MTH3008 Weekly Problems 11

Original Documents: Problem Sheet / My Handwritten Solutions

Vibes: Whole-module revision in five problems. Two warm-ups (rewriting a vector identity in suffix notation; re-deriving $\nabla \times \nabla f=0$); a transformation-rule recipe for an arbitrary mixed rank-5 tensor; a full dual-basis exercise (cross products, components, transformation matrix, and second-rank tensor in the new system); and a 2D coordinate-system grind (basis, metric, arc length, both kinds of Christoffel symbols, three Riemann components). The 2D system in 11.5 is just flat ${\mathbb{R}}^2$ in disguise (set $X=e^x\sin \theta$, $Y=e^x\cos \theta$), so all Riemann components vanish.

Used Techniques:

• Suffix-symmetric / antisymmetric contraction: ${ϵ}_{ijk}{\partial }_j{\partial }_{k}f=0$ since ${ϵ}_{ijk}$ is antisymmetric in $j,k$ and ${\partial }_j{\partial }_k$ is symmetric.
• Transformation rule slot-by-slot: each free index gets one $L$, using $L_{i^{\prime }}^j$ for lower-position indices and $L_j^{i^{\prime }}$ for upper-position indices.
• Dual basis recipe: ${\mathbf{e}}^i=({\mathbf{e}}_j\times {\mathbf{e}}_k)/V$ for cyclic $(i,j,k)$, where $V={\mathbf{e}}_1\cdot ({\mathbf{e}}_2\times {\mathbf{e}}_3)$.
• Component formulae: $A_i=\mathbf{A}\cdot {\mathbf{e}}_i$ (covariant), $A^i=\mathbf{A}\cdot {\mathbf{e}}^i$ (contravariant).
• Matrix form of the rank-2 transformation: $A_{ik}^{\prime }=L_{i^{\prime }}^{\ell }{L}_{k^{\prime }}^m{A}_{\ell m}$ becomes $A^{\prime }=LA{L}^T$.
• Diagonal metric Christoffel shortcuts: nonzero first-kind symbols are ${\Gamma }_{iik}={\Gamma }_{iki}=\frac{1}{2}{\partial }_k{g}_{ii}$, ${\Gamma }_{ijj}=-\frac{1}{2}{\partial }_i{g}_{jj}$, plus the diagonal ${\Gamma }_{iii}=\frac{1}{2}{\partial }_i{g}_{ii}$.
• Riemann antisymmetry: $R_{ijk}^r=-R_{ikj}^r$, so $R_{ijj}^r=0$ for any $r,i,j$.

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11.1. Vector Equation in Suffix Notation

Question

Write the vector equation

$$\mathbf{u}+(\mathbf{a}\cdot \mathbf{b})\mathbf{v}=|\mathbf{a}{|}^2(\mathbf{b}\cdot \mathbf{v})\mathbf{a}$$

in suffix notation.

Translate each piece using Suffix Notation and the Kronecker Delta / dot-product identity $\mathbf{x}\cdot \mathbf{y}=x_j{y}_j$:

• $\mathbf{u}\to u_i$ (free index $i$).
• $(\mathbf{a}\cdot \mathbf{b})\mathbf{v}\to (a_j{b}_j)v_i$ - the dot product is a scalar with dummy index $j$, multiplying the free $v_i$.
• $|\mathbf{a}{|}^2=\mathbf{a}\cdot \mathbf{a}\to a_k{a}_k$ - new dummy index $k$ to keep it disjoint from $j$.
• $(\mathbf{b}\cdot \mathbf{v})\mathbf{a}\to (b_m{v}_m)a_i$ - another fresh dummy $m$.

$$\boxed{u_i+(a_j{b}_j)v_i=(a_k{a}_k)(b_m{v}_m)a_i.}$$

Use distinct dummies

Each contracted pair must use a different letter from the others. Re-using $j$ for $\mathbf{b}\cdot \mathbf{v}$ on the right would silently force $a_j{b}_j=b_j{v}_j$, which is nonsense.

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11.2. Curl of a Gradient

Question

Let $f$ be a scalar field. Using suffix notation, evaluate $\nabla \times (\nabla f)$.

