Tensors chapter 1 — Vishal Rao

#Chapter 1: Tensors

In an \(n\)-dimentional space a tensor of rank \('r'\) is a mathematical object that has \('r'\) indices and \(n^r\) componente and obey certain transformation rules.

[the word "components" is written in below the line, as an insertion]

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  \draw[->] (0,0) -- (0,1.7) node[left] {$z$};
  \draw[->] (0,0) -- (1.8,0) node[right] {$y$};
  \draw[->] (0,0) -- (-0.9,-1.1) node[below] {$x$};
  \draw[->] (0,0) -- (0.35,1.0);
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\[ \vec{A} = x\hat{i} + y\hat{j} + z\hat{k} \qquad \text{here } (\vec A = \vec r) \]

components in basis \((\hat i, \hat j, \hat k)\)

vector can also define area ,

\[ \vec{dA} = |dA|\,\hat n \]

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  \draw[->] (0.55,0.5) -- (1.7,0.6) node[right] {$\hat n$};
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\((\text{area} = dA)\)

scalar

To specify temp of water

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  \draw (0,1.0) -- (0.12,0) -- (0.95,0) -- (1.07,1.0);
  \draw (0.15,0.72) -- (0.95,0.72);
  \draw (0.28,0.42) -- (0.72,0.42);
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we don't need any basis vector.

\(0\) Basis vector / component

vector :-

Per component we require one basis vector.

\[ \vec A = 3\hat i + 4\hat j \]

for each component we require only one basis vector.

How will you define stress at point \(O\).

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  \fill (2.0,0.72) circle (0.05);
  \node at (2.35,0.72) {$O$};
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we can cut surface in 3 ways at point \('O'\)

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  % first cut
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    \draw[->] (0.6,0.35) -- (0.6,-0.75) node[below] {$\sigma_{xx}$};
  \end{scope}
  % second cut
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    \draw[->] (0.45,0.35) -- (1.7,0.35) node[right] {$\sigma_{yy}$};
    \draw[->] (0.45,0.35) -- (0.45,-0.8) node[below] {$\sigma_{yx}$};
  \end{scope}
  % third cut
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    \fill (0.45,0.35) circle (0.04);
    \draw[->] (0.45,0.35) -- (0.45,1.4) node[above right] {$\sigma_{zz}$};
    \draw[->] (0.45,0.35) -- (1.7,0.35) node[right] {$\sigma_{zy}$};
    \draw[->] (0.45,0.35) -- (0.45,-0.8) node[below] {$\sigma_{zx}$};
  \end{scope}
\end{tikzpicture}

first \('x'\) indicate dir\(^n\) of area & second indices indicate dir\(^n\) of stsers/pressure.

\[ \sigma = \begin{bmatrix} \sigma_{xx} & \sigma_{xy} & \sigma_{xz} \\ \sigma_{yx} & \sigma_{yy} & \sigma_{yz} \\ \sigma_{zx} & \sigma_{zy} & \sigma_{zz} \end{bmatrix} \]

[we need 2 basis vectors to write one component.]

(arrows under the columns of the matrix) All components with \(\sigma\) along \(x\)-dirn \(\big|\) \(\left(\sigma \text{ along } 'y'\right)\) \(\big|\) \(\left(\sigma \text{ along } z\right)\)

[we can't add \(\sigma_{xx} + \sigma_{yx} + \sigma_{zx}\) just because all are in same dir\(^n\) but these componente have different nature]

longitudnal stress \(= \sigma_{xx}\)

shear stress \(= \sigma_{yx}\)

To specify piezoelectricity & stiffness we require 3 basis vector per component & 4 basis vector per component respectively.

So, For a tensor of rank \('r'\) in \('n'\) dimentional space has \(n^r\) components and require \('r'\) basis to define one component.

Summary

\[ \begin{aligned} \text{scalar} \quad &-\quad \text{Tensor of rank } '0' \qquad &&(3^0 = 1 \text{ component})\\ \text{vector} \quad &-\quad \text{Tensor of rank } '1' \qquad &&(3^1 = 3 \text{ components})\\ \text{stress} \quad &-\quad \text{Tensor of rank } '2' \qquad &&(3^2 = 9 \text{ components})\\ & && (4^2 = 16 \text{ components in 4-D} \cdots)\\ \text{piezoelectricity} \quad &-\quad \text{Tensor of rank } '3' \qquad &&(\text{In 3D has } 3^3 = 27 \text{ components})\\ \text{stiffness} \quad &-\quad \text{Tensor of rank } '4' \qquad &&(\text{In 3D has } 3^4 = 81 \text{ components}) \end{aligned} \]

[the word "Tensor" is written and struck out on the line below "vector"]

In form of tensor, any quantity is represented as

\[ T_i \qquad \text{or} \qquad T^i \]

component \(\big|\) basis vector (also called indices)

\(\left\{T^i = T \text{ super } i \ ;\quad T_i = T \text{ sub } i\right.\)

Representation of Tensor quantity.

\(\rightarrow\) superscript represents contravariant tensor represented as column vector

\[ p^i = \left(p^i\right)_{i=1,2,3} = \begin{pmatrix} p^1 \\ p^2 \\ p^3 \end{pmatrix} \]

\(\rightarrow\) A tensor quantity with subscript Represents covariant tensor with representation of row vector

\[ L_i = \left(L_i\right)_{i=1,2,3} = \begin{pmatrix} L_1 & L_2 & L_3\end{pmatrix} \]

\(\rightarrow\) A tensor of second rank is a matrix

\[ M_i^{\,j} = \begin{bmatrix} M^1_{\ 1} & M^1_{\ 2} & M^1_{\ 3} \\ M^2_{\ 1} & M^2_{\ 2} & M^2_{\ 3} \\ M^3_{\ 1} & M^3_{\ 2} & M^3_{\ 3} \end{bmatrix} \]

\(\rightarrow\) Any twice repeated index in single term is summed over.

\[ \sum_{i=1}^{3} a_i x_i = a_1x_1 + a_2x_2 + a_3x_3 \] \[ \sum a_i x_i = a_i x_i \qquad (\text{einstien summation convention}) \] \[ a_{ij}x_j = a_{i1}x_1 + a_{i2}x_2 + a_{i3}x_3 \] \[ a^i_{\ i} = a^1_{\ 1} + a^2_{\ 2} + a^3_{\ 3} \qquad ; \qquad a^i x_i = a^1x_1 + a^2x_2 + \cdots + a^n x_n \]

\(\rightarrow\) Any index repeated twice in a single term is called dummy index.

\[ a_{ij}x_j \qquad (j = \text{dummy index}) \]

dummy index can be replaced with other index of same range.

\[ a_{ij}x_j = a_{ik}x_k = a_{i\ell}x_\ell \ \neq\ a_{ii}x_i \]

range should be same \(\qquad (j = 1,2,3\ ,\quad k = 1,2,3\ ,\quad \ell = 1,2,3\ )\)

\(\rightarrow\) Any index occuring once in single term is called free index.

\[ a_{ij}x_j \qquad (i - \text{free index}) \]

\((i = 1,2,3)\)

index \('i'\) can have one out of 3 values.

\[ a_{ij}x_j = a_{1j}x_j = a_{2j}x_j = a_{3j}x_j \]

here

\[ a_{1j}x_j = a_{11}x_1 + a_{12}x_2 + a_{13}x_3 \] \[ a_{2j}x_j = a_{21}x_1 + a_{22}x_2 + a_{23}x_3 \] \[ a_{3j}x_j = a_{31}x_1 + a_{32}x_2 + a_{33}x_3 \] \[ \boxed{a_{ij}x_j} \]

\(j = \) dummy index \(\qquad\qquad\) \(i = \) free index

\(-\) Repeats twice in single term. \(\qquad\) \(-\) occurs once in single term

\(-\) Represents sum over range \(\qquad\) \(-\) do not represent summation

\(-\) can be replaced by other index \(\qquad\) \(-\) can't be replaced by other index.

\(-\) Index can not occur more than 2 times.

\[ \left\{ \begin{aligned} &X_{ij}b_j \quad \checkmark\\ &X_{ik}b_k \quad \checkmark\\ &X_{ii}b_i \quad \times \end{aligned}\right. \]

while counting indices, count superscript & subscript together.

\[ X^j_{\ j} \qquad j = \text{dummy index} \] \[ X^j_{\ j} = \ \ast \ \text{matrix} \ \text{(?)} \] \[ X_i^{\ jj} \qquad \begin{aligned} &i = \text{free index}\\ &j = \text{dummy index} \end{aligned} \]

\(\rightarrow\) In an equation involving einstien notation free indices on both sides must match

(1) \(\quad x_i = a_{ij}b_j \qquad \checkmark\)

(2) \(\quad x_i = a_{jj} \qquad (\times) \qquad (\text{invalid eq}^n)\)

(3) \(\quad a_i = A_{ki}B_{kj}x_j \qquad (\checkmark)\)

(4) \(\quad x_j = A_{ik}v_k \qquad (\times) \qquad (\text{invalid eq}^n)\)

\(-\) Contravariant Tensor & covariant tensor

Tensor \(\vec{AB}\) remains unchanged under general rotation about origin.

In general, in coordinate transformations component of tensor transforms in two manner contravariant and covariant.

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\[ \vec{AB} = 3\hat i + 2\hat j \] \[ \hat i' = 2\hat i \] \[ \hat j' = \hat j \]

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\[ \vec{AB} = 1.5\,\hat i' + 3\hat j' \]

components of tensor \(\vec{AB}\) are changing inverse of change of basis vectors.

so, we call these tensor as contravariant tensor.

If Basis is doubled \(\rightarrow\) component halved.

We represent such quantities as

\[ x^i,\ v^i,\ a^i \ \cdots \]

displacement, velocity, accelaration are contravariant vectors.

