was stable, [itex]V(a)[/itex] is just a constant, the [itex](\frac{∂V}{∂r})_{r=a}=0[/itex] because it was a stable that point, and you only have the 2nd derivative term...

    [itex] V(r+a)= V(a)+ \frac{r^{2}}{2} (\frac{∂^{2}V}{∂r^{2}})_{r=a}+O(r^{3})[/itex]
    So everything, no matter what kind of force you have, at small displacements works like the Hook's law (harmonic oscillator): the potential has the [itex]r^{2}[/itex] dependence.
    Last edited: Jan 27, 2014
  4. Jan 27, 2014 #3
    This doesn't explain the fact that a larger force is required to keep a spring stretched at greater length, does it? Two particles attracted to each other are still easier to pull apart when they are far away from each other, but a spring is opposite that.
  5. Jan 27, 2014 #4


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    But in the string the particle of the one edge does not interact with the particle on the other edge... each is interacting with their neighbors in the way I explained.
  6. Jan 27, 2014 #5
    The attractive force is not the only force acting on the particles.
    You cannot have an equilibrium with attraction only.
    If the atoms get too close to each other there is a repulsion force.
    The equilibrium distance between particles is given by the balance of the attraction and repulsion.
    If you stretch the crystal, the distance between particles increases and both attraction and repulsion force decreases. But the repulsion decreases much faster with distance so the net effect is an attraction towards the equilibrium position.

    See. for example. Van der Waals or ionic crystals, for specific examples of how the forces depend on distance.
    http://physics.unl.edu/tsymbal/teaching/SSP-927/Section 03_Crystal_Binding.pdf
    Last edited: Jan 27, 2014
  7. Jan 27, 2014 #6


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    Yes. Each particles sits in a 'potential well' and the restoring force for small perturbations (together or apart) is proportional to the displacement. Billions of atoms (in line), each one moving by minute distances,means that the restoring force is linear with overall large distortion of the bulk metal.
  8. Jan 27, 2014 #7
    This makes good sense.
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