MTH3008 Lecture 21

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Continuing the revision sweep from Lecture 20 (Chapters 1-4): today consolidates Chapter 5 (dual bases, covariant/contravariant components of vectors and second-rank tensors), Chapter 6 (tensors in generalised coordinate systems and symmetries) on a worked example. The unifying technique is building the dual basis from cross products, then reading off components by dotting with the appropriate basis. Tensor algebra, the metric tensor & arc length, Christoffel symbols, Ricci’s theorem, and the Riemann-Christoffel tensor are deferred to lecture 22.

Generalised Coordinate Systems

A generalised coordinate system does not necessarily have an orthogonal basis. Switching between two such systems is more involved than rotating an orthonormal frame - rather than a single rotation matrix $L$ with $L^T=L^{-1}$, we now need separate rules for covariant and contravariant components. The starting point is the dual basis.

Dual Bases

Definition

Two bases ${\mathbf{e}}_1,{\mathbf{e}}_2,{\mathbf{e}}_3$ and ${\mathbf{e}}^1,{\mathbf{e}}^2,{\mathbf{e}}^3$ are dual if

$${\mathbf{e}}_i\cdot {\mathbf{e}}^k={\delta }_i^k=\{\begin{array}{ll}1& i=k\\ 0& i\ne k.\end{array}$$

Method of Dual Basis

Given the original basis ${\mathbf{e}}_i$, the dual basis is obtained from cross products:

$$\boxed{{\mathbf{e}}^i=\frac{{\mathbf{e}}_j\times {\mathbf{e}}_k}{{\mathbf{e}}_i\cdot ({\mathbf{e}}_j\times {\mathbf{e}}_k)}=\frac{{\mathbf{e}}_j\times {\mathbf{e}}_k}{V},}$$

where $(i,j,k)$ is a cyclic permutation of $(1,2,3)$ and $V={\mathbf{e}}_1\cdot ({\mathbf{e}}_2\times {\mathbf{e}}_3)$ is the volume of the parallelepiped spanned by ${\mathbf{e}}_1,{\mathbf{e}}_2,{\mathbf{e}}_3$. The reverse formula ${\mathbf{e}}_i=({\mathbf{e}}^j\times {\mathbf{e}}^k)/V^{\prime }$ recovers the original basis from its dual.

Example: Constructing the Dual Basis

Example

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

$${\mathbf{e}}_1=2{\mathbf{i}}_1+2{\mathbf{i}}_2+{\mathbf{i}}_3,{\mathbf{e}}_2={\mathbf{i}}_1+{\mathbf{i}}_2,{\mathbf{e}}_3={\mathbf{i}}_1+2{\mathbf{i}}_2+4{\mathbf{i}}_3.$$

Find the dual basis ${\mathbf{e}}^i$.

Cross products.

$${\mathbf{e}}_2\times {\mathbf{e}}_3=(4,-4,1{)}^T,{\mathbf{e}}_3\times {\mathbf{e}}_1=(-6,7,-2{)}^T,{\mathbf{e}}_1\times {\mathbf{e}}_2=(-1,1,0{)}^T.$$

Volume. $V={\mathbf{e}}_3\cdot ({\mathbf{e}}_1\times {\mathbf{e}}_2)=(1,2,4)\cdot (-1,1,0)=-1+2+0=1$.

Dual basis vectors.

$$\boxed{{\mathbf{e}}^1=(4,-4,1{)}^T,{\mathbf{e}}^2=(-6,7,-2{)}^T,{\mathbf{e}}^3=(-1,1,0{)}^T.}$$

Covariant and Contravariant Components of a Vector

Two Expansions

A vector $\mathbf{A}$ has two natural expansions - one against each basis:

$$\mathbf{A}=A^i{\mathbf{e}}_i\text{(contravariant components, summed against }{\mathbf{e}}_i),$$ $$\mathbf{A}=A_i{\mathbf{e}}^i\text{(covariant components, summed against }{\mathbf{e}}^i).$$

The placement of the index encodes the basis: upper index $\leftrightarrow$ original basis ${\mathbf{e}}_i$ (contravariant), lower index $\leftrightarrow$ dual basis ${\mathbf{e}}^i$ (covariant).

