MTH3008 Lecture 20

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We’ve completed all seven chapters of the course. Today is a synthesis revision: we collect the headline results of Chapters 1-4 (suffix notation, Kronecker delta, alternating tensor, vector differential operators, local coordinate transformations, and tensors in orthogonal Cartesian systems), and work through two assessment-style problems - one from a past coursework, one from a past portfolio test - to consolidate the techniques.

Suffix Notation, Kronecker Delta, and Alternating Tensor

Vector vs. Suffix Notation

Suffix notation is local - it refers to a single component of a tensor. Vector notation is global - it refers to the whole object.

Object	Vector notation	Suffix notation
Vector	$\mathbf{v}=(v_1,v_2,v_3)$	$v_i$
Scalar multiple	$\lambda \mathbf{v}=(\lambda v_1,\lambda v_2,\lambda v_3)$	$\lambda v_i$

A free index appears exactly once in each term and labels a component (e.g. $v_i$ - three values $v_1,v_2,v_3$). A dummy index appears exactly twice in a term and indicates summation under the Einstein convention. The dot product is the canonical example:

$$\mathbf{a}\cdot \mathbf{b}=a_1{b}_1+a_2{b}_2+a_3{b}_3=\sum _{i=1}^3{a}_i{b}_i=a_i{b}_i.$$

Kronecker Delta

$${\delta }_{ij}=\{\begin{array}{ll}1& \text{if }i=j\\ 0& \text{if }i\ne j.\end{array}$$

Key properties:

• Substitution tensor: ${\delta }_{ij}{a}_j=a_i$ (the repeated index gets “swallowed”). Likewise ${\delta }_{ji}{a}_i=a_j$.
• Dot product via $\delta$: $\mathbf{a}\cdot \mathbf{b}={\delta }_{ij}{a}_i{b}_j$.

Alternating Tensor

$${ϵ}_{ijk}=\{\begin{array}{ll}0& \text{if any of }i,j,k\text{ are equal}\\ +1& \text{if }(i,j,k)\in \{(1,2,3),(2,3,1),(3,1,2)\}\\ -1& \text{if }(i,j,k)\in \{(1,3,2),(2,1,3),(3,2,1)\}.\end{array}$$

Symmetries:

• Cyclic: ${ϵ}_{ijk}={ϵ}_{jki}={ϵ}_{kij}$.
• Antisymmetric on swap: ${ϵ}_{ijk}=-{ϵ}_{jik}$ (and similarly for any other index swap).

Applications:

$$(\mathbf{a}\times \mathbf{b}{)}_i={ϵ}_{ijk}{a}_j{b}_k,|M|={ϵ}_{ijk}{M}_{1i}{M}_{2j}{M}_{3k}={ϵ}_{ijk}{M}_{i1}{M}_{j2}{M}_{k3},\mathbf{a}\cdot (\mathbf{b}\times \mathbf{c})={ϵ}_{ijk}{a}_i{b}_j{c}_k.$$

Practice: 2023-24 Coursework

Coursework 23-24

Let $\mathbf{a},\mathbf{b},\mathbf{c},\mathbf{d}$ be vectors. Use suffix notation to show

$$((\mathbf{a}\times \mathbf{b})\times \mathbf{c})\cdot \mathbf{d}=-(\mathbf{a}\cdot \mathbf{d})(\mathbf{b}\cdot \mathbf{c})+(\mathbf{a}\cdot \mathbf{c})(\mathbf{b}\cdot \mathbf{d}).$$

Solution. Write $\mathbf{u}=\mathbf{a}\times \mathbf{b}$, so $u_m={ϵ}_{mpq}{a}_p{b}_q$. Then $(\mathbf{u}\times \mathbf{c}{)}_i={ϵ}_{imn}{u}_m{c}_n={ϵ}_{imn}{ϵ}_{mpq}{a}_p{b}_q{c}_n$, and contracting with $\mathbf{d}$:

$$((\mathbf{a}\times \mathbf{b})\times \mathbf{c})\cdot \mathbf{d}={ϵ}_{imn}{ϵ}_{mpq}{a}_p{b}_q{c}_n{d}_i.$$

Using the cyclic symmetry ${ϵ}_{imn}=-{ϵ}_{min}$, then applying the $ϵ$-$\delta$ identity ${ϵ}_{min}{ϵ}_{mpq}={\delta }_{ip}{\delta }_{nq}-{\delta }_{iq}{\delta }_{np}$:

$${ϵ}_{imn}{ϵ}_{mpq}=-({\delta }_{ip}{\delta }_{nq}-{\delta }_{iq}{\delta }_{np})={\delta }_{iq}{\delta }_{np}-{\delta }_{ip}{\delta }_{nq}.$$

