Exercises — Christoffel symbols — intro

The minus sits on ${\partial }_m{g}_{ij}$ — the derivative with respect to the ==dummy index $m$==, the same $m$ that is tied to the inverse metric $g^{km}$.

Why it matters: the two plus terms differentiate $g$ with respect to the free lower indices $i$ and $j$; the minus term differentiates with respect to the summed index. Mixing them up flips signs and produces wrong "fictitious forces." Mnemonic: "half-inverse, two plus, one minus."

The upper index $k$ takes $2$ values. The lower pair $(i,j)$ is symmetric, so instead of $2\times 2=4$ ordered pairs we count unordered pairs: $(1,1),(1,2),(2,2)$ = $3$ choices.

Total independent components $=2\times 3=\boxed{6}$.

(In $n$ dimensions the count is $n\cdot \frac{n(n+1)}{2}$.)

True. Every term in the master formula is a derivative $\partial g$. If each $g_{ij}$ is constant then ${\partial }_i{g}_{mj}={\partial }_j{g}_{mi}={\partial }_m{g}_{ij}=0$, so the whole bracket is $0$ and ${\Gamma }_{ij}^k=0$ for all indices.

This is exactly the Cartesian case $g_{ij}={\delta }_{ij}$ from Worked Example 2 in the parent note.

Only $g_{\theta \theta }=r^2$ depends on a coordinate; its only nonzero derivative is ${\partial }_r{g}_{\theta \theta }=2r$. Everything else has zero derivative.

${\Gamma }_{rr}^r$: $\frac{1}{2}{g}^{rr}({\partial }_r{g}_{rr}+{\partial }_r{g}_{rr}-{\partial }_r{g}_{rr})=\frac{1}{2}(1)(0)=0$.

${\Gamma }_{r\theta }^r$: $\frac{1}{2}{g}^{rr}({\partial }_r{g}_{r\theta }+{\partial }_{\theta }{g}_{rr}-{\partial }_r{g}_{r\theta })=0$.

${\Gamma }_{\theta \theta }^r$: $\frac{1}{2}{g}^{rr}({\partial }_{\theta }{g}_{r\theta }+{\partial }_{\theta }{g}_{r\theta }-{\partial }_r{g}_{\theta \theta })=\frac{1}{2}(1)(0+0-2r)=\boxed{-r}$.

${\Gamma }_{rr}^{\theta }$: $\frac{1}{2}{g}^{\theta \theta }({\partial }_r{g}_{\theta r}+{\partial }_r{g}_{\theta r}-{\partial }_{\theta }{g}_{rr})=0$.

${\Gamma }_{r\theta }^{\theta }$: $\frac{1}{2}{g}^{\theta \theta }({\partial }_r{g}_{\theta \theta }+{\partial }_{\theta }{g}_{\theta r}-{\partial }_{\theta }{g}_{r\theta })=\frac{1}{2}\cdot \frac{1}{r^2}\cdot 2r=\boxed{\frac{1}{r}}$.

${\Gamma }_{\theta \theta }^{\theta }$: $\frac{1}{2}{g}^{\theta \theta }({\partial }_{\theta }{g}_{\theta \theta }+{\partial }_{\theta }{g}_{\theta \theta }-{\partial }_{\theta }{g}_{\theta \theta })=\frac{1}{2}{g}^{\theta \theta }(0)=0$.

Nonzero: ${\Gamma }_{\theta \theta }^r=-r$ and ${\Gamma }_{r\theta }^{\theta }={\Gamma }_{\theta r}^{\theta }=\frac{1}{r}$. All four others vanish.

Inverse metric: $g^{\theta \theta }=1/a^2$, $g^{\varphi \varphi }=1/(a^2{\sin }^2\theta )$. The only coordinate-dependent component is $g_{\varphi \varphi }=a^2{\sin }^2\theta$, with ${\partial }_{\theta }{g}_{\varphi \varphi }=2a^2\sin \theta \cos \theta$.

