ExamplesA

Section 3.3 ExamplesA

Example 3.3.1.

Show that a scalar is invariant under any coordinate transformation.

Solution.

We have the transformation equations for contravariant and covariant tensor as

\begin{equation*} \bar{A}^{i}=\frac{\partial \bar{x}^{i}}{\partial x^{j}}A^{j} \end{equation*}

and

\begin{equation*} \bar{A}_{i}=\frac{\partial x^{j}}{\partial \bar{x}^{i}}A_{j} \end{equation*}

now scalar is a tensor of rank zero, so set $i=j=0\text{,}$ we get from above equations, $A = \bar{A}$ i.e., scalar is invarient under any coordinate transformation.

Example 3.3.2.

show that in a cartensian coordinate system, the contravarient and covarient components of a vector are identical.

Solution.

Let $A^{i}$ and $A_{i}$ are the contravarient and covarient components of a vector respectively in a cartesian coordinates, then they are related by $A_{i}=g_{ij}A^{j}=g_{ji}A^{j}\text{.}$ Here,

\begin{equation*} \left\{g_{ij} = \frac{\partial \bar{x}^{j}}{\partial x^{i}}\cdot \frac{\partial \bar{x}^{i}}{\partial x^{j}}\right. \end{equation*}

and in cartesian coordinate $h_{1}= h_{2} = h_{3} =1\text{.}$ Therefore, $g_{11} =g_{22} =g_{33}=1 $ in orthogonal system. For a cartesian system $g_{ii}=1$ [if $i=j$] and $g_{ij} =0$ [if $i \neq j$]. Hence, $A_{i}=g_{ii}A^{i}$ and $A_{i}=A^{i}\text{,}$ i. e., covarient component is similar to contravarient components.

Example 3.3.3.

Show that velocity and acceleration are contravarient vectors and that of the gradient of a scalar field is a covarient vector.

Solution.

Let the transformation eqn. as

\begin{equation} \,d\bar{x}^{i}=\frac{\partial \bar{x}^{i}}{\partial x^{j}}\,dx^{j} \tag{3.3.1} \end{equation}

Dividing by $\,dt\text{,}$ we get -

\begin{equation} \frac{\,d\bar{x}^{i}}{\,dt}=\frac{\partial \bar{x}^{i}}{\partial x^{j}}\frac{dx^{j}}{\,dt} \tag{3.3.2} \end{equation}

But,

\begin{equation*} \frac{\,d\bar{x}^{i}}{\,dt} = \bar{v}^{i} \end{equation*}

and

\begin{equation*} \frac{\,dx^{j}}{\,dt} = \bar{v}^{j} \end{equation*}

\begin{equation} \therefore \bar{v}^{i} = \frac{\,d\bar{x}^{i}}{\partial x^{j}}v^{j} \tag{3.3.3} \end{equation}

Velocity $v^{j}$ is a contravarient vector because (3.3.3) is a transformation rule for contravarient tensor. Again, differentiating (3.3.3), w. r. t. $'t'\text{,}$ we get -

\begin{equation*} \frac{\,d\bar{v}^{i}}{\,dt} = \frac{\,d\bar{x}^{i}}{\partial x^{j}}\frac{\,dv^{j}}{\,dt} \end{equation*}

\begin{equation} \therefore \bar{a}^{i} = \frac{\,d\bar{x}^{i}}{\partial x^{j}}a^{j} \tag{3.3.4} \end{equation}

hence acceleration is also a contravarient vector.

Note: The coordinates $x^{j}$ in $\frac{\,dx^{j}}{\,dt}$ are the coordinates of a particle in motion, while the coefficients $\frac{\,d\bar{x}^{i}}{\partial x^{j}}$ only denotes a relation between two coordinate systems, which is independent of time.
Let $\phi = \phi(x^{j})$ be a scalar field. The functional form of scalar field is same under coordinate transformations, hence

\begin{equation*} \phi(x^{j}) = \bar{\phi}(\bar{x}^{i}) = \phi(\bar{x}^{i}) \end{equation*}