Computing components

Dotting both expansions against ${\mathbf{e}}_i$ or ${\mathbf{e}}^i$ and using duality ${\mathbf{e}}_i\cdot {\mathbf{e}}^k={\delta }_i^k$:

$$A_i=\mathbf{A}\cdot {\mathbf{e}}_i,A^i=\mathbf{A}\cdot {\mathbf{e}}^i.$$

Example: Components of $\mathbf{J}=(2,-1,0{)}^T$

Covariant components $J_i=\mathbf{J}\cdot {\mathbf{e}}_i$:

$$J_1=(2,-1,0)\cdot (2,2,1)=2,J_2=(2,-1,0)\cdot (1,1,0)=1,J_3=(2,-1,0)\cdot (1,2,4)=0.$$

Contravariant components $J^i=\mathbf{J}\cdot {\mathbf{e}}^i$:

$$J^1=(2,-1,0)\cdot (4,-4,1)=12,J^2=(2,-1,0)\cdot (-6,7,-2)=-19,J^3=(2,-1,0)\cdot (-1,1,0)=-3.$$

Verification. Both expansions reproduce $\mathbf{J}$:

$$2{\mathbf{e}}^1+{\mathbf{e}}^2+0{\mathbf{e}}^3=2(4,-4,1)+(-6,7,-2)=(2,-1,0)=\mathbf{J}.✓$$ $$12{\mathbf{e}}_1-19{\mathbf{e}}_2-3{\mathbf{e}}_3=(24,24,12)+(-19,-19,0)+(-3,-6,-12)=(2,-1,0)=\mathbf{J}.✓$$

Transformation Rule

Transformation matrix in generalised coordinates

$$L_{i^{\prime }}^k={\mathbf{e}}_i^{\prime }\cdot {\mathbf{e}}^k,L_j^{i^{\prime }}={\mathbf{e}}_j\cdot {\mathbf{e}}^{\prime i}.$$

The components of $\mathbf{A}$ transform between systems via:

$$\boxed{A_i^{\prime }=L_{i^{\prime }}^j{A}_j\text{(covariant)},A^{\prime i}=L_j^{i^{\prime }}{A}^j\text{(contravariant)}.}$$

Cartesian special case

When both bases are orthonormal, ${\mathbf{e}}^k={\mathbf{e}}_k$ and ${\mathbf{e}}^{\prime i}={\mathbf{e}}_i^{\prime }$, so $L_{i^{\prime }}^k=L_k^{i^{\prime }}=L_{ik}$ collapses to the single rotation matrix from mth3008 lecture 6.

Raising and Lowering with the Metric Tensor

Within a single coordinate system, covariant and contravariant components are linked by the metric:

$$\boxed{A_i=g_{ik}{A}^k,A^i=g^{ik}{A}_k,}$$

where

$$g_{ik}=g_{ki}={\mathbf{e}}_i\cdot {\mathbf{e}}_k,g^{ik}=g^{ki}={\mathbf{e}}^i\cdot {\mathbf{e}}^k.$$

So $g_{ik}$ lowers an index and $g^{ik}$ raises one - see Index Raising and Lowering.

Covariant and Contravariant Components of a Second-Rank Tensor

The Four Flavours

A second-rank tensor has four kinds of components:

• covariant $A_{ik}$,
• contravariant $A^{ik}$,
• mixed $A_i^{\cdot k}$ and $A_{\cdot k}^i$.

