MTH3008 Lecture 19

Quote

…

Last time we introduced the Riemann-Christoffel Tensor and showed that a space is Euclidean if and only if $R_{ijk}^r=0$. Today we continue Chapter 7 by introducing the Ricci Tensor - a contraction of the Riemann-Christoffel tensor that gives a simpler curvature summary - and then run a full revision-style example: computing the Ricci tensor in cylindrical coordinates from scratch.

Reminder: Christoffel Symbols and the Riemann-Christoffel Tensor

We are working in a generalised coordinate system with Local Basis ${\mathbf{e}}_1,{\mathbf{e}}_2,{\mathbf{e}}_3$ varying from point to point, ${\mathbf{e}}_j={\mathbf{e}}_j(x^1,x^2,x^3)$, with dual basis ${\mathbf{e}}^j$ and metric coefficients $g_{ij}={\mathbf{e}}_i\cdot {\mathbf{e}}_j$, $g^{ij}={\mathbf{e}}^i\cdot {\mathbf{e}}^j$.

The Christoffel Symbols of the first and second kind are computed from the metric as

$${\Gamma }_{ijk}=\frac{1}{2}\left(\frac{\partial g_{ij}}{\partial x^k}+\frac{\partial g_{ik}}{\partial x^j}-\frac{\partial g_{jk}}{\partial x^i}\right),{\Gamma }_{jk}^i=\frac{1}{2}{g}^{i\ell }\left(\frac{\partial g_{\ell j}}{\partial x^k}+\frac{\partial g_{\ell k}}{\partial x^j}-\frac{\partial g_{jk}}{\partial x^{\ell }}\right),$$

and are symmetric in their last two indices: ${\Gamma }_{ijk}={\Gamma }_{ikj}$ and ${\Gamma }_{jk}^i={\Gamma }_{kj}^i$.

By Ricci’s Theorem (the metric is covariantly constant), we have the useful identity

$$\frac{\partial g_{ik}}{\partial x^{\ell }}={\Gamma }_{ik\ell }+{\Gamma }_{ki\ell },$$

and in orthogonal coordinates the off-diagonal version ${\Gamma }_{ik\ell }=-{\Gamma }_{ki\ell }$ holds.

The Riemann-Christoffel Tensor is defined by $R_{ijk}^r{A}_r={\nabla }_j{\nabla }_k{A}_i-{\nabla }_k{\nabla }_j{A}_i$, which expands to

$$\boxed{R_{ijk}^r=\frac{\partial {\Gamma }_{ki}^r}{\partial x^j}-\frac{\partial {\Gamma }_{ij}^r}{\partial x^k}+{\Gamma }_{ik}^p{\Gamma }_{pj}^r-{\Gamma }_{ij}^p{\Gamma }_{pk}^r.}$$

It is a tensor, and a space is Euclidean iff $R_{ijk}^r\equiv 0$.

The Ricci Tensor

Definition

The Ricci tensor is the contraction of the Riemann-Christoffel tensor on the upper index and the second lower index:

$$\boxed{R_{ij}=R_{ikj}^k.}$$

Substituting the explicit formula for the Riemann-Christoffel tensor and renaming indices,

$$R_{ij}=R_{ikj}^k=\frac{\partial {\Gamma }_{ji}^k}{\partial x^k}-\frac{\partial {\Gamma }_{ik}^k}{\partial x^j}+{\Gamma }_{ij}^p{\Gamma }_{pk}^k-{\Gamma }_{ik}^p{\Gamma }_{pj}^k.$$

Note

$R_{ij}$ is symmetric: $R_{ij}=R_{ji}$. This means the Ricci tensor in 3D has at most six independent components, not nine.

The Ricci tensor inherits tensor character from the Riemann-Christoffel tensor (the contraction of a tensor is a tensor). It is the “summary” curvature object used in general relativity - Einstein’s field equations relate $R_{ij}$ (and its trace, the scalar curvature) to the stress-energy of matter.

