14.2: C.2- 1st-order isotropic tensors
- Page ID
- 18092
A 1st-order tensor is a vector. If the vector \(\vec{v}\) is isotropic, then under an infinitesimal rotation
\[v_{i}^{\prime}=v_{j} C_{j i}=v_{j}\left(\delta_{j i}+r_{j i}\right)=v_{i}+v_{j} r_{j i}=v_{i}. \nonumber \]
Therefore,
\[v_{j} r_{j i}=0. \nonumber \]
This represents three algebraic equations, one for each value of \(i\):
\[\begin{array}{l}
v_{1} r_{11}+v_{2} r_{21}+v_{3} r_{31}=0 \\
v_{1} r_{12}+v_{2} r_{22}+v_{3} r_{32}=0 \\
v_{1} r_{13}+v_{2} r_{23}+v_{3} r_{33}=0.
\end{array}\label{eqn:1} \]
Because \(\underset{\sim}{r}\) is antisymmetric, \(r_{11}\) = \(r_{22}\) = \(r_{33}\) = 0, removing one term from each equation.
Now consider the first equation of Equation \(\ref{eqn:1}\):
\[v_{2} r_{21}+v_{3} r_{31}=0.\label{eqn:2} \]
Here is a crucial point: if \(\vec{v}\) is isotropic, then Equation \(\ref{eqn:1}\) must be true for all antisymmetric matrices \(\underset{\sim}{r}\), i.e., regardless of the values of \(r_{21}\) and \(r_{31}\). The only way this can be true is if \(v_2\) = 0 and \(v_3\) = 0. The same considerations applied to the second equation of Equation \(\ref{eqn:1}\) tell us that \(v_1\) must also be zero, hence the only isotropic 1st-order tensor is the trivial case
\[\vec{v}=0. \nonumber \]