Question #102643
Define covariant and contravariant tensors of rank two. What is the rank of a scalar
tensor? Show that the Kronecker delta function is a mixed tensor of rank 2
1
Expert's answer
2020-02-19T05:14:25-0500

Contravariant tensor of the second rank:-

If n 2 quantities A q s  in a coordinate system (x 1,x 2,........x n)are related to n 2 other quantities Aˉ\bar{A}pr in another coordinate system(x1ˉ\bar{x^1},x2ˉ\bar{x^2} ,............xnˉ\bar{x^n}) by the transformation equations

Aˉ\bar{A} pr= s=1n\displaystyle\sum_{s=1}^n q=1n\displaystyle\sum_{q=1}^n xpˉxq{\partial\bar{x^p} \above{2pt}\partial x^q } xrˉxs{\partial\bar{x^r} \above{2pt}\partial x^s } A q s p,r=1,2,..........,n

or, by our conventions,

Aˉ\bar{A} pr=xpˉxq{\partial\bar{x^p} \above{2pt}\partial x^q } xrˉxs{\partial\bar{x^r} \above{2pt}\partial x^s }

they are called components of a contravariant tensor of the second rank (or of rank two).

Covariant tensor of the second rank :-

If n 2 quantities A q s in a coordinate system (x 1,x 2,.......x n) are related to n 2 other quantities Aˉ\bar{A} pr in another coordinate system (x1ˉ\bar{x^1},.........,xnˉ\bar{x^n} ) by the transformation equations

Aˉ\bar{A} pr= s=1n\displaystyle\sum_{s=1}^n q=1n\displaystyle\sum_{q=1}^n xqxpˉ{\partial x^q\above{2pt}\partial\bar{x^p} } xsxrˉ{\partial x^s \above{2pt}\partial\bar{x^r} } A q s p,r=1,2,........,n

or, by our conventions,

Aˉ\bar{A} pr=xqxpˉ{\partial x^q \above{2pt}\partial \bar{x^p} } xsxrˉ{\partial x^s \above{2pt}\partial\bar{x^r} }

they are called components of a covariant tensor of the second rank.

Show that the Kronecker delta function is a mixed tensor of rank 2

The Kronecker delta or Kronecker tensor, written, δqp\delta^p_q and defined by

δqp\delta^p_q ={0if p=/q1if p=q\begin{cases} 0 &\text{if } p{=}\mathllap{/\,}q \\ 1 &\text{if } p=q \end{cases}

is a mixed tensor of rank 2,justify the notation used.

Proof:-

we must prove that

δkjˉ\bar{δ^j_k} =ujˉup{\partial\bar{u^j} \above{2pt}\partial u^p } uqukˉ{\partial u^q \above{2pt} \partial\bar{u^k} } δqp\delta^p_q

By definition

δqpˉ=δqp=\bar{δ^p_q}=\delta^p_q= {0if p=/q1if p=q\begin{cases} 0 &\text{if } p{=}\mathllap{/\,}q \\ 1 &\text{if } p=q \end{cases}

Coordinates ujˉ\bar{u ^j} are functions of coordinates u p which are in turn functions of coordinates ukˉ\bar{u ^k} .Then,by chain rule,

ujˉupupukˉ=ujˉukˉ=δkj{\partial \bar{u ^j} \above{2pt}\partial u^p }{\partial u^p \above{2pt}\partial \bar{u ^k} }={\partial \bar{u ^j} \above{2pt}\partial \bar{u ^k} }=\delta^j_k

Now we must prove that the right side of the transformation equations of the mixed tensor above equals δkjˉ\bar{\delta^j_k } .It does:

ujˉupuqukˉδqp{\partial \bar{u ^j} \above{2pt}\partial u^p }{\partial u^q\above{2pt}\partial \bar{u^k} } \delta^p_q =ujˉupupukˉ{\partial \bar{u ^j} \above{2pt} \partial u^p}{\partial u^p \above{2pt}\partial \bar{u ^k} } =ujˉukˉ=δkj=δkjˉ{\partial \bar{u^j} \above{2pt}\partial \bar{u^k} }=\delta^j_k=\bar{\delta^j_k}

Rank of Scalar tensor:-

In fact tensors are merely a generalization of scalars and vectors; a scalar is a zero rank tensor, and a vector is a first rank tensor. The rank (or order) of a tensor is defined by the number of directions (and hence the dimensionality of the array) required to describe it.

A tensor is a mathematical representation of a scalar (tensor of rank zero), a vector(rank 1)

scalar=rank 0->magnitude ,no direction.

vector=rank 1->magnitude and direction.



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