plane unit vectors normal and tangential to the dislocation line. From equation follow $${\boldsymbol{\nu}}=\frac{1}{\sqrt{l_{,1}^2+l_{,2}^2}}(l_{,1}\mathbf{e}_1+l_{,2}\mathbf{e}_2), \quad {\boldsymbol{\tau}}=\frac{1}{\sqrt{l_{,1}^2+l_{,2}^2}}(-l_{,2}\mathbf{e}_1+l_{,1}\mathbf{e}_2),$$ where the comma before an index denotes the partial derivative with respect to the corresponding coordinate. Note that ${\boldsymbol{\nu}}$, ${\boldsymbol{\tau}}$, $\mathbf{m}$ form a right-handed basis vectors of the three-dimensional space (see Fig. \[fig:loop\]).

![A dislocation loop in the chosen coordinate system[]{data-label="fig:loop"}](loop){height="6.5cm"}

@Ortiz99 introduced the resultant Burgers vector of excess dislocations, whose lines cross the area $\mathcal{A}$ in the reference configuration, in the following way $$\label{eq:2.2}
\mathbf{b}_r=\oint_{\mathcal{C}} \mathbf{F}^p\cdot d\mathbf{x} ,$$ where $\mathcal{C}$ is the close contour surrounding $\mathcal{A}$. @Le2014nonlinear have shown that, in the continuum limit, when the atomic distance goes to zero at the fixed sizes of the representative volume element and the fixed density of dislocations per area of unit cell, integral gives the total closure failure induced by $\mathbf{F}^p$ which must be equal to the resultant Burgers vector. It is natural to assume $\mathbf{F}^p$ continuously differentiable in this continuum limit, so, applying Stoke’s theorem we get from $$\mathbf{b}_r=- \int _{\mathcal{A}}(\mathbf{F}^p\times \nabla ) \cdot \mathbf{n}da,$$ where $\times $ denotes the vector product, $da$ the surface element, and $\mathbf{n}$ the unit vector normal to $\mathcal{A}$. This legitimates the introduction of the dislocation density tensor 