MTH3008 Weekly Problems 5

Original Documents: Problem Sheet / Provided Solutions

Vibes: Index-gymnastics + “differentiate the position vector” to get bases/metrics, then use orthogonality to simplify everything.

Used Techniques:

• Tensor transformation law (free vs dummy indices).
• Orthogonal matrix identity $L_{ij}{L}_{kj}={\delta }_{ik}$ (aka “two $L$‘s give a delta”).
• Contraction/trace as “sum over a repeated index” + use $\delta$ to collapse sums.
• For generalised coordinates: ${\mathbf{e}}_i=\partial \mathbf{r}/\partial x^i$, $g_{ij}={\mathbf{e}}_i\cdot {\mathbf{e}}_j$, and (orthogonal case) $d{s}^2=\sum (h_{i}d{x}^i{)}^2$ with $h_i=\sqrt{g_{ii}}$.
• Raise/lower with metric in an orthogonal basis via $A_i=g_{ii}{A}^i$ and $g^{ii}=1/g_{ii}$.

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5.1. Transformation Rules for Rank Four and Six Tensors

Question

Write down the transformation rule for:

1. A tensor of rank four.
2. A tensor of rank six.

Using the lecture rule “one $L$” for each free index, we have:

1. Rank 4:

$$T_{ijk\ell }^{\prime }=L_{im}{L}_{jn}{L}_{kp}{L}_{\ell q}{T}_{mnpq}.$$

1. Rank 6:

$$S_{ijk\ell mn}^{\prime }=L_{ia}{L}_{jb}{L}_{kc}{L}_{\ell d}{L}_{me}{L}_{nf}{S}_{abcdef}.$$

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5.2. Tensor Contraction to Rank Two

Question

Show that if $Q_{ijk\ell }$ is a tensor of rank four, then $Q_{ijj\ell }$ is a tensor of rank two.

Start from the rank-4 rule and then contract over the repeated $j$:

$$Q_{ijj\ell }^{\prime }=L_{im}{L}_{jn}{L}_{jp}{L}_{\ell q}{Q}_{mnpq}.$$

Because $L$ is an orthogonal transformation, $L_{jn}{L}_{jp}={\delta }_{np}$, so:

$$Q_{ijj\ell }^{\prime }=L_{im}{L}_{\ell q}{Q}_{mnnq}.$$

This is exactly the rank-2 transformation law in the free indices $i,\ell$, hence $Q_{ijj\ell }$ is a rank-2 tensor.

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5.3. Trace of a Second-Rank Tensor

Question

Show that if $T_{ij}$ is a tensor, then $T_{ii}$ is a scalar.

Use the rank-2 transformation rule and contract:

$$T_{ii}^{\prime }=L_{im}{L}_{in}{T}_{mn}.$$

Orthogonality gives $L_{im}{L}_{in}={\delta }_{mn}$, so $T_{ii}^{\prime }={\delta }_{mn}{T}_{mn}=T_{mm}$, i.e. the trace is unchanged by the transformation.

Therefore, $T_{ii}$ is a scalar (rank 0 tensor).

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5.4. Basis Vectors and Metric Tensor in Polar Coordinates

Question

Consider the 2D polar coordinates $(x^1,x^2)=(r,\theta )$ with $\mathbf{r}=r\cos \theta {\mathbf{i}}_1+r\sin \theta {\mathbf{i}}_2$.

1. Find ${\mathbf{e}}_1,{\mathbf{e}}_2$.
2. Find $g_{ij}$ and $g^{ij}$.
3. Give $ds$ and the metric coefficients $h_1,h_2$.

1. Basis vectors come from ${\mathbf{e}}_i=\partial \mathbf{r}/\partial x^i$:

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

1. Metric is $g_{ij}={\mathbf{e}}_i\cdot {\mathbf{e}}_j$:

$$[g_{ij}]=\left(\begin{array}{cc}{\mathbf{e}}_1\cdot {\mathbf{e}}_1& {\mathbf{e}}_1\cdot {\mathbf{e}}_2\\ {\mathbf{e}}_2\cdot {\mathbf{e}}_1& {\mathbf{e}}_2\cdot {\mathbf{e}}_2\end{array}\right)=\left(\begin{array}{cc}1& 0\\ 0& r^2\end{array}\right),$$