Goal: Compute the Ricci Tensor in Cylindrical Coordinates

Example

Consider the three-dimensional coordinate system $(x^1,x^2,x^3)=(\rho ,\theta ,z)$ with position vector

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

Find the Ricci tensor $R_{ij}$ in this coordinate system.

This is a single revision exercise that touches every concept of the previous weeks. The strategy works backwards along the dependency chain:

$$\text{basis vectors }{\mathbf{e}}_i\to \text{metric }{g}_{ij},g^{ij}\to \text{Christoffel symbols }{\Gamma }_{jk}^i\to \text{Riemann-Christoffel }{R}_{ijk}^r\to \text{Ricci }{R}_{ij}.$$

Step 1: Basis Vectors

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

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

Step 2: Metric Coefficients

Off-diagonal entries. Direct dot products give

$$g_{12}={\mathbf{e}}_1\cdot {\mathbf{e}}_2=-\rho \cos \theta \sin \theta +\rho \cos \theta \sin \theta =0,$$ $$g_{13}={\mathbf{e}}_1\cdot {\mathbf{e}}_3=0,g_{23}={\mathbf{e}}_2\cdot {\mathbf{e}}_3=0.$$

By symmetry, all $g_{ij}=0$ for $i\ne j$ - the system is orthogonal.

Diagonal entries.

$$g_{11}={\mathbf{e}}_1\cdot {\mathbf{e}}_1={\cos }^2\theta +{\sin }^2\theta =1,$$ $$g_{22}={\mathbf{e}}_2\cdot {\mathbf{e}}_2={\rho }^2{\sin }^2\theta +{\rho }^2{\cos }^2\theta ={\rho }^2,$$ $$g_{33}={\mathbf{e}}_3\cdot {\mathbf{e}}_3=1.$$

So the metric matrices are

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

Note

$g_{22}={\rho }^2\ne 1$, so the system is orthogonal but not orthonormal. The arc length element is $(ds{)}^2=(d\rho {)}^2+(\rho d\theta {)}^2+(dz{)}^2$, with scale factors $h_1=1,h_2=\rho ,h_3=1$.

Step 3: Christoffel Symbols of the First Kind

The only nonzero metric derivative is $\partial g_{22}/\partial x^1=\partial {\rho }^2/\partial \rho =2\rho$. Plugging into

$${\Gamma }_{ijk}=\frac{1}{2}\left(\frac{\partial g_{ij}}{\partial x^k}+\frac{\partial g_{ik}}{\partial x^j}-\frac{\partial g_{jk}}{\partial x^i}\right),$$

the only surviving symbols are those with two indices equal to $2$ and one index equal to $1$. Computing the first one:

$${\Gamma }_{122}=\frac{1}{2}\left(\frac{\partial g_{12}}{\partial x^2}+\frac{\partial g_{12}}{\partial x^2}-\frac{\partial g_{22}}{\partial x^1}\right)=-\frac{1}{2}(2\rho )=-\rho .$$

By Ricci’s theorem in orthogonal coordinates, ${\Gamma }_{212}=-{\Gamma }_{122}=\rho$. By the symmetry of ${\Gamma }_{ijk}$ in $(j,k)$, ${\Gamma }_{221}={\Gamma }_{212}=\rho$. All other ${\Gamma }_{ijk}=0$:

$$\boxed{{\Gamma }_{122}=-\rho ,{\Gamma }_{212}={\Gamma }_{221}=\rho ,{\Gamma }_{ijk}=0\text{otherwise}.}$$

Step 4: Christoffel Symbols of the Second Kind

Using ${\Gamma }_{jk}^i=g^{i\ell }{\Gamma }_{\ell jk}$ with the diagonal $g^{ij}$:

$${\Gamma }_{22}^1=g^{11}{\Gamma }_{122}=1\cdot (-\rho )=-\rho ,$$ $${\Gamma }_{12}^2={\Gamma }_{21}^2=g^{22}{\Gamma }_{221}=\frac{1}{{\rho }^2}\cdot \rho =\frac{1}{\rho }.$$

