MTH3008 Weekly Problems 10

Original Documents: mth3008 weekly problem sheet 10.pdf / mth3008 weekly problem sheet 10 handwritten solutions.pdf

Vibes: Three Ricci-tensor calculations followed by a five-part revision sweep through the whole course (basis vectors, orthogonality, covariant components, metric, Christoffel symbols, Riemann-Christoffel tensor) on one bizarre coordinate system parametrising flat space. Every “compute the Ricci tensor” reduces to: list nonzero $\Gamma$, plug into $R_{ij}={\partial }_k{\Gamma }_{ji}^k-{\partial }_j{\Gamma }_{ik}^k+{\Gamma }_{ij}^p{\Gamma }_{pk}^k-{\Gamma }_{ik}^p{\Gamma }_{pj}^k$. The revision exercises 10.4-10.8 all sit on the same coordinate system, so the work compounds.

Used Techniques:

• For an orthogonal diagonal metric, only ${\Gamma }_{iik}={\Gamma }_{iki}=\frac{1}{2}{\partial }_k{g}_{ii}$ and ${\Gamma }_{ijj}=-\frac{1}{2}{\partial }_i{g}_{jj}$ survive (the four cases of Christoffel Symbols).
• Raise with $g^{ii}=1/g_{ii}$ on a diagonal metric.
• Ricci formula: $R_{ij}={\partial }_k{\Gamma }_{ji}^k-{\partial }_j{\Gamma }_{ik}^k+{\Gamma }_{ij}^p{\Gamma }_{pk}^k-{\Gamma }_{ik}^p{\Gamma }_{pj}^k$ (sum over $k$ and $p$).
• Antisymmetry $R_{ijk}^r=-R_{ikj}^r$ kills any component with the last two indices equal - instantly $R_{ijj}^r=0$.
• When $\mathbf{r}$ is a smooth invertible map into ${\mathbb{R}}^3$ with non-singular Jacobian, the induced metric is flat and all $R_{ijk}^r=0$ - verify by computing one component.

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10.1. Ricci Tensor for a 2D Exponential Metric

Question

Given the 2D metric $[g_{ij}]=\left(\begin{array}{cc}1& 0\\ 0& e^{2x}\end{array}\right)$ in coordinates $(x^1,x^2)=(x,y)$, calculate the components of the Ricci tensor $R_{ij}$.

Inverse metric. $g^{11}=1$, $g^{22}=e^{-2x}$.

Nonzero metric derivative. Only $\partial g_{22}/\partial x=2e^{2x}$.

Christoffel symbols (first kind). From the four-case table for orthogonal coordinates, the only surviving symbols are those with two indices $=2$ and one index $=1$:

$${\Gamma }_{122}=-\frac{1}{2}{\partial }_1{g}_{22}=-e^{2x},{\Gamma }_{212}={\Gamma }_{221}=\frac{1}{2}{\partial }_1{g}_{22}=e^{2x}.$$

Christoffel symbols (second kind).

$$\boxed{{\Gamma }_{22}^1=-e^{2x},{\Gamma }_{12}^2={\Gamma }_{21}^2=e^{-2x}\cdot e^{2x}=1.}$$

All others vanish.

Ricci tensor. Apply $R_{ij}={\partial }_k{\Gamma }_{ji}^k-{\partial }_j{\Gamma }_{ik}^k+{\Gamma }_{ij}^p{\Gamma }_{pk}^k-{\Gamma }_{ik}^p{\Gamma }_{pj}^k$.

For $R_{11}$: ${\partial }_k{\Gamma }_{11}^k=0$ (${\Gamma }_{11}^k=0$ for all $k$). ${\Gamma }_{1k}^k={\Gamma }_{12}^2=1$, so ${\partial }_1(1)=0$. Third term: ${\Gamma }_{11}^p=0$, gives $0$. Fourth term: ${\Gamma }_{1k}^p{\Gamma }_{p1}^k$ - only nonzero when $p=k=2$, giving ${\Gamma }_{12}^2{\Gamma }_{21}^2=1$. So $R_{11}=0-0+0-1=-1$.

