Worked examples — Christoffel symbols — intro

This page is a drill. The parent note gave you the one formula you must memorise. Here we throw every kind of input at it — flat and curved, diagonal and off-diagonal metrics, zero cases, degenerate cases, a word problem, and an exam trap — so that after this page no scenario can surprise you.

Everything rests on the single boxed formula from the parent:

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The scenario matrix

Before working anything, let us list every distinct thing that can go wrong or change. Each worked example below is tagged with the cell it lands in.

Cell	What it tests	Example
A. Flat / trivial	constant metric $\Rightarrow$ all $\Gamma =0$	Ex 1
B. Diagonal, one non-constant entry	the classic polar radial symbol	Ex 2
C. Diagonal, sign of the derivative	when a $\Gamma$ comes out positive vs negative	Ex 3
D. Off-diagonal metric	$g^{km}$ mixes indices; two terms in the sum survive	Ex 4
E. Coordinate-dependent everywhere (sphere)	multiple non-zero $\Gamma$, latitude sign flips	Ex 5
F. Degenerate / limiting input	what happens as $r\to 0$ (coordinate breakdown)	Ex 6
G. Word problem (physics)	read a "fictitious force" off a $\Gamma$	Ex 7
H. Exam twist (is-it-a-tensor trap)	conformal rescale, symmetry check	Ex 8

The goal: after Ex 8 you should be able to place any new problem into one of these rows and know which trick it wants.

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Ex 1 — Cell A: the flat sanity check

Forecast: guess the answer before reading on. What is the derivative of a constant?

1. Write the formula with $k=x,i=j=y$. ${\Gamma }_{yy}^x=\frac{1}{2}{g}^{xm}({\partial }_y{g}_{my}+{\partial }_y{g}_{my}-{\partial }_m{g}_{yy}).$ Why this step? We are just substituting the requested indices into the machine; nothing clever yet.

2. Note every metric component is a constant number. Since $g_{yy}=1$ everywhere, ${\partial }_m{g}_{yy}=0$. Since $g_{my}$ is constant, ${\partial }_y{g}_{my}=0$. Why this step? The formula only ever contains derivatives of the metric. A constant metric has all derivatives zero — so the entire bracket collapses.

3. Conclude. ${\Gamma }_{yy}^x=\frac{1}{2}{g}^{xm}\cdot 0=\boxed{0}.$ The same argument kills every Christoffel symbol here.

Verify: units/logic — basis vectors $\hat{x},\hat{y},\hat{z}$ point the same way at every point, so they cannot "turn," so their turn-rate cards ($\Gamma$) must all read zero. Consistent. ✔

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Ex 2 — Cell B: the classic polar radial symbol

Forecast: which single derivative in the formula can possibly be non-zero here?

Look at the figure: the arrows ${\mathbf{e}}_r$ (burnt orange) point outward and ${\mathbf{e}}_{\theta }$ (teal) point sideways. As you swing $\theta$, notice ${\mathbf{e}}_{\theta }$ swings inward toward the origin — its change has a component along $-{\mathbf{e}}_r$. That inward lean is precisely ${\Gamma }_{\theta \theta }^r$.

1. Substitute $k=r,i=j=\theta$. ${\Gamma }_{\theta \theta }^r=\frac{1}{2}{g}^{rm}({\partial }_{\theta }{g}_{m\theta }+{\partial }_{\theta }{g}_{m\theta }-{\partial }_m{g}_{\theta \theta }).$ Why this step? We are targeting one specific symbol — the "output basis" is $r$ (upper index $k=r$) and the "which basis vector, in which direction" pair is $\theta ,\theta$ — so we place those exact labels into the machine before touching any numbers.

2. Kill the sum over $m$. The metric is diagonal, so $g^{rm}$ is non-zero only for $m=r$, giving $g^{rr}=1$. Why this step? Diagonal inverse metric means "only the matching index survives" — this is what makes diagonal problems fast. ${\Gamma }_{\theta \theta }^r=\frac{1}{2}(1)({\partial }_{\theta }{g}_{r\theta }+{\partial }_{\theta }{g}_{r\theta }-{\partial }_r{g}_{\theta \theta }).$

3. Evaluate each derivative. $g_{r\theta }=0$ so its $\theta$-derivative is $0$. And $g_{\theta \theta }=r^2$, so ${\partial }_r{g}_{\theta \theta }=2r$. ${\Gamma }_{\theta \theta }^r=\frac{1}{2}(0+0-2r)=\boxed{-r}.$ Why this step? The minus sign lives on the ${\partial }_m$ term (the one tied to $g^{km}$) — remember the mnemonic "two plus, one minus."

Verify: the geometric picture said the change of ${\mathbf{e}}_{\theta }$ leans toward $-{\mathbf{e}}_r$ (a negative radial component), matching the minus sign. Physically this is the centripetal term $-r{\dot{\theta }}^2$. ✔

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Ex 3 — Cell C: watch the sign — a positive Christoffel symbol

Forecast: will this be positive or negative? Guess, then check.

1. Substitute $k=\theta ,i=r,j=\theta$. ${\Gamma }_{r\theta }^{\theta }=\frac{1}{2}{g}^{\theta m}({\partial }_r{g}_{m\theta }+{\partial }_{\theta }{g}_{mr}-{\partial }_m{g}_{r\theta }).$ Why this step? This time the "output basis" is $\theta$ (upper index $k=\theta$) and the direction/vector pair is $r,\theta$; we must feed those precise slots in — the whole exercise is to see how a different index choice changes the surviving derivative and thus the sign.

2. Diagonal $\Rightarrow$ only $m=\theta$ survives: $g^{\theta \theta }=1/r^2$. ${\Gamma }_{r\theta }^{\theta }=\frac{1}{2}\cdot \frac{1}{r^2}({\partial }_r{g}_{\theta \theta }+{\partial }_{\theta }{g}_{\theta r}-{\partial }_{\theta }{g}_{r\theta }).$ Why this step? Exactly as in Ex 2: the inverse metric is diagonal, so $g^{\theta m}$ vanishes unless $m=\theta$. Dropping the $m=r$ term is legitimate only because $g^{\theta r}=0$ here — for an off-diagonal metric (Ex 4) we could not.

3. Evaluate. ${\partial }_r{g}_{\theta \theta }=2r$; the other two involve $g_{\theta r}=0$, so are $0$. ${\Gamma }_{r\theta }^{\theta }=\frac{1}{2}\cdot \frac{1}{r^2}\cdot 2r=\boxed{\frac{1}{r}}.$ Why this step? Here the non-zero derivative sits in a plus slot (it is ${\partial }_i{g}_{mj}$, not ${\partial }_m{g}_{ij}$), so the answer is positive — the opposite sign to Ex 2 even though it came from the same ${\partial }_r{r}^2=2r$.

Verify: as $r$ grows, $1/r$ shrinks — far from the origin the coordinate lines are "less curved," so the twist-rate drops. Sensible. This is the Coriolis-like term. ✔

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Ex 4 — Cell D: an off-diagonal metric (the sum really sums)

Forecast: the inverse metric now has off-diagonal entries, so the $m$-sum has two terms. But look hard at the metric first — is anything actually varying?

1. Write the sum explicitly over $m=1,2$. ${\Gamma }_{22}^1=\frac{1}{2}\left[g^{11}\right(2{\partial }_2{g}_{12}-{\partial }_1{g}_{22})+g^{12}(2{\partial }_2{g}_{22}-{\partial }_2{g}_{22}\left)\right].$ Why this step? With an off-diagonal $g^{12}$, we must not drop the $m=