All others vanish:

$$\boxed{{\Gamma }_{22}^1=-\rho ,{\Gamma }_{12}^2={\Gamma }_{21}^2=\frac{1}{\rho },{\Gamma }_{jk}^i=0\text{otherwise}.}$$

Step 5: Sample Riemann-Christoffel Component

To illustrate the calculation, compute $R_{121}^2$:

$$R_{121}^2=\frac{\partial {\Gamma }_{11}^2}{\partial x^2}-\frac{\partial {\Gamma }_{12}^2}{\partial x^1}+{\Gamma }_{11}^p{\Gamma }_{p2}^2-{\Gamma }_{12}^p{\Gamma }_{p1}^2.$$

Since ${\Gamma }_{11}^2=0$, the first term drops. Among the products, only $p=2$ in the last sum survives (${\Gamma }_{12}^2{\Gamma }_{21}^2$), so

$$R_{121}^2=-\frac{\partial }{\partial \rho }\left(\frac{1}{\rho }\right)-\frac{1}{\rho }\cdot \frac{1}{\rho }=\frac{1}{{\rho }^2}-\frac{1}{{\rho }^2}=0.$$

Note

Cylindrical coordinates parametrise ordinary Euclidean ${\mathbb{R}}^3$, so we expect every Riemann-Christoffel component to vanish. The example confirms this for one representative component, and the same cancellation pattern works for all others.

Step 6: Ricci Tensor

Since the indices range over $\{1,2,3\}$ but ${\Gamma }^3$ symbols all vanish, the contraction $R_{ij}=R_{ikj}^k$ collapses to

$$R_{ij}=R_{i1j}^1+R_{i2j}^2.$$

The $R_{i1j}^1$ piece. Expanding,

$$R_{i1j}^1=\frac{\partial {\Gamma }_{1i}^1}{\partial x^j}-\frac{\partial {\Gamma }_{ij}^1}{\partial x^1}+{\Gamma }_{i1}^p{\Gamma }_{pj}^1-{\Gamma }_{ij}^p{\Gamma }_{p1}^1.$$

The terms with ${\Gamma }_{1i}^1$ and ${\Gamma }_{p1}^1$ vanish (no nonzero ${\Gamma }^1$ has a $1$ in the lower indices). What remains is

$$R_{i1j}^1=-\frac{\partial {\Gamma }_{ij}^1}{\partial x^1}+{\Gamma }_{i1}^2{\Gamma }_{2j}^1.$$

For $i=1$ or $j=1$, both terms vanish. For $i=j=2$,

$$R_{212}^1=-\frac{\partial {\Gamma }_{22}^1}{\partial \rho }+{\Gamma }_{21}^2{\Gamma }_{22}^1=-\frac{\partial (-\rho )}{\partial \rho }+\frac{1}{\rho }(-\rho )=1-1=0.$$

So $R_{i1j}^1=0$ for all $i,j$.

The $R_{i2j}^2$ piece. Expanding,

$$R_{i2j}^2=\frac{\partial {\Gamma }_{2i}^2}{\partial x^j}-\frac{\partial {\Gamma }_{ij}^2}{\partial x^2}+{\Gamma }_{i2}^p{\Gamma }_{pj}^2-{\Gamma }_{ij}^p{\Gamma }_{p2}^2.$$

The term $\partial {\Gamma }_{ij}^2/\partial x^2=\partial {\Gamma }_{ij}^2/\partial \theta =0$ since none of the ${\Gamma }^2$ symbols depend on $\theta$. For $(i,j)\in \{(1,2),(2,1)\}$,

$$R_{i2j}^2={\Gamma }_{i2}^1{\Gamma }_{1j}^2+{\Gamma }_{i2}^2{\Gamma }_{2j}^2-{\Gamma }_{ij}^1{\Gamma }_{12}^2-{\Gamma }_{ij}^2{\Gamma }_{22}^2=0$$

(every product term either has a vanishing factor or cancels). The remaining case $i=j=2$:

$$R_{222}^2={\Gamma }_{22}^1{\Gamma }_{12}^2-{\Gamma }_{22}^1{\Gamma }_{12}^2=0.$$

Together with $R_{121}^2=0$ from Step 5, $R_{i2j}^2=0$ for all $i,j$.

