Question

In equation (5.16), show that if ${\mathbf{T}}_{\mathbf{jk}}$is a tensor (that is, not a pseudotensor), then ${\mathbf{V}}_{\mathbf{i}}$is a pseudovector (axial vector). Also show that if ${\mathbf{V}}_{\mathbf{i}}$is a pseudotensor, then ${\mathbf{T}}_{\mathbf{jk}}$is a vector (true or polar vector). You know that if role="math" localid="1659251751142" ${\mathbf{V}}_{\mathbf{i}}$is a cross product of polar vectors, then it is a pseudovector. Is its dual ${\mathbf{T}}_{\mathbf{jk}}$a tensor or a pseudotensor?

Short answer

The required proofs are shown below.

Step-by-step solution

Given information.

Physics definitions are given.

Definition of a tensor.

Tensors, like scalars and vectors, are mathematical constructs that can be used to describe physical qualities. Tensors are just a combination of scalars and vectors, with a scalar being a zero rank tensor and a vector being a first rank tensor.

Define a tensor or a pseudo tensor.

Define a rank two tensor T . Write its transformation.

$T_{\alpha \beta }^{\text{'}}=k{a}_{\alpha \iota }{a}_{\beta j}{T}_{ij}$

Simplify equations further.

Simplify equations if${\mathrm{V}}_{\mathrm{i}}=\frac{1}{2}{\mathrm{\epsilon }}_{\mathrm{ijk}}{\mathrm{T}}_{\mathrm{jk}}$.

$$\begin{gathered} {\mathrm{V}}_{\propto }^{\text{'}}=\frac{1}{2}\left(\mathrm{detA}\right){\mathrm{a}}_{\mathrm{\alpha i}}{\mathrm{a}}_{\mathrm{\beta j}}{\mathrm{a}}_{\mathrm{\gamma k}}{\mathrm{\epsilon }}_{\mathrm{ijk}}{\mathrm{ka}}_{\mathrm{\beta p}}{\mathrm{a}}_{\mathrm{\gamma q}}{\mathrm{T}}_{\mathrm{pq}} \\ =\frac{1}{2}\mathrm{k}\left(\mathrm{detA}\right){\mathrm{a}}_{\mathrm{\alpha i}}{\mathrm{\epsilon }}_{\mathrm{ijk}}{\mathrm{T}}_{\mathrm{jk}} \\ =\mathrm{k}\left(\mathrm{detA}\right){\mathrm{a}}_{\mathrm{\alpha i}}{\mathrm{V}}_{\mathrm{i}} \end{gathered}$$

Make conclusions.

If T is a pseudo tensor, write the conclusions.

$$\begin{gathered} k=detA \\ k\left(detA\right)=1 \end{gathered}$$

This implies that V is a vector.

If T is a tensor, write the conclusions.

$$\begin{gathered} K=1 \\ K\left(detA\right)=detA \end{gathered}$$

This implies that V is a pseudo vector.