MTH3008 Weekly Problems 6

Original Documents: mth3008 weekly problem sheet 6.pdf / mth3008 weekly problem sheet 6 handwritten solutions.pdf / mth3008 weekly problem sheet 6 solutions.pdf

Vibes: Heavy mix of index-chasing proofs and grindy coordinate-change computations - for proofs, lean on the chain rule, $L_j^{i^{\prime }}{L}_{k^{\prime }}^j={\delta }_{k^{\prime }}^{i^{\prime }}$, and the Quotient Rule; for computation, grind through $P^{\prime }=LP{L}^T$ with dual bases.

Used Techniques:

• Multi-variable chain rule on $\partial x^k/\partial x^{\prime i}$ applied to identity maps.
• Tensor transformation laws for covariant, contravariant, and Mixed Components rank-2 tensors.
• Associated Tensors using the Metric Tensor $g_{ik}$ and $g^{ik}$.
• Dual-basis construction: ${\mathbf{e}}^i=({\mathbf{e}}_j\times {\mathbf{e}}_k)/V$, then ${\mathbf{e}}^i\cdot {\mathbf{e}}_k={\delta }_k^i$.
• Matrix form of covariant transformation: $P^{\prime }=LP{L}^T$, with $L$ having rows ${\mathbf{e}}_i$ in Cartesian components.

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6.1. Composition Of Coordinate Transformation Matrices

Question

Using the alternative definitions of the transformation coefficients $L^k{}_{i^{\prime }}$ and $L^{i^{\prime }}{}_k$, namely

$$L^k{}_{i^{\prime }}=\frac{\partial x^k}{\partial x^{\prime i}},L^{i^{\prime }}{}_k=\frac{\partial x^{\prime i}}{\partial x^k},$$

show that

$$L^i{}_{k^{\prime }}{L}^{k^{\prime }}{}_j={\delta }^i{}_j,$$

where ${\delta }^i{}_j$ is the usual Kronecker delta.

Apply the chain rule to the identity map $x^i=x^i(x^{\prime }(x))$:

$$\frac{\partial x^i}{\partial x^j}=\sum _k\frac{\partial x^i}{\partial x^{\prime k}}\frac{\partial x^{\prime k}}{\partial x^j}=L^i{}_{k^{\prime }}{L}^{k^{\prime }}{}_j.$$

But $\partial x^i/\partial x^j={\delta }_j^i$ (coordinates are independent), so

$$\boxed{L^i{}_{k^{\prime }}{L}^{k^{\prime }}{}_j={\delta }^i{}_j.}$$

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6.2. Contraction Of Covariant And Contravariant Tensors

Question

Let $A_{ik\ell }$ be a covariant tensor of order $3$ and $B^{pqmn}$ a contravariant tensor of order $4$. Prove that $A_{ik\ell }{B}^{k\ell mn}$ is a mixed tensor of order $3$ with one covariant and two contravariant indices.

Under a coordinate change, $A_{ik\ell }$ picks up three covariant $L$ factors and $B^{pqmn}$ picks up four contravariant ones:

$$A_{ik\ell }^{\prime }=L^a{}_{i^{\prime }}{L}^b{}_{k^{\prime }}{L}^c{}_{{\ell }^{\prime }}{A}_{abc},B^{\prime pqmn}=L^{p^{\prime }}{}_r{L}^{q^{\prime }}{}_s{L}^{m^{\prime }}{}_t{L}^{n^{\prime }}{}_u{B}^{rstu}.$$

Form the claimed contraction:

$$A_{ik\ell }^{\prime }{B}^{\prime k\ell mn}=L^a{}_{i^{\prime }}{L}^b{}_{k^{\prime }}{L}^c{}_{{\ell }^{\prime }}{A}_{abc}\cdot L^{k^{\prime }}{}_r{L}^{{\ell }^{\prime }}{}_s{L}^{m^{\prime }}{}_t{L}^{n^{\prime }}{}_u{B}^{rstu}.$$

Collapse the pairs using 6.1: $L^b{}_{k^{\prime }}{L}^{k^{\prime }}{}_r={\delta }_r^b$ and $L^c{}_{{\ell }^{\prime }}{L}^{{\ell }^{\prime }}{}_s={\delta }_s^c$:

$$=L^a{}_{i^{\prime }}{L}^{m^{\prime }}{}_t{L}^{n^{\prime }}{}_u{A}_{abc}{\delta }_r^b{\delta }_s^c{B}^{rstu}=L^a{}_{i^{\prime }}{L}^{m^{\prime }}{}_t{L}^{n^{\prime }}{}_u\underset{\text{contracted object}}{\underbrace{A_{abc}{B}^{bcmn}}}$$

That is exactly the transformation law of a $(1,2)$-type tensor: one covariant factor $L^a{}_{i^{\prime }}$ and two contravariant factors $L^{m^{\prime }}{}_t,L^{n^{\prime }}{}_u$. Hence $\boxed{A_{ik\ell }{B}^{k\ell mn}}$ is a mixed rank-3 tensor with one lower and two upper indices. ✓

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6.3. Valid Relations Between Associated Tensors

Question

For each of the following proposed relations, decide whether it is correct.