For $R_{22}$:

• ${\partial }_k{\Gamma }_{22}^k$: only ${\partial }_1{\Gamma }_{22}^1={\partial }_x(-e^{2x})=-2e^{2x}$.
• ${\partial }_2{\Gamma }_{2k}^k={\partial }_y({\Gamma }_{22}^2)={\partial }_y(0)=0$.
• ${\Gamma }_{22}^p{\Gamma }_{pk}^k$: $p=1$: ${\Gamma }_{22}^1\cdot {\Gamma }_{1k}^k=(-e^{2x})(1)=-e^{2x}$.
• ${\Gamma }_{2k}^p{\Gamma }_{p2}^k$: $(p,k)=(1,2)$: ${\Gamma }_{22}^1{\Gamma }_{12}^2=(-e^{2x})(1)=-e^{2x}$. $(p,k)=(2,1)$: ${\Gamma }_{21}^2{\Gamma }_{22}^1=(1)(-e^{2x})=-e^{2x}$. Sum $-2e^{2x}$.

$R_{22}=-2e^{2x}-0+(-e^{2x})-(-2e^{2x})=-e^{2x}$.

For $R_{12}$ (and $R_{21}=R_{12}$ by symmetry): every term evaluates to zero (the ${\Gamma }_{12}^p=0$ kills the third term; the fourth requires $p,k$ both giving nonzero factors, which doesn’t happen for the cross indices).

$$\boxed{R_{11}=-1,R_{22}=-e^{2x},R_{12}=R_{21}=0.}$$

Sanity check

The metric $d{s}^2=d{x}^2+e^{2x}d{y}^2$ is the hyperbolic plane, with Gaussian curvature $K=-1$. In 2D, $R_{ij}=K{g}_{ij}$, giving $R_{11}=-1\cdot 1=-1$ and $R_{22}=-1\cdot e^{2x}=-e^{2x}$. ✓

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10.2. Ricci Tensor in Spherical Coordinates

Question

Consider 3D space in spherical coordinates with arc length $d{s}^2=d{r}^2+r^{2}d{\theta }^2+r^2{\sin }^2\theta d{\varphi }^2$. Compute the Ricci tensor components $R_{kk}$ for $(x^1,x^2,x^3)=(r,\theta ,\varphi )$.

Metric. $g_{11}=1$, $g_{22}=r^2$, $g_{33}=r^2{\sin }^2\theta$, off-diagonal zero. Inverse $g^{11}=1$, $g^{22}=1/r^2$, $g^{33}=1/(r^2{\sin }^2\theta )$.

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

Christoffel symbols (second kind). Using ${\Gamma }_{jk}^i=g^{ii}{\Gamma }_{ijk}$ on a diagonal metric and the four-case table:

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

All others zero.

Ricci tensor. Spherical coordinates are a non-singular smooth chart of Euclidean ${\mathbb{R}}^3$, so the space is flat and all components must vanish. Verifying $R_{11}$:

$$R_{11}={\partial }_k{\Gamma }_{11}^k-{\partial }_1{\Gamma }_{1k}^k+{\Gamma }_{11}^p{\Gamma }_{pk}^k-{\Gamma }_{1k}^p{\Gamma }_{p1}^k.$$

• ${\Gamma }_{11}^k=0$ for all $k$, so first term is $0$.
• ${\Gamma }_{1k}^k={\Gamma }_{12}^2+{\Gamma }_{13}^3=2/r$. So $-{\partial }_r(2/r)=2/r^2$.
• ${\Gamma }_{11}^p=0$, third term $=0$.
• ${\Gamma }_{1k}^p{\Gamma }_{p1}^k$: nonzero only for $(p,k)=(2,2)$: ${\Gamma }_{12}^2{\Gamma }_{21}^2=1/r^2$, and $(p,k)=(3,3)$: ${\Gamma }_{13}^3{\Gamma }_{31}^3=1/r^2$. Sum $2/r^2$.

