MTH3008 Weekly Problems 8

Original Documents: mth3008 weekly problem sheet 8.pdf / mth3008 weekly problem sheet 8 handwritten solutions.pdf / mth3008 weekly problem sheet 8 solutions.pdf

Vibes: Christoffel-symbol computation at scale. Once the four orthogonal-coordinate cases from 8.1 are internalised, the later problems reduce to “find the nonzero metric derivatives, drop them into the right slot, raise with $g^{ii}$.” The final problem 8.5 is a standalone proof showing the covariant derivative really is a tensor.

Used Techniques:

• First-kind Christoffel formula: ${\Gamma }_{ijk}=\frac{1}{2}({\partial }_k{g}_{ij}+{\partial }_j{g}_{ik}-{\partial }_i{g}_{jk})$.
• Ricci’s Theorem: orthogonal ⇒ only $(i=j=k),(i=j\ne k),(i\ne j=k)$ and distinct-index cases matter.
• Raising the first index with $g^{ii}=1/g_{ii}$ (diagonal metric).
• Local Basis from ${\mathbf{e}}_i=\partial \mathbf{r}/\partial x^i$, then $g_{ij}={\mathbf{e}}_i\cdot {\mathbf{e}}_j$.
• Christoffel Symbols of ${\Gamma }_{jk}^i$ + identity $L_r^{k^{\prime }}{L}_{k^{\prime }}^s={\delta }_r^s$ for the tensor-character proof.

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8.1. Christoffel Symbols in Orthogonal Coordinates

Question

Simplify ${\Gamma }_{ijk}=\frac{1}{2}({\partial }_j{g}_{ik}+{\partial }_k{g}_{ij}-{\partial }_i{g}_{kj})$ when $g_{ij}=0$ for $i\ne j$, in the cases: (1) $i=j=k$; (2) $i=j\ne k$; (3) $i\ne j=k$; (4) all distinct.

For an orthogonal metric, every $g_{ij}$ with $i\ne j$ vanishes, and so do all of its derivatives. Evaluating the three terms in each case:

(1) $i=j=k$. All three terms are $\partial g_{ii}/\partial x^i$:

$$\boxed{{\Gamma }_{iii}=\frac{1}{2}\frac{\partial g_{ii}}{\partial x^i}.}$$

(2) $i=j\ne k$. $g_{ik}=g_{ki}=0$, so only the middle term survives:

$$\boxed{{\Gamma }_{iik}=\frac{1}{2}\frac{\partial g_{ii}}{\partial x^k}.}$$

(3) $i\ne j=k$. $g_{ij}=0$, so only the third term (with a minus) survives:

$$\boxed{{\Gamma }_{ijj}=-\frac{1}{2}\frac{\partial g_{jj}}{\partial x^i}.}$$

(4) All distinct. Every $g$ in the three terms is off-diagonal, so:

$$\boxed{{\Gamma }_{ijk}=0.}$$

These four shortcuts are used in every subsequent orthogonal-coordinate problem.

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8.2. Christoffel Symbols in Spherical Coordinates

Question

In $(r,\phi ,\theta )$ with $g_{11}=1$, $g_{22}=r^2$, $g_{33}=r^2{\sin }^2\phi$ (off-diagonal entries zero), find all ${\Gamma }_{ijk}$ and ${\Gamma }_{jk}^i$.

Nonzero metric derivatives. $\partial g_{22}/\partial r=2r$, $\partial g_{33}/\partial r=2r{\sin }^2\phi$, $\partial g_{33}/\partial \phi =2r^2\sin \phi \cos \phi$.

First-kind symbols via 8.1 cases:

• ${\Gamma }_{122}=-\frac{1}{2}\partial g_{22}/\partial r=-r$.
• ${\Gamma }_{133}=-\frac{1}{2}\partial g_{33}/\partial r=-r{\sin }^2\phi$.
• ${\Gamma }_{233}=-\frac{1}{2}\partial g_{33}/\partial \phi =-r^2\sin \phi \cos \phi$.
• ${\Gamma }_{221}={\Gamma }_{212}=\frac{1}{2}\partial g_{22}/\partial r=r$.
• ${\Gamma }_{331}={\Gamma }_{313}=\frac{1}{2}\partial g_{33}/\partial r=r{\sin }^2\phi$.
• ${\Gamma }_{332}={\Gamma }_{323}=\frac{1}{2}\partial g_{33}/\partial \phi =r^2\sin \phi \cos \phi$.

${\Gamma }_{iii}$ - all zero (diagonal metric derivatives w.r.t. own coord vanish). Fully-distinct case - zero.