${\Gamma }_{\varphi \varphi }^{\theta }$: $k=\theta ,i=j=\varphi$. $\frac{1}{2}{g}^{\theta \theta }({\partial }_{\varphi }{g}_{\theta \varphi }+{\partial }_{\varphi }{g}_{\theta \varphi }-{\partial }_{\theta }{g}_{\varphi \varphi })=\frac{1}{2}\cdot \frac{1}{a^2}(0+0-2a^2\sin \theta \cos \theta )=\boxed{-\sin \theta \cos \theta }.$

${\Gamma }_{\theta \varphi }^{\varphi }$: $k=\varphi ,i=\theta ,j=\varphi$. $\frac{1}{2}{g}^{\varphi \varphi }({\partial }_{\theta }{g}_{\varphi \varphi }+{\partial }_{\varphi }{g}_{\varphi \theta }-{\partial }_{\varphi }{g}_{\theta \varphi })=\frac{1}{2}\cdot \frac{1}{a^2{\sin }^2\theta }\cdot 2a^2\sin \theta \cos \theta =\boxed{\cot \theta }.$

With ${\Gamma }_{\theta \theta }^r=-r$: $\ddot{r}+(-r){\dot{\theta }}^2=0⟹\ddot{r}=r{\dot{\theta }}^2.$ Wait — the full geodesic sum is $\ddot{r}+{\Gamma }_{\theta \theta }^r{\dot{\theta }}^2=\ddot{r}-r{\dot{\theta }}^2=0$, so $\boxed{\ddot{r}-r{\dot{\theta }}^2=0}.$ Interpretation: this is Newton's "no force, straight line" written in polar coordinates. The term $-r{\dot{\theta }}^2$ is the centripetal acceleration — the piece you'd call a "fictitious force" in a rotating description. It appears purely because the basis vectors ${\hat{e}}_r,{\hat{e}}_{\theta }$ rotate; there is no real force. The Christoffel symbol is that fictitious term.

Nonzero polar symbols: ${\Gamma }_{\theta \theta }^r=-r$, ${\Gamma }_{r\theta }^{\theta }={\Gamma }_{\theta r}^{\theta }=1/r$. All others zero.

• ${\partial }_r{\Gamma }_{\theta \theta }^r={\partial }_r(-r)=-1$.
• ${\partial }_{\theta }{\Gamma }_{r\theta }^r={\partial }_{\theta }(0)=0$.
• ${\Gamma }_{rm}^r{\Gamma }_{\theta \theta }^m$: sum over $m$. ${\Gamma }_{rr}^r=0,{\Gamma }_{r\theta }^r=0$, so this whole term is $0$.
• ${\Gamma }_{\theta m}^r{\Gamma }_{r\theta }^m$: sum over $m$. Only ${\Gamma }_{\theta \theta }^r=-r$ is nonzero with $m=\theta$, paired with ${\Gamma }_{r\theta }^{\theta }=1/r$. Term $=(-r)(1/r)=-1$.

Assemble: $R_{\theta r\theta }^r=(-1)-0+0-(-1)=-1+1=\boxed{0}.$

Moral: the plane is flat. Nonzero $\Gamma$'s (from polar coordinates) canceled perfectly inside the curvature combination. $\Gamma \ne 0$ never means curved — only $\partial \Gamma +\Gamma \Gamma$, done invariantly, detects curvature.

Components are constant, so every ${\partial }_j{V}^k=0$. Only the $\Gamma$ terms survive.

${\nabla }_{\theta }{V}^{\theta }={\partial }_{\theta }{V}^{\theta }+{\Gamma }_{i\theta }^{\theta }{V}^i$. Sum over $i$: $=0+{\Gamma }_{r\theta }^{\theta }{V}^r+{\Gamma }_{\theta \theta }^{\theta }{V}^{\theta }=\frac{1}{r}\cdot 1+0\cdot 0=\boxed{\frac{1}{r}}.$

${\nabla }_{\theta }{V}^r={\partial }_{\theta }{V}^r+{\Gamma }_{i\theta }^r{V}^i$. Sum over $i$: $=0+{\Gamma }_{r\theta }^r{V}^r+{\Gamma }_{\theta \theta }^r{V}^{\theta }=0\cdot 1+(-r)\cdot 0=\boxed{0}.$

Interpretation: the field $V^r=1$ is the outward unit radial field ${\hat{e}}_r$. As you swing in the $\theta$ direction, ${\hat{e}}_r$ physically rotates into the ${\hat{e}}_{\theta }$ direction. The covariant derivative sees this real turning even though the number $V^r$ never changes. The nonzero ${\nabla }_{\theta }{V}^{\theta }=1/r$ records exactly that rotation. Constant components ≠ constant vector on a curvilinear grid.

In the $x$-chart, $d{s}^2=d{x}^2$, so $\Gamma =0$ (constant metric).