Recall $[\nabla f{]}_k={\partial }_{k}f$ and $[\nabla \times \mathbf{u}{]}_i={ϵ}_{ijk}{\partial }_j{u}_k$. Combining:

$$[\nabla \times (\nabla f){]}_i={ϵ}_{ijk}{\partial }_j{\partial }_{k}f.$$

The mixed partial ${\partial }_j{\partial }_{k}f={\partial }_k{\partial }_{j}f$ is symmetric in $j,k$ (Schwarz/Clairaut), while ${ϵ}_{ijk}=-{ϵ}_{ikj}$ is antisymmetric in $j,k$. The contraction of a symmetric and antisymmetric tensor over the same pair vanishes:

$${ϵ}_{ijk}{\partial }_j{\partial }_{k}f=\frac{1}{2}({ϵ}_{ijk}+{ϵ}_{ikj}){\partial }_j{\partial }_{k}f=0.$$ $$\boxed{\nabla \times (\nabla f)=0.}$$

The sym/antisym argument in one line

Swap the dummies $j\leftrightarrow k$: ${ϵ}_{ijk}{\partial }_j{\partial }_{k}f={ϵ}_{ikj}{\partial }_k{\partial }_{j}f=-{ϵ}_{ijk}{\partial }_j{\partial }_{k}f$, so ${ϵ}_{ijk}{\partial }_j{\partial }_{k}f=0$.

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11.3. Transformation Laws

Question

Write the transformation law for the following tensors. (1) A rank-5 tensor in Cartesian coordinates. (2) A rank-5 contravariant tensor in generalised coordinates. (3) The rank-5 mixed tensor $T_{\cdot \cdot mn\cdot }^{k\ell \cdot \cdot p}$ in generalised coordinates.

The recipe: one $L$ per free index, with the kind of $L$ matching the index position.

(1) Cartesian, rank 5. Index position is irrelevant; one rotation matrix $L_{ij}$ does everything.

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

(2) Contravariant rank 5, generalised. Each upper index uses $L_j^{i^{\prime }}$:

$$\boxed{T^{\prime ijk\ell m}=L_p^{i^{\prime }}{L}_q^{j^{\prime }}{L}_r^{k^{\prime }}{L}_s^{{\ell }^{\prime }}{L}_t^{m^{\prime }}{T}^{pqrst}.}$$

(3) Mixed rank 5. $T_{\cdot \cdot mn\cdot }^{k\ell \cdot \cdot p}$ has positions 1,2,5 upper and positions 3,4 lower. Match each:

• Position 1 (upper $k$): $L_a^{k^{\prime }}$.
• Position 2 (upper $\ell$): $L_b^{{\ell }^{\prime }}$.
• Position 3 (lower $m$): $L_{m^{\prime }}^c$.
• Position 4 (lower $n$): $L_{n^{\prime }}^d$.
• Position 5 (upper $p$): $L_e^{p^{\prime }}$.

$$\boxed{T_{\cdot \cdot mn\cdot }^{\prime k\ell \cdot \cdot p}=L_a^{k^{\prime }}{L}_b^{{\ell }^{\prime }}{L}_{m^{\prime }}^c{L}_{n^{\prime }}^d{L}_e^{p^{\prime }}{T}_{\cdot \cdot cd\cdot }^{ab\cdot \cdot e}.}$$

The dummy indices $a,b,c,d,e$ on the original $T$ sit in the same positions (1,2 upper, 3,4 lower, 5 upper) as the primed indices on the result.

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11.4. Dual Basis & Tensor Transformation

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

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

(1) Dual Basis Vectors

Question

Compute ${\mathbf{e}}^1,{\mathbf{e}}^2,{\mathbf{e}}^3$.