Transformation Rules

Each free index transforms by one factor of $L$, with the position of the index dictating which $L$ to use:

$$A_{ik}^{\prime }=L_{i^{\prime }}^{\ell }{L}_{k^{\prime }}^m{A}_{\ell m},A^{\prime ik}=L_{\ell }^{i^{\prime }}{L}_m^{k^{\prime }}{A}^{\ell m},$$ $$A_i^{\prime \cdot k}=L_{i^{\prime }}^{\ell }{L}_m^{k^{\prime }}{A}_{\ell }^{\cdot m},A_{\cdot k}^{\prime i}=L_{\ell }^{i^{\prime }}{L}_{k^{\prime }}^m{A}_{\cdot m}^{\ell }.$$

Matrix Form

A clever rewrite turns the index-laden formula into a matrix product. For the covariant case:

$$A_{ik}^{\prime }=L_{i^{\prime }}^{\ell }{L}_{k^{\prime }}^m{A}_{\ell m}=L_{i^{\prime }}^{\ell }{A}_{\ell m}(L^T{)}_m^{k^{\prime }}=(LA{L}^T{)}_{ik}.$$

So $A^{\prime }=LA{L}^T$. The same trick works for $G=[g^{ik}]$ raising/lowering:

$$A^{\prime ik}=g^{i\ell }{g}^{km}{A}_{\ell m}^{\prime }=(G{A}^{\prime }{G}^T{)}_{ik},A_i^{\prime \cdot k}=(A^{\prime }{G}^T{)}_{ik},A_{\cdot k}^{\prime i}=(G{A}^{\prime }{)}_{ik}.$$

Example: Transforming $[A_{ik}]$

Example

Continuing with $K^{\prime }$ from above, take the second-order tensor of $K$ with components

$$[A_{ik}]=[A^{ik}]=[A_i^{\cdot k}]=[A_{\cdot k}^i]=\left(\begin{array}{ccc}1& 0& 1\\ -1& 0& 0\\ 0& 1& 0\end{array}\right).$$

Express its covariant components in $K^{\prime }$.

Transformation matrix. Reading off coefficients from ${\mathbf{e}}_i=L_{ij}{\mathbf{i}}_j$:

$$L=\left(\begin{array}{ccc}2& 2& 1\\ 1& 1& 0\\ 1& 2& 4\end{array}\right).$$

Covariant components. $A^{\prime }=LA{L}^T$:

$$A^{\prime }=\left(\begin{array}{ccc}2& 2& 1\\ 1& 1& 0\\ 1& 2& 4\end{array}\right)\left(\begin{array}{ccc}1& 0& 1\\ -1& 0& 0\\ 0& 1& 0\end{array}\right)\left(\begin{array}{ccc}2& 1& 1\\ 2& 1& 2\\ 1& 0& 4\end{array}\right)=\left(\begin{array}{ccc}0& 1& 2\\ 0& 0& 1\\ -2& 4& 1\end{array}\right)\left(\begin{array}{ccc}2& 1& 1\\ 2& 1& 2\\ 1& 0& 4\end{array}\right)=\left(\begin{array}{ccc}4& 1& 11\\ 1& 0& 4\\ 5& 2& 14\end{array}\right).$$

Metric in $K^{\prime }$. Using $g^{ik}={\mathbf{e}}^i\cdot {\mathbf{e}}^k$ with the dual basis from above:

$$G=[g^{ik}]=\left(\begin{array}{ccc}22& -54& -8\\ -54& 89& 13\\ -8& 13& 2\end{array}\right).$$

Contravariant components. $A^{\prime ik}=(G{A}^{\prime }G{)}_{ik}$:

$$[A^{\prime ik}]=\left(\begin{array}{ccc}56& -84& -13\\ -86& 128& 20\\ -17& 26& 4\end{array}\right).$$

Mixed components.

$$[A_i^{\prime \cdot k}]=A^{\prime }G=\left(\begin{array}{ccc}-10& 16& 3\\ 1& -2& 0\\ -55& 90& 14\end{array}\right),[A_{\cdot k}^{\prime i}]=G{A}^{\prime }=\left(\begin{array}{ccc}38& 17& 35\\ -62& -28& -56\\ -9& -4& -8\end{array}\right).$$

