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Why do materials extend proportionally to the force exerted on them (Hooke's law)? I thought that

  1. when materials are compressed or extended under force, their atoms become closer or further apart;
  2. the inter-atomic forces are essentially electrostatic;
  3. electrostatic forces (and indeed most other forces) follow an inverse-square law;

So I would naively conclude that springs should follow an inverse square law. But clearly in most situations the law is linear. Where is my logic flawed?

Qmechanic
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Sanjay Manohar
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    Hooke's law is an idealization or approximation for potentials near their equilibrium value. It is not a fundamental law. – BioPhysicist Jul 22 '18 at 13:48
  • Sure - but the question is, why does approximate to a linear, as opposed to an inverse square law? – Sanjay Manohar Jul 22 '18 at 13:52
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    So you are asking why the entire system has some equilibrium position that we can Taylor expand the potential about to get a quadratic approximation of that potential? – BioPhysicist Jul 22 '18 at 14:15
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    Possible duplicates: https://physics.stackexchange.com/q/408987/2451 , https://physics.stackexchange.com/q/159021/2451 and links therein. – Qmechanic Jul 22 '18 at 15:16
  • In the limit of very small changes of a variable, just about all continuous functions show a linear dependence with respect to the variable. Try doing a back-of-the-envelope calculation of how much you think that inter-atomic distances change when a typical spring is compressed. The percentage change is really, really small. –  Jul 22 '18 at 16:43
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    Possible duplicate of Can Hooke's law be derived? – M. Enns Jul 22 '18 at 21:09
  • @SanjayManohar The inverse square law has no equilibrium point. Point charges beginning at rest in such a potential are accelerated and travel to infinite distance. – Ryan Thorngren Jul 22 '18 at 23:56

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When one considers a relatively small segment of a curve, most appear approximately linear when one zooms in far enough. Most materials are not truly linear when stretched a considerable distance, but for small deviations about the equilibrium some are approximately linear. This can be seen when one considers Taylor Series expansions of functions.

The inverse square law for electrostatics can be expanded as a Taylor Series as well, and it contains a linear component as can be seen here:

$$\frac{1}{(r-r_0)^2}=\frac{1}{r_0^2}\left[1+2\left(\frac{r}{r_0}\right)+3\left(\frac{r}{r_0}\right)^2+4\left(\frac{r}{r_0}\right)^3+\cdots\right]\,.$$

For small deviations $\Delta r\approx0$, the inverse square law is approximately linear since the higher-order terms $\mathcal{O}[(\Delta r)^2]$ fall off very quickly. This can be seen using the Binomial approximation $(1+x)^n\approx1+nx$ where $x\approx0$. In this case we have:

$$(\Delta r-r_0)^{-2}=\frac{1}{r_0^2}\left(1-\frac{\Delta r}{r_0}\right)^{-2}\approx\frac{1}{r_0^2}\left[1-(-2)\left(\frac{\Delta r}{r_0}\right)\right]=\frac{1}{r_0^2}\left[1+\frac{2\Delta r}{r_0}\right]\,.$$

Of course this is far too simple to describe the more complex material effects of crystal lattices being stretched from equilibrium, but this is the general idea of how many materials can exhibit linear relationships between force and distance for small deviations from equilibrium.

BoisterousLemma
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  • Thanks - amazing - so Hooke's law is actually incredibly general, because any force-vs-distance relation will look linear over tiny displacements -- which it will be if there are a very large number of atoms. It works more like a 'law of large numbers'... – Sanjay Manohar Jul 23 '18 at 15:21
  • is r extension or absolute length. If it is absolute length then why do the higher order terms drop off, r/r0 even fro small delta r will still be aproximately 1. I understand why they drop off is r is delta r (i.e. extension).? – MooseBoost Jan 28 '20 at 12:58