Cross products.

$${\mathbf{e}}_2\times {\mathbf{e}}_3=(0,1,-1)\times (1,-1,1)=((1)(1)-(-1)(-1),(-1)(1)-(0)(1),(0)(-1)-(1)(1))=(0,-1,-1).$$ $${\mathbf{e}}_3\times {\mathbf{e}}_1=(1,-1,1)\times (1,0,1)=((-1)(1)-(1)(0),(1)(1)-(1)(1),(1)(0)-(-1)(1))=(-1,0,1).$$ $${\mathbf{e}}_1\times {\mathbf{e}}_2=(1,0,1)\times (0,1,-1)=((0)(-1)-(1)(1),(1)(0)-(1)(-1),(1)(1)-(0)(0))=(-1,1,1).$$

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

Dual basis ${\mathbf{e}}^i=({\mathbf{e}}_j\times {\mathbf{e}}_k)/V$ for cyclic $(i,j,k)$:

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

Check. ${\mathbf{e}}_1\cdot {\mathbf{e}}^1=(1,0,1)\cdot (0,1,1)=1✓$, ${\mathbf{e}}_2\cdot {\mathbf{e}}^2=(0,1,-1)\cdot (1,0,-1)=1✓$, ${\mathbf{e}}_3\cdot {\mathbf{e}}^3=(1,-1,1)\cdot (1,-1,-1)=1✓$, all off-diagonals zero.

(2) Covariant Components of $\mathbf{V}={\mathbf{i}}_1-2{\mathbf{i}}_3$

Question

Compute the covariant components of $\mathbf{V}=(1,0,-2{)}^T$.

$V_i=\mathbf{V}\cdot {\mathbf{e}}_i$:

$$V_1=(1,0,-2)\cdot (1,0,1)=1+0-2=-1,$$ $$V_2=(1,0,-2)\cdot (0,1,-1)=0+0+2=2,$$ $$V_3=(1,0,-2)\cdot (1,-1,1)=1+0-2=-1.$$ $$\boxed{V_1=-1,V_2=2,V_3=-1.}$$

(3) Contravariant Components of $\mathbf{V}={\mathbf{i}}_1-2{\mathbf{i}}_3$

Question

Compute the contravariant components.

$V^i=\mathbf{V}\cdot {\mathbf{e}}^i$:

$$V^1=(1,0,-2)\cdot (0,1,1)=0+0-2=-2,$$ $$V^2=(1,0,-2)\cdot (1,0,-1)=1+0+2=3,$$ $$V^3=(1,0,-2)\cdot (1,-1,-1)=1+0+2=3.$$ $$\boxed{V^1=-2,V^2=3,V^3=3.}$$

Sanity check via reconstruction. $V_i{\mathbf{e}}^i=-1(0,1,1)+2(1,0,-1)+(-1)(1,-1,-1)=(2-1,-1+1,-1-2+1)=(1,0,-2)=\mathbf{V}✓$.

(4) Transformation Matrix

Question

Find $L=[L_{i^{\prime }}^j]$.

$L_{i^{\prime }}^j={\mathbf{e}}_i^{\prime }\cdot {\mathbf{e}}^j$, with $K^{\prime }$ the primed system. In $K$ Cartesian, ${\mathbf{e}}^j={\mathbf{i}}_j$, so $L_{i^{\prime }}^j={\mathbf{e}}_i\cdot {\mathbf{i}}_j$ - i.e. just the components of ${\mathbf{e}}_i$. Reading off:

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

(5) Covariant Components of $[P_{ik}]$ in $K^{\prime }$

Question

Express the covariant components of the rank-2 tensor of $K$ with components

$$[P_{ik}]=[P^{ik}]=[P_i^{\cdot k}]=[P_{\cdot k}^i]=\left(\begin{array}{ccc}0& 1& 0\\ -1& 0& -1\\ 0& 0& 1\end{array}\right)$$

in $K^{\prime }$.

Use the rank-2 transformation $P_{ik}^{\prime }=L_{i^{\prime }}^{\ell }{L}_{k^{\prime }}^m{P}_{\ell m}=(LP{L}^T{)}_{ik}$.

Step 1: $LP$.

