Similar questions have been already asked here and here, but I'm still confused. I think of a wire of a material, say aluminum, as formed by stacking layers of atoms. Suppose that the distance between layers at equilibrium is $r_0$, and that $\Delta r := r-r_0$ is the displacement from equilibrium. I suppose that the restoring force between two consecutive layers is $F = -K'\Delta r$; I neglect the interaction between distant layers.

I assume that $K'$ depends on the area $A$ of the layers as $K' = KA$. Hence, the force needed to stretch the wire is $F_t = KAN\Delta r$, where $N$ is the number of layers. We have that the initial length of the wire is $L_0 = Nr_0$ and the extension is $\Delta L = N\Delta r$. Therefore, Young's modulus $E$ should be $E = (F_t/A)/(\Delta L/L_0) = L_0K$.

In my derivation $K$ is the fundamental quantity, so $E$ should depend on the length of the sample, but I know that $E$ is the intrinsic property of the metal. What am I missing? It's hard to imagine that $K \propto \frac{1}{L_0}$.

Think of two wires of aluminum, the transversal area of both 1 mm$^2$, the length of the sample A 10 cm, and the length of the sample B 1 m. Suppose we stretch both by 1 cm, so $\Delta L/L_0 = $ 0.1 (A) and 0.01 (B). Since $E$ is constant, I have to apply more pressure to stretch (A); however, if $K$ were the fundamental quantity the force should be equal in both cases. The reason for the latter is that, even though the strain is different, in (A) the number of layers opposing to the force is fewer but $\Delta r/r_0 = \Delta L/L_0$ is larger, and in (B) the number of layers opposing to the force is larger but $\Delta r/r_0 = \Delta L/L_0$ is smaller; both effects offset, and the force is the same.

I'd appreciate any clarification.


1 Answer 1


The forces in the layers do not add up. Essentially, you have a chain of springs - if you stretch the chain on both ends, the force in each spring is the same. Otherwise, the point between the first and the second spring would be puled towards the centre, since the force in that direction would be $N-1$ times the one pulling towards the end.

Forces do add up when the springs are in parallel -- you have correcty taken that into account by assumeing the force is proportional to the area.

  • $\begingroup$ To compute the force my reasoning was: the potential energy of the configuration is $V = \frac{1}{2}\sum_{\textrm{layers}}K'(\Delta r)^2$, so shouldn't be the magnitude of the force to stretch it $|F| = |V'| = \sum_{\textrm{layers}}K'\Delta r$? My physics is slippery, but your suggestion sounds fine, I get $F_t = KA\Delta r$, and so $E = Kr_0$, is it right? $\endgroup$
    – user90189
    Apr 2, 2020 at 14:11
  • 1
    $\begingroup$ @user90189 right. If you chain several springs, the combined displacement-based spring constant (derived from $F/\Delta l$ gets smaller, but the strain-based modulus ($F/\epsilon$) stays constant $\endgroup$
    – Toffomat
    Apr 2, 2020 at 17:34

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