now under frame rotation

\begin{equation*} \frac{\partial\phi}{\partial x^{j}} = \frac{\partial\phi}{\partial \bar{x}^{i}}\frac{\partial\bar{x}^{i}}{\partial x^{j}} \end{equation*}

\begin{equation*} \frac{\partial\phi}{\partial x^{j}} = \frac{\partial\phi}{\partial \bar{x}^{i}}\frac{\partial\bar{x}^{i}}{\partial x^{j}} \end{equation*}

or,

\begin{equation} \quad A_{j} = \frac{\partial\bar{x}^{i}}{\partial x^{j}}\bar{A_{i}} \tag{3.3.5} \end{equation}

i.e., the (3.3.5) shows the transformation rule for covarient tensor. If we suppose, $A^{j} = \frac{\partial\phi}{\partial x^{j}}$ and $\bar{A^{i}} = \frac{\partial\phi}{\partial \bar{x}^{i}}$ then $A^{j}=\frac{\partial\bar{x^{i}}}{\partial x^{j}}\bar{A^{i}}$ which is not the transformation rule. Hence the gradient of a scalar field is not a contravarient vector.

Example 3.3.4.

Show that the contraction of the outer product of the tensors $A^{p}$ and $B_{q}$ is an invariant.

Solution.

Since $A^{p}$ and $B_{q}$ are tensors, we have,

\begin{equation*} \bar{A}^{j} = \frac{\partial\bar{x}^{j}}{\partial x^{p}}A^{p} \end{equation*}

and

\begin{equation*} \bar{B}_{k} = \frac{\partial x^{q}}{\partial \bar{x}^{k}}B_{q} \end{equation*}

therefore, the outer product is

\begin{equation*} \bar{A}^{j}\bar{B}_{k} = \frac{\partial\bar{x}^{j}}{\partial x^{p}}\frac{\partial x^{q}}{\partial \bar{x}^{k}}A^{p}B_{q} \end{equation*}

by contraction, put $j=k\text{,}$ we have

\begin{equation*} \bar{A}^{j}\bar{B}_{j} = \frac{\partial\bar{x}^{j}}{\partial x^{p}}\frac{\partial x^{q}}{\partial \bar{x}^{j}}A^{p}B_{q} = \frac{\partial x^{q}}{\partial x^{p}}A^{p}B_{q} = \delta^{q}_{p}A^{p}B_{q} = A^{p}B_{q} = \text{invarient}. \end{equation*}

Example 3.3.5.

If $A^{ij}$ is symmetric tensor in a reference system then it is symmetric in all other reference system.

Solution.

Let $A^{ij}$ is a components of symmetric tensor then $A^{ij} = A^{ji}\text{.}$ If $\bar{A}^{pq}$ be the components in barred system, then

\begin{equation*} \bar{A}^{pq}= \frac{\partial\bar{x}^{p}}{\partial x^{i}}\frac{\partial \bar{x}^{q}}{\partial \bar{x}^{j}}A^{ij} \end{equation*}

on interchanging the indices $p\text{,}$ $q$ and $i\text{,}$ $j\text{,}$ we have

\begin{equation*} \bar{A}^{qp}= \frac{\partial\bar{x}^{q}}{\partial x^{i}}\frac{\partial \bar{x}^{p}}{\partial x^{j}}A^{ji} \end{equation*}

\begin{equation*} \therefore \bar{A}^{qp} = \bar{A}^{pq} \quad \text{proved}. \end{equation*}

Example 3.3.6.

The symmetric or antisymmetric property may not be true for the interchange of contravariant and covariant index in all system, i.e., $A^{p} _{q} \neq A^{q} _{p}\text{.}$

Solution.

