# How local is the stress tensor?

I am confused by the definition of the stress tensor in a crystal (let's say a semi-conductor), I don't see how it could be "more local" than over an unit cell. I know that in field theory the stress tensor, or more precisely the whole stress-energy tensor, is local function, a function of the point $\vec{x}$ but I don't understand the interpretation you can give to shearing a point. Plus, in the lattice case I stressed (pun semi-intended), you shouldn't be able to look at displacements or deformations on smallest scales than your elementary cell?

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## 2 Answers

An interesting question.

You are right, the stress in a crystalline solid, or any solid, is treated by engineers as a macroscopic property of matter assuming matter is a continuous medium. It is given in terms of the external forces acting on the solid per unit area at some direction. Hence the distinction of $\sigma_{xy}, \sigma_{yz}$ etc. This goes with the definition of stress

$\sigma_{ij}={\frac {dF_i}{dA_j}}$

where $dF_i, dA_j$ are Cartesian tensors (not like the general contravariant and covariant tensors in general tensor analysis as in GR.) This derivative is meaningful locally and, as said above, treats the solid as a continuous medium.

In order to define stress in terms of the crystal structure you can relate it via the strain tensor, which can be defined in terms of relative infinitesimal displacements of the atoms in the cell. There are a couple of ways of doing this, one of which is the Lagrangian description. In this, the coordinates $(x_1,x_2,x_3)$ of the atoms in the unrestrained state are taken as the independent variables, while $(u_1,u_2,u_3)$ are the relative displacements of the atoms, and they are the dependent variables. This leads to the following definition for the strain tensor (Langrangian strain)

$\eta_{ij}={\frac 1 2}( {\frac {\partial u_i}{\partial xj}}+{\frac {\partial u_j}{\partial x_i}}+\Sigma {\frac {\partial u_r}{\partial x_i}} {\frac {\partial u_r}{\partial x_j}} )$

where the summation is over the index $r$. Note that this is a function of the coordinates of the atoms in the lattice, so $\eta_{ij}$ is a locally defined quantity. Macroscopically strain and stress relate via Young’s modulus $E$ in the relation $\sigma=\epsilon E$ (Hooke’s law,) which would be ok for an isotropic material, in which direction of application of the force is irrelevant. For anisotropic materials, such as crystalline solids in condensed matter physics, this relation is generalised to the following

$\sigma_{ij}=\Sigma C_{ijkl}\eta_{kl}$

and the summation is assumed over the $kl$ indices. The coefficients $C_{ijkl}$ are the second order ‘elastic constants’ or elastic stiffness coefficients of the material, and they are fourth order tensors, defined as derivatives of the elastic energy of the material, i.e. the potential energy of the atoms in the crystal lattice due to their relative displacements. I think this is what TMS meant in his comment. I hope this helps to understand the notion of locality of stress in crystalline solids.

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You right, you can't, when we defining stress-energy tensor like that we suppose that we have everywhere solid, continuous matter inside the object.

And when we want to incorporate the discrete structure of crystal, we usually doing that in a "smooth" way by using the function of Radial distribution function of atoms in crystal cells.

This can be justified by noting that we are usually interested in the potentials between the atoms when we try to study physical properties of the crystal, not the atom's locations itself, which in its turn can't be determined precisely as you know and makes no sense.

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I don't know the term "radial destitution". Maybe autocorrect played a trick on you? – Rafael Reiter Feb 10 at 9:26
Yes, corrected thx. – TMS Feb 10 at 10:07
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