MTH3008 Weekly Problems 4

Original Documents: Problem Sheet / Provided Solutions

Vibes: Dual-basis grind, then satisfying metric/index-raising consistency checks.

Used Techniques:

• Dual basis via ${\mathbf{e}}^i=\frac{{\mathbf{e}}_j\times {\mathbf{e}}_k}{V}$ with $V={\mathbf{e}}_1\cdot ({\mathbf{e}}_2\times {\mathbf{e}}_3)$ (cyclic $i,j,k$).
• Covariant/contravariant components: $A_i=\mathbf{A}\cdot {\mathbf{e}}_i$, $A^i=\mathbf{A}\cdot {\mathbf{e}}^i$.
• Expansion coefficients and transform rule: $L_{m^{\prime }}^n={\mathbf{e}}_m^{\prime }\cdot {\mathbf{i}}_n$, $B_i^{\prime }=L_{i^{\prime }}^j{B}_j$.
• Metric: $g_{ik}={\mathbf{e}}_i\cdot {\mathbf{e}}_k$, and raising/lowering via $A_i=g_{ik}{A}^k$, $A^i=g^{ik}{A}_k$.

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4.1. Dual Basis and Components for Oblique System

Question

1. Given an orthogonal coordinate system $K$ with orthonormal basis ${\mathbf{i}}_1,{\mathbf{i}}_2,{\mathbf{i}}_3$, consider a new coordinate system $K^{\prime }$ with basis vectors ${\mathbf{e}}_1={\mathbf{i}}_1+{\mathbf{i}}_2-{\mathbf{i}}_3$, ${\mathbf{e}}_2={\mathbf{i}}_1+{\mathbf{i}}_2$, ${\mathbf{e}}_3={\mathbf{i}}_1-{\mathbf{i}}_3$. Find a basis ${\mathbf{e}}^1,{\mathbf{e}}^2,{\mathbf{e}}^3$ dual to ${\mathbf{e}}_1,{\mathbf{e}}_2,{\mathbf{e}}_3$.
2. Find the covariant components of the vector joining the origin to the point $(2,0,-1)$.
3. Find the contravariant components of the vector joining the origin to the point $(2,0,-1)$.

1) Dual basis. Using the dual-basis formula with $V={\mathbf{e}}_1\cdot ({\mathbf{e}}_2\times {\mathbf{e}}_3)=1$, the dual basis is:

$${\mathbf{e}}^1=\left(\begin{array}{c}-1\\ 1\\ -1\end{array}\right),{\mathbf{e}}^2=\left(\begin{array}{c}1\\ 0\\ 1\end{array}\right),{\mathbf{e}}^3=\left(\begin{array}{c}1\\ -1\\ 0\end{array}\right).$$

Let $\mathbf{A}=(2,0,-1{)}^T=2{\mathbf{i}}_1-{\mathbf{i}}_3$.

2) Covariant components $A_i=\mathbf{A}\cdot {\mathbf{e}}_i$:

$$(A_1,A_2,A_3)=(3,2,3).$$

3) Contravariant components $A^i=\mathbf{A}\cdot {\mathbf{e}}^i$:

$$(A^1,A^2,A^3)=(-1,1,2).$$

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4.2. Dual Basis and Components

Question

1. Given an orthogonal coordinate system $K$ with orthonormal basis ${\mathbf{i}}_1,{\mathbf{i}}_2,{\mathbf{i}}_3$, consider a new coordinate system $K^{\prime }$ with basis vectors ${\mathbf{e}}_1={\mathbf{i}}_1+2{\mathbf{i}}_2+4{\mathbf{i}}_3$, ${\mathbf{e}}_2={\mathbf{i}}_2$, ${\mathbf{e}}_3={\mathbf{i}}_1+2{\mathbf{i}}_2+5{\mathbf{i}}_3$. Find a basis ${\mathbf{e}}^1,{\mathbf{e}}^2,{\mathbf{e}}^3$ dual to ${\mathbf{e}}_1,{\mathbf{e}}_2,{\mathbf{e}}_3$.
2. Find the covariant components of the vector joining the origin to the point $(1,1,1)$.
3. Find the contravariant components of the vector joining the origin to the point $(1,1,1)$.