So

$$((\mathbf{a}\times \mathbf{b})\times \mathbf{c})\cdot \mathbf{d}=({\delta }_{iq}{\delta }_{np}-{\delta }_{ip}{\delta }_{nq})a_p{b}_q{c}_n{d}_i=a_n{b}_i{c}_n{d}_i-a_i{b}_n{c}_n{d}_i.$$

Recognising the dot products: $a_n{c}_n=\mathbf{a}\cdot \mathbf{c}$, $b_i{d}_i=\mathbf{b}\cdot \mathbf{d}$, $a_i{d}_i=\mathbf{a}\cdot \mathbf{d}$, $b_n{c}_n=\mathbf{b}\cdot \mathbf{c}$:

$$\boxed{((\mathbf{a}\times \mathbf{b})\times \mathbf{c})\cdot \mathbf{d}=(\mathbf{a}\cdot \mathbf{c})(\mathbf{b}\cdot \mathbf{d})-(\mathbf{a}\cdot \mathbf{d})(\mathbf{b}\cdot \mathbf{c}).}$$

Vector Differential Operators

The standard trio in Cartesian coordinates:

$$[\nabla f{]}_i=\frac{\partial f}{\partial x_i},\nabla \cdot \mathbf{u}=\frac{\partial u_j}{\partial x_j},[\nabla \times \mathbf{u}{]}_i={ϵ}_{ijk}\frac{\partial u_k}{\partial x_j}.$$

Compact derivative shorthand

Many calculations use ${\partial }_j=\partial /\partial x_j$. Then divergence is ${\partial }_j{u}_j$ and curl is ${ϵ}_{ijk}{\partial }_j{u}_k$.

Position Vector

Let $\mathbf{r}=(x_1,x_2,x_3)$ with magnitude $r=|\mathbf{r}|=\sqrt{x_1^2+x_2^2+x_3^2}$. The standard results are:

$$\nabla r=\frac{\mathbf{r}}{r},\nabla \cdot \mathbf{r}=3,\nabla \times \mathbf{r}=0.$$

These are easy to derive in suffix notation:

• Gradient. $\partial r/\partial x_i=x_i/r$, since $r^2=x_j{x}_j$ gives $2r(\partial r/\partial x_i)=2x_i$.
• Divergence. $\partial x_j/\partial x_j={\delta }_{jj}=3$.
• Curl. ${ϵ}_{ijk}{\partial }_j{x}_k={ϵ}_{ijk}{\delta }_{jk}=0$ (sum of an antisymmetric and symmetric tensor).

Local Coordinate Transformation

Bases and Coordinate Systems

A basis ${\mathbf{v}}_1,{\mathbf{v}}_2,{\mathbf{v}}_3$ is a set of vectors that is

• linearly independent: the only ${\lambda }_1{\mathbf{v}}_1+{\lambda }_2{\mathbf{v}}_2+{\lambda }_3{\mathbf{v}}_3=0$ has ${\lambda }_1={\lambda }_2={\lambda }_3=0$;
• spanning: every vector $\mathbf{v}$ is some linear combination ${\mu }_1{\mathbf{v}}_1+{\mu }_2{\mathbf{v}}_2+{\mu }_3{\mathbf{v}}_3$.

A coordinate system is orthogonal if ${\mathbf{v}}_i\cdot {\mathbf{v}}_j=0$ whenever $i\ne j$. It is orthonormal if additionally $|{\mathbf{v}}_i|=1$ for all $i$. The standard Cartesian basis $\hat{\mathbf{i}}=(1,0,0),\hat{\mathbf{j}}=(0,1,0),\hat{\mathbf{k}}=(0,0,1)$ is the canonical orthonormal basis.

Coordinate Transformation Matrix

Switching between two coordinate systems with bases ${\mathbf{e}}_1,{\mathbf{e}}_2,{\mathbf{e}}_3$ (for $(x_1,x_2,x_3)$) and ${\mathbf{e}}_1^{\prime },{\mathbf{e}}_2^{\prime },{\mathbf{e}}_3^{\prime }$ (for $(x_1^{\prime },x_2^{\prime },x_3^{\prime })$) is governed by the matrix

$$\boxed{L_{ij}={\mathbf{e}}_i^{\prime }\cdot {\mathbf{e}}_j.}$$

Component-wise, $x_i^{\prime }=L_{ij}{x}_j$, i.e. coordinates transform via $L$.