A tensor whose components change in same manner as basis change under given transformation.

\[ \nabla F = \frac{\partial F}{\partial x}\hat i + \frac{\partial F}{\partial y}\hat j + \frac{\partial F}{\partial z}\hat k \] \[ \hat i' = 2\hat i \] \[ \hat j' = 2\hat j \] \[ \hat k' = 2\hat k \]

\(\partial x = \) contravariant

\[ \partial x' = \frac{\partial x}{2}\ ,\qquad \partial y' = \frac{\partial y}{2}\ ,\qquad \partial z' = \frac{\partial z}{2} \] \[ \nabla F = \frac{\partial F}{\partial x'}\hat i' + \frac{\partial F}{\partial y'}\hat j' + \frac{\partial F}{\partial z'}\hat k' \]

The quantities \(\frac{\partial F}{\partial x'}, \frac{\partial F}{\partial y'}, \frac{\partial F}{\partial z'}\) also doubles just like basis vectors, such vectors are called as covariant tensors.

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[If \('a'\) is contravariant vector, its components are calculated from parallel projection.]

\[ \vec a = a^1\hat e_1 + a^2\hat e_2 \]

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[If \(\vec a\) is covariant tensor, its components are calculated from perpendicular projection.]

\[ \vec a = a_1\hat e_1 + a_2\hat e_2 \]

(1) For covariant tensor :-

suppose \(F(x)\) is scalar field defined in a coordinate system \((x^i)\)

\[ F(x) = F(x^1, x^2, x^3 \cdots x^N) \] \[ \nabla F = \left(\frac{\partial F}{\partial x^1},\ \frac{\partial F}{\partial x^2},\ \frac{\partial F}{\partial x^3} \cdots \frac{\partial F}{\partial x^n}\right) \]

where \(u_i = \frac{\partial F}{\partial x^i}\) \(\qquad (i = 1,2,3\cdots n)\) \(\quad (1 \le i \le n)\)

\[ u_i = \frac{\partial F}{\partial x^i} \]

represents \(i^{th}\) component of \(\nabla F\).

used as subsmpt as \(\nabla F\) is covarient tensor.

\[ \nabla F_i = \left(\frac{\partial F}{\partial x^i}\right) = u_i \]

If the coordinate system transformed to new coordinate system \((\bar x^i)\) gradient is defined as

\[ \nabla F = \left(\frac{\partial F}{\partial \bar x^1},\ \frac{\partial F}{\partial \bar x^2},\ \frac{\partial F}{\partial \bar x^3} \cdots \frac{\partial F}{\partial \bar x^n}\right) \]

where,

\[ \bar u_i = \frac{\partial F}{\partial \bar x^i} \qquad (1 \le i \le N) \]

represents \(i^{th}\) component of \(\nabla F\).

\[ \bar u_i = \frac{\partial F}{\partial \bar x^i} = \frac{\partial F}{\partial \bar x^i}\left(\frac{\partial x^r}{\partial x^r}\right) \] \[ = \left(\frac{\partial F}{\partial x^r}\right)\frac{\partial x^r}{\partial \bar x^i} = \frac{\partial F}{\partial x^1}\frac{\partial x^1}{\partial \bar x^i} + \frac{\partial F}{\partial x^2}\frac{\partial x^2}{\partial \bar x^i} + \cdots \] \[ \boxed{\ \bar u_i = u_r\,\frac{\partial x^r}{\partial \bar x^i}\ } \]

where \(\left(u_r = \frac{\partial F}{\partial x^r}\right)\)

\[ \bar u_i = u_r\frac{\partial x^r}{\partial \bar x^i} \qquad \begin{aligned} \bar u_i &= \text{component of } \nabla F \text{ in } \bar x^i\\ u_r &= \underline{\qquad\qquad} \text{ in } x^i \end{aligned} \]

\(\frac{\partial x^r}{\partial \bar x^i}\) represents transformation from \((x^i\) to \(\bar x^i)\) coordinate system

Def\(^n\) :-

covariant 1st rank tensor :-

The vector field \('u'\) is said to be covariant first rank tensor, if its component \(u_i\) in \((x^i)\) coordinate system and \(\bar u_i\) in \((\bar x^i)\) coordinate system are related by following Transformation.

\[ \bar u_i = u_r\frac{\partial x^r}{\partial \bar x^i} \qquad (1 \le i \le n) \]

(2) Transformation for contravariant tensor :-

suppose \(v(x) = v(x^1, x^2 \cdots x^n)\) is velocity defined in coordinate system \((x^i)\) \(\quad\) [\((n\)-d space)]

\[ \vec v = \left(\frac{\partial x^1}{\partial t},\ \frac{\partial x^2}{\partial t},\ \frac{\partial x^3}{\partial t},\ \cdots \frac{\partial x^n}{\partial t}\right) \]

where

\[ v^i = \frac{\partial x^i}{\partial t} \qquad (1 \le i \le n) \]

If coordinate system is transformed to new coordinate system, \((\bar x^i)\)

\[ \vec v = \left(\frac{\partial \bar x^1}{\partial t},\ \frac{\partial \bar x^2}{\partial t},\ \frac{\partial \bar x^3}{\partial t},\ \cdots \frac{\partial \bar x^n}{\partial t}\right) \] \[ \bar v^i = \frac{\partial \bar x^i}{\partial t} \] \[ = \frac{\partial \bar x^i}{\partial t}\ \frac{\partial x^r}{\partial x^r} \] \[ = \frac{\partial \bar x^i}{\partial x^r}\left(\frac{\partial x^r}{\partial t}\right) = \frac{\partial \bar x^i}{\partial x^1}\frac{\partial x^1}{\partial t} + \frac{\partial \bar x^i}{\partial x^2}\frac{\partial x^2}{\partial t} + \cdots \] \[ \boxed{\ \bar v^i = v^r\,\frac{\partial \bar x^i}{\partial x^r}\ } \qquad v^r = \frac{\partial x^r}{\partial t} \]

(ex)

\[ \bar v^1 = v^r\frac{\partial \bar x^1}{\partial x^r} = v^1\frac{\partial \bar x^1}{\partial x^1} + v^2\frac{\partial \bar x^1}{\partial x^2} + v^3\frac{\partial \bar x^1}{\partial x^3} + \cdots \]

Summary

\[ \bar v^i = v^r\frac{\partial \bar x^i}{\partial x^r} \qquad \begin{aligned} \bar v^i &= \text{component of } \vec v \text{ in } \bar x^i\\ v^r &= \underline{\qquad\qquad} \text{ in } x^i \end{aligned} \]

Def\(^n\) :- The vector field \('v'\) is said to be contravariant 1st rank tensor if its components \(v^i\) in \((x^i)\) coordinate system and \(\bar v^i\) in \(\bar x^i\) coordinate system are related by following set of transformation.

\[ \bar v^i = v^r\frac{\partial \bar x^i}{\partial x^r} \qquad (1 \le i \le n) \]

Summary :-

\[ \begin{array}{ll} \underline{\text{covariant}} & \underline{\text{contravarient}}\\ (\text{in same manner}) & (\text{in opposite manner}) \end{array} \]

\(-\) Basis & components transform in same manner \(\qquad\) \(-\) Basis and components transform in opposite manner

\(-\) Indices are in subscript of component \((T_i)\) \(\qquad\) \(-\) indices are superscript of tensor quantity \((T^i)\)

\[ \vec a = a_1\hat e_1 + a_2\hat e_2 \qquad\qquad \vec a = a^1\hat e_1 + a^2\hat e_2 \]

ex \(\ \left. \nabla F\right)_i = u_i = \frac{\partial F}{\partial x^i}\) \(\qquad\qquad\) ex displacement, velocity, accelaration, \(\cdots\)

\(-\) Transformation rule \(\qquad\qquad\) \(-\) Transformation rule

\[ \left\{\bar u_i = u_r\frac{\partial x^r}{\partial \bar x^i}\right\} \qquad\qquad \left\{\bar v^i = v^r\frac{\partial \bar x^i}{\partial x^r}\right\} \]

(Q.1) expand \(a_{ij}(b_i + c_j + d_k)\)

\((\times)\)

\[ \left\{ \begin{aligned} &= a_{ij}b_i + a_{ij}c_j + a_{ij}d_k\\ &= \left(a_{1j}b_1 + a_{2j}b_2 + a_{3j}b_3\right) + \left(a_{i1}c_1 + a_{i2}c_2 + a_{i3}c_3\right) + a_{ij}d_k \end{aligned}\right. \]

Brackets in Einstien notation

\[ a_{ij}(b_i + c_j + d_k) \]

(step-1) Combine terms outside parantheses with each term inside seperately.

\[ \left(a_{ij}b_i\right) + \left(a_{ij}c_j\right) + \left(a_{ij}d_k\right) \] \[ \begin{array}{lll} i=2 & i=1 & i=1\\ j=1 & j=2 & j=1,\ k>1 \end{array} \] \[ \left.\begin{aligned} i_{max} &= 2\\ j_{max} &= 2 \end{aligned}\right\} i,j = \text{dummy} \] \[ k_{max} = 1 \qquad \longrightarrow\ k = \text{free indices} \]

(1)

\[ a_{ij}(x_i + y_j) \neq a_{ij}x_i + a_{ij}y_j \]

LHS : \(a_{ij}(x_i + y_j)\) \(\qquad\) \(\left.\begin{aligned} i_{max} &= 2\\ j_{max} &= 2\end{aligned}\right\} i,j = \text{dummys}\)

RHS \(\quad a_{ij}x_i + a_{ij}y_j\)

[under the first term: \(i = \) dumy, \(j = \) free ; under the second term: \(j = \) dummy, \(i = \) free]

RHS is inconsistent with LHS

(2)

\[ a_{ij}x_iy_j \neq a_{ij}x_jy_i \qquad (i,j = \text{dumy both sides}) \]

but coeff of \(x_1y_2 = \left\{\begin{aligned} a_{12}&\quad \text{LHS}\\ a_{21}&\quad \text{RHS}\end{aligned}\right.\)

(3)

\[ (a_{ij}+a_{ji})\,x_iy_i \ \neq\ 2a_{ij}x_iy_j \]

\((i,j = \) dummy both sides but\()\) \(\underline{(a_{ij}+a_{ji})}\) .

