$$\mathbf{T}=-\mathbf{F}^p\times \nabla .$$ For the plastic deformation taken from $$\mathbf{T}=-\mathbf{F}^p\times \nabla =\mathbf{s}\otimes (\nabla \beta \times \mathbf{m}).$$ If we choose now an infinitesimal area $da$ with the unit normal vector ${\boldsymbol{\tau}}$, then the resultant Burgers vector of all excess dislocations, whose dislocation lines cross this area at right angle is given by $$\mathbf{b}_r=\mathbf{T}\cdot {\boldsymbol{\tau}}\, da=-\mathbf{s}\, (\nabla \beta \cdot {\boldsymbol{\nu}})da=-\mathbf{s}\, \partial _\nu \beta \, da.$$ This resultant Burgers vector can be decomposed into the sum of two vectors $$\mathbf{b}_r=\mathbf{b}_{r\perp}+\mathbf{b}_{r\parallel}=-({\boldsymbol{\nu}}s_\nu +{\boldsymbol{\tau}}s_\tau )\partial _\nu \beta\, da,$$ where $s_\nu =\mathbf{s}\cdot {\boldsymbol{\nu}}=\nu _1$ and $s_\tau =\mathbf{s}\cdot {\boldsymbol{\tau}}=\tau _1$ are the projections of the slip vector onto the normal and tangential direction to the dislocation line, respectively. This allows us to define two scalar densities (or the numbers of excess dislocations per unit area) of edge and screw dislocations $$\label{eq:2.3a}
\begin{split}
\rho_\perp =\frac{|\mathbf{b}_{r\perp}|}{b}=\frac{1}{b}|s_\nu \partial _\nu \beta| =\frac{1}{b}\left| \frac{l_{,1}(\beta _{,1}l_{,1}+\beta _{,2}l_{,2})}{l_{,1}^2+l_{,2}^2}\right| , 
\\
\rho_\parallel =\frac{|\mathbf{b}_{r\parallel}|}{b}=\frac{1}{b}|s_\tau \partial _\nu \beta|=\frac{1}{b}\left| \frac{l_{,2}(\beta _{,1}l_{,1}+\beta _{,2}l_{,2})}{l_{,1}^2+l_{,2}^2}\right| ,
\end{split}$$ with $b$ the magnitude of Burgers vector. We see that the three-dimensional dislocation densities $\rho_\perp $ and $\rho_\parallel $ depend on both the gradient of the plastic slip and the gradient of the loop function $l(\mathbf{x})$ through the vectors ${\boldsymbol{\nu}}$ and ${\boldsymbol{\tau}}$.

Consider now the 