Properties of $L$

• $L_{ij}$ equals the cosine of the angle between ${\mathbf{e}}_i^{\prime }$ and ${\mathbf{e}}_j$.
• For two orthonormal bases, $L$ is orthogonal: $L^{T}L=L{L}^T=I$. Equivalently, $L_{ij}{L}_{kj}={\delta }_{ik}=L_{ji}{L}_{jk}$.
• The inverse transformation is $x_i=L_{ji}{x}_j^{\prime }$, so $\partial x_i^{\prime }/\partial x_j=L_{ij}$ and $\partial x_i/\partial x_j^{\prime }=L_{ji}$ (see Jacobian).

Example: Constructing $L$

Example

Let $K$ be Cartesian with orthonormal basis ${\mathbf{i}}_1,{\mathbf{i}}_2,{\mathbf{i}}_3$, and let $K^{\prime }$ have basis

$${\mathbf{e}}_1={\mathbf{i}}_1+{\mathbf{i}}_2+{\mathbf{i}}_3,{\mathbf{e}}_2=2{\mathbf{i}}_2,{\mathbf{e}}_3={\mathbf{i}}_1+3{\mathbf{i}}_2.$$

Find the rotation matrix $L=[L_{ij}]$.

Method 1 (read off coefficients). Since ${\mathbf{i}}_1,{\mathbf{i}}_2,{\mathbf{i}}_3$ is the Cartesian basis,

$$\left(\begin{array}{c}{\mathbf{e}}_1\\ {\mathbf{e}}_2\\ {\mathbf{e}}_3\end{array}\right)=\left(\begin{array}{ccc}1& 1& 1\\ 0& 2& 0\\ 1& 3& 0\end{array}\right)\left(\begin{array}{c}{\mathbf{i}}_1\\ {\mathbf{i}}_2\\ {\mathbf{i}}_3\end{array}\right)⟹L=\left(\begin{array}{ccc}1& 1& 1\\ 0& 2& 0\\ 1& 3& 0\end{array}\right).$$

Method 2 (definition). Compute each $L_{ij}={\mathbf{e}}_i\cdot {\mathbf{i}}_j$:

$$L_{11}=1,L_{12}=1,L_{13}=1,L_{21}=0,L_{22}=2,L_{23}=0,L_{31}=1,L_{32}=3,L_{33}=0.$$

Warning

The basis $K^{\prime }$ here is not orthonormal (e.g. $|{\mathbf{e}}_1|=\sqrt{3}\ne 1$, and ${\mathbf{e}}_1\cdot {\mathbf{e}}_3=4\ne 0$), so the resulting $L$ is not an orthogonal matrix. The "$L^{T}L=I$" property assumes a rotation between two orthonormal bases. The two methods above still give the correct change-of-basis matrix, but the orthogonality identity does not hold here.

Tensors in Orthogonal (Cartesian) Coordinates

Formal Definition of a Vector

A quantity $v_i$ is a vector if its components transform under rotation as

$$\boxed{v_i^{\prime }=L_{ij}{v}_j.}$$

This is the transformation rule of a vector. Together with $L^T=L^{-1}$, it gives the orthogonality identity

$$L_{ij}{L}_{kj}={\delta }_{ik}=L_{ji}{L}_{jk}.$$

See also Basis Transformation.

Tensor Transformation Rule

A quantity is a tensor if each free index transforms according to the vector rule. Examples:

• Rank 2: $T_{ij}^{\prime }=L_{im}{L}_{jn}{T}_{mn}$
• Rank 3: $T_{ijk}^{\prime }=L_{im}{L}_{jn}{L}_{kp}{T}_{mnp}$
• Rank 4: $T_{ijk\ell }^{\prime }=L_{im}{L}_{jn}{L}_{kp}{L}_{\ell q}{T}_{mnpq}$

Definition

The rank (or order) of a tensor is the number of free indices.

Practice: 2023-24 Portfolio Test

Portfolio Test 23-24

Suppose $T_{ijk\ell m}$ is a rank-five tensor. Using the formal definition (the transformation rule), show that $T_{ijkjk}$ is a rank-one tensor.

