Answer to Question #102643 in Linear Algebra for BIVEK SAH

Question

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

Answer 1

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 $\bar{A}$^pr in another coordinate system($\bar{x^1}$,$\bar{x^2}$ ,............$\bar{x^n}$) by the transformation equations

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

or, by our conventions,

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

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 $\bar{A}$ _pr in another coordinate system ($\bar{x^1}$,.........,$\bar{x^n}$ ) by the transformation equations

$\bar{A}$ _pr= $\displaystyle\sum_{s=1}^n$ $\displaystyle\sum_{q=1}^n$ ${\partial x^q\above{2pt}\partial\bar{x^p} }$ ${\partial x^s \above{2pt}\partial\bar{x^r} }$ A _{q s} p,r=1,2,........,n

or, by our conventions,

$\bar{A}$ _pr=${\partial x^q \above{2pt}\partial \bar{x^p} }$ ${\partial x^s \above{2pt}\partial\bar{x^r} }$

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, $\delta^p_q$ and defined by

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

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

Proof:-

we must prove that

$\bar{δ^j_k}$ =${\partial\bar{u^j} \above{2pt}\partial u^p }$ ${\partial u^q \above{2pt} \partial\bar{u^k} }$ $\delta^p_q$

By definition

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

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

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

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

${\partial \bar{u ^j} \above{2pt}\partial u^p }{\partial u^q\above{2pt}\partial \bar{u^k} } \delta^p_q$ =${\partial \bar{u ^j} \above{2pt} \partial u^p}{\partial u^p \above{2pt}\partial \bar{u ^k} }$ =${\partial \bar{u^j} \above{2pt}\partial \bar{u^k} }=\delta^j_k=\bar{\delta^j_k}$

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.