Can we derive Hooke's law from the theory of elasticity? I know it is not a fundamental law and therefore can be derived from more basic considerations.


2 Answers 2


Yes, we can derive Hooke's Law from more basic continuum conditions, provided that the material be stable and at equilibrium, so that the strain energy is smoothly minimized with respect to the distance between atoms. (This energy is sometimes called the pair potential and is modeled using functions such as the Lennard-Jones potential.)

Consider just a pair of atoms. For slight positional deviations $\delta$ (corresponding to the small-strain assumption of linear elasticity) around the smooth energy minimum $E(0)$, the energy $E$, regardless of its true functional form, can be expanded using a Taylor series:


where the prime notation denotes derivatives with respect to position. Now let's set our energy reference to $E(0)=0$, note that $E^\prime(0)=0$ because we're at an energy minimum, and drop all but the very next term. This gives


which describes the energy of an idealized spring with spring constant $E^{\prime\prime}$ and displacement $\delta$; the derivative of this equation with respect to position provides the restoring force, which is

$$F(\delta)\approx E^{\prime\prime}(0)\delta.$$

Now define stress as $\sigma=F/A$ and engineering strain as $\epsilon=\delta/L$ and we have $$\sigma=\frac{LE^{\prime\prime}(0)}{A}\epsilon$$ or $$\sigma=Y\epsilon,$$

which is simple Hooke's Law for a corresponding elastic modulus $Y$.

  • $\begingroup$ You actually derived Hooke's law from Lennard-Jones potential and then you derived elasticity from Hooke's law. $\endgroup$
    – Diracology
    Commented May 29, 2018 at 18:53
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    $\begingroup$ I specifically noted that one does not need to use the Lennard-Jones potential or any specific potential; it's sufficient to assume that the substance is initially at a stable equilibrium. That's the beauty of a Taylor series expansion; it proves that every smooth minimum looks like a parabola from up close. A parabolic energy response is equivalent to a spring-like response. $\endgroup$ Commented May 29, 2018 at 18:59
  • $\begingroup$ That derivation works for materials that doesn't have any imperfections. Like perfect crystals. $\endgroup$
    – nicoguaro
    Commented May 29, 2018 at 19:10
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    $\begingroup$ It works pretty well until vacancies/impurities start to make up a fair fraction of the atomic volume fraction. Small amounts of imperfections in crystals generally have much more of an effect on strength than on stiffness. $\endgroup$ Commented May 29, 2018 at 19:36
  • $\begingroup$ Great derivation! I have problems with the step "... provides the restoring force". You do not seem to differentiate w.r.t. the position, otherwise. $E''(0)$ would get an additional prime. Instead, you seem to divide the equation by $\frac{1}{2}\delta$ to get $$\frac{2E(\delta)}{\delta} \approx E''(0)\delta$$ at the LHS and then use $$\frac{2E(\delta)}{\delta} = F(\delta) \tag{$\star$}$$. But why does this last equation $(\star)$ hold? $\endgroup$ Commented Nov 15, 2022 at 13:00

I like this particular derivation because it avoids the requirement to actually specify inter-atomic forces, such as the coulomb force and the Pauli force. Technically, however, you will have one relationship for compression and another for tension and therefore two elastic constants. On the macro-scale this method will yield the classic work and virtual work functions structural engineers are so familiar with. This is an excellent derivation.

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    $\begingroup$ This does not provide an answer to the question. Once you have sufficient reputation you will be able to comment on any post; instead, provide answers that don't require clarification from the asker. - From Review $\endgroup$
    – jng224
    Commented Nov 24, 2021 at 14:42
  • $\begingroup$ It's not true that the derivation yields one relation for compression and another for tension. As I noted elsewhere, every smooth minimum looks like a symmetric parabola up close. $E^{\prime\prime}(0)$ provides a stiffness for both compression and tension. And indeed, we experimentally measure the same elastic modulus for small deformations of either contraction or elongation. $\endgroup$ Commented Oct 25, 2023 at 3:54

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