Tensors in Generalised Coordinate Systems

Higher-Rank Transformation

The pattern from the rank-1 and rank-2 cases generalises: each index transforms individually, with the kind of $L$ matching the position of the index. For a mixed rank-5 tensor:

$$\boxed{A_{ij\cdot \cdot m}^{\prime \cdot \cdot k\ell \cdot }=L_{i^{\prime }}^a{L}_{j^{\prime }}^b{L}_c^{k^{\prime }}{L}_d^{{\ell }^{\prime }}{L}_{m^{\prime }}^e{A}_{ab\cdot \cdot e}^{\cdot \cdot cd\cdot }.}$$

How to read this

• Original (left-side) indices keep their position.
• Each $L$ has one primed index matching the original tensor index, one unprimed dummy index matching the right-side tensor.
• Lower-position indices use $L_{i^{\prime }}^j$ (covariant rule); upper-position indices use $L_j^{i^{\prime }}$ (contravariant rule).

Relations Between Components of a Second-Rank Tensor

Index raising/lowering with $g_{ik}$, $g^{ik}$ links all four kinds:

$$A_{ik}=g_{i\ell }{g}_{km}{A}^{\ell m}=g_{k\ell }{A}_i^{\cdot \ell }=g_{i\ell }{A}_{\cdot k}^{\ell },$$ $$A^{ik}=g^{i\ell }{g}^{km}{A}_{\ell m}=g^{i\ell }{A}_{\ell }^{\cdot k}=g^{k\ell }{A}_{\cdot \ell }^i,$$ $$A_i^{\cdot k}=g^{k\ell }{A}_{i\ell }=g_{i\ell }{A}^{\ell k},A_{\cdot k}^i=g^{i\ell }{A}_{\ell k}=g_{k\ell }{A}^{i\ell }.$$

Symmetries

Cartesian Coordinates

In Cartesian (or orthogonal) coordinates, the index position is irrelevant, so symmetry is just an index-swap relation:

Definition (Cartesian)

$T_{ij}$ is symmetric if $T_{ij}=T_{ji}$; antisymmetric if $T_{ij}=-T_{ji}$.

For higher rank, symmetry/antisymmetry can apply to any pair of indices. Example: the Kronecker Delta is symmetric (${\delta }_{ij}={\delta }_{ji}$); the Alternating Tensor is antisymmetric in any pair (${ϵ}_{ijk}=-{ϵ}_{jik}$ etc.).

Generalised Coordinates

In generalised coordinates the index position matters, so symmetry only applies to pairs of indices in the same position:

Definition (generalised)

$A_{ik}^{\cdot \cdot \ell }$ is symmetric in $i,k$ if $A_{ik}^{\cdot \cdot \ell }=A_{ki}^{\cdot \cdot \ell }$.

$B_{\cdot \cdot \ell }^{ik}$ is antisymmetric in $i,k$ if $B_{\cdot \cdot \ell }^{ik}=-B_{\cdot \cdot \ell }^{ki}$.

You cannot meaningfully ask whether $A_i^{\cdot k}$ is symmetric in $i,k$, because the two indices live in different positions and so transform with different $L$‘s.

Where We Came From

This lecture covered the headline mechanics of Chapters 5-6:

• Chapter 5 (lectures 7-10): dual basis, contravariant components of a vector, metric tensor and raising/lowering, Quotient Rule.
• Chapter 6 (lectures 11-14): second-rank tensor flavours, generalised-coordinate tensor transformations, symmetries.

Lecture 22 will close out the revision with tensor algebra (Tensor Addition, Outer Product, Contraction), metric tensor and arc length, Christoffel Symbols & Ricci’s Theorem, and the Riemann-Christoffel Tensor. The mock exam will be solved alongside.

Mock Exam

The mock exam is to be solved before lecture 22 - the in-class walkthrough is most useful as a check on attempted work, not as first exposure.