And since the basis is orthogonal, $g^{11}=1/g_{11}$, $g^{22}=1/g_{22}$:

$$[g^{ij}]=\left(\begin{array}{cc}1& 0\\ 0& \frac{1}{r^2}\end{array}\right).$$

1. Arc length satisfies $d{s}^2=g_{ij}d{x}^{i}d{x}^j$, and for orthogonal systems $d{s}^2=(h_{1}dr{)}^2+(h_{2}d\theta {)}^2$ with $h_i=\sqrt{g_{ii}}$:

$$d{s}^2=d{r}^2+r^{2}d{\theta }^2,h_1=1,h_2=r.$$

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5.5. Basis Vectors and Metric Coefficients in Spherical Coordinates

Question

In spherical coordinates $(x^1,x^2,x^3)=(r,\varphi ,\theta )$ with $\mathbf{r}=r\sin \varphi \cos \theta {\mathbf{i}}_1+r\sin \varphi \sin \theta {\mathbf{i}}_2+r\cos \varphi {\mathbf{i}}_3$, find the basis vectors and the metric coefficients.

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

$${\mathbf{e}}_1=\frac{\partial \mathbf{r}}{\partial r}=\sin \varphi \cos \theta {\mathbf{i}}_1+\sin \varphi \sin \theta {\mathbf{i}}_2+\cos \varphi {\mathbf{i}}_3,$$ $${\mathbf{e}}_2=\frac{\partial \mathbf{r}}{\partial \varphi }=r\cos \varphi \cos \theta {\mathbf{i}}_1+r\cos \varphi \sin \theta {\mathbf{i}}_2-r\sin \varphi {\mathbf{i}}_3,$$ $${\mathbf{e}}_3=\frac{\partial \mathbf{r}}{\partial \theta }=-r\sin \varphi \sin \theta {\mathbf{i}}_1+r\sin \varphi \cos \theta {\mathbf{i}}_2.$$

Then $g_{ij}={\mathbf{e}}_i\cdot {\mathbf{e}}_j$, and this coordinate system is orthogonal so $g_{ij}=0$ for $i\ne j$:

$$g_{11}=1,g_{22}=r^2,g_{33}=r^2{\sin }^2\varphi ,$$

so the metric coefficients $h_i=\sqrt{g_{ii}}$ are:

$$h_1=1,h_2=r,h_3=r\sin \varphi .$$

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5.6. Antisymmetry of a Second-Order Tensor

Question

Given vectors $\mathbf{A},\mathbf{B}$ with components $A_i,B_i$, show that $C_{ik}=A_i{B}_k-A_k{B}_i$ is antisymmetric.

Compute $C_{ki}=A_k{B}_i-A_i{B}_k=-(A_i{B}_k-A_k{B}_i)=-C_{ik}$, so $C_{ik}=-C_{ki}$.

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5.7. Zero Entries of a Mixed-Symmetry Third-Rank Tensor

Question

$T_{ijk}$ is symmetric in its last two indices and antisymmetric in its first two indices. Show $T_{ijk}=0$ for all $i,j,k$.

From symmetry in the last two indices, $T_{ijk}=T_{ikj}$.

Apply antisymmetry in the first two indices to $T_{ikj}$: $T_{ikj}=-T_{kij}$, hence $T_{ijk}=-T_{kij}$.

But the same symmetry/antisymmetry reasoning (swap $j\leftrightarrow k$ where allowed, then $i\leftrightarrow k$) also gives $T_{ijk}=T_{kij}$, so $T_{ijk}=-T_{ijk}$ and therefore $T_{ijk}=0$.

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5.8. Transformation to Orthogonal Coordinates and Metric Tensors

Question

Coordinates $q^1,q^2,q^3$ are related to orthogonal coordinates $x^1,x^2,x^3$ (with orthonormal basis ${\mathbf{i}}_1,{\mathbf{i}}_2,{\mathbf{i}}_3$) by $q^1=x^1+x^2$, $q^2=x^1-x^2$, $q^3=2x^3$.

1. Find ${\mathbf{e}}_1,{\mathbf{e}}_2,{\mathbf{e}}_3$ and show orthogonal.
2. Find $g_{ik}$ and $g^{ik}$.
3. Give $ds$.
4. Find covariant/contravariant components of $\mathbf{A}=2{\mathbf{i}}_1$, $\mathbf{B}={\mathbf{i}}_1+{\mathbf{i}}_2$, $\mathbf{C}=2{\mathbf{i}}_1-3{\mathbf{i}}_3$ w.r.t. ${\mathbf{e}}_1,{\mathbf{e}}_2,{\mathbf{e}}_3$, without computing the dual basis.