(1)

\[ a_{ij}(x_j + y_j) = a_{ij}x_j + a_{ij}y_j \]

(2)

\[ a_{ij}x_iy_j = a_{ij}y_jx_i \]

(3)

\[ a_{ij}x_ix_j = a_{ji}x_ix_j \equiv a_{ji}x_jx_i \]

(4)

\[ (a_{ij}+a_{ji})x_ix_j \equiv 2a_{ij}x_ix_j \]

(5)

\[ (a_{ij}-a_{ji})x_ix_j \equiv 0 \]

\[ \delta_{ij} \equiv \delta^i_j \equiv \delta^{ij} \equiv \begin{cases} 1 & i = j\\ 0 & i \neq j\end{cases} \] \[ \delta_{ij} = \delta_{ji} \qquad (\forall\ i,j) \]

example \(\quad\) Simplify \(\ \delta_{ij}x_ix_j\)

\[ \delta_{ij}x_ix_j = \delta_{ii}x_ix_i = \delta_{jj}x_jx_j \] \[ \delta_{ij}x_ix_j = x_ix_i = x_jx_j \qquad \checkmark \]

linear transformations

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\begin{tikzpicture}[scale=1.0]
  \draw[->] (0,0) -- (0,2.6);
  \draw[->] (0,0) -- (-1.2,0);
  \draw[->] (0,0) -- (-0.7,2.6);
  \draw[->] (0,0) -- (-0.5,-0.5);
  \draw[->] (0,0) -- (1.3,1.4) node[above left] {$x$};
  \draw[->] (0,0) -- (1.5,1.35) node[above right] {$x'$};
  \draw[->] (0,0) -- (3.6,0) node[right] {$c$};
  \draw[->] (0,0) -- (3.4,0.9) node[right] {$c'$};
\end{tikzpicture}

[linear \(-\) scaling, rotation]

Suppose we have vector \('x'\) in coordinate system \(c\), If we do linear transformation of coordinate system to \(c'\), then vector \(x\) will look like \(x'\) in new \(c'\) coordinate system.

\[ x' = L\,x \]

\(\rightarrow\) matrix \((2\times 2)\) Transformation matrix

General Transformation :

\usetikzlibrary{arrows.meta,decorations.pathreplacing,calc,angles,quotes,positioning,patterns}
\begin{tikzpicture}[scale=1.0]
  \draw[->] (0,0) -- (0,2.4);
  \draw[->] (0,0) -- (3.6,0) node[right] {$c$};
  \draw[->] (0,0) -- (1.4,1.5) node[above right] {$x$};
  \draw[->] (-0.9,-0.7) .. controls (-1.2,0.8) .. (-0.7,1.9);
  \draw[->] (-1.4,-0.8) .. controls (0.6,-0.4) and (2.2,0.9) .. (3.3,1.0) node[right] {$c'$};
\end{tikzpicture}

New vector will look like \(x'\) in new coordinate system \(c'\),

\[ x' = f(x) = T(\vec x) \]

(or)

\[ x'_i = T_p(x_1,x_2,x_3 \cdots x_n) \]

A bijective transformation is one to one, every \('x'\) in original coordinate system coorespondes to unique \(x'\) in the image coordinate system & every \(x'\) in image coordinate system corresponds to a unique \(x\) in the original coordinate system.

\usetikzlibrary{arrows.meta,decorations.pathreplacing,calc,angles,quotes,positioning,patterns}
\begin{tikzpicture}[scale=1.0]
  \draw plot [smooth cycle] coordinates {(0,0) (0.9,0.9) (0.4,1.9) (-0.7,1.6) (-1.0,0.7)};
  \draw plot [smooth cycle] coordinates {(5.2,0) (6.2,0.8) (5.9,2.0) (4.7,1.8) (4.3,0.8)};
  \node at (-0.2,-0.6) {$c$};
  \node at (5.3,-0.6) {$c'$};
  \node at (-0.55,1.35) {$x_1$}; \fill (-0.25,1.35) circle (0.03);
  \node at (-0.55,0.75) {$x_2$}; \fill (-0.25,0.75) circle (0.03);
  \node at (-0.55,0.2) {$x_3$}; \fill (-0.25,0.2) circle (0.03);
  \node at (5.3,1.5) {$x_1'$};
  \node at (5.3,1.0) {$x_2'$};
  \node at (5.3,0.5) {$x_3'$};
  \draw[->] (-0.2,1.4) .. controls (2.0,1.9) .. (5.05,1.5);
  \draw[->] (-0.2,0.8) .. controls (2.0,1.2) .. (5.05,1.0);
  \draw[->] (-0.2,0.25) .. controls (2.0,0.7) .. (5.05,0.5);
  \node at (2.4,2.0) {one$-$one};
\end{tikzpicture}

The image of \(x'\) of bijective transformation \('T'\) represents a set of admissible coordinates for \(x\)

\(\left(\begin{aligned}\text{collection of } x_i &\text{ represents } c\\ "\qquad\quad x'_i &\quad " \qquad c'\end{aligned}\right)\)

If \(T\) is linear \(x_i\) coordinate system is called affine coordinate system. If \(T\) is non-linear then \(\{x_i\}\) makes up curvilinear coordinate systems ex, polar, cylindrical, spherical coordinate system.

Suppose

\[ w = f(u_1, u_2, u_3 \cdots u_n) \]

where \(u_i = u_i(x_1, x_2 \cdots x_m)\)

Then we know that

\[ \frac{\partial w}{\partial x_j} = \frac{\partial f}{\partial u_1}\frac{\partial u_1}{\partial x_j} + \frac{\partial f}{\partial u_2}\frac{\partial u_2}{\partial x_j} + \frac{\partial f}{\partial u_3}\frac{\partial u_3}{\partial x_j} + \cdots \]

\[ \left(\frac{\partial w}{\partial x_j} = \frac{\partial f}{\partial u_i}\frac{\partial u_i}{\partial x_j}\right) \quad \longleftarrow \ \text{einstiens summation convention} \]

New notation :- A vector in \(\mathbb{R}^n\)

\usetikzlibrary{arrows.meta,decorations.pathreplacing,calc,angles,quotes,positioning,patterns}
\begin{tikzpicture}[scale=1.0]
  \draw (0,-0.2) -- (0,1.6);
  \draw (0,-0.2) -- (2.6,-0.2);
  \fill (0.25,0.15) circle (0.04); \node[below] at (0.25,0.1) {$P$};
  \fill (1.3,0.95) circle (0.04); \node[above] at (1.3,1.0) {$Q$};
  \draw (0.25,0.15) -- (1.3,0.95);
  \fill (1.15,0.15) circle (0.04); \node[right] at (1.2,0.15) {$A$};
\end{tikzpicture}

\[ \overline{PQ} = \begin{pmatrix} x^1 \\ x^2 \\ x^3 \\ \vdots \\ x^N\end{pmatrix} \]

[use parantheses for power]

[\((a_i)^2 = a_i\,a_i\)]

[\((a^i)^2 = a^i \times a^i\)]

coordinate of \('A'\) in \(\mathbb{R}^N\) \(=\)

\[ \begin{pmatrix} a^1 \\ a^2 \\ a^3 \\ \vdots \\ a^N\end{pmatrix} \]

Ractangular coordinates :-

\usetikzlibrary{arrows.meta,decorations.pathreplacing,calc,angles,quotes,positioning,patterns}
\begin{tikzpicture}[scale=1.0]
  \draw (0,-0.2) -- (0,1.9);
  \draw (-0.6,0) -- (3.0,0);
  \node[below left] at (0,0) {$O$};
  \fill (1.1,1.05) circle (0.04); \node[above left] at (1.1,1.05) {$B$};
  \draw[decorate,decoration={brace,amplitude=4pt}] (1.25,1.05) -- (1.25,0.05);
  \node[right] at (1.35,0.55) {$x^2$};
  \draw[decorate,decoration={brace,mirror,amplitude=4pt}] (0,-0.25) -- (1.1,-0.25);
  \node[below] at (0.55,-0.35) {$x^1$};
\end{tikzpicture}

\[ B = \begin{pmatrix} x^1 \\ x^2\end{pmatrix} \]

A coordinate system is said to be ractangular if distance b/w two points \(C(x^1x^2 \cdots x^N)\) & \(D(y^1\,y^2 \cdots y^N)\) is given by

\[ \sqrt{(x^1-y^1)^2 + (x^2-y^2)^2 + (x^3-y^3)^2 + \cdots} \]

\[ \sqrt{\delta_{ij}\,\Delta x^i \Delta x^j} \]

\[ \sqrt{\Delta x^i \Delta x^i} \qquad\qquad \Delta x^i = (x^i - y^i) \]

suppose:

\(P\) is a point in a coordinate system \((x^i)\) in \(\mathbb{R}^n\) given by \(P: (x^1, x^2 \cdots , x^n)\)
\((\bar x^i)\) is another coordinate system in \(\mathbb{R}^n\) such that the coordinates of \('p'\) in this system are \((\bar x^1\ \bar x^2 \cdots \bar x^n)\)
suppose that \(\bar x^i = \bar x^i(x^1\ x^2\ x^3 \cdots x^N)\) [\(F\) denotes \(N\) such transformations]. If functions \(\bar x^i(x^1 x^2 \cdots x^n)\) are all real valued, have continuous 2nd order partial derivatives every where and are all invertible then \(F\) is called coordinate transformation.