Second-kind symbols via ${\Gamma }_{jk}^i=g^{i\ell }{\Gamma }_{\ell kj}$ with $g^{11}=1$, $g^{22}=1/r^2$, $g^{33}=1/(r^2{\sin }^2\phi )$:

$$\boxed{{\Gamma }_{22}^1=-r,{\Gamma }_{33}^1=-r{\sin }^2\phi ,{\Gamma }_{12}^2={\Gamma }_{21}^2=\frac{1}{r},{\Gamma }_{33}^2=-\sin \phi \cos \phi ,{\Gamma }_{13}^3={\Gamma }_{31}^3=\frac{1}{r},{\Gamma }_{23}^3={\Gamma }_{32}^3=\cot \phi .}$$

All others vanish.

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8.3. Exotic Polar-Type Coordinates in 2D

Question

For $\mathbf{r}={\rho }^2\cos (2\theta ){\mathbf{i}}_1+{\rho }^2\sin (2\theta ){\mathbf{i}}_2$ with $(x^1,x^2)=(\rho ,\theta )$, find the basis vectors, metric (covariant and contravariant), and all Christoffel symbols of both kinds.

1) Basis vectors.

$${\mathbf{e}}_1=\frac{\partial \mathbf{r}}{\partial \rho }=2\rho \cos (2\theta ){\mathbf{i}}_1+2\rho \sin (2\theta ){\mathbf{i}}_2,{\mathbf{e}}_2=\frac{\partial \mathbf{r}}{\partial \theta }=-2{\rho }^2\sin (2\theta ){\mathbf{i}}_1+2{\rho }^2\cos (2\theta ){\mathbf{i}}_2.$$

2) Metric. Dot products collapse via ${\cos }^2+{\sin }^2=1$:

$$g_{11}=4{\rho }^2,g_{22}=4{\rho }^4,g_{12}={\mathbf{e}}_1\cdot {\mathbf{e}}_2=0.$$

Orthogonal, so $\boxed{[g_{ij}]=\left(\begin{array}{cc}4{\rho }^2& 0\\ 0& 4{\rho }^4\end{array}\right)}$ and $\boxed{[g^{ij}]=\left(\begin{array}{cc}1/(4{\rho }^2)& 0\\ 0& 1/(4{\rho }^4)\end{array}\right)}$.

3) First-kind Christoffel symbols. Nonzero metric derivatives: $\partial g_{11}/\partial \rho =8\rho$, $\partial g_{22}/\partial \rho =16{\rho }^3$. Apply 8.1:

$${\Gamma }_{111}=\frac{1}{2}\partial g_{11}/\partial \rho =4\rho ,$$ $${\Gamma }_{221}={\Gamma }_{212}=\frac{1}{2}\partial g_{22}/\partial \rho =8{\rho }^3,$$ $${\Gamma }_{122}=-\frac{1}{2}\partial g_{22}/\partial \rho =-8{\rho }^3.$$

All others zero.

4) Second-kind Christoffel symbols. Raise with $g^{ii}=1/g_{ii}$:

$$\boxed{{\Gamma }_{11}^1=\frac{1}{\rho },{\Gamma }_{22}^1=-2\rho ,{\Gamma }_{12}^2={\Gamma }_{21}^2=\frac{2}{\rho }.}$$

All others zero.

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8.4. Time-Dependent 3D Coordinate System

Question

For $\mathbf{r}=(x^2-y^2){\mathbf{i}}_1+2xy{\mathbf{i}}_2+t{\mathbf{i}}_3$ with $(x^1,x^2,x^3)=(x,y,t)$, find the basis vectors, metric, and all Christoffel symbols.

1) Basis.

$${\mathbf{e}}_1=2x{\mathbf{i}}_1+2y{\mathbf{i}}_2,{\mathbf{e}}_2=-2y{\mathbf{i}}_1+2x{\mathbf{i}}_2,{\mathbf{e}}_3={\mathbf{i}}_3.$$

2) Metric. ${\mathbf{e}}_1\cdot {\mathbf{e}}_2=-4xy+4xy=0$, so orthogonal. Magnitudes give:

$$\boxed{[g_{ij}]=\left(\begin{array}{ccc}4(x^2+y^2)& 0& 0\\ 0& 4(x^2+y^2)& 0\\ 0& 0& 1\end{array}\right),[g^{ij}]=\left(\begin{array}{ccc}1/(4(x^2+y^2))& 0& 0\\ 0& 1/(4(x^2+y^2))& 0\\ 0& 0& 1\end{array}\right).}$$

3) First-kind Christoffel symbols. Derivatives: $\partial g_{11}/\partial x=\partial g_{22}/\partial x=8x$, $\partial g_{11}/\partial y=\partial g_{22}/\partial y=8y$; all $t$-derivatives and all derivatives of $g_{33}$ vanish.

Applying 8.1 case by case:

$${\Gamma }_{111}=4x,{\Gamma }_{222}=4y,{\Gamma }_{112}={\Gamma }_{121}=4y,{\Gamma }_{221}={\Gamma }_{212}=4x,{\Gamma }_{122}=-4x,{\Gamma }_{211}=-4y.$$

All others zero (any Christoffel with a $3$ anywhere involves only $g_{33}=1$ or derivatives ${\partial }_t$, both trivial).