If $A^{i} _{j}= A^{j} _{i}\text{,}$ then it is not necessary that $\bar{A}^{q} _{p} = \bar{A}^{p} _{q}\text{.}$ Suppose $A^{i} _{j}$ is a component of a mixed tensor in unbarred system such that $A^{i} _{j}= A^{j} _{i}\text{.}$ Then in barred system, we have -

\begin{equation} \bar{A}^{p} _{q} = \frac{\partial\bar{x}^{p}}{\partial x^{i}} \frac{\partial x^{j}}{\partial \bar{x}^{q}}A^{i} _{j} \tag{3.3.6} \end{equation}

Interchange $p{,}q$ and $i{,}j\text{,}$ we get -

\begin{equation} \bar{A}^{q} _{p} = \frac{\partial\bar{x}^{q}}{\partial x^{j}} \frac{\partial x^{i}}{\partial \bar{x}^{p}}A^{j} _{i} = \frac{\partial\bar{x}^{q}}{\partial x^{j}}\frac{\partial x^{i}}{\partial \bar{x}^{p}}A^{i} _{j} \tag{3.3.7} \end{equation}

from eqns. (3.3.6) and (3.3.7),

\begin{equation*} \because \quad \frac{\partial\bar{x}^{q}}{\partial x^{j}} \neq \frac{\partial x^{j}}{\partial \bar{x}^{q}} \end{equation*}

\begin{equation*} \frac{\partial x^{i}}{\partial \bar{x}^{p}} \neq \frac{\partial\bar{x}^{p}}{\partial x^{i}} \end{equation*}

Hence,

\begin{equation*} \bar{A}^{p} _{q} \neq \bar{A}^{q} _{p}. \end{equation*}

Thus the symmetric propeties are not preserved in contravariant and covariant of different systems.

Example 3.3.7.

Show that $\frac{\partial A_{p}}{\partial x^{q}}$ is not a tensor even though $A_{p}$ is a covarient tensor of rank one.

Solution.

we have -

\begin{equation*} \bar{A_{j}} = \frac{\partial x^{p}}{\partial \bar{x}^{j}}A_{p} \end{equation*}

Differentiating w. r. t. $\bar{x}^{k}$ we get -

\begin{equation*} \frac{\partial \bar{A_{j}}}{\bar{x}^{k}} = \frac{\partial x^{p}}{\partial \bar{x}^{j}}.\frac{\partial A_{p}}{\partial \bar{x}^{k}} + \frac{\partial^{2} x^{p}}{\partial \bar{x}^{k} \partial \bar{x}^{j}}A_{p} \end{equation*}

\begin{equation*} = \frac{\partial x^{p}}{\partial \bar{x}^{j}}.\frac{\partial A_{p}}{\partial x^{q}} \frac{\partial x_{q}}{\partial \bar{x}^{k}}+\frac{\partial^{2} x^{p}}{\partial \bar{x}^{k} \partial \bar{x}^{j}}A_{p} \end{equation*}

\begin{equation*} = \frac{\partial x^{p}}{\partial \bar{x}^{j}}.\frac{\partial x_{q}}{\partial \bar{x}^{k}} \frac{\partial A_{p}}{\partial x^{q}}+\frac{\partial^{2} x^{p}}{\partial \bar{x}^{k} \partial \bar{x}^{j}}A_{p} \end{equation*}

because of the presence of second term on the right $\frac{\partial A_{p}}{\partial x^{q}}$ does not transform like a tensor.

Example 3.3.8.

Prove that the contraction of the tensor $A^{p} _{q}$ is a scalar or invarient.

Solution.

We have

\begin{equation*} \bar{A}^{j} _{k} = \frac{\partial\bar{x}^{j}}{\partial x^{p}} \frac{\partial x^{q}}{\partial \bar{x}^{k}} A^{p} _{q} \end{equation*}

put $j=k\text{,}$ then

\begin{equation*} \bar{A}^{j} _{j} = \frac{\partial\bar{x}^{j}}{\partial x^{p}} \frac{\partial x^{q}}{\partial \bar{x}^{j}} A^{p} _{q} = \delta_{p}^{q}A^{p} _{q} = A^{p} _{p}. \end{equation*}

Thus $\bar{A}^{j} _{j} = A^{p} _{p}$ which means $A^{p} _{p}$ must be an invarient.

Example 3.3.9.

Show that the inner product of any $2^{nd}$ rank tensors is a tensor of rank two.

Solution.