Transform. With $x=\sqrt{\tilde{x}}$, $\frac{dx}{d\tilde{x}}=\frac{1}{2\sqrt{\tilde{x}}}$. The metric transforms (see Tensor Transformation Laws): $\tilde{g}=g{\left(\frac{dx}{d\tilde{x}}\right)}^2=1\cdot \frac{1}{4\tilde{x}}=\frac{1}{4\tilde{x}}.$ Inverse: ${\tilde{g}}^{-1}=4\tilde{x}$. In 1D the master formula reads $\tilde{\Gamma }=\frac{1}{2}{\tilde{g}}^{-1}{\partial }_{\tilde{x}}\tilde{g}$ (the three bracket terms collapse to one since all indices equal): $\tilde{\Gamma }=\frac{1}{2}(4\tilde{x})\cdot {\partial }_{\tilde{x}}\left(\frac{1}{4\tilde{x}}\right)=\frac{1}{2}(4\tilde{x})\cdot \left(-\frac{1}{4{\tilde{x}}^2}\right)=-\frac{1}{2\tilde{x}}\ne 0.$ So $\boxed{\tilde{\Gamma }=-\frac{1}{2\tilde{x}}}$. A perfectly flat line acquired a nonzero Christoffel symbol purely from the coordinate stretch. This is the non-tensor behaviour: had $\Gamma$ been a tensor, $\Gamma =0$ in one chart would force $\tilde{\Gamma }=0$ in every chart. It doesn't.

(a) Metric: $g_{rr}=1,g_{\theta \theta }=f(r{)}^2,g_{r\theta }=0$; inverse $g^{rr}=1,g^{\theta \theta }=1/f^2$. Only ${\partial }_r{g}_{\theta \theta }=2f{f}^{\prime }$ is nonzero.

${\Gamma }_{\theta \theta }^r=\frac{1}{2}{g}^{rr}(-{\partial }_r{g}_{\theta \theta })=\frac{1}{2}(1)(-2f{f}^{\prime })=\boxed{-f{f}^{\prime }}.$

${\Gamma }_{r\theta }^{\theta }=\frac{1}{2}{g}^{\theta \theta }\left({\partial }_r{g}_{\theta \theta }\right)=\frac{1}{2}\cdot \frac{1}{f^2}\cdot 2f{f}^{\prime }=\boxed{\frac{f^{\prime }}{f}}.$

All others vanish (check: every remaining bracket contains only derivatives of constants or of $g_{rr}=1$). This recovers polar results with $f=r$: $-f{f}^{\prime }=-r$, $f^{\prime }/f=1/r$. ✓

(b) With $f=\sin r$, $f^{\prime }=\cos r$: ${\Gamma }_{\theta \theta }^r=-\sin r\cos r,{\Gamma }_{r\theta }^{\theta }=\frac{\cos r}{\sin r}=\cot r.$ (These match the sphere result of Problem 2.2 with $a=1$ and $\theta \to r$: $-\sin \theta \cos \theta$ and $\cot \theta$. ✓)

(c) Radial geodesic: $\ddot{r}+{\Gamma }_{\theta \theta }^r{\dot{\theta }}^2=0$, i.e. $\boxed{\ddot{r}-f{f}^{\prime }{\dot{\theta }}^2=0}\text{(general)},\ddot{r}-\sin r\cos r{\dot{\theta }}^2=0\text{(sphere)}.$ The term $-f{f}^{\prime }{\dot{\theta }}^2$ is the generalised centripetal acceleration: motion "wants" to curve toward or away from the axis at a rate set by how fast the circle radius $f(r)$ changes. On the sphere near the equator ($r=\pi /2$) we get $\cos r=0$, so the term vanishes — great circles round the equator need no radial pull, exactly as expected.

Can you, from memory: write the master formula? ::: ${\Gamma }_{ij}^k=\frac{1}{2}{g}^{km}({\partial }_i{g}_{mj}+{\partial }_j{g}_{mi}-{\partial }_m{g}_{ij})$ Name the two nonzero polar symbols? ::: ${\Gamma }_{\theta \theta }^r=-r,{\Gamma }_{r\theta }^{\theta }=1/r$ Explain why flat polar space has nonzero $\Gamma$? ::: The polar grid's basis arrows rotate; $\Gamma$ tracks that, but the Riemann combination still cancels to zero.