1) Dual basis. Here $V=1$, and the dual basis is:

$${\mathbf{e}}^1=\left(\begin{array}{c}5\\ 0\\ -1\end{array}\right),{\mathbf{e}}^2=\left(\begin{array}{c}-2\\ 1\\ 0\end{array}\right),{\mathbf{e}}^3=\left(\begin{array}{c}-4\\ 0\\ 1\end{array}\right).$$

Let $\mathbf{B}=(1,1,1{)}^T$.

2) Covariant components $B_i=\mathbf{B}\cdot {\mathbf{e}}_i$:

$$(B_1,B_2,B_3)=(7,1,8).$$

3) Contravariant components $B^i=\mathbf{B}\cdot {\mathbf{e}}^i$:

$$(B^1,B^2,B^3)=(4,-1,-3).$$

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4.3. Dual Basis and Components for Non-Orthogonal Basis

Question

1. Given an orthogonal coordinate system $K$ with orthonormal basis ${\mathbf{i}}_1,{\mathbf{i}}_2,{\mathbf{i}}_3$, consider a new coordinate system $K^{\prime }$ with basis vectors ${\mathbf{e}}_1={\mathbf{i}}_1+2{\mathbf{i}}_2$, ${\mathbf{e}}_2=2{\mathbf{i}}_1+2{\mathbf{i}}_2+4{\mathbf{i}}_3$, ${\mathbf{e}}_3=2{\mathbf{i}}_3$. Find a basis ${\mathbf{e}}^1,{\mathbf{e}}^2,{\mathbf{e}}^3$ dual to ${\mathbf{e}}_1,{\mathbf{e}}_2,{\mathbf{e}}_3$.
2. Find the covariant components of the vector $\mathbf{C}=2{\mathbf{i}}_2+2{\mathbf{i}}_3$.
3. Find the contravariant components of the vector $\mathbf{C}=2{\mathbf{i}}_2+2{\mathbf{i}}_3$.

1) Dual basis. Using $V={\mathbf{e}}_3\cdot ({\mathbf{e}}_1\times {\mathbf{e}}_2)=-4$, we get:

$${\mathbf{e}}^1=\left(\begin{array}{c}-1\\ 1\\ 0\end{array}\right),{\mathbf{e}}^2=\left(\begin{array}{c}1\\ -\frac{1}{2}\\ 0\end{array}\right),{\mathbf{e}}^3=\left(\begin{array}{c}-2\\ 1\\ \frac{1}{2}\end{array}\right).$$

Let $\mathbf{C}=(0,2,2{)}^T$.

2) Covariant components $C_i=\mathbf{C}\cdot {\mathbf{e}}_i$:

$$(C_1,C_2,C_3)=(4,12,4).$$

3) Contravariant components $C^i=\mathbf{C}\cdot {\mathbf{e}}^i$:

$$(C^1,C^2,C^3)=(2,-1,3).$$

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4.4. Expansion Coefficients and Metric Tensor

Question

1. Given an orthogonal coordinate system $K$ with orthonormal basis ${\mathbf{i}}_1,{\mathbf{i}}_2,{\mathbf{i}}_3$, consider the vector $\mathbf{B}={\mathbf{i}}_1+3{\mathbf{i}}_2$. Let $K^{\prime }$ be a new coordinate system with basis vectors ${\mathbf{e}}_1^{\prime }={\mathbf{i}}_1$, ${\mathbf{e}}_2^{\prime }={\mathbf{i}}_1-{\mathbf{i}}_2$, ${\mathbf{e}}_3^{\prime }={\mathbf{i}}_1+{\mathbf{i}}_2+{\mathbf{i}}_3$. Work out the expansion coefficients $L_{i^{\prime }}^1,L_{i^{\prime }}^2,L_{i^{\prime }}^3$ for each $i=1,2,3$.
2. Compute the covariant components $B_i^{\prime }$ of $\mathbf{B}$ in the coordinate system $K^{\prime }$.
3. Find a basis dual to ${\mathbf{e}}_1^{\prime },{\mathbf{e}}_2^{\prime },{\mathbf{e}}_3^{\prime }$.
4. Compute the contravariant components $B^{\prime i}$ of $\mathbf{B}$ in the coordinate system $K^{\prime }$.
5. Let $g_{ik}={\mathbf{e}}_i^{\prime }\cdot {\mathbf{e}}_k^{\prime }$. Compute these $g_{ik}$ for all $i,k=1,2,3$.
6. Using your answers to parts 2 and 5, work out the contravariant components $B^{\prime i}$ of $\mathbf{B}$ in $K^{\prime }$ in another way.