Let us consider $A^{p}_{r}$ and $B^{qs}$ are any tensors of rank two, then -

\begin{equation*} \bar{A}^{j} _{k} = \frac{\partial\bar{x}^{j}}{\partial x^{p}} \frac{\partial x^{r}}{\partial \bar{x}^{k}} A^{p} _{r} \end{equation*}

and

\begin{equation*} \bar{B}^{lm}= \frac{\partial\bar{x}^{l}}{\partial x^{q}}. \frac{\partial\bar{x}^{m}}{\partial x^{s}} B^{qs} \end{equation*}

multiplying them, we get -

\begin{equation*} \bar{A}^{j} _{k}\bar{B}^{lm} = \frac{\partial\bar{x}^{j}}{\partial x^{p}} \frac{\partial x^{r}}{\partial \bar{x}^{k}}\frac{\partial\bar{x}^{l}}{\partial x^{q}}. \frac{\partial\bar{x}^{m}}{\partial x^{s}} A^{p} _{r}B^{qs} \end{equation*}

set $k=l,$ we get -

\begin{equation*} \bar{A}^{j} _{k}\bar{B}^{km} = \frac{\partial\bar{x}^{j}}{\partial x^{p}} \frac{\partial x^{r}}{\partial \bar{x}^{k}}\frac{\partial\bar{x}^{k}}{\partial x^{q}}. \frac{\partial\bar{x}^{m}}{\partial x^{s}} A^{p} _{r}B^{qs} \end{equation*}

\begin{equation*} = \frac{\partial\bar{x}^{j}}{\partial x^{p}} \delta_{q}^{r} \frac{\partial\bar{x}^{m}}{\partial x^{s}} A^{p} _{r}B^{qs} \end{equation*}

\begin{equation*} = \frac{\partial\bar{x}^{j}}{\partial x^{p}} \frac{\partial\bar{x}^{m}}{\partial x^{s}} A^{p} _{q}B^{qs} \end{equation*}

\begin{equation*} = \frac{\partial\bar{x}^{j}}{\partial x^{p}} \frac{\partial\bar{x}^{m}}{\partial x^{s}} C^{ps} \end{equation*}

which is a second rank tensor.

Example 3.3.10.

Show that the inner product of a $2^{nd}$ tensor and a $3^{rd}$ rank tensor is a tensor of rank three.

Solution.

Let us consider $A^{p}_{r}$ and $B^{qs}_{t}$ are any tensors of rank two and three, respectively then -

\begin{equation*} \bar{A}^{j} _{k} = \frac{\partial\bar{x}^{j}}{\partial x^{p}} \frac{\partial x^{r}}{\partial \bar{x}^{k}} A^{p} _{r} \end{equation*}

and

\begin{equation*} \bar{B}^{lm} _{n}= \frac{\partial\bar{x}^{l}}{\partial x^{q}}. \frac{\partial\bar{x}^{m}}{\partial x^{s}}. \frac{\partial x^{t}}{\partial \bar{x}^{n}}B^{qs} _{t} \end{equation*}

multiplying, we get -

\begin{equation*} \bar{A}^{j} _{k}\bar{B}^{lm} _{n} = \frac{\partial\bar{x}^{j}}{\partial x^{p}} \frac{\partial x^{r}}{\partial \bar{x}^{k}}\frac{\partial\bar{x}^{l}}{\partial x^{q}}. \frac{\partial\bar{x}^{m}}{\partial x^{s}}. \frac{\partial x^{t}}{\partial \bar{x}^{n}}A^{p} _{r}B^{qs} _{t} \end{equation*}

put $j=n\text{,}$

\begin{equation*} \bar{A}^{j} _{k}\bar{B}^{lm} _{j} = \frac{\partial\bar{x}^{j}}{\partial x^{p}} \frac{\partial x^{r}}{\partial \bar{x}^{k}}\frac{\partial\bar{x}^{l}}{\partial x^{q}}. \frac{\partial\bar{x}^{m}}{\partial x^{s}}. \frac{\partial x^{t}}{\partial \bar{x}^{j}}A^{p} _{r}B^{qs} _{t} \end{equation*}

\begin{equation*} = \delta_{p}^{t} \frac{\partial x^{r}}{\partial \bar{x}^{k}}\frac{\partial\bar{x}^{l}}{\partial x^{q}} \frac{\partial\bar{x}^{m}}{\partial x^{s}} A^{p} _{r}B^{qs} _{t} \end{equation*}