Invert the coordinate map:

$$x^1=\frac{q^1+q^2}{2},x^2=\frac{q^1-q^2}{2},x^3=\frac{q^3}{2}.$$

Write the position vector $\mathbf{r}=x^1{\mathbf{i}}_1+x^2{\mathbf{i}}_2+x^3{\mathbf{i}}_3$ in terms of $q$:

$$\mathbf{r}=\frac{q^1+q^2}{2}{\mathbf{i}}_1+\frac{q^1-q^2}{2}{\mathbf{i}}_2+\frac{q^3}{2}{\mathbf{i}}_3.$$

1. Basis vectors ${\mathbf{e}}_i=\partial \mathbf{r}/\partial q^i$:

$${\mathbf{e}}_1=\frac{1}{2}({\mathbf{i}}_1+{\mathbf{i}}_2),{\mathbf{e}}_2=\frac{1}{2}({\mathbf{i}}_1-{\mathbf{i}}_2),{\mathbf{e}}_3=\frac{1}{2}{\mathbf{i}}_3.$$

Orthogonality is by dot products ${\mathbf{e}}_1\cdot {\mathbf{e}}_2={\mathbf{e}}_1\cdot {\mathbf{e}}_3={\mathbf{e}}_2\cdot {\mathbf{e}}_3=0$.

1. Metric $g_{ik}={\mathbf{e}}_i\cdot {\mathbf{e}}_k$ is diagonal (orthogonal basis):

$$[g_{ik}]=\left(\begin{array}{ccc}\frac{1}{2}& 0& 0\\ 0& \frac{1}{2}& 0\\ 0& 0& \frac{1}{4}\end{array}\right),[g^{ik}]=\left(\begin{array}{ccc}2& 0& 0\\ 0& 2& 0\\ 0& 0& 4\end{array}\right)(\text{since }{g}^{ii}=1/g_{ii}).$$

1. Arc length:

$$d{s}^2=g_{ik}d{q}^{i}d{q}^k=\frac{1}{2}(d{q}^1{)}^2+\frac{1}{2}(d{q}^2{)}^2+\frac{1}{4}(d{q}^3{)}^2=(h_{1}d{q}^1{)}^2+(h_{2}d{q}^2{)}^2+(h_{3}d{q}^3{)}^2,$$

with $h_1=h_2=1/\sqrt{2}$, $h_3=1/2$.

1. Contravariant components come from solving $\mathbf{V}=V^i{\mathbf{e}}_i$, then covariant are $V_i=g_{ii}{V}^i$ (no dual basis needed).

• $\mathbf{A}=2{\mathbf{i}}_1$: $A^1=2,A^2=2,A^3=0$ and $A_1=1,A_2=1,A_3=0$.
• $\mathbf{B}={\mathbf{i}}_1+{\mathbf{i}}_2$: $B^1=2,B^2=0,B^3=0$ and $B_1=1,B_2=0,B_3=0$.
• $\mathbf{C}=2{\mathbf{i}}_1-3{\mathbf{i}}_3$: $C^1=2,C^2=2,C^3=-6$ and $C_1=1,C_2=1,C_3=-\frac{3}{2}$.

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5.9. Kronecker Delta from Transformation Coefficients

Question

With $L_{i^{\prime }}^j=\partial x^j/\partial x^{\prime i}$ and $L_j^{i^{\prime }}=\partial x^{\prime i}/\partial x^j$, show that ${\delta }_i^j=L_i^{k^{\prime }}{L}_{k^{\prime }}^j$.

Use the multi variable chain rule on the identity map $x^j=x^j(x^{\prime }(x))$:

$$\frac{\partial x^j}{\partial x^i}=\frac{\partial x^j}{\partial x^{\prime k}}\frac{\partial x^{\prime k}}{\partial x^i}=L_{k^{\prime }}^j{L}_i^{k^{\prime }}.$$

But $\partial x^j/\partial x^i={\delta }_i^j$, so ${\delta }_i^j=L_i^{k^{\prime }}{L}_{k^{\prime }}^j$ (just reorder the scalar factors).