If \((x^i)\) being inputs of \(F\) are ractangular coordinates then \((\bar x^i)\) is

Affine if \(F\) is linear
curvilinear if \(F\) is not-linear

\(-\) examples of curvilinear coordinates

(1) polar coordinates

\usetikzlibrary{arrows.meta,decorations.pathreplacing,calc,angles,quotes,positioning,patterns}
\begin{tikzpicture}[scale=1.0]
  \draw[->] (0,-0.1) -- (0,2.0);
  \draw[->] (-0.3,0) -- (2.8,0) node[below] {$x^1$};
  \fill (1.6,1.2) circle (0.04); \node[above right] at (1.6,1.2) {$P(x^1,x^2)$};
  \draw (0,0) -- (1.6,1.2);
  \node[above left] at (0.75,0.6) {$\bar x^1$};
  \draw (0.5,0) arc (0:37:0.5);
  \node at (0.72,0.2) {$\bar x^2$};
  \draw[dashed] (1.6,1.2) -- (1.6,0);
  \node[right] at (1.65,0.6) {$x^2$};
\end{tikzpicture}

\[ \left.\begin{aligned} \bar x^1 &= \sqrt{(x^1)^2 + (x^2)^2}\\[4pt] \bar x^2 &= \arctan\left(\frac{x^2}{x^1}\right) \end{aligned}\right\} T_p \]

\(T_p\) is coordinate transformation

\(-\) As inputs of & outputs of \(T\) are real numbers

\(-\) \(T\) has or functions of \(T\) has continuous 2nd partial derivative everywhere (except origin)

\(-\) \(T_p\) is invertible

\[ \left.\begin{aligned} x^1 &= \bar x^1\cos(\bar x^2)\\ x^2 &= \bar x^1\sin(\bar x^2) \end{aligned}\right\} T_p^{-1} \]

eq\(^n\) of \(T_p\) are non-linear so \(T_p\) represents curvilinear coordinate transformation

(2) cylindrical coordinates :

\usetikzlibrary{arrows.meta,decorations.pathreplacing,calc,angles,quotes,positioning,patterns}
\begin{tikzpicture}[scale=1.0]
  \draw[->] (0,-0.2) -- (0,2.2);
  \draw[->] (0,0) -- (2.9,0);
  \draw[->] (0,0) -- (-1.1,-1.0);
  \fill (1.1,1.5) circle (0.04); \node[above right] at (1.1,1.5) {$P(x^1,x^2,x^3)$};
  \draw[dashed] (1.1,1.5) -- (1.1,0.0);
  \draw (0,0) -- (1.1,0.0);
  \node[above] at (0.75,0.02) {$\bar x^1$};
  \draw[decorate,decoration={brace,amplitude=4pt}] (-0.15,1.5) -- (-0.15,0.0);
  \node[left] at (-0.3,0.75) {$\bar x^3$};
  \node at (0.55,0.62) {$\bar x^2$};
  \node at (-0.75,-0.75) {$\bar x^2$};
\end{tikzpicture}

\[ \left.\begin{aligned} \bar x^1 &= \sqrt{(x^1)^2 + (x^2)^2}\\[4pt] \bar x^2 &= \arctan\left(\frac{x^2}{x^1}\right)\\[4pt] \bar x^3 &= x^3 \end{aligned}\right\} T_p \] \[ T_p^{-1} \Rightarrow \left\{ \begin{aligned} x^1 &= \bar x^1\cos(\bar x^2)\\ x^2 &= \bar x^1\sin(\bar x^2)\\ x^3 &= \bar x^3 \end{aligned}\right. \]

[Non-linear so represents Curvilnear Coordinate System.]

(3) spherical coordinate system :-

\usetikzlibrary{arrows.meta,decorations.pathreplacing,calc,angles,quotes,positioning,patterns}
\begin{tikzpicture}[scale=1.0]
  \draw[->] (0,-0.2) -- (0,2.3);
  \draw[->] (0,0) -- (3.0,0);
  \draw[->] (0,0) -- (-1.2,-1.1);
  \fill (1.2,1.6) circle (0.04); \node[above right] at (1.2,1.6) {$P(x^1,x^2,x^3)$};
  \draw[dashed] (0,0) -- (1.2,1.6);
  \node[left] at (0.55,1.0) {$\bar x^1$};
  \draw (0,0.7) arc (90:53:0.7);
  \node at (0.28,0.62) {$\bar x^2$};
  \draw[dashed] (1.2,1.6) -- (1.2,-0.4);
  \draw[dashed] (0,0) -- (-0.7,-0.65);
  \draw[dashed] (-0.7,-0.65) -- (1.2,-0.4);
  \node[below left] at (-0.35,-0.35) {$\bar x^3$};
\end{tikzpicture}

\[ \left.\begin{aligned} \bar x^1 &= \sqrt{(x^1)^2 + (x^2)^2 + (x^3)^2}\\[6pt] \bar x^2 &= \cos^{-1}\left(\frac{x^3}{\sqrt{(x^1)^2+(x^2)^2+(x^3)^2}}\right)\\[6pt] \bar x^3 &= \tan^{-1}\left(\frac{x^2}{x^1}\right) \end{aligned}\right\} T_p \] \[ T_p^{-1} \Rightarrow \left\{ \begin{aligned} x^1 &= \bar x^1\sin(\bar x^2)\cos(\bar x^3)\\ x^2 &= \bar x^1\sin(\bar x^2)\sin(\bar x^3)\\ x^3 &= \bar x^1\cos(\bar x^2) \end{aligned}\right. \]

\(T_p\) & \(T_p^{-1}\) have non-linear functions so \(T_p\) & \(T_p^{-1}\) represents curvilinear coordinate transformation.

A Tensor whose components change in contravarient/covariant manner is said to be contravariant/Covariant tensor.

contravariant change :- suppose we have 2-D non-orthogonal coordinate system.

\usetikzlibrary{arrows.meta,decorations.pathreplacing,calc,angles,quotes,positioning,patterns}
\begin{tikzpicture}[scale=1.0]
  \draw[->] (0,0) -- (3.4,0) node[right] {$x^1$};
  \draw[->] (0,0) -- (1.4,2.2) node[right] {$x^2$};
  \draw[->] (0,0) -- (0.28,0.44) node[left] {$\vec e_2$};
  \draw[->] (0,0) -- (0.8,0) node[below] {$\vec e_1$};
  \draw[->] (0,0) -- (2.7,1.6) node[right] {$\vec a$};
  \draw[->] (0,0) -- (1.15,1.8);
  \node at (0.72,1.15) {$a^2$};
  \draw[decorate,decoration={brace,mirror,amplitude=4pt}] (0,-0.45) -- (2.3,-0.45);
  \node[below] at (1.1,-0.55) {$a^1$};
\end{tikzpicture}

\[ \vec a = a^1\vec e_1 + a^2\vec e_2 \] \(a^1 = \) no. of \(e_1\) req. \(a^2 = \) no. of \(e_2\) requred.

what happens to \(\vec a\) if \(\vec{e_1'} = \alpha\vec{e_1}\) & \(\vec{e_2'} = \beta\vec{e_2}\)

\(\vec a\) does not change but its components will change

\usetikzlibrary{arrows.meta,decorations.pathreplacing,calc,angles,quotes,positioning,patterns}
\begin{tikzpicture}[scale=1.0]
  \draw[->] (0,0) -- (3.4,0) node[right] {$x^1$};
  \draw[->] (0,0) -- (1.4,2.2) node[right] {$x^2$};
  \draw[->] (0,0) -- (0.55,0.87) node[left] {$\bar e_2$};
  \draw[->] (0,0) -- (1.6,0) node[below] {$\bar e_1$};
  \draw[->] (0,0) -- (2.7,1.4) node[right] {$\vec a$};
\end{tikzpicture}

\[ \vec a = \bar a^1\bar e_1 + \bar a^2\bar e_2 \] \[ \bar a^1 = \frac{a^1}{\alpha} \qquad\qquad \bar a^2 = \frac{a^2}{\beta} \]

So components \(\bar a^1\) & \(\bar a^2\) are changing in contrary to \(\bar e_1\) & \(\bar e_2\).

For representation purpose we use superscripts to distinguish from covariant quantities.

Covariant change :-

\usetikzlibrary{arrows.meta,decorations.pathreplacing,calc,angles,quotes,positioning,patterns}
\begin{tikzpicture}[scale=1.0]
  \draw[->] (0,0) -- (3.4,0) node[right] {$x^1$};
  \draw[->] (0,0) -- (1.3,2.1) node[right] {$x^2$};
  \draw[->] (0,0) -- (0.28,0.45) node[left] {$\vec e_2$};
  \draw[->] (0,0) -- (0.7,0) node[below] {$\vec e_1$};
  \draw[->] (0,0) -- (2.3,1.5) node[right] {$\vec b$};
  \draw[->] (0,0) -- (1.0,1.6);
  \node at (0.62,1.0) {$b^2$};
  \draw[dashed] (2.3,1.5) -- (1.3,0.0);
  \draw[dashed] (2.3,1.5) -- (2.3,0.0);
  \draw[decorate,decoration={brace,mirror,amplitude=3pt}] (0,-0.28) -- (1.3,-0.28);
  \node[below] at (0.65,-0.36) {$b^1$};
  \draw[decorate,decoration={brace,mirror,amplitude=3pt}] (0,-0.85) -- (2.3,-0.85);
  \node[below] at (1.15,-0.93) {$b_1$};
\end{tikzpicture}

we can write \(\vec b\) as \(\vec b = b^1e_1 + b^2e_2\) in contravariant components.

We can also write \(\vec b = b_1e_1 + b_2e_2\)

where \(b_1\) & \(b_2\) are covariant components.

\[ \vec b = b_1\vec e_1 + b_2\vec e_2 \]

where

\[ b_1 = \vec b\cdot e_1 = (b^1e_1 + b^2e_2)\cdot e_1 = b^1(e_1\cdot e_1) + b^2(e_2\cdot e_1) \] \[ b_2 = \vec b\cdot e_2 = (b^1e_1 + b^2e_2)\cdot e_2 = b^1(e_1\cdot e_2) + b^2(e_2\cdot e_2) \]

If we change our basis \(e_1' = \alpha e_1\) & \(e_2' = \beta e_2\) then

\[ \vec b = b_1'e_1' + b_2'e_2' \]

where

\[ b_1' = \vec b\cdot e_1' = \vec b\cdot(\alpha e_1) = \alpha\, b_1 \] \[ b_2' = \vec b\cdot e_2' = \vec b\cdot(\beta e_2) = \beta\, b_2 \]

so we can see that components \(b_1'\) & \(b_2'\) also change in same manner as \(e_1'\) & \(e_2'\) are changing so such components are called as covariant components.

Caveats :-

vectors with covariant components are covectors (different from regular vectors)
co-vectors are better thought as functions
co-vectors are usually specified with dual vectors as basis instead of conventional basis vectors.

