Let a and b be vectors with components ai and bi· and A be a tensor with components aij. Show that aibi and aii are scalar invariants.

Question

Let a and b be vectors with components [latex]a_{i}\ and b_{i}[/latex]· and A be a tensor with components [latex]a_{ij}[/latex] . Show that [latex]a_{i}b_{i} [/latex] and [latex]a_{ii}[/latex] are scalar invariants.

Accepted Answer

If [latex]a^{,}_{i}[/latex] and [latex]b_{i}[/latex] are components of a and b in the [latex]x^{,}_{i}[/latex] system, we have, by the transformation rule of vector components,

[latex]a^{,}_{i}b^{,}_{i}=(\alpha _{ip}a_{p})(\alpha _{iq}b_{q})=(\alpha _{ip}\alpha _{iq})a_{p}a_{q}=\delta _{pq}a_{p}b_{p}=a_{i}b_{i}[/latex] (2.3.19)

Thus, [latex]a_{i}b_{i}[/latex] has the same value in all coordinate systems; it is therefore a scalar invariant. Note that this scalar invariant is actually the scalar product of a and b, namely a · b.
If [latex]a^{,}_{ij}[/latex] are the components of A in the [latex]x^{,}_{i}[/latex] system, we have, by the transformation rule of tensor components of order 2, [latex]a^{,}_{ij}=\alpha _{ip}\alpha _{jq}a_{pq}[/latex]. Hence

[latex]a^{,}_{ii}=\alpha _{ip}\alpha _{iq}a_{pq}=\delta _{pq}a_{pq}=a_{pp}=a_{ii}[/latex] (2.3.20)

Thus, [latex]a_{ii}[/latex] has the same value in all coordinate systems; it is therefore a scalar invariant. This invariant is called the trace of A, denoted by tr A.

Question 2.3.3: Let a and b be vectors with components ai and bi· and A be a...

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