\begin{equation*} =\frac{\partial x^{r}}{\partial \bar{x}^{k}}\frac{\partial\bar{x}^{l}}{\partial x^{q}}. \frac{\partial\bar{x}^{m}}{\partial x^{s}}A^{p} _{r}B^{qs} _{p} = \frac{\partial x^{r}}{\partial \bar{x}^{k}}\frac{\partial\bar{x}^{l}}{\partial x^{q}} \frac{\partial\bar{x}^{m}}{\partial x^{s}}C^{qs} _{r} \end{equation*}

i.e., $A^{p} _{r}B^{qs} _{t}$ is a tensor of rank three.

Example 3.3.11.

If a tensor $A^{pqr} _{st}$ is a symmetric w.r.t. indices $p$ and $q$ in one coordinate system then show that it remains symmetric w.r.t. $p$ and $q$ in any other coordinate systems.

Solution.

Since only indices $p$ and $q$ are involved, we shall prove that the results for $B^{pq}\text{.}$ If $B^{pq}$ is symmetric then $B^{pq}= B^{qp}$ and

\begin{equation*} \bar{B}^{jk} =\frac{\partial\bar{x}^{j}}{\partial x^{p}}\frac{\partial\bar{x}^{k}}{\partial x^{q}}B^{pq} = \frac{\partial\bar{x}^{k}}{\partial x^{q}}\frac{\partial\bar{x}^{j}}{\partial x^{p}}B^{qp} = \bar{B}^{kj} \end{equation*}

hence $B_{qp}$ remains symmetric in the $\bar{x}^{i}$ coordinate system. If $B^{pq}$ is skew - symmetric, then $B^{pq} = - B^{qp}$ and

\begin{equation*} \bar{B}^{jk} =\frac{\partial\bar{x}^{j}}{\partial x^{p}}\frac{\partial\bar{x}^{k}}{\partial x^{q}}B^{pq} =- \frac{\partial\bar{x}^{k}}{\partial x^{q}}\frac{\partial\bar{x}^{j}}{\partial x^{p}}B^{qp} = -\bar{B}^{kj} \end{equation*}

hence $B_{pq}$ remains skew - symmetric in $\bar{x}^{i}$ system.

Example 3.3.12.

Prove that $\delta^{p}_{q}$ is a mixed tensor of rank two.

Solution.

If $\delta^{p}_{q}$ is a mixed tensor it must transform as

\begin{equation*} \bar{\delta}^{j}_{k} = \frac{\partial\bar{x}^{j}}{\partial x^{p}}\frac{\partial x^{q}}{\partial \bar{x}^{k}}\delta^{p}_{q} = \frac{\partial\bar{x}^{j}}{\partial x^{p}}\frac{\partial x^{p}}{\partial \bar{x}^{k}} = \frac{\partial\bar{x}^{j}}{\partial \bar{x}^{k}} = \delta^{j}_{k}. \end{equation*}

\begin{equation*} \because \quad \bar{\delta}^{j}_{k} = \delta^{j}_{k} = \begin{cases} 1,& \text{if } j = k \\ 0, & \text{if } j \neq k \end{cases} \end{equation*}

Therefore, $\delta^{p}_{q}$ is a mixed tensor of rank two.

Example 3.3.13.

Prove $\delta_{ii} = 3\text{.}$

Solution.

\begin{equation*} \delta_{ii} = \delta_{11} +\delta_{22}+\delta_{33}= 3$. \end{equation*}

Example 3.3.14.

Prove $\delta_{ik}\omega_{ik} = \omega_{ii}\text{.}$

Solution.

In $\delta_{ik}\omega_{ik}$ both $i$ and $k$ are dummy and they are to be summed over from 1 to 3. As $\delta_{ik} = 1$ only if $i = k$ and $\delta_{ik} = 0$ for $i \neq k\text{.}$ So only terms for $i=k = 1,2,3$ will contribute and thus

\begin{equation*} \delta_{ik}\omega_{ik} = \omega_{kk}= \omega_{11}+\omega_{22}+\omega_{33}=\omega_{ii}. \end{equation*}

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