Skip to main content Skip to navigation

Dislocations 2: Equations of Motion

To understand the motion of dislocations, one needs to understand the mechanics of strain. Suppose one has a volume V with boundary surface A deep inside a material. If the content exerts a force per unit volume $\mathbf{f}$ on the boundary A, then the total force for each component $f_i$ from $\mathbf{f}=(f_1,f_2,f_3)$ will equal $\int_V f_i dV$. If the volume V is static, one must apply an qual and opposite force on boundary A. Let this opposite force per unit area have components $-\sigma_{i,k}$. Then the total force applied to the surface is

$\int_A -\mathbf{\sigma_i}\cdot \mathbf{\nu}dx_1 dx_2 = \int_A \sum_{1 \leq k \leq 3}-\sigma_{i,k}d\nu_{k}$

where $\mathbf{\nu}$ is the normal to the surface at the point of integration. It must be equal to the total force in the volume and by Green’s integral theorem, one can replace the circle integral with a volume integral giving

$\int_V f_i dV = \sum_k \int_V -\frac{d\sigma_{i,k}}{dx_k}dV = \sum_k \int_A -\sigma_{i,k}d\nu_k \Rightarrow \sum_k \frac{d\sigma_{i,k}}{dx_k}+f_i=0$

Now consider the work done by deforming the volume during a thermodynamically reversible process. If the displacement is $\delta u_k$, then the work done is

$\delta R = \sum_i f_i \delta u_i = \sum_i \sum_k \frac{d \sigma_{i,k}}{dx_k}\delta u_i$


$\int_V \delta R dV = \sum_i \sum_k \int_V \frac{d\sigma_{i,k}}{x_k}\delta u_i dV = \sum_i \sum_k \int_A \sigma_{i,k}\delta u_i d\nu_k-\int_V \sigma_{i,k}\frac{d \delta u_i}{d x_k}dV$

by partial integration. If the deformation is local, the first boundary term will vanish as the surface is taken large/far enough. Now the variation $\delta$ is essentially a differentiation in an undetermined parameter so we assume it can be interchanged with the differentiation $\frac{d}{dx_k}$. Instead of summing over $\frac{d u_i}{dx_k}$, one can sum over $\frac{du_i}{x_k}+\frac{u_k}{x_i}$ and take halve its value, so

$\sum_i \sum_k \sigma{i,k}\frac{d \delta u_i}{d x_k} = \sum_i \sum_k \frac{1}{2}\sigma_{i,k}(\frac{d \delta u_i}{d\x_k}+\frac{d \delta u_k}{x_i})$.

Introducing the stress tensor $u_{i,k}=\frac{1}{2}(\frac{du_i}{dx_k}+\frac{du_k}{dx_i})$ (see next paragraph for a motivation), one arrives at $\delta R = \sigma_{i,k} \delta u_{i,k}$. From thermodynamics,

$dE = TdS-PdV = TdS+\sigma_{i,k}\delta u_{i,k}$


dF = -SdT + \sigma_{i,k} \delta u_{i,k}



with F the free energy. Physically, it is clear that if there is no stress, there is no force per area $\sigma_{i,k}$ needed to counter the stress. On the other hand, if the stress $u_{i,k}$ is big, the stress should also be big. Hence one can expand

$F=c_0 + c_1 u_{i,k}+c_2 u_{i,k}^2+...$


\sigma_{i,k} = c_1 + 2c_2 u_{i,k}+...

From the discussion above,


therefore $c_1 = 0$. One can assume Hooke`s law, making the force dependent on stress or deviation from equilibrium upto linear order. Then \sigma{i,k} = 2 c_2 u_{i,k}. Inserting the definition of the stress tensor back in the equation, combining with

$\sum_k \frac{\sigma_{i,k}}{x_k}+f_i=0$

and adopting the Einsitein summation convention in order to drop the sum signs, one obtains

$c_{i,j,k,l}\frac{\partial^2 u_{k}}{\partial x_j \partial x_l} + f_k=0$

If $f_k$ is constant, we have the static case and we can solve the poisson equation. If we want dynamics, we set the force per unit volume equal to $\rho_0 \frac{d^2 u_k}{dt^2}$ with $\rho_0$ the mass density giving the equivalent of newton’s equation of motion. The resulting equation is a wave equation.

Now a short explanation about the stress tensor. Suppose that without strain/no deformation, we can define a coordinate system \{ x_1, x_2, x_3 \}. After applying strain, the material is deformed and the coordinate axes shift so that for any position \{ x_1, x_2, x_3 \} \mapsto \{ y_1, y_2, y_3\}. Define u_i = y_i - x_i. Then for any infinitesimal displacement dx_i in the old coordinates, the new displacement is

dy_i=dx_i+du_i=dx_i+\sum_k \frac{\partial u_i}{\partial x_k}dx_k

and if the total distance used to be dl=\sqrt{dx_1^2+dx_2^2+dx_3^2}, the new distance will be

dl_{new}=dl+\sum_{i,k,l} 2\frac{\partial u_i}{\partial x_k}dx_i dx_k + \frac{\partial u_i}{\partial x_k}\frac{\partial u_i}{\partial u_l}dx_k dx_l=dl+\sum_{i,k,l} 2\frac{\partial u_i}{\partial x_k}dx_i dx_k+ \mathbf{O}(2).

Now we can play the same trick with the summation again summing

\sum_{i,k} \frac{\partial u_i}{\partial x_k} = \sum_{i,k} \frac{1}{2}(\frac{\partial u_i}{\partial x_k}+\frac{\partial u_k}{\partial x_i}) =\sum_{i,k}u_{i,k}

so that

dl_{new}-dl=2u_{i,k}dx_i dx_k

The stress tensor is thus the first-order correction in the length of a line between two points in the material after deformation.