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

Step 2: $(LP)L^T$. With $L^T=L$ (it’s symmetric here),

$$(LP)L^T=\left(\begin{array}{ccc}0& 1& 1\\ -1& 0& -2\\ 1& 1& 2\end{array}\right)\left(\begin{array}{ccc}1& 0& 1\\ 0& 1& -1\\ 1& -1& 1\end{array}\right)=\left(\begin{array}{ccc}1& 0& 0\\ -3& 2& -3\\ 3& -1& 2\end{array}\right).$$ $$\boxed{[P_{ik}^{\prime }]=\left(\begin{array}{ccc}1& 0& 0\\ -3& 2& -3\\ 3& -1& 2\end{array}\right).}$$

Why $L^T=L$ here

The basis ${\mathbf{e}}_1=(1,0,1),{\mathbf{e}}_2=(0,1,-1),{\mathbf{e}}_3=(1,-1,1)$ produces an $L$ that happens to be symmetric. This is a coincidence of the chosen ${\mathbf{e}}_i$, not a general property - $K^{\prime }$ is non-orthonormal (e.g. ${\mathbf{e}}_1\cdot {\mathbf{e}}_1=2$, ${\mathbf{e}}_1\cdot {\mathbf{e}}_2=-1$), so $L$ is not an orthogonal matrix.

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11.5. 2D Coordinate System with $\mathbf{r}=e^x\sin \theta {\mathbf{i}}_1+e^x\cos \theta {\mathbf{i}}_2$

Coordinates $(x^1,x^2)=(x,\theta )$.

(1) Basis Vectors

Question

Compute ${\mathbf{e}}_1,{\mathbf{e}}_2$.

Local basis ${\mathbf{e}}_i=\partial \mathbf{r}/\partial x^i$:

$${\mathbf{e}}_1=\frac{\partial \mathbf{r}}{\partial x}=e^x\sin \theta {\mathbf{i}}_1+e^x\cos \theta {\mathbf{i}}_2=e^x(\sin \theta ,\cos \theta {)}^T,$$ $${\mathbf{e}}_2=\frac{\partial \mathbf{r}}{\partial \theta }=e^x\cos \theta {\mathbf{i}}_1-e^x\sin \theta {\mathbf{i}}_2=e^x(\cos \theta ,-\sin \theta {)}^T.$$

(The system is two-dimensional - the question’s mention of ${\mathbf{e}}_3$ is a typo from a 3D template.)

(2) Metric Tensor

Question

Compute $g_{ik}$.

$g_{ik}={\mathbf{e}}_i\cdot {\mathbf{e}}_k$:

$$g_{11}=e^{2x}({\sin }^2\theta +{\cos }^2\theta )=e^{2x},g_{22}=e^{2x}({\cos }^2\theta +{\sin }^2\theta )=e^{2x},$$ $$g_{12}=g_{21}=e^{2x}(\sin \theta \cos \theta -\cos \theta \sin \theta )=0.$$ $$\boxed{[g_{ik}]=e^{2x}\left(\begin{array}{cc}1& 0\\ 0& 1\end{array}\right)=e^{2x}{\delta }_{ik}.}$$

So the system is orthogonal (off-diagonals zero), with both scale factors equal: $h_1=h_2=e^x$. The induced metric is conformally flat.

(3) Arc Length

Question

Describe the arc-length element in terms of the metric coefficients.

$$d{s}^2=g_{ik}d{x}^{i}d{x}^k=g_{11}d{x}^2+g_{22}d{\theta }^2,$$ $$\boxed{d{s}^2=e^{2x}(d{x}^2+d{\theta }^2).}$$

(4) Christoffel Symbols Of the First Kind

Question

Determine ${\Gamma }_{ijk}$.

${\Gamma }_{ijk}=\frac{1}{2}({\partial }_j{g}_{ik}+{\partial }_k{g}_{ij}-{\partial }_i{g}_{jk})$.