\(-\) Note: we have learned that contravariant vector \('v'\) transform under change of coordinate system \((x^i \rightarrow \bar x^i)\) as

\[ \bar v^i = v^r\,\frac{\partial \bar x^i}{\partial x^r} \qquad 1 \le i \le N \]

\(-\) Tangent vector of parametric curve \((x^1, x^2, x^3 \cdots x^N)\) (where \(x^i = x^i(t)\))

\[ \theta^i = \frac{dx^i}{dt} \]

is also contravariant vector / contravariant tensor of rank 1

under coordinate transformation \(x^i \longrightarrow \bar x^i\)

\[ \bar\theta^i = \frac{d\bar x^i}{dt} \qquad\qquad \bar x^i = \bar x^i(x^1, x^2 \cdots x^N) \] \[ \bar\theta^i = \frac{\partial \bar x^i}{\partial x^1}\frac{\partial x^1}{\partial t} + \frac{\partial \bar x^i}{\partial x^2}\frac{\partial x^2}{\partial t} + \cdots \] \[ \bar\theta^i = \frac{\partial \bar x^i}{\partial x^r}\left(\frac{\partial x^r}{\partial t}\right) \] \[ \boxed{\ \bar\theta^i = \theta^r\,\frac{\partial \bar x^i}{\partial x^r}\ } \]

\(\theta^i\) transform like contravariant tensor of rank 1 so it's an contravariant vector.

we have seen that covariant vector (Tensors of rank 1) transform \(v_i \rightarrow \bar v_i\) under coordinate transformation \(x^i \longrightarrow \bar x^i\).

\[ \bar v_i = v_r\,\frac{\partial x^r}{\partial \bar x^i} \]

To know how tensors of higher rank transform, In order to define higher rank tensors, we start with assumption.

[the word "components" is inserted above "tensors", and "tensor" is inserted above "rank"]

Assumptions :- suppose that \((w/V)\) is a matrix field (i.e. \((w/V)\) provides us matrix for every point in space) composed of \(n\times n\) scalar field (functions) defined

over region 'u' in \(\mathbb{R}^n\).

Assume that in \(\{x^i\}\) coordinate system, the components of V are \(V^{ij}\). Assume also that after a coordinate transformation \(T: \bar{x}^i = \bar{x}^i(x^1 x^2 - . x^n)\) that takes us to the \(\bar{x}^i\) coordinate system, the components of V becomes \(\bar{V}^{ij}\).

matrix is not a tensor. A tensor is an object that is invariant under change of coordinate systems with components that change according to a special set of mathematical formulae

A matrix field V is a contravariant tensor of rank 2 if its components \(V^{ij}\) in \(\{x^i\}\) coordinate system transform to \(\bar{V}^{ij}\) in \(\{\bar{x}^i\}\) coordinate system, such that

\[ \bar{V}^{ij} = V^{rs}\,\frac{\partial \bar{x}^i}{\partial x^r}\,\frac{\partial \bar{x}^j}{\partial x^s} \]

[\(i,j\) -- free indices]

[\(r,s\) -- dummy indices.]

\[ (1 \leq i,j,r,s \leq n) \]

A matrix field 'W' is covariant tensor of rank-2 if its components \(W_{ij}\) in \(\{x^i\}\) coordinate system & \(\bar{W}_{ij}\) in \(\{\bar{x}^i\}\) coordinate system obey

\[ \bar{W}_{ij} = W_{rs}\,\frac{\partial x^r}{\partial \bar{x}^i}\,\frac{\partial x^s}{\partial \bar{x}^j} \]

suppose that \(Z\) is a matrix field composed of \(n\times n\) scalar fields defined over region \(u\) in \(\mathbb{R}^n\). Assume that in \(\{x^i\}\) coordinate system, the components of \(Z\) are \(Z^i_j\), which becomes \(\bar{Z}^i_j\) after coordinate transformation \(T: \bar{x}^i = \bar{x}^i(x^1 x^2 . x^N)\)

then \(Z\) (is matrix field) is mixed tensor of rank-2 of its components \(Z^i_j\) in \(x^i\) coordinate system.

& \(\bar{Z}^i_j\) in \(\bar{x}^i\) coordinate system obey

\[ \bar{Z}^i_j = Z^r_s\,\frac{\partial \bar{x}^i}{\partial x^r}\,\frac{\partial x^s}{\partial \bar{x}^j} \]

Suppose \(V\) is tensor with components

\[ V^{i_1 i_2 i_3 - i_p}_{j_1 j_2 -- j_q} \]

then total rank of \(V\) is \((m = p+q)\).

we call \(V\) as \((p,q)\) tensor.

[\(p\) -- contravariant rank ; \(q\) -- covariant rank.]

ex, If \(A\) is a tensor with components

\[ A^{ij} \quad \text{then } A \text{ is } (2,0) \text{ tensor} \] \[ A_{ij} \quad \text{then } A \text{ is } (0,2) \text{ tensor} \] \[ A^i_j \quad \text{-----------} \quad (1,1) \text{ tensor.} \]

Assumption :- suppose \(V\) is an array field (It returns 1D/2D/3D -- mD array) For each point in \(\mathbb{R}^n\)) composed of \(n^m\) scalar fields (functions) defined over a region 'u' in \(\mathbb{R}^n\). Assume that in the \((x^i)\) coordinate system components of \(V\) are \(V^{i_1 i_2 -- i_p}_{j_1 j_2 - j_q}\) where \((p+q=m)\). Assume also that after a coordinate transformation \(T: \bar{x}^i = \bar{x}^i(x^1 x^2 - x^N)\) that takes us to \((\bar{x}^i)\) coordinate system components of \(\bar{V}\) becomes.

\[ \bar{V}^{i_1 i_2 -- i_p}_{j_1 j_2 - j_q} \]

In order to be array field \(\bar{V}\) to be tensor of rank \(m = p+q\) with contravariant rank \(p\) & covariant rank 'q', then its components \(\bar{V}^{i_1 i_2 -- i_p}_{j_1 j_2 - j_q}\) in \((\bar{x}^i)\) coordinate system & \(V^{i_1 i_2 - i_p}_{j_1 j_2 -- j_q}\) in \((x^i)\) coordinate system should obey

\[ \bar{V}^{i_1 i_2 -- i_p}_{j_1 j_2 -- j_q} = V^{r_1 r_2 -- r_p}_{s_1 s_2 -- s_q}\ \frac{\partial \bar{x}^{i_1}}{\partial x^{r_1}}\frac{\partial \bar{x}^{i_2}}{\partial x^{r_2}}\frac{\partial \bar{x}^{i_3}}{\partial x^{r_3}}\cdots\frac{\partial x^{s_1}}{\partial \bar{x}^{j_1}}\frac{\partial x^{s_2}}{\partial \bar{x}^{j_2}}\frac{\partial x^{s_q}}{\partial \bar{x}^{j_q}} \] \[ \left(1 \leq i,j,r,s \leq N\right) \]

[\(i\)'s & \(j\)'s are free indices]

[\(r\)'s & \(s\) are dummy indices.]

suppose that 'S' & 'T' are tensors given by

\[ S \equiv \left( S^{i_1 i_2 -- i_p}_{j_1 j_2 --- j_q} \right) \ , \qquad T \equiv \left( T^{k_1 k_2 - k_r}_{l_1 l_2 -- l_s} \right) \]

\(S\) is \((p,q)\) tensor , \(T\) is \((r,s)\) tensor.

summation :- \[ \left.\begin{array}{c} p = r \\ q = s \end{array}\right\}\ \text{contravariant \& covariant rank should be same.} \] \[ S+T = \left( S^{i_1 i_2 --- i_p}_{j_1 j_2 - j_q} + T^{k_1 k_2 -- k_p}_{l_1 l_2 - l_q} \right) \] Then \((S+T)\) is tensor of rank same as \(S\) & \(T\). Proof :- Assume that we change coordinates \(x^i \rightarrow \bar{x}^i\) Then \[ \bar{S}^{i_1 i_2 -- i_p}_{j_1 j_2 - j_q} = S^{\alpha_1 \alpha_2 -- \alpha_p}_{\beta_1 \beta_2 - \beta_q}\ \frac{\partial \bar{x}^{i_1}}{\partial x^{\alpha_1}}\frac{\partial \bar{x}^{i_2}}{\partial x^{\alpha_2}}\cdots\frac{\partial \bar{x}^{i_p}}{\partial x^{\alpha_p}}\ \frac{\partial x^{\beta_1}}{\partial \bar{x}^{j_1}}\frac{\partial x^{\beta_2}}{\partial \bar{x}^{j_2}}\cdots\frac{\partial x^{\beta_q}}{\partial \bar{x}^{j_q}} \qquad (1) \] Similarly \[ \bar{T}^{i_1 i_2 -- i_p}_{j_1 j_2 -- j_q} = T^{\alpha_1 \alpha_2 - \alpha_p}_{\beta_1 \beta_2 - \beta_q}\ \frac{\partial \bar{x}^{i_1}}{\partial x^{\alpha_1}}\frac{\partial \bar{x}^{i_2}}{\partial x^{\alpha_2}}\cdots\frac{\partial \bar{x}^{i_p}}{\partial x^{\alpha_p}}\ \frac{\partial x^{\beta_1}}{\partial \bar{x}^{j_1}}\frac{\partial x^{\beta_2}}{\partial \bar{x}^{j_2}}\cdots\frac{\partial x^{\beta_q}}{\partial \bar{x}^{j_q}} \qquad (2) \] \((1) \pm (2) \Rightarrow\) \[ \bar{S}^{i_1 i_2 - i_p}_{j_1 j_2 - j_q} \pm \bar{T}^{i_1 i_2\ i_p}_{j_1 j_2 - j_q} = \left( S^{\alpha_1 \alpha_2 - \alpha_p}_{\beta_1 \beta_2 - \beta_q} \pm T^{\alpha_1 \alpha_2 - \alpha_p}_{\beta_1 \beta_2 - \beta_q} \right) C \] \[ C = \frac{\partial \bar{x}^{i_1}}{\partial x^{\alpha_1}}\frac{\partial \bar{x}^{i_2}}{\partial x^{\alpha_2}}\cdots\frac{\partial \bar{x}^{i_p}}{\partial x^{\alpha_p}}\cdot\frac{\partial x^{\beta_1}}{\partial \bar{x}^{j_1}}\frac{\partial x^{\beta_2}}{\partial \bar{x}^{j_2}}\cdots\frac{\partial x^{\beta_q}}{\partial \bar{x}^{j_q}} \] [\(\downarrow\) Transformation law for \((S \pm T)\)] \(S+T\) is also tensor of same rank as \(S\) & \(T\) \((p,q)\) \[ \left(\begin{array}{l} S, T \text{ are } (p,q) \text{ tensor} \\ (S \pm T) \text{ is } (p,q) \text{ tensor} \end{array}\right) \]

(2) Scalar multiplication :-

Suppose \(A\) is a \((p,q)\) tensor. If \(\phi\) is a scalar then \(\phi A\) is found by

\[ \phi A^{i_1 i_2 -- i_p}_{j_1 j_2 -- j_q} \]

i.e. multiplying each component \(A^{i_1 i_2 - i_p}_{j_1 j_2 - j_q}\) by \(\phi\).