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Pre-Lecture Notes from University Notes

• Chapter 5-6 revision focusing on dual bases, vector & tensor components in generalised coordinates, and symmetries
• Generalised coordinate systems: not necessarily orthogonal; transitioning between them needs more than one rotation matrix
• Dual basis ${\mathbf{e}}^i$: defined by ${\mathbf{e}}_i\cdot {\mathbf{e}}^k={\delta }_i^k$; constructed via ${\mathbf{e}}^i=({\mathbf{e}}_j\times {\mathbf{e}}_k)/V$ for cyclic $(i,j,k)$, with $V={\mathbf{e}}_1\cdot ({\mathbf{e}}_2\times {\mathbf{e}}_3)$
• Worked example: dual basis for ${\mathbf{e}}_1=2{\mathbf{i}}_1+2{\mathbf{i}}_2+{\mathbf{i}}_3,{\mathbf{e}}_2={\mathbf{i}}_1+{\mathbf{i}}_2,{\mathbf{e}}_3={\mathbf{i}}_1+2{\mathbf{i}}_2+4{\mathbf{i}}_3$ gives $V=1$ and ${\mathbf{e}}^1=(4,-4,1),{\mathbf{e}}^2=(-6,7,-2),{\mathbf{e}}^3=(-1,1,0)$
• Covariant component $A_i=\mathbf{A}\cdot {\mathbf{e}}_i$ (expand against dual basis); contravariant component $A^i=\mathbf{A}\cdot {\mathbf{e}}^i$ (expand against original basis); index position encodes which basis is used
• Worked example: $\mathbf{J}=(2,-1,0{)}^T$ has covariant components $(2,1,0)$ and contravariant components $(12,-19,-3)$; both expansions reproduce $\mathbf{J}$
• Transformation matrices: $L_{i^{\prime }}^j={\mathbf{e}}_i^{\prime }\cdot {\mathbf{e}}^j$ (covariant rule), $L_j^{i^{\prime }}={\mathbf{e}}_j\cdot {\mathbf{e}}^{\prime i}$ (contravariant rule)
• Rules: $A_i^{\prime }=L_{i^{\prime }}^j{A}_j$ for covariant, $A^{\prime i}=L_j^{i^{\prime }}{A}^j$ for contravariant
• Raise/lower with metric: $A_i=g_{ik}{A}^k$, $A^i=g^{ik}{A}_k$, where $g_{ik}={\mathbf{e}}_i\cdot {\mathbf{e}}_k$ and $g^{ik}={\mathbf{e}}^i\cdot {\mathbf{e}}^k$
• Second-rank tensors: four flavours $A_{ik},A^{ik},A_i^{\cdot k},A_{\cdot k}^i$; each transforms by one $L$ per free index according to its position
• Matrix form via $A^{\prime }=LA{L}^T$, $A^{\prime ik}=(G{A}^{\prime }G{)}_{ik}$, $A_i^{\prime \cdot k}=(A^{\prime }G{)}_{ik}$, $A_{\cdot k}^{\prime i}=(G{A}^{\prime }{)}_{ik}$ - turns index gymnastics into 3×3 matrix products
• Worked example: a $3\times 3$ tensor in $K$ transformed to $K^{\prime }$, all four flavours computed by matrix products
• Higher-rank generalisation: each index transforms individually with one $L$; new (primed) indices on $L$ go with the original tensor’s index positions, dummy (unprimed) indices on $L$ contract with the right-side tensor
• Symmetries (Cartesian): $T_{ij}=\pm T_{ji}$; can apply to any pair of indices for higher-rank tensors
• Symmetries (generalised): only meaningful for pairs of indices in the same position - can’t sensibly ask if $A_i^{\cdot k}$ is symmetric in $i,k$
• Mock exam: to be attempted before lecture 22; class walkthrough only as effective as the prep work
• Next lecture: tensor algebra, arc length & metric tensor, Christoffel symbols & Ricci’s theorem, Riemann-Christoffel tensor (Chapters 6-7 closeout)