1. $T^p{}_q=g^r{}_p{T}_{rq}$.
2. $S^{pq}=g^{rp}{g}^{sq}{S}_{rs}$.
3. $W^p{}_{\cdot rs}=g^s{}_q{W}^{pq}{}_{\cdot \cdot s}$.
4. $V^{qm}{}_{\cdot tk}{}^{\cdot \cdot n}=g^p{}_k{g}^s{}_n{g}^r{}_m{V}^{q\cdot st}{}_{\cdot r}{}^{\cdot \cdot p}$.

Key fact: the mixed metric $g^r{}_p={\mathbf{e}}^r\cdot {\mathbf{e}}_p={\delta }_p^r$, so contracting with $g^r{}_p$ renames an index without changing its level.

(1) Invalid. $g^r{}_p{T}_{rq}={\delta }_p^r{T}_{rq}=T_{pq}$, which has both indices lowered. But $T^p{}_q$ has $p$ raised. Index positions mismatch between the two sides, so the equation cannot hold as an identity between tensors.

(2) Valid. This is the standard “raise both indices” recipe: $S^{pq}=g^{rp}{g}^{sq}{S}_{rs}$ with $g^{rp}$ and $g^{sq}$ the contravariant metric. Each metric factor raises one covariant dummy ($r$ or $s$) into the corresponding free upper index ($p$ or $q$). See Index Raising and Lowering.

(3) Invalid. Two problems. First, $g^s{}_q={\delta }_q^s$ just renames $q\to s$. Second, after contraction, the RHS carries $s$ both as an upper dummy (from the renamed $q$) and as a lower dummy (inherited), so $s$ is doubly repeated - a violation of summation-convention rules. The LHS has $s$ as a free lower index, so the structures disagree.

(4) Invalid. Every factor on the right is a mixed metric $g^{\cdot }{}_{\cdot }={\delta }_{\cdot }^{\cdot }$, so the RHS reduces by relabelling to $V^{q\cdot nt}{}_{\cdot m\cdot \cdot k}$ - same entries, but with different index positions from the LHS $V^{qm}{}_{\cdot tk}{}^{\cdot \cdot n}$. The positions are what distinguish associated tensors, so the equality fails.

Why the mixed metric isn't useful for raising/lowering

Only $g_{ik}$ (fully covariant) lowers, and only $g^{ik}$ (fully contravariant) raises. The mixed $g^i{}_k={\delta }_k^i$ is inert - it never changes a component, only relabels.

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6.4. Changing Tensor Components Under A Non-Orthonormal Basis

Question

In Cartesian $K$, $[P_{ik}]=[P^{ik}]=[P_i^{\cdot k}]=[P_{\cdot k}^i]=\left(\begin{array}{ccc}1& 1& -1\\ 2& 3& 0\\ 0& -2& 1\end{array}\right)$. New basis: ${\mathbf{e}}_1={\mathbf{i}}_1$, ${\mathbf{e}}_2={\mathbf{i}}_1-{\mathbf{i}}_2$, ${\mathbf{e}}_3={\mathbf{i}}_1+{\mathbf{i}}_2+2{\mathbf{i}}_3$. Find the dual basis and compute $P_{i^{\prime }{k}^{\prime }}$, $P^{i^{\prime }{k}^{\prime }}$, $P_{i^{\prime }}^{\cdot k^{\prime }}$.

1) Dual basis via ${\mathbf{e}}^i=({\mathbf{e}}_j\times {\mathbf{e}}_k)/V$ with $V={\mathbf{e}}_1\cdot ({\mathbf{e}}_2\times {\mathbf{e}}_3)$.

