# How can Young modulus depend on the geometry of the item & Lennard Jones (question related to an older one)

In the question Young's modulus and geometry of test material, the answer was that it cannot play any role. Nonetheless, for the Lennard-Jones solid, in many places the Young modulus is given as : $$Y=\frac{dF/A}{dr/r_0}=-\frac{r_0}{A}\frac{d^2 U}{dr^2}$$, which seems to depend independently on $$A$$, which is the area on which the force is acting, and $$U(r)$$ which is the Lennard-Jones potential. How can that be ?

• The constant A in the right side is not the area but one of the constants that define the L-J potential. – thermomagnetic condensed boson Nov 19 '18 at 15:23
• @coniferous_smellerULPBG-W8ZgjR It's not; please see my answer. – Chemomechanics Nov 19 '18 at 20:18

## 1 Answer

As you note, Young's modulus is a property of the material and shouldn't depend on the cross-sectional area. Implicit in your derivation of $$Y$$ from the L-J potential, however, is that you're applying a force to change a single bond length (this force is $$dU/dr$$). But Young's modulus doesn't directly describe forces; it couples the strain with the stress. To turn the force into a stress, we need to divide by an area ($$A$$ or $$r_0^2$$).

This may seem odd because you say that the stiffness seems to depend on the potential and on the test material's geometry, but this framing isn't quite accurate, is it? The stiffness depends on the (1) curvature of the potential with respect to the bond spacing and (2) the bond spacing itself. Any geometry concerns are at the bond scale, not the test sample scale.

• thx ! So the actual Young modulus reads : $Y=-\frac{1}{r_0}\frac{d^2U}{dr^2}$ ? – J.A Nov 20 '18 at 9:26
• If you idealize the crystal as simple cubic. – Chemomechanics Nov 20 '18 at 13:23