Step 7: Conclusion

Adding the two pieces,

$$\boxed{R_{ij}=R_{i1j}^1+R_{i2j}^2=0\text{for all }i,j.}$$

The Ricci tensor vanishes identically - exactly as expected, since cylindrical coordinates parametrise flat Euclidean ${\mathbb{R}}^3$.

Important

$R_{ij}=0$ is necessary but not sufficient for a space to be Euclidean. The Ricci tensor is a contraction of $R_{ijk}^r$ and so loses information; a space can have $R_{ij}=0$ without being flat (Ricci-flat manifolds appear in vacuum solutions of general relativity). The Riemann-Christoffel tensor, not the Ricci tensor, is the definitive curvature test.

---

Pre-Lecture Notes from mth3008 lecture notes 19.pdf

• Setup: generalised coordinates with local basis ${\mathbf{e}}_j(x^1,x^2,x^3)$; Christoffel symbols computed from the metric; symmetric in last two lower indices
• Reminder: Ricci’s theorem $\Rightarrow$ $\partial g_{ik}/\partial x^{\ell }={\Gamma }_{ik\ell }+{\Gamma }_{ki\ell }$; in orthogonal coordinates ${\Gamma }_{ik\ell }=-{\Gamma }_{ki\ell }$
• Riemann-Christoffel tensor: $R_{ijk}^r{A}_r={\nabla }_j{\nabla }_k{A}_i-{\nabla }_k{\nabla }_j{A}_i$, coordinate formula $R_{ijk}^r={\partial }_j{\Gamma }_{ki}^r-{\partial }_k{\Gamma }_{ij}^r+{\Gamma }_{ik}^p{\Gamma }_{pj}^r-{\Gamma }_{ij}^p{\Gamma }_{pk}^r$; Euclidean iff $R_{ijk}^r=0$
• Ricci tensor: definition $R_{ij}=R_{ikj}^k$ (contraction); coordinate formula $R_{ij}={\partial }_k{\Gamma }_{ji}^k-{\partial }_j{\Gamma }_{ik}^k+{\Gamma }_{ij}^p{\Gamma }_{pk}^k-{\Gamma }_{ik}^p{\Gamma }_{pj}^k$; symmetric tensor
• Worked example: cylindrical coordinates $(\rho ,\theta ,z)$, position $\mathbf{r}=\rho \cos \theta {\mathbf{i}}_1+\rho \sin \theta {\mathbf{i}}_2+z{\mathbf{i}}_3$
• Dependency chain: basis ${\mathbf{e}}_i$ $\to$ metric $g_{ij}$ $\to$ Christoffel ${\Gamma }_{jk}^i$ $\to$ Riemann-Christoffel $R_{ijk}^r$ $\to$ Ricci $R_{ij}$
• Cylindrical results: orthogonal but not orthonormal; $g_{11}=g_{33}=1$, $g_{22}={\rho }^2$; only nonzero Christoffel symbols are ${\Gamma }_{22}^1=-\rho$, ${\Gamma }_{12}^2={\Gamma }_{21}^2=1/\rho$; all $R_{ijk}^r=0$, hence $R_{ij}=0$ - confirming Euclidean character
• Next lecture: revision of Chapters 1-4 (suffix and vector notation, Kronecker delta, alternating tensor, vector differential operators, local coordinate transformation, tensors in orthogonal Cartesian coordinates)