Let $\mathbf{B}=(1,3,0{)}^T$ in the orthonormal basis ${\mathbf{i}}_n$.

1) Expansion coefficients. Using $L_{m^{\prime }}^n={\mathbf{e}}_m^{\prime }\cdot {\mathbf{i}}_n$, the rows are just the Cartesian components of ${\mathbf{e}}_m^{\prime }$:

$$(L_{1^{\prime }}^n)=(1,0,0),(L_{2^{\prime }}^n)=(1,-1,0),(L_{3^{\prime }}^n)=(1,1,1).$$

2) Covariant components. Using $B_i^{\prime }=L_{i^{\prime }}^j{B}_j$:

$$(B_1^{\prime },B_2^{\prime },B_3^{\prime })=(1,-2,4).$$

3) Dual basis. From the cross-product construction, the dual basis is:

$${\mathbf{e}}^{\prime 1}=\left(\begin{array}{c}1\\ 1\\ -2\end{array}\right),{\mathbf{e}}^{\prime 2}=\left(\begin{array}{c}0\\ -1\\ 1\end{array}\right),{\mathbf{e}}^{\prime 3}=\left(\begin{array}{c}0\\ 0\\ 1\end{array}\right).$$

4) Contravariant components. Using $B^{\prime i}=\mathbf{B}\cdot {\mathbf{e}}^{\prime i}$:

$$(B^{\prime 1},B^{\prime 2},B^{\prime 3})=(4,-3,0).$$

5) Metric tensor. By definition $g_{ik}={\mathbf{e}}_i^{\prime }\cdot {\mathbf{e}}_k^{\prime }$, so:

$$(g_{ik})=\left(\begin{array}{ccc}1& 1& 1\\ 1& 2& 0\\ 1& 0& 3\end{array}\right).$$

6) Raise the index using $g^{ik}$. Since $g^{ik}$ is the inverse of $g_{ik}$ (contravariant metric), here:

$$(g^{ik})=(g_{ik}{)}^{-1}=\left(\begin{array}{ccc}6& -3& -2\\ -3& 2& 1\\ -2& 1& 1\end{array}\right),$$

And then $B^{\prime i}=g^{ik}{B}_k^{\prime }$ gives $(B^{\prime 1},B^{\prime 2},B^{\prime 3})=(4,-3,0)$.

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4.5. Dual Basis and Components from Past Exam

Question

Question from Final Exam 23-24.

In this question, denote by $K$ the Cartesian coordinate system with vector basis ${\mathbf{i}}_1,{\mathbf{i}}_2,{\mathbf{i}}_3$. Denote by $K^{\prime }$ the coordinate system with vector basis ${\mathbf{e}}_1,{\mathbf{e}}_2,{\mathbf{e}}_3$ given by ${\mathbf{e}}_1=2{\mathbf{i}}_1+{\mathbf{i}}_3$, ${\mathbf{e}}_2={\mathbf{i}}_2$, ${\mathbf{e}}_3={\mathbf{i}}_1-{\mathbf{i}}_2+{\mathbf{i}}_3$.

1. Find the dual basis ${\mathbf{e}}^1,{\mathbf{e}}^2,{\mathbf{e}}^3$.
2. Find the covariant and contravariant components of the vector $\mathbf{V}=3{\mathbf{i}}_1-2{\mathbf{i}}_2+{\mathbf{i}}_3$ with respect to ${\mathbf{e}}_1,{\mathbf{e}}_2,{\mathbf{e}}_3$ and ${\mathbf{e}}^1,{\mathbf{e}}^2,{\mathbf{e}}^3$.

1) Dual basis. Since $V={\mathbf{e}}_1\cdot ({\mathbf{e}}_2\times {\mathbf{e}}_3)=1$, we get:

$${\mathbf{e}}^1=\left(\begin{array}{c}1\\ 0\\ -1\end{array}\right),{\mathbf{e}}^2=\left(\begin{array}{c}-1\\ 1\\ 2\end{array}\right),{\mathbf{e}}^3=\left(\begin{array}{c}-1\\ 0\\ 2\end{array}\right).$$

2) Components of $\mathbf{V}=(3,-2,1{)}^T$.

• Covariant: $V_i=\mathbf{V}\cdot {\mathbf{e}}_i\Rightarrow (V_1,V_2,V_3)=(7,-2,6)$.
• Contravariant: $V^i=\mathbf{V}\cdot {\mathbf{e}}^i\Rightarrow (V^1,V^2,V^3)=(2,-3,-1)$.