$R_{11}=0+2/r^2+0-2/r^2=0$. Similar cancellations give $R_{22}=R_{33}=0$.

$$\boxed{R_{kk}=0\text{for }k=1,2,3.}$$

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10.3. Ricci Tensor for a Triple-Exponential Metric

Question

Given $g_{ij}=\mathrm{diag}(e^{2u},e^{2v},e^{2w})$ in coordinates $(x^1,x^2,x^3)=(u,v,w)$, find the Ricci tensor components $R_{ij}$.

Inverse metric. $g^{11}=e^{-2u}$, $g^{22}=e^{-2v}$, $g^{33}=e^{-2w}$.

Key observation. Each diagonal $g_{ii}$ depends only on the corresponding coordinate $x^i$. So ${\partial }_k{g}_{ii}=0$ unless $k=i$. Hence the only nonzero first-kind Christoffel symbols are the diagonal ones:

$${\Gamma }_{111}=\frac{1}{2}{\partial }_u{e}^{2u}=e^{2u},{\Gamma }_{222}=e^{2v},{\Gamma }_{333}=e^{2w}.$$

Second kind.

$$\boxed{{\Gamma }_{11}^1=e^{-2u}\cdot e^{2u}=1,{\Gamma }_{22}^2=1,{\Gamma }_{33}^3=1.}$$

All others zero.

Ricci tensor. This metric is $d{s}^2=e^{2u}d{u}^2+e^{2v}d{v}^2+e^{2w}d{w}^2=d{X}^2+d{Y}^2+d{Z}^2$ under the substitution $X=e^u,Y=e^v,Z=e^w$. So the space is flat. Verifying $R_{11}$:

• ${\partial }_k{\Gamma }_{11}^k={\partial }_1{\Gamma }_{11}^1={\partial }_u(1)=0$.
• ${\Gamma }_{1k}^k={\Gamma }_{11}^1=1$. ${\partial }_1(1)=0$.
• ${\Gamma }_{11}^p{\Gamma }_{pk}^k$: only $p=1$ gives a nonzero factor ${\Gamma }_{11}^1=1$, and ${\Gamma }_{1k}^k=1$. Product $1$.
• ${\Gamma }_{1k}^p{\Gamma }_{p1}^k$: only $(p,k)=(1,1)$: ${\Gamma }_{11}^1{\Gamma }_{11}^1=1$.

$R_{11}=0-0+1-1=0$. Symmetric arguments give $R_{22}=R_{33}=0$, and off-diagonals vanish trivially.

$$\boxed{R_{ij}=0\text{for all }i,j.}$$

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Revision Exercises (10.4-10.8)

The remaining problems all use the same coordinate system. Define

$$f(\theta )=\sin \theta +\cos \theta ,f^{\prime }(\theta )=\cos \theta -\sin \theta .$$

Useful identities: $f^2=1+\sin 2\theta$, $(f^{\prime }{)}^2=1-\sin 2\theta$, $f{f}^{\prime }=\cos 2\theta$, $f^2+(f^{\prime }{)}^2=2$, $f^{\prime \prime }=-f$.

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10.4. Basis Vectors and Orthogonality

Question

Coordinate system $(x^1,x^2,x^3)=(\varphi ,\theta ,z)$ with position vector

$$\mathbf{r}=\cos \varphi (\sin \theta +\cos \theta ){\mathbf{i}}_1+\sin \varphi (\sin \theta +\cos \theta ){\mathbf{i}}_2+z{\mathbf{i}}_3.$$

(1) Find basis vectors ${\mathbf{e}}_1,{\mathbf{e}}_2,{\mathbf{e}}_3$. (2) Show this system is orthogonal.