Then \(\phi A\) is also \((p,q)\) tensor.

Proof :- \(A\) transform like -- (If it is a \((p,q)\) tensor)

\[ \bar{A}^{i_1 i_2 -- i_p}_{j_1 j_2 - j_q} = A^{\alpha_1 \alpha_2 -- \alpha_p}_{\beta_1 \beta_2 -- \beta_q}\left(\frac{\partial \bar{x}^{i_1}}{\partial x^{\alpha_1}}\frac{\partial \bar{x}^{i_2}}{\partial x^{\alpha_2}}\cdots\frac{\partial \bar{x}^{i_p}}{\partial x^{\alpha_p}}\right)\left(\frac{\partial x^{\beta_1}}{\partial \bar{x}^{j_1}}\frac{\partial x^{\beta_2}}{\partial \bar{x}^{j_2}}\cdots\frac{\partial x^{\beta_q}}{\partial \bar{x}^{j_q}}\right) \]

we can multiply each side by \(\phi\) (scalar)

\[ \phi\,\bar{A}^{i_1 i_2 - i_p}_{j_1 j_2 - j_q} = \phi\,A^{\alpha_1 \alpha_2 - \alpha_p}_{\beta_1 \beta_2 - \beta_q}\left(\frac{\partial \bar{x}^{i_1}}{\partial x^{\alpha_1}}\frac{\partial \bar{x}^{i_2}}{\partial x^{\alpha_2}}\cdots\frac{\partial \bar{x}^{i_p}}{\partial x^{\alpha_p}}\right)\left(\frac{\partial x^{\beta_1}}{\partial \bar{x}^{j_1}}\frac{\partial x^{\beta_2}}{\partial \bar{x}^{j_2}}\cdots\frac{\partial x^{\beta_q}}{\partial \bar{x}^{j_q}}\right) \]

since \(\phi\) is scalar \(\left(\bar{\phi} = \phi\right)\)

\[ \bar{\phi}\,\bar{A}^{i_1 i_2 -- i_p}_{j_1 j_2 - j_q} = \phi\,A^{\alpha_1 \alpha_2 - \alpha_p}_{\beta_1 \beta_2 -- \beta_q}\left(\frac{\partial \bar{x}^{i_1}}{\partial x^{\alpha_1}}\frac{\partial \bar{x}^{i_2}}{\partial x^{\alpha_2}}\cdots\frac{\partial \bar{x}^{i_p}}{\partial x^{\alpha_p}}\right)\left(\frac{\partial x^{\beta_1}}{\partial \bar{x}^{j_1}}\frac{\partial x^{\beta_2}}{\partial \bar{x}^{j_2}}\cdots\frac{\partial x^{\beta_q}}{\partial \bar{x}^{j_q}}\right) \]

So this implies \((\phi A)\) also obey transformation law of that of \(A\) i.e. \((p,q)\) tensor. So \((\phi A)\) is also a \((p,q)\) tensor.

(3) Linear combinations :-

If \((T_1\ T_2\ T_3 -- T_u)\) are tensors of same contravariant & covariant rank i.e. \((p,q)\) tensor

then, \(\left(\lambda_1 T_1 + \lambda_2 T_2 + \lambda_3 T_3 - \lambda_u T_u\right)\) is also tensor of same contravariant & covariant rank.

i.e.

\[ \left(\lambda_1 T_1 + \lambda_2 T_2 + \cdots \lambda_u T_u\right) = \lambda_i T_i \ \text{ is also } (p,q) \text{ tensor.} \] \[ \left(\text{if } T_i \text{ is } (p,q) \text{ tensor}\right). \] \[ (1 \leq i \leq u). \]

Tensor operations :-

suppose 'S' & 'T' are tensors given by:

\[ S \text{ is } (p,q) \text{ tensor} \ : \ S = \left( S^{i_1 i_2 -- i_p}_{j_1 j_2 - j_q} \right) \] \[ T \text{ is } (r,s) \text{ tensor} \ : \ T = \left( T^{k_1 k_2 - k_r}_{l_1 l_2 -- l_s} \right) \]

i.e

\[ \bar{S}^{i_1 i_2 -- i_p}_{j_1 j_2 - j_q} = S^{\alpha_1 \alpha_2 -- \alpha_p}_{\beta_1 \beta_2 - \beta_q}\left(\frac{\partial \bar{x}^{i_1}}{\partial x^{\alpha_1}}\frac{\partial \bar{x}^{i_2}}{\partial x^{\alpha_2}}\cdots\frac{\partial \bar{x}^{i_p}}{\partial x^{\alpha_p}}\right)\left(\frac{\partial x^{\beta_1}}{\partial \bar{x}^{j_1}}\frac{\partial x^{\beta_2}}{\partial \bar{x}^{j_2}}\cdots\frac{\partial x^{\beta_q}}{\partial \bar{x}^{j_q}}\right) \] \[ \bar{T}^{k_1 k_2 -- k_r}_{l_1 l_2 -- l_s} = T^{a_1 a_2 -- a_p}_{b_1 b_2 -- b_s}\left(\frac{\partial \bar{x}^{k_1}}{\partial x^{a_1}}\frac{\partial \bar{x}^{k_2}}{\partial x^{a_2}}\cdots\frac{\partial \bar{x}^{k_r}}{\partial x^{a_r}}\right)\left(\frac{\partial x^{b_1}}{\partial \bar{x}^{l_1}}\frac{\partial x^{b_2}}{\partial \bar{x}^{l_2}}\cdots\frac{\partial x^{b_s}}{\partial \bar{x}^{l_s}}\right) \]

Contraction :-

suppose \(S\) is \((p,q)\) tensor such that

\[ S = \left( S^{i_1 i_2 -- i_p}_{j_1 j_2 - j_q} \right) \]

The contraction of \(S\) w.r.t a contravariant index \(i_f\) and covariant index \(j_g\) (where \((1 \leq f \leq p)\) & \((1 \leq g \leq q)\)) is given by setting \(\left(i_f = j_g = u\right)\) and evaluating

\[ \left( S' \equiv S^{i_1 i_2 -- i_f -- i_p}_{j_1 j_2 -- u - j_q} \right) \]

[\(S'\) is \((p-1, q-1)\) tensor]

example :- Suppose we have \((2,2)\) tensor \(A = A^{i_1 i_2}_{j_1 j_2}\)

contraction of \(A\) relative to \(i_2\) & \(j_1\) \(\left(A' : A^{i_1 u}_{u j_2}\right)\)

\(A' = \) contraction of \(A\) w.r.t \(i_2\) & \(j_1\)

\[ A' = A^{i,u}_{u,j_2} \]

[\(\leftarrow\) (repeated index gets summed over)]

in \(\mathbb{R}^n\),

\[ A' = A^{i,1}_{1,j_2} + A^{i,2}_{2,j_2} + A^{i,3}_{3,j_2} + \cdots + A^{i,n}_{n,j_2} \]

in 3-D \(\left(\mathbb{R}^3\right.\),

\[ A' = A^{i,1}_{1,j_2} + A^{i,2}_{2,j_2} + A^{i,3}_{3,j_2} \]

[\((1,1)\) tensor]

\(\therefore\) in going from \(A \longrightarrow A'\)

contravariant & covariant rank reduces by 1.

So \(A'\) is \((1,1)\) tensor.

Proof :-

For \(S\) is a \((p,q)\) tensor, we know its transformation law,

\[ \bar{S}^{i_1 i_2 - i_p}_{j_1 j_2 - j_q} = S^{\alpha_1 \alpha_2 -- \alpha_f - \alpha_p}_{\beta_1 \beta_2 -- \beta_g -- \beta_q}\ \frac{\partial \bar{x}^{i_1}}{\partial x^{\alpha_1}}\frac{\partial \bar{x}^{i_2}}{\partial x^{\alpha_2}}\cdots\frac{\partial \bar{x}^{i_f}}{\partial x^{\alpha_f}}\cdot\frac{\partial \bar{x}^{i_p}}{\partial x^{\alpha_p}}\ \frac{\partial x^{\beta_1}}{\partial \bar{x}^{j_1}}\frac{\partial x^{\beta_2}}{\partial \bar{x}^{j_2}}\cdots\frac{\partial x^{\beta_g}}{\partial \bar{x}^{j_g}}\cdots\frac{\partial x^{\beta_q}}{\partial \bar{x}^{j_q}} \qquad (1) \] \[ \left(i_f = j_g = u\right) \] \[ \bar{S}^{i_1 i_2 - u\ i_p}_{j_1 j_2 - u\ j_q} = S^{\alpha_1 \alpha_2 -- \alpha_p}_{\beta_1 \beta_2 -- \beta_q}\left(\frac{\partial \bar{x}^{i_1}}{\partial x^{\alpha_1}}\frac{\partial \bar{x}^{i_2}}{\partial x^{\alpha_2}}\cdots\frac{\partial \bar{x}^{i_p}}{\partial x^{\beta_p}}\right)\left(\frac{\partial \bar{x}^{u}}{\partial x^{\alpha_f}}\frac{\partial x^{\beta_g}}{\partial \bar{x}^{u}}\right)\left(\frac{\partial x^{\beta_1}}{\partial \bar{x}^{j_1}}\frac{\partial x^{\beta_2}}{\partial \bar{x}^{j_2}}\cdots\frac{\partial x^{\beta_q}}{\partial \bar{x}^{j_q}}\right) \qquad (2) \]

we know that from reverse chain rule

\[ \left(\frac{\partial \bar{x}^u}{\partial x^{\alpha_f}}\ \frac{\partial x^{\beta_g}}{\partial \bar{x}^u}\right) = \frac{\partial x^{\beta_g}}{\partial x^{\alpha_f}} = \delta^{\beta_g}_{\alpha_f} \] \[ \delta^{\beta_g}_{\alpha_f} = \begin{cases} 1 & \beta_g = \alpha_f \\ 0 & \beta_g \neq \alpha_f \end{cases} \]

So, eqn (2) becomes transformation law for \((p-1, q-1)\) tensor

so \(S'\) represents \((p-1, q-1)\) tensor.

