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Question 11.10: A transport equation such as Ohm’s law (11.74),∇ϕ = −ρ · jq,...

A transport equation such as Ohm’s law (11.74),

\nabla \varphi = −ρ (s, n_A, q) · j_q                                     (11.74)

\nabla \varphi = −ρ · j_q

relates two vectors, which are the conductive electric current density j_q and electric potential gradient \nabla \varphi , through a linear application, which is the electric resistivity ρ.
Mathematically, a vector is a rank-1 tensor and a linear application between two vectors is a rank-2 tensor.

a) Show that the electric resistivity ρ can be decomposed into the sum of a symmetric part ρ^s and an antisymmetric part ρ^α .
b) Show that the antisymmetric part ρ^α has a contribution to the transport that can be written as,

\nabla^α \varphi = −ρ^a (\hat{u} × j_q)

where \nabla^α \varphi is the antisymmetric part of the electric potential gradient and \hat{u} is a unit axial vector.
The decomposition and the expression for the antisymmetric part of the electric potential gradient is a general result that applies for any empirical linear relation between a current density vector and a generalised force vector.

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a) The components of the symmetric rank-2 resistivity tensor are written as,

\rho ^s_{ij} = \frac{1}{2} (\rho _{ij} + \rho _{ji}) .

The components of the antisymmetric rank-2 resistivity tensor are given by,

\rho ^α_{ij} = \frac{1}{2} (\rho _{ij} + \rho _{ji}) .

The components of the rank-2 resistivity tensor are written as,

\rho _{ij} = \frac{1}{2} (\rho _{ij} + \rho _{ji}) + \frac{1}{2} (\rho _{ij} + \rho _{ji}) .

which implies that,

\rho _{ij} = \rho ^s_{ij} + \rho ^α_{ij} .

Thus, the conductivity tensor ρ is the sum of the symmetric rank-2 conductivity tensor ρ^s and the antisymmetric rank-2 conductivity tensor ρ^α ,

ρ = ρ^s + ρ^α

a) The linear application ρ^a · j_q can be written in components as,

\Biggl(\begin{matrix} 0 & \rho ^\alpha _{12} & \rho ^\alpha _{13} \\ -\rho ^\alpha _{12} & 0 & \rho ^\alpha _{23} \\ -\rho ^\alpha _{13} & -\rho ^\alpha _{23} & 0 \end{matrix}\Biggr) \Biggl(\begin{matrix} j_{q1} \\ j_{q2} \\ j_{q3} \end{matrix}\Biggr) =\Biggl(\begin{matrix} \rho ^\alpha _{12}j_{q2} + \rho ^\alpha _{13}j_{q3} \\ -\rho ^\alpha _{12}j_{q1} + \rho ^\alpha _{23}j_{q3}\\ -\rho ^\alpha_{13}j_{q1} – \rho ^\alpha _{23}j_{q2}\end{matrix}\Biggr) 

The vector product ρ^α \hat{u} × j_q is written in components as,

\Biggl(\begin{matrix} \rho ^\alpha \hat{u}_1\\ \rho ^\alpha \hat{u}_2 \\ \rho ^\alpha \hat{u}_3\end{matrix} \Biggr) × \Biggl(\begin{matrix} j_{q1} \\ j_{q2}\\ j_{q3}\end{matrix} \Biggr) = \Biggl(\begin{matrix}ρ^α \hat{u}_2 j_{q3} − ρ^α \hat{u}_3 j_{q2} \\ρ^α \hat{u}_3 j_{q1} − ρ^α \hat{u}_1 j_{q3}\\ ρ^α \hat{u}_1 j_{q2}− ρ^α \hat{u}_2 j_{q1}\end{matrix} \Biggr)

The identification of the components of these two vectors yields,

\hat{u}_1 = – \frac{ρ^α_{23}}{ρ^α}                         \hat{u}_2 = – \frac{ρ^α_{13}}{ρ^α}                           \hat{u}_3 = – \frac{ρ^α_{12}}{ρ^α}

Since the vector \hat{u} is a unit axial vector,

\hat{u}^2  = \hat{u}^2_1 + \hat{u}^2_2 + \hat{u}^2_3 = \frac{1}{(ρ^α)^2} \biggl((\rho ^\alpha _{23})^2 + (\rho ^\alpha _{13})^2 + (\rho ^\alpha _{13})^2\biggr) =1

which implies that,

\rho ^\alpha = \sqrt{(\rho ^\alpha _{23})^2 + (\rho ^\alpha _{13})^2 + (\rho ^\alpha _{12})^2}

Thus, the antisymmetric part of the electric potential gradient is given by,

\nabla ^\alpha \varphi = – \rho ^\alpha .j_q = -\rho ^\alpha (\hat{u}\times j_q )

where the unit axial vector \hat{u} is written in components as,

\hat{u} = \frac{1}{ρ_α}  \Biggl(\begin{matrix} −ρ^α_{23} \\ ρ^α_{13} \\ −ρ^α_{12} \end{matrix} \Biggr)

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