4) Second-kind Christoffel symbols. Raise with $g^{11}=g^{22}=1/(4(x^2+y^2))$:

$$\boxed{{\Gamma }_{11}^1=\frac{x}{x^2+y^2},{\Gamma }_{12}^1={\Gamma }_{21}^1=\frac{y}{x^2+y^2},{\Gamma }_{22}^1=-\frac{x}{x^2+y^2},}$$ $$\boxed{{\Gamma }_{11}^2=-\frac{y}{x^2+y^2},{\Gamma }_{12}^2={\Gamma }_{21}^2=\frac{x}{x^2+y^2},{\Gamma }_{22}^2=\frac{y}{x^2+y^2}.}$$

All others zero.

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8.5. Tensorial Nature of the Covariant Derivative

Question

Given ${\Gamma }^{i^{\prime }}{}_{j^{\prime }{k}^{\prime }}=L^{i^{\prime }}{}_{\ell }{L}^m{}_{j^{\prime }}{L}^n{}_{k^{\prime }}{\Gamma }^{\ell }{}_{mn}+L^{i^{\prime }}{}_m{L}^n{}_{k^{\prime }}\frac{\partial L^m{}_{j^{\prime }}}{\partial x^n}$, show that the covariant derivative of a covariant vector is a rank-2 covariant tensor.

We want: $A_{i^{\prime },j^{\prime }}=L^m{}_{i^{\prime }}{L}^n{}_{j^{\prime }}{A}_{m,n}$.

Start from $A_{i^{\prime },j^{\prime }}=\frac{\partial A_{i^{\prime }}}{\partial x^{j^{\prime }}}-{\Gamma }_{i^{\prime }{j}^{\prime }}^{k^{\prime }}{A}_{k^{\prime }}$ with $A_{i^{\prime }}=L^k{}_{i^{\prime }}{A}_k$.

Step 1 - derivative of the transformed component, using the product rule and the chain rule on $\partial /\partial x^{j^{\prime }}=L^n{}_{j^{\prime }}\partial /\partial x^n$:

$$\frac{\partial A_{i^{\prime }}}{\partial x^{j^{\prime }}}=L^k{}_{i^{\prime }}{L}^n{}_{j^{\prime }}\frac{\partial A_k}{\partial x^n}+A_k\cdot L^n{}_{j^{\prime }}\frac{\partial L^k{}_{i^{\prime }}}{\partial x^n}.$$

Step 2 - Christoffel contribution. Using $A_{k^{\prime }}=L^s{}_{k^{\prime }}{A}_s$ and the given transformation of ${\Gamma }_{i^{\prime }{j}^{\prime }}^{k^{\prime }}$, multiply and collapse $L^{k^{\prime }}{}_r{L}^s{}_{k^{\prime }}={\delta }_r^s$ (the identity from 6.1):

$${\Gamma }_{i^{\prime }{j}^{\prime }}^{k^{\prime }}{A}_{k^{\prime }}=L^m{}_{i^{\prime }}{L}^n{}_{j^{\prime }}{\Gamma }^r{}_{mn}{A}_r+L^n{}_{j^{\prime }}\frac{\partial L^r{}_{i^{\prime }}}{\partial x^n}{A}_r.$$

Step 3 - subtract and cancel. Combining Steps 1 and 2:

$$A_{i^{\prime },j^{\prime }}=L^k{}_{i^{\prime }}{L}^n{}_{j^{\prime }}\frac{\partial A_k}{\partial x^n}+A_k{L}^n{}_{j^{\prime }}\frac{\partial L^k{}_{i^{\prime }}}{\partial x^n}-L^m{}_{i^{\prime }}{L}^n{}_{j^{\prime }}{\Gamma }^r{}_{mn}{A}_r-L^n{}_{j^{\prime }}\frac{\partial L^r{}_{i^{\prime }}}{\partial x^n}{A}_r.$$

The second and fourth terms cancel exactly (they differ only in dummy-index name $k\leftrightarrow r$). What remains is

$$A_{i^{\prime },j^{\prime }}=L^m{}_{i^{\prime }}{L}^n{}_{j^{\prime }}\left(\frac{\partial A_m}{\partial x^n}-{\Gamma }^r{}_{mn}{A}_r\right)=L^m{}_{i^{\prime }}{L}^n{}_{j^{\prime }}{A}_{m,n}.$$

$\boxed{A_{i^{\prime },j^{\prime }}=L^m{}_{i^{\prime }}{L}^n{}_{j^{\prime }}{A}_{m,n}}$ is the rank-2 covariant transformation law. ✓

The cancellation is exactly why Christoffel symbols exist

The non-tensorial extra term in $\Gamma$‘s transformation is precisely what cancels the non-tensorial extra term in the derivative $\partial A_{i^{\prime }}/\partial x^{j^{\prime }}$. Together they reconstruct a tensor - which is the whole point of Covariant Differentiation.