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

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

Check: ${\mathbf{e}}^i\cdot {\mathbf{e}}_k={\delta }_k^i$ for all pairs. ✓

2) Transformation matrices. Writing each basis vector as a row in Cartesian components:

$$L=[L^{\ell }{}_{i^{\prime }}]=\left(\begin{array}{ccc}1& 0& 0\\ 1& -1& 0\\ 1& 1& 2\end{array}\right),M^{\ast }=[L^{i^{\prime }}{}_{\ell }]=\left(\begin{array}{ccc}1& 1& -1\\ 0& -1& \frac{1}{2}\\ 0& 0& \frac{1}{2}\end{array}\right).$$

Covariant: $P_{i^{\prime }{k}^{\prime }}^{\prime }=L^{\ell }{}_{i^{\prime }}{L}^m{}_{k^{\prime }}{P}_{\ell m}=(LP{L}^T{)}_{i^{\prime }{k}^{\prime }}$:

$$LP=\left(\begin{array}{ccc}1& 1& -1\\ -1& -2& -1\\ 3& 0& 1\end{array}\right),[P_{i^{\prime }{k}^{\prime }}^{\prime }]=LP{L}^T=\boxed{\left(\begin{array}{ccc}1& 0& 0\\ -1& 1& -5\\ 3& 3& 5\end{array}\right)}.$$

Contravariant: $P^{\prime i^{\prime }{k}^{\prime }}=L^{i^{\prime }}{}_{\ell }{L}^{k^{\prime }}{}_m{P}^{\ell m}=(M^{\ast }P(M^{\ast }{)}^T{)}_{i^{\prime }{k}^{\prime }}$:

$$M^{\ast }P=\left(\begin{array}{ccc}3& 6& -2\\ -2& -4& \frac{1}{2}\\ 0& -1& \frac{1}{2}\end{array}\right),[P^{\prime i^{\prime }{k}^{\prime }}]=\boxed{\left(\begin{array}{ccc}11& -7& -1\\ -\frac{13}{2}& \frac{17}{4}& \frac{1}{4}\\ -\frac{3}{2}& \frac{5}{4}& \frac{1}{4}\end{array}\right)}.$$

Mixed: $P_{i^{\prime }}^{\prime \cdot k^{\prime }}=L^{\ell }{}_{i^{\prime }}{L}^{k^{\prime }}{}_m{P}_{\ell }^{\cdot m}=(LP(M^{\ast }{)}^T{)}_{i^{\prime }{k}^{\prime }}$:

$$[P_{i^{\prime }}^{\prime \cdot k^{\prime }}]=\boxed{\left(\begin{array}{ccc}3& -\frac{3}{2}& -\frac{1}{2}\\ -2& \frac{3}{2}& -\frac{1}{2}\\ 2& \frac{1}{2}& \frac{1}{2}\end{array}\right)}.$$

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6.5. Tensor Component Transformation With A Different Basis Change

Question

In Cartesian $K$, $[P_{ik}]=[P^{ik}]=[P_i^{\cdot k}]=[P_{\cdot k}^i]=\left(\begin{array}{ccc}2& 0& 0\\ 2& 0& -1\\ 1& 0& 0\end{array}\right)$. New basis: ${\mathbf{e}}_1={\mathbf{i}}_1-{\mathbf{i}}_2$, ${\mathbf{e}}_2={\mathbf{i}}_2$, ${\mathbf{e}}_3={\mathbf{i}}_1+2{\mathbf{i}}_3$. Find the dual basis and compute $P_{i^{\prime }{k}^{\prime }}$, $P^{i^{\prime }{k}^{\prime }}$, $P_{i^{\prime }}^{\cdot k^{\prime }}$.

1) Dual basis. ${\mathbf{e}}_2\times {\mathbf{e}}_3=(2,0,-1)$, ${\mathbf{e}}_3\times {\mathbf{e}}_1=(2,2,-1)$, ${\mathbf{e}}_1\times {\mathbf{e}}_2=(0,0,1)$, $V=2$:

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

Verify ${\mathbf{e}}^i\cdot {\mathbf{e}}_k={\delta }_k^i$. ✓

2) Transformation matrices.

$$L=\left(\begin{array}{ccc}1& -1& 0\\ 0& 1& 0\\ 1& 0& 2\end{array}\right),M^{\ast }=\left(\begin{array}{ccc}1& 0& -\frac{1}{2}\\ 1& 1& -\frac{1}{2}\\ 0& 0& \frac{1}{2}\end{array}\right).$$

Covariant $LP{L}^T$:

$$LP=\left(\begin{array}{ccc}0& 0& 1\\ 2& 0& -1\\ 4& 0& 0\end{array}\right),[P_{i^{\prime }{k}^{\prime }}^{\prime }]=\boxed{\left(\begin{array}{ccc}0& 0& 2\\ 2& 0& 0\\ 4& 0& 4\end{array}\right)}.$$