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4.6. Vector Triple Product Identity Proof

Question

Consider dual bases ${\mathbf{e}}_1,{\mathbf{e}}_2,{\mathbf{e}}_3$ and ${\mathbf{e}}^1,{\mathbf{e}}^2,{\mathbf{e}}^3$, and denote

$$V={\mathbf{e}}_1\cdot ({\mathbf{e}}_2\times {\mathbf{e}}_3),V^{\prime }={\mathbf{e}}^1\cdot ({\mathbf{e}}^2\times {\mathbf{e}}^3).$$

Show, using the vector triple product formula, that $V^{\prime }=\frac{1}{V}$.

Using ${\mathbf{e}}^1=\frac{{\mathbf{e}}_2\times {\mathbf{e}}_3}{V}$, ${\mathbf{e}}^2=\frac{{\mathbf{e}}_3\times {\mathbf{e}}_1}{V}$, ${\mathbf{e}}^3=\frac{{\mathbf{e}}_1\times {\mathbf{e}}_2}{V}$, we have:

$$V^{\prime }=\frac{{\mathbf{e}}_2\times {\mathbf{e}}_3}{V}\cdot \left(\frac{{\mathbf{e}}_3\times {\mathbf{e}}_1}{V}\times \frac{{\mathbf{e}}_1\times {\mathbf{e}}_2}{V}\right)=\frac{1}{V^3}({\mathbf{e}}_2\times {\mathbf{e}}_3)\cdot (({\mathbf{e}}_3\times {\mathbf{e}}_1)\times ({\mathbf{e}}_1\times {\mathbf{e}}_2)).$$

Apply the vector triple product identity and use $({\mathbf{e}}_3\times {\mathbf{e}}_1)\perp {\mathbf{e}}_1$ to drop one term, giving $({\mathbf{e}}_3\times {\mathbf{e}}_1)\times ({\mathbf{e}}_1\times {\mathbf{e}}_2)=V{\mathbf{e}}_1$.

Hence, $V^{\prime }=\frac{1}{V^3}({\mathbf{e}}_2\times {\mathbf{e}}_3)\cdot (V{\mathbf{e}}_1)=\frac{1}{V^2}{\mathbf{e}}_1\cdot ({\mathbf{e}}_2\times {\mathbf{e}}_3)=\frac{1}{V}$.

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4.7. Scalar Triple Product in Components

Question

See lecture.

Express the scalar triple product $(\mathbf{A}\times \mathbf{B})\cdot \mathbf{C}$ in terms of the covariant and contravariant components of $\mathbf{A},\mathbf{B},\mathbf{C}$ with respect to dual bases. You may use the equality

$$\mathbf{A}\times \mathbf{B}:C^i=A_j{B}_k-A_k{B}_j\text{(with i,j,k cyclic)},{\mathbf{e}}_j\times {\mathbf{e}}_k=V{\mathbf{e}}^i,$$

where $V={\mathbf{e}}_1\cdot ({\mathbf{e}}_2\times {\mathbf{e}}_3)$.

Let $\mathbf{D}=\mathbf{A}\times \mathbf{B}$ and $E=(\mathbf{A}\times \mathbf{B})\cdot \mathbf{C}=\mathbf{D}\cdot \mathbf{C}$.

From the lecture cross-product derivation, if $\mathbf{A}=A^j{\mathbf{e}}_j$, $\mathbf{B}=B^k{\mathbf{e}}_k$, then $\mathbf{D}=D^i{\mathbf{e}}_i$ with $D^i=(A^j{B}^k-A^k{B}^j)$ (cyclic $i,j,k$) and ${\mathbf{e}}_j\times {\mathbf{e}}_k=V{\mathbf{e}}^i$.

Therefore, one clean component form is

$$E=D^i{C}_i,$$

And you can rewrite $C_i=g_{i\ell }{C}^{\ell }$ (or $D^i=g^{i\ell }{D}_{\ell }$) to produce mixed covariant/contravariant versions as needed.

The provided-solutions style final expression (in terms of a metric contraction) is:

$$E=\frac{g_{i\ell }}{V}(A_j{B}_k-A_k{B}_j)C_{\ell }\text{(with i,j,k cyclic)},$$

With analogous variants obtained by raising/lowering indices.