(1) Basis vectors. Using ${\mathbf{e}}_i=\partial \mathbf{r}/\partial x^i$ with $f=\sin \theta +\cos \theta$:

$$\boxed{{\mathbf{e}}_1=-f\sin \varphi {\mathbf{i}}_1+f\cos \varphi {\mathbf{i}}_2,{\mathbf{e}}_2=f^{\prime }\cos \varphi {\mathbf{i}}_1+f^{\prime }\sin \varphi {\mathbf{i}}_2,{\mathbf{e}}_3={\mathbf{i}}_3.}$$

(2) Orthogonality.

$${\mathbf{e}}_1\cdot {\mathbf{e}}_2=-f{f}^{\prime }\sin \varphi \cos \varphi +f{f}^{\prime }\cos \varphi \sin \varphi =0,$$

${\mathbf{e}}_1\cdot {\mathbf{e}}_3=0$ (no ${\mathbf{i}}_3$ component in ${\mathbf{e}}_1$), ${\mathbf{e}}_2\cdot {\mathbf{e}}_3=0$. Orthogonal. $■$

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10.5. Covariant Components of $\mathbf{A}=-{\mathbf{i}}_1+{\mathbf{i}}_2$

Question

Find the covariant components $A_i=\mathbf{A}\cdot {\mathbf{e}}_i$ of $\mathbf{A}=-{\mathbf{i}}_1+{\mathbf{i}}_2$ with respect to the basis of 10.4.

$$A_1=\mathbf{A}\cdot {\mathbf{e}}_1=(-1)(-f\sin \varphi )+(1)(f\cos \varphi )=f(\sin \varphi +\cos \varphi ),$$ $$A_2=\mathbf{A}\cdot {\mathbf{e}}_2=(-1)(f^{\prime }\cos \varphi )+(1)(f^{\prime }\sin \varphi )=f^{\prime }(\sin \varphi -\cos \varphi ),$$ $$A_3=\mathbf{A}\cdot {\mathbf{e}}_3=0.$$ $$\boxed{A_1=(\sin \theta +\cos \theta )(\sin \varphi +\cos \varphi ),A_2=(\cos \theta -\sin \theta )(\sin \varphi -\cos \varphi ),A_3=0.}$$

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10.6. Metric Coefficients and Arc Length

Question

Compute the metric coefficients (scale factors) and the components of the covariant metric tensor $g_{ii}$ for the coordinate system of 10.4.

Using $g_{ii}={\mathbf{e}}_i\cdot {\mathbf{e}}_i$:

$$g_{11}=f^2{\sin }^2\varphi +f^2{\cos }^2\varphi =f^2=1+\sin 2\theta ,$$ $$g_{22}=(f^{\prime }{)}^2{\cos }^2\varphi +(f^{\prime }{)}^2{\sin }^2\varphi =(f^{\prime }{)}^2=1-\sin 2\theta ,$$ $$g_{33}=1.$$

Off-diagonals zero (confirmed in 10.4). Scale factors $h_i=\sqrt{g_{ii}}$:

$$\boxed{h_1=|f|,h_2=|f^{\prime }|,h_3=1.}$$

Arc length element:

$$\boxed{d{s}^2=f^{2}d{\varphi }^2+(f^{\prime }{)}^{2}d{\theta }^2+d{z}^2=(1+\sin 2\theta )d{\varphi }^2+(1-\sin 2\theta )d{\theta }^2+d{z}^2.}$$

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10.7. Christoffel Symbols

Question

Determine all Christoffel symbols of the first and second kind for the coordinate system of 10.4.

Inverse metric. $g^{11}=1/f^2$, $g^{22}=1/(f^{\prime }{)}^2$, $g^{33}=1$.

Nonzero metric derivatives. Both $g_{11}=f^2$ and $g_{22}=(f^{\prime }{)}^2$ depend only on $\theta =x^2$:

$${\partial }_2{g}_{11}=2f{f}^{\prime },{\partial }_2{g}_{22}=2f^{\prime }{f}^{\prime \prime }=-2f{f}^{\prime }.$$

(Using $f^{\prime \prime }=-f$.) Note $f{f}^{\prime }=\cos 2\theta$.