Contravariant $M^{\ast }P(M^{\ast }{)}^T$:

$$M^{\ast }P=\left(\begin{array}{ccc}\frac{3}{2}& 0& 0\\ \frac{7}{2}& 0& -1\\ \frac{1}{2}& 0& 0\end{array}\right),[P^{\prime i^{\prime }{k}^{\prime }}]=\boxed{\left(\begin{array}{ccc}\frac{3}{2}& \frac{3}{2}& 0\\ 4& 4& -\frac{1}{2}\\ \frac{1}{2}& \frac{1}{2}& 0\end{array}\right)}.$$

Mixed $LP(M^{\ast }{)}^T$:

$$[P_{i^{\prime }}^{\prime \cdot k^{\prime }}]=\boxed{\left(\begin{array}{ccc}-\frac{1}{2}& -\frac{1}{2}& \frac{1}{2}\\ \frac{5}{2}& \frac{5}{2}& -\frac{1}{2}\\ 4& 4& 0\end{array}\right)}.$$

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6.6. Transforming Components Using Metric And Dual Basis

Question

In Cartesian $K$, $[B_{ik}]=[B^{ik}]=[B_i^{\cdot k}]=[B_{\cdot k}^i]=\left(\begin{array}{ccc}0& 1& 0\\ -1& 0& 1\\ 0& 0& 1\end{array}\right)$. New basis: ${\mathbf{e}}_1={\mathbf{i}}_1-{\mathbf{i}}_2$, ${\mathbf{e}}_2={\mathbf{i}}_1+4{\mathbf{i}}_2+{\mathbf{i}}_3$, ${\mathbf{e}}_3={\mathbf{i}}_1+3{\mathbf{i}}_2+{\mathbf{i}}_3$.

(1) Compute $B_{i^{\prime }{k}^{\prime }}$. (2) Compute the dual basis. (3) Compute $B^{i^{\prime }{k}^{\prime }}$ using the Metric Tensor.

1) Covariant components. $L=\left(\begin{array}{ccc}1& -1& 0\\ 1& 4& 1\\ 1& 3& 1\end{array}\right)$:

$$LB=\left(\begin{array}{ccc}1& 1& -1\\ -4& 1& 5\\ -3& 1& 4\end{array}\right),[B_{i^{\prime }{k}^{\prime }}^{\prime }]=LB{L}^T=\boxed{\left(\begin{array}{ccc}0& 4& 3\\ -5& 5& 4\\ -4& 5& 4\end{array}\right)}.$$

2) Dual basis. ${\mathbf{e}}_2\times {\mathbf{e}}_3=(1,0,-1)$, ${\mathbf{e}}_3\times {\mathbf{e}}_1=(1,1,-4)$, ${\mathbf{e}}_1\times {\mathbf{e}}_2=(-1,-1,5)$, $V=1$:

$$\boxed{{\mathbf{e}}^1=(1,0,-1),{\mathbf{e}}^2=(1,1,-4),{\mathbf{e}}^3=(-1,-1,5).}$$

Check ${\mathbf{e}}^i\cdot {\mathbf{e}}_k={\delta }_k^i$ entry-by-entry. ✓

3) Contravariant components. Two equivalent routes:

Route A (via the metric). $[g_{i^{\prime }{k}^{\prime }}]=[{\mathbf{e}}_i\cdot {\mathbf{e}}_k]=\left(\begin{array}{ccc}2& -3& -2\\ -3& 18& 14\\ -2& 14& 11\end{array}\right)$; invert to get $[g^{i^{\prime }{k}^{\prime }}]$; then $B^{\prime i^{\prime }{k}^{\prime }}=g^{i^{\prime }{\ell }^{\prime }}{g}^{k^{\prime }{m}^{\prime }}{B}_{{\ell }^{\prime }{m}^{\prime }}^{\prime }$.

Route B (direct, using the dual basis). Use $M^{\ast }=\left(\begin{array}{ccc}1& 0& -1\\ 1& 1& -4\\ -1& -1& 5\end{array}\right)$ and $B^{\prime i^{\prime }{k}^{\prime }}=M^{\ast }B(M^{\ast }{)}^T$:

$$M^{\ast }B=\left(\begin{array}{ccc}0& 1& -1\\ -1& 1& -3\\ 1& -1& 4\end{array}\right),[B^{\prime i^{\prime }{k}^{\prime }}]=\boxed{\left(\begin{array}{ccc}1& 5& -6\\ 2& 12& -15\\ -3& -16& 20\end{array}\right)}.$$

Both routes yield the same answer - Route B avoids inverting a $3\times 3$ matrix.