Solution. Apply the rank-five rule:

$$T_{ijk\ell m}^{\prime }=L_{ip}{L}_{jq}{L}_{kr}{L}_{\ell s}{L}_{mt}{T}_{pqrst}.$$

Setting $\ell =j$ and $m=k$ on both sides (i.e. contracting the second-and-fourth and third-and-fifth indices),

$$T_{ijkjk}^{\prime }=L_{ip}{L}_{jq}{L}_{kr}{L}_{js}{L}_{kt}{T}_{pqrst}.$$

Group the $L$‘s with shared dummy indices: $L_{jq}{L}_{js}$ and $L_{kr}{L}_{kt}$. By the orthogonality identity,

$$L_{jq}{L}_{js}={\delta }_{qs},L_{kr}{L}_{kt}={\delta }_{rt}.$$

Substituting,

$$T_{ijkjk}^{\prime }=L_{ip}{\delta }_{qs}{\delta }_{rt}{T}_{pqrst}=L_{ip}{T}_{pqrqr}.$$

Defining $S_p=T_{pqrqr}$ (the contracted object on the right), we have

$$\boxed{T_{ijkjk}^{\prime }=L_{ip}{S}_p=L_{ip}{T}_{pqrqr}.}$$

This is exactly the rank-one transformation rule. So $T_{ijkjk}$ is a rank-one tensor (a vector). $■$

Why this works

Each pair of contracted indices contributes a factor $L\cdot L^T=I$, removing two free indices and two factors of $L$. A rank-$n$ tensor with $p$ contracted index-pairs contracts down to a rank-$(n-2p)$ tensor. Here $n=5,p=2$, so the result has rank $5-4=1$.

Where We Came From

This lecture covered the foundational mechanics of Chapters 1-4. The course’s natural progression from here:

• Chapter 5 (lectures 7-10): dual basis ${\mathbf{e}}^i$, Covariant and Contravariant Components of vectors and second-rank tensors, the Metric Tensor $g_{ij}$, Quotient Rule.
• Chapter 6 (lectures 11-14): tensors in generalised (curvilinear) coordinates, Symmetry and Antisymmetry, tensor algebra (Tensor Addition, Outer Product, Contraction).
• Chapter 7 (lectures 15-19): Local Basis, Covariant Differentiation, Christoffel Symbols, Ricci’s Theorem, Riemann-Christoffel Tensor, Ricci Tensor.

Lectures 21-22 (if delivered) continue the revision through these later chapters; the final exam covers the whole course with emphasis on Chapters 5-7.

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Pre-Lecture Notes from mth3008 lecture notes 20.pdf

• Chapters 1-4 revision - single sweep through the foundational notation and tensors in Cartesian systems
• Suffix notation: local (one component) vs. vector notation (whole object); free vs. dummy indices; Einstein summation
• Kronecker delta ${\delta }_{ij}$ as substitution tensor: ${\delta }_{ij}{a}_j=a_i$; dot product $\mathbf{a}\cdot \mathbf{b}={\delta }_{ij}{a}_i{b}_j$
• Alternating tensor ${ϵ}_{ijk}$: cyclic invariant, sign-flips on swap; cross product $(\mathbf{a}\times \mathbf{b}{)}_i={ϵ}_{ijk}{a}_j{b}_k$, determinant, scalar triple product
• $ϵ$-$\delta$ identity: ${ϵ}_{ijk}{ϵ}_{ipq}={\delta }_{jp}{\delta }_{kq}-{\delta }_{jq}{\delta }_{kp}$, the workhorse of suffix-notation algebra (Coursework 23-24)
• Differential operators: gradient $[\nabla f{]}_i={\partial }_{i}f$, divergence ${\partial }_j{u}_j$, curl ${ϵ}_{ijk}{\partial }_j{u}_k$; standard position-vector results $\nabla r=\mathbf{r}/r$, $\nabla \cdot \mathbf{r}=3$, $\nabla \times \mathbf{r}=0$
• Coordinate systems: basis = linearly independent + spanning; orthogonal vs. orthonormal
• Transformation matrix $L_{ij}={\mathbf{e}}_i^{\prime }\cdot {\mathbf{e}}_j$; orthogonality $L^{T}L=I$ (assumes orthonormal bases), equivalently $L_{ij}{L}_{kj}={\delta }_{ik}$; coordinates transform via $x_i^{\prime }=L_{ij}{x}_j$
• Worked example: transformation matrix between Cartesian ${\mathbf{i}}_{1,2,3}$ and a general (non-orthonormal) basis - computed two ways, by reading off coefficients and by definition
• Vector definition (formal): quantity with $v_i^{\prime }=L_{ij}{v}_j$
• Tensor definition (formal): each free index obeys the vector rule; rank = number of free indices; rank-$n$ tensor with $p$ contracted index-pairs becomes rank-$(n-2p)$ (Portfolio Test 23-24)
• Next lecture: Chapter 5 revision - dual basis, covariant/contravariant components, tensors in generalised coordinates, symmetries