# 14.2: C.2- 1st-order isotropic tensors

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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

$\vec{v}=0. \nonumber$

This page titled 14.2: C.2- 1st-order isotropic tensors is shared under a CC BY-NC 4.0 license and was authored, remixed, and/or curated by Bill Smyth via source content that was edited to the style and standards of the LibreTexts platform; a detailed edit history is available upon request.