(2) Inner product :-

suppose \(S\) and \(T\) are two tensors given by

\[ S = \left( S^{i_1 i_2 -- i_p}_{j_1 j_2 --- j_q} \right) \quad \text{and} \quad T = \left( T^{k_1 k_2 -- k_r}_{l_1 l_2 -- l_s} \right) \]

To find inner product \(S\cdot T\), choose a contravariant index on \(S\) \((i_f)\) and covariant index \((l_g)\) on \(T\).

let \(\left(i_f = l_g = u\right)\) then

\[ S \circ T = \left( S^{i_1 i_2 -- u - i_p}_{j_1 j_2 --- j_q}\ T^{k_1 k_2 -- k_r}_{l_1 l_2 - u - l_s} \right) \]

\(S\cdot T\) is \((p+r-1,\ q+s-1)\) tensor

or total rank of \(S\cdot T = m = p+r-1+q+s-1\)

\[ = p+r+q+s-2 \]

Proof can be found similer previous method.

ex (1) Evaluate \(u\cdot v\) where, \(V = \left(V^i\right)\) and \(u = \left(u_j\right)\)

\[ V\cdot u = V^s u_s \qquad (i=j=s) \]

i.e., \(\left(\text{in } \mathbb{R}^3\right)\)

\[ V\cdot u = V^1 u_1 + V^2 u_2 + V^3 u_3 \]

\(V, u\) & tensors following

\[ \bar{V}^i = V^r\,\frac{\partial \bar{x}^i}{\partial x^r} \] \[ \bar{u}_j = u_s\,\frac{\partial x^s}{\partial \bar{x}^j} \]

for inner product \((i=j=k)\)

\[ \bar{V}^k \bar{u}_k = \left( V^r\,\frac{\partial \bar{x}^k}{\partial x^r} \right)\left( u_s\,\frac{\partial x^s}{\partial \bar{x}^k} \right) \] \[ = V^r u_s\ \frac{\partial \bar{x}^k}{\partial x^r}\frac{\partial x^s}{\partial \bar{x}^k} \] \[ = V^r u_s\ \frac{\partial x^s}{\partial x^r} \] \[ = V^r u_s\ \delta^s_r \]

[\(\dfrac{\partial x^s}{\partial x^r} = \delta^s_r\)]

\[ \left(\bar{V}^k \bar{u}_k = V^s u_s\right) \]

inner product of contravariant and covariant vector is invariant / scalar.

\[ m = \text{rank of } S\cdot T\ |\ V\cdot u = p+q+r+s-2 \] \[ = 1+1-2 = \underline{0} \]

So \(\left(V\cdot u = \text{scalar}\right)\) i.e. tensor of rank \(\underline{0}\).

(3) Outer product :-

Suppose \(S\) & \(T\) are tensors given by

\[ S = \left( S^{i_1 i_2 -- i_p}_{j_1 j_2 -- j_q} \right) \quad \& \quad T = \left( T^{k_1 k_2 -- k_r}_{l_1 l_2 -- l_s} \right) \]

'S' is \((p,q)\) tensor & 'T' is \((r,s)\) tensor.

Then the outer product / direct product / tensor product

\[ S \otimes T \equiv \left( S^{i_1 i_2 - i_p}_{j_1 j_2 - j_q}\cdot T^{k_1 k_2 - k_r}_{l_1 l_2 - l_s} \right) \]

\(S\otimes T\) is tensor whose components are product of each component of \(S\) & each component of \(T\).

\(S\otimes T\) is \((p+r,\ q+s)\) tensor.

\(S\otimes T\) is tensor of rank \((m = p+r+q+s)\).

example (1) Find outer product of \(a = \begin{pmatrix} a^1 \\ a^2 \\ a^3 \end{pmatrix}\) & \(b = \begin{pmatrix} b_1 & b_2 & b_3 \end{pmatrix}\)

\[ a \otimes b = \begin{pmatrix} a^1 b_1 & a^1 b_2 & a^1 b_3 \\ a^2 b_1 & a^2 b_2 & a^2 b_3 \\ a^3 b_1 & a^3 b_2 & a^3 b_3 \end{pmatrix} \]

[\(\rightarrow\) mixed tensor of rank 2.]

\[ \downarrow \] \[ a^i \otimes b_j \]

[(Schaum's outlines)]

(1) (i)

\[ a_1 x_1 + a_2 x_2 + \cdots + a_n x_n = \sum_{i=1}^{N} a_i x_i \equiv a_i x_i \]

(Repeated indices gets summed over)

(ii)

\[ a_{ii} x_K = a_{11} x_K + a_{22} x_K + a_{33} x_K + \cdots + a_{nn} x_n \]

(iii)

\[ a_{ij} x_j = a_{i1} x_1 + a_{i2} x_2 + a_{i3} x_3 + \cdots + a_{in} x_n \]

repeated indices \(=\) dummy indices

non-repeated indices \(=\) free indices.

Note

\[ a_{ij} x_j \neq a_{kj} x_j \qquad (\text{unless } i=k) \]

(iv)

\[ y_i = a_{ir} x_r \] \[ y_i = a_{i1} x_1 + a_{i2} x_2 + a_{i3} x_3 \]

[tensor of rank 1]

[If \(\left(y_i \in \mathbb{R}^3\right)\)]

\[ y_1 = a_{11} x_1 + a_{12} x_2 + a_{13} x_3 \] \[ y_2 = a_{21} x_1 + a_{22} x_2 + a_{23} x_3 \] \[ y_3 = a_{31} x_1 + a_{32} x_2 + a_{33} x_3 \] \[ \left(x_j\right)\left(y_i\right) = \text{covariant tensor of rank 1} \] \[ a_{ir} = \text{covariant tensor of rank 2} \] \[ \begin{pmatrix} y_1 \\ y_2 \\ y_3 \end{pmatrix} = \begin{pmatrix} a_{11} & a_{12} & a_{13} \\ a_{21} & a_{22} & a_{23} \\ a_{31} & a_{32} & a_{33} \end{pmatrix}\begin{pmatrix} x_1 \\ x_2 \\ x_3 \end{pmatrix} \] \[ \boxed{\,y_i = a_{ir} x_r\,} \]

Any expression involving a twice repeated index (occuring twice as superscript, or twice as subscript, or once a superscript and once a subscript) shall automatically stand for its sum over the values \(1,2,3, --- n\). of repeated index.

\[ a^i_i = a^1_1 + a^2_2 + a^3_3 + \cdots + a^n_n \]

for \(a^i_j\) to be \((1,1)\) tensor in n-D. \(\left(\mathbb{R}^n\right)\)

\[ a_{ii} = a_{11} + a_{22} + a_{33} + \cdots + a_{nn} \] \[ a^{jj} = a^{11} + a^{22} + a^{33} + \cdots + a^{nn} \] \[ a^i x^i = a^1 x^1 + a^2 x^2 + \cdots + a^n x^n \]

If \(a^i x^i\) is an tensor of rank 0. (scalar).

\[ Q = b_{ij} y_i x_j \qquad \text{where } y_i = a_{ij} x_j \]

But 'j' is duplicated so

we replace \(j \rightarrow r\) (different index) to substitute \(y_i\) in \(Q\).

\[ Q = b_{ij} y_i x_j = b_{ij}\left(a_{ir} x_r\right)x_j \] \[ Q = a_{ir} b_{ij} x_r x_j \]

ex If \(y_i = a_{ij}x_j\) then express quadratic form

\[ Q = g_{ij} y_i y_j \quad \text{in terms of } x-\text{variables.} \] \[ Q = g_{ij} y_i y_j \]

where \(y_i = a_{ir} x_r\) (\(j\) is replaced with \(r\)) as \(j\) is duplicated

\[ y_j = a_{js} x_s \] \[ Q = g_{ij}\left(a_{ir}x_r\right)\left(a_{js}x_s\right) = \left(g_{ij}a_{ir}a_{js}\right)x_r x_s \]

(or)

\[ Q = h_{rs} x_r x_s \]

where

\[ h_{rs} = g_{ij} a_{ir} a_{js} \]

(\(i\) & \(j\) gets summed over).

A much used symbol in tensor calculus has the effect of annihilating the off-diagonal terms in a double summation.

kronecker delta

\[ \delta_{ij} = \delta^{ij} = \delta^i_j = \begin{cases} 1 & i=j \\ 0 & i \neq j \end{cases} \]

clearly \(\left(\delta_{ij} = \delta_{ji} \quad \forall\ i\ \&\ j\right)\).

ex (1)

\[ \delta_{ij} x_i x_j = ? \qquad \text{for } 1 \leq i,j \leq 3 \] \[ \delta_{ij} x_i x_j = \delta_{i1} x_i x_1 + \delta_{i2} x_i x_2 + \delta_{i3} x_i x_3 \] \[ = \left(\delta_{11}x_1x_1 + \underbrace{\delta_{21}x_2x_1}_{0} + \underbrace{\delta_{31}x_3x_1}_{0}\right) + \left(\delta_{22}x_2x_2 + \underbrace{\delta_{12}x_1x_2}_{0} + \underbrace{\delta_{32}x_3x_2}_{0}\right) + \left(\underbrace{\delta_{13}x_1x_3}_{0} + \underbrace{\delta_{23}x_2x_3}_{0} + \delta_{33}x_3x_3\right) \] \[ \delta_{ij} x_i x_j = x_1^2 + x_2^2 + x_3^2 = x_i x_i = x_j x_j \] \[ \delta^r_j a_{ir} x_i = a_{ij} x_i \] \[ \delta^r_j = \begin{cases} 1 & r=j \\ 0 & r \neq j \end{cases} \]

ex 2) Suppose \(T^i = g^i_r a_{rs} y_s\) , \(y_i = b_{ir} x_r\) , \(a_{ir} b_{rj} = \delta_{ij}\)

Find \(T^i\) in terms of \(x_p\).

\[ T^i = g^i_r a_{rs} y_s \]

[\(y_s = b_{sr} x_r\) but (\(r\) is duplicated) so \(y_s = b_{sj} x_j\)]

\[ T^i = g^i_r \underline{a_{rs} b_{sj}}\, x_j \] \[ = g^i_r\, \delta_{rj}\, x_j \] \[ \left(T^i = g^i_r x_r\right) \]

\[ a_{ij}\left(x_i + y_j\right) \neq a_{ij}x_i + a_{ij}y_j \] \[ a_{ij} x_i y_j \neq a_{ij} y_i x_j \] \[ \left(a_{ij} + a_{ji}\right)x_i y_j \neq 2a_{ij}x_i x_j \]

Repeated indices :-

use summation convention to write following

(a)

\[ a_{11}b_{11} + a_{21}b_{12} + a_{31}b_{13} + a_{41}b_{14} = a_{i1}b_{1i} \]

\(a_{i1}b_{1i}\) is tensor of rank 1 in \(\underline{\mathbb{R}^4}\). \((n=4)\)

(b)

\[ a_{11}b_{11} + a_{12}b_{12} + a_{13}b_{13} + a_{14}b_{14} + a_{15}b_{15} + a_{16}b_{16} \] \[ = a_{1i}b_{1i} \qquad \text{in } \mathbb{R}^6 \qquad (n=6) \]

(c)

\[ C^i_{11} + C^i_{22} + C^i_{33} + C^i_{44} + C^i_{55} + C^i_{66} + C^i_{77} + C^i_{88} = C^i_{jj} \]

_Tensor in \(\mathbb{R}^8\) \((n=8)\)_

(1.2) use summation convention to write following systems.