Christoffel symbols of the first kind. Apply the four-case table:

• ${\partial }_2{g}_{11}\ne 0$ contributes:
• ${\Gamma }_{112}={\Gamma }_{121}=\frac{1}{2}{\partial }_2{g}_{11}=f{f}^{\prime }=\cos 2\theta$,
• ${\Gamma }_{211}=-\frac{1}{2}{\partial }_2{g}_{11}=-f{f}^{\prime }=-\cos 2\theta$.
• ${\partial }_2{g}_{22}\ne 0$ contributes:
• ${\Gamma }_{222}=\frac{1}{2}{\partial }_2{g}_{22}=-f{f}^{\prime }=-\cos 2\theta$.

All others vanish.

Christoffel symbols of the second kind. Using ${\Gamma }_{jk}^i=g^{ii}{\Gamma }_{ijk}$ (no sum, diagonal):

$${\Gamma }_{12}^1={\Gamma }_{21}^1=\frac{f{f}^{\prime }}{f^2}=\frac{f^{\prime }}{f},{\Gamma }_{11}^2=-\frac{f{f}^{\prime }}{(f^{\prime }{)}^2}=-\frac{f}{f^{\prime }},{\Gamma }_{22}^2=-\frac{f{f}^{\prime }}{(f^{\prime }{)}^2}=-\frac{f}{f^{\prime }}.$$

In closed form:

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

All others zero.

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10.8. Riemann-Christoffel Tensor

Question

Determine $R_{122}^1$ and the general $R_{ijk}^r$ for the coordinate system of 10.4.

Antisymmetry shortcut. $R_{ijk}^r=-R_{ikj}^r$ (interchanging the last two indices flips sign), so any component with $j=k$ is automatically zero. In particular:

$$\boxed{R_{122}^1=0\text{(by antisymmetry, since the last two indices are both 2).}}$$

General $R_{ijk}^r$. The position vector $\mathbf{r}(\varphi ,\theta ,z)$ is a smooth map ${\mathbb{R}}^3\to {\mathbb{R}}^3$ with non-singular Jacobian on the open set where $f(\theta )\ne 0$ and $f^{\prime }(\theta )\ne 0$. So this is a coordinate change on flat Euclidean space, and the induced metric is flat: all components $R_{ijk}^r=0$.

Verification via $R_{212}^1$. This is the key 2D-style component to check (since the $z$-direction decouples). With $r=1,i=2,j=1,k=2$:

$$R_{212}^1={\partial }_1{\Gamma }_{22}^1-{\partial }_2{\Gamma }_{21}^1+{\Gamma }_{22}^p{\Gamma }_{p1}^1-{\Gamma }_{21}^p{\Gamma }_{p2}^1.$$

• ${\Gamma }_{22}^1=0$, so ${\partial }_1(0)=0$.

• ${\Gamma }_{21}^1=f^{\prime }/f$. Using $f^{\prime \prime }=-f$:

  $${\partial }_{\theta }\left(\frac{f^{\prime }}{f}\right)=\frac{f^{\prime \prime }f-(f^{\prime }{)}^2}{f^2}=\frac{-f^2-(f^{\prime }{)}^2}{f^2}=-\frac{2}{f^2}.$$

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So $-{\partial }_2{\Gamma }_{21}^1=2/f^2$.

• ${\Gamma }_{22}^p{\Gamma }_{p1}^1$: only $p=2$ gives a nonzero factor: ${\Gamma }_{22}^2{\Gamma }_{21}^1=(-f/f^{\prime })(f^{\prime }/f)=-1$.
• ${\Gamma }_{21}^p{\Gamma }_{p2}^1$: only $p=1$ contributes: ${\Gamma }_{21}^1{\Gamma }_{12}^1=(f^{\prime }/f{)}^2$.

Combining:

$$R^1\ast 212=0+\frac{2}{f^2}+(-1)-\frac{(f^{\prime }{)}^2}{f^2}=\frac{2-(f^{\prime }{)}^2}{f^2}-1.$$

Using $2-(f^{\prime }{)}^2=2-(1-\sin 2\theta )=1+\sin 2\theta =f^2$:

$$R^1\ast 212=\frac{f^2}{f^2}-1=1-1=0.✓$$ $$\boxed{R_{ijk}^r=0\text{for all }r,i,j,k-\text{the space is flat (Euclidean) in disguise.}}$$