Nonzero metric derivatives. Since $g_{ik}=e^{2x}{\delta }_{ik}$ depends only on $x=x^1$:

$${\partial }_1{g}_{11}=2e^{2x},{\partial }_1{g}_{22}=2e^{2x},\text{all }{\partial }_2{g}_{ik}=0,{\partial }_i{g}_{12}=0.$$

Going through all eight cases (in 2D, $2^3=8$ symbols):

• ${\Gamma }_{111}=\frac{1}{2}{\partial }_1{g}_{11}=e^{2x}$,
• ${\Gamma }_{112}={\Gamma }_{121}=\frac{1}{2}{\partial }_2{g}_{11}=0$,
• ${\Gamma }_{122}=-\frac{1}{2}{\partial }_1{g}_{22}=-e^{2x}$,
• ${\Gamma }_{211}=-\frac{1}{2}{\partial }_2{g}_{11}=0$,
• ${\Gamma }_{212}={\Gamma }_{221}=\frac{1}{2}{\partial }_1{g}_{22}=e^{2x}$,
• ${\Gamma }_{222}=\frac{1}{2}{\partial }_2{g}_{22}=0$.

$$\boxed{{\Gamma }_{111}=e^{2x},{\Gamma }_{122}=-e^{2x},{\Gamma }_{212}={\Gamma }_{221}=e^{2x},\text{all others }=0.}$$

(5) Christoffel Symbols Of the Second Kind

Question

Determine ${\Gamma }_{jk}^i$.

For a diagonal metric, $g^{ii}=1/g_{ii}=e^{-2x}$ (no sum), and ${\Gamma }_{jk}^a=g^{ai}{\Gamma }_{ijk}=g^{aa}{\Gamma }_{ajk}$. Multiplying the first-kind table by $e^{-2x}$:

$${\Gamma }_{11}^1=e^{-2x}\cdot e^{2x}=1,{\Gamma }_{22}^1=e^{-2x}\cdot (-e^{2x})=-1,$$ $${\Gamma }_{12}^2={\Gamma }_{21}^2=e^{-2x}\cdot e^{2x}=1.$$ $$\boxed{{\Gamma }_{11}^1=1,{\Gamma }_{22}^1=-1,{\Gamma }_{12}^2={\Gamma }_{21}^2=1,\text{all others }=0.}$$

(6) Riemann-Christoffel Components

Question

Compute $R_{111}^1,R_{121}^2,R_{222}^1$.

Antisymmetry shortcut. $R_{ijk}^r=-R_{ikj}^r$, so any component with the last two indices equal vanishes. Therefore

$$R_{111}^1=0,R_{222}^1=0\text{(by antisymmetry, j=k).}$$

Direct computation of $R_{121}^2$. Using

$$R_{ijk}^r={\partial }_j{\Gamma }_{ik}^r-{\partial }_k{\Gamma }_{ij}^r+{\Gamma }_{ik}^p{\Gamma }_{pj}^r-{\Gamma }_{ij}^p{\Gamma }_{pk}^r,$$

with $r=2,i=1,j=2,k=1$:

• ${\partial }_2{\Gamma }_{11}^2={\partial }_{\theta }(0)=0$.
• $-{\partial }_1{\Gamma }_{12}^2=-{\partial }_x(1)=0$.
• ${\Gamma }_{11}^p{\Gamma }_{p2}^2$: $p=1$ gives ${\Gamma }_{11}^1{\Gamma }_{12}^2=1\cdot 1=1$; $p=2$ gives ${\Gamma }_{11}^2{\Gamma }_{22}^2=0$. Sum $=1$.
• ${\Gamma }_{12}^p{\Gamma }_{p1}^2$: $p=1$ gives ${\Gamma }_{12}^1{\Gamma }_{11}^2=0\cdot 0=0$; $p=2$ gives ${\Gamma }_{12}^2{\Gamma }_{21}^2=1\cdot 1=1$. Sum $=1$.

$$R_{121}^2=0-0+1-1=0.$$ $$\boxed{R_{111}^1=R_{121}^2=R_{222}^1=0.}$$

Why everything vanishes

Set $X=e^x\sin \theta$, $Y=e^x\cos \theta$. Then $X^2+Y^2=e^{2x}$, so $\mathbf{r}=(X,Y)$ is just flat 2D Euclidean space in disguised coordinates (a logarithmic-polar chart). Smooth and invertible on the open set where $e^x\ne 0$ (always true), so the induced metric is flat and all $R_{ijk}^r=0$. Computing one nonzero-looking component (here $R_{121}^2$) confirms this.