(a)

\[ \left.\begin{array}{l} C_{11}x_1 + C_{12}x_2 + C_{13}x_3 = 2 \\ C_{21}x_1 + C_{22}x_2 + C_{23}x_3 = -3 \\ C_{31}x_1 + C_{32}x_2 + C_{33}x_3 = 5 \end{array}\right\} \qquad \begin{array}{l} C_{1j}x_j = 2 \\ C_{2j}x_j = -3 \\ C_{3j}x_j = 5 \end{array} \] \[ \boxed{\ C_{ij}x_j = d_i \quad \text{where } \left(d_1 = 2,\ d_2 = -3,\ d_3 = 5\right)\ } \]

(b)

\[ a^1_j x_1 + a^2_j x_2 + a^3_j x_3 + a^4_j x_4 = b_j \qquad (j=1,2) \] \[ a^i_j x_i = b_j \]

[\(j\) = free index ; \(i\) = dummy index]

[\(j = 1,2\) ; \(i = 1,2,3,4\)]

(1.3) write out explicit summations

\(C_i\left(x_i + y_i\right)\) & \(C_j x_j + C_k y_k\) \((n=4)\) \[ C_i\left(x_i+y_i\right) = C_1\left(x_1+y_1\right) + C_2\left(x_2+y_2\right) + C_3\left(x_3+y_3\right) + C_4\left(x_4+y_4\right) \] \[ = \left(C_1x_1 + C_2x_2 + C_3x_3 + C_4x_4\right) + \left(C_1y_1 + C_2y_2 + C_3y_3 + C_4y_4\right) \] \[ = C_j x_j + C_k y_k \] \[ \boxed{\ C_i\left(x_i+y_i\right) = C_j x_j + C_k y_k\ } \] Double sum

(1.4) If \(n=3\), expand \(Q = a^{ij}x_i x_j\)

\[ Q = a^{1j}x_1 x_j + a^{2j}x_2 x_j + a^{3j}x_3 x_j \] \[ = \left(a^{11}x_1x_1 + a^{12}x_1x_2 + a^{13}x_1x_3\right) + \left(a^{21}x_2x_1 + a^{22}x_2x_2 + a^{23}x_2x_3\right) + \left(a^{31}x_3x_1 + a^{32}x_3x_2 + a^{33}x_3x_3\right) \]

(1.5) use summation convention to write the following & state \(n = ?\)

(a)

\[ \left(a_{11}b_{11} + a_{21}b_{12} + a_{31}b_{13}\right) + \left(a_{12}b_{21} + a_{22}b_{22} + a_{32}b_{23}\right) + \left(a_{13}b_{31} + a_{23}b_{32} + a_{33}b_{33}\right) \] \[ = a_{i1}b_{1i} + a_{i2}b_{2i} + a_{i3}b_{3i} \] \[ = a_{ij}b_{ji} \qquad (n=3) \]

(b)

\[ \left( g^1_{11} + g^1_{12} + g^1_{21} + g^1_{22} + g^2_{11} + g^2_{12} + g^2_{21} + g^2_{22} \right) \] \[ = g^1_{1i}C_i + g^1_{2i}C_i + g^2_{1i}C_i + g^2_{2i}C_i \] \[ = \left(g^1_{ji}C_j\right)C_i + \left(g^2_{ji}C_j\right)C_i \]

[\(C_i, C_j = 1\) \(\forall\ i,j = 1,2\)]

\[ = \left(g^k_{ji}C_k\right)C_j C_i = g^k_{ji}C_k C_j C_i \] \[ \left(C_k = C_j = C_i = 1\right) \qquad \left(i,j,k = 1,2\right) \quad \text{i.e } \underline{(n=2)} \]

(1.6) For \(n=2\) write out explicitly the tripple summation

\[ C_{rst}\,x^r y^s z^t \] \[ C_{rst}\,x^r y^s z^t = C_{rs1}x^r y^s z^1 + C_{rs2}x^r y^s z^2 \] \[ = C_{r11}x^r y^1 z^1 + C_{r21}x^r y^2 z^1 + C_{r12}x^r y^1 z^2 + C_{r22}x^r y^2 z^2 \] \[ = C_{111}x^1y^1z^1 + C_{211}x^2y^1z^1 + C_{121}x^1y^2z^1 + C_{221}x^2y^2z^1 \] \[ + C_{112}x^1y^1z^2 + C_{212}x^2y^1z^2 + C_{122}x^1y^2z^2 + C_{222}x^2y^2z^2 \]

(1.7) show that \(a_{ij}x_i x_j = 0\) if \(a_{ij} = i-j\)

\[ (i-j)x_i x_j = (i-1)x_i x_1 + (i-2)x_i x_2 \] \[ = (2-1)x_2x_1 + (1-1)x_1x_1 + (1-2)x_1x_2 + (2-2)x_2x_2 \] \[ = x_2x_1 + 0 + -x_1x_2 + 0 \] \[ = 0 \]

(1.8) If \(a_{ij}\) are const. calculate partial derivatives

\[ \frac{\partial}{\partial x_K}\left(a_{ij}x_i x_j\right) = a_{ij}\frac{\partial}{\partial x_K}x_i x_j \] \[ = a_{ij}\,\partial_{x_K}\,x_i x_j \] \[ = a_{ij}\left(\delta_{iK}x_j + \delta_{Kj}x_i\right) \] \[ = a_{ij}\delta_{iK}x_j + a_{ij}\delta_{Kj}x_i \] \[ = a_{Kj}x_j + a_{iK}x_i \]

Since \(i,j\) has same range,

\[ \frac{\partial}{\partial x_K}\left(a_{ij}x_i x_j\right) = \left(a_{Ki} + a_{iK}\right)x_i \]

(1.9) Substitutions, kronecker delta :-

express \(b^{ij}y_i y_j\) in Terms of \(x\)-variables, if \(y_i = C_{ij}x_j\)

& \(b^{ij}C_{jk} = \delta^i_k\)

\[ b^{ij}y_i y_j = ? \] \[ y_i = C_{ij}x_j \qquad (\text{but } j \text{ is duplicated}) \]

replace \(j \rightarrow r\)

\[ y_i = C_{ir}x_r \] \[ y_j = C_{jk}x_k \] \[ b^{ij}\left(C_{ir}x_r\right)\left(C_{jk}x_k\right) = \underbrace{b^{ij}C_{ir}}\,C_{jk}\,x_r x_k \] \[ = \delta^j_r\,C_{jk}\,x_r x_k \] \[ = C_{rk}\,x_r x_k \] \[ b^{ij}y_i y_j = C_{rk}x_r x_k \] \[ = C_{ij}x_i x_j \]

[\((i,j,r,k)\) has same range]

\[ \left(1 \leq r,k,i,j \leq n\right) \]

(1.11) If \(a_{ij} = a_{ji}\)

\[ \frac{\partial^2}{\partial x_K \partial x_\ell}\left(a_{ij}x_i x_j\right) \] \[ = \frac{\partial}{\partial x_K}\left(\frac{\partial}{\partial x_\ell}\left(a_{ij}x_i x_j\right)\right) \] \[ = \frac{\partial}{\partial x_K}\left(a_{ij}\,\delta_{i\ell}\,x_j + a_{ij}\,\delta_{j\ell}\,x_i\right) \] \[ = \frac{\partial}{\partial x_K}\left(a_{\ell j}x_j + a_{i\ell}x_i\right) \] \[ = \frac{\partial}{\partial x_K}\left(\left(a_{\ell i} + a_{i\ell}\right)x_i\right) \] \[ = \left(a_{\ell i} + a_{i\ell}\right)\delta_{iK} \] \[ = a_{\ell K} + a_{K\ell} = 2a_{K\ell} \]

(1.12) consider a system of linear eqn of form \(y^i = a^{ij}x_j\) and suppose that \(\left(b_{ij}\right)\) is matrix of numbers such that \(\forall\ i,j\) \(b_{ir}a^{rj} = \delta^j_i\) (\(b_{ij}\) is inverse of \(a^{ij}\))

\[ y^i = a^{ij}x_j \] \[ b_{ki}y^i = b_{ki}a^{ij}x_j = \delta^j_k x_j \] \[ b_{ji}y^i = x_j \] \[ x_j = b_{ji}y^i \]

\[ x_i = b_{ij}y^j \] \[ b_{ki}y^i = x_k \]

\[ \left(x_i = b_{ij}y^j\right) \]

[Both represents same sol\(^n\).]

(1.13) show that,

\[ a_{ijk}\left(x_i + y_j\right)z_k \neq a_{ijk}x_i z_k + a_{ijk}y_j z_k \]

[\(i,j,k\) = dummy indices ,]

[\(\uparrow\) \(j\) = free ; \(\uparrow\) \(i\) = free]

So not equal.

Supplementary problems will be addressed in revision (in the end of all 13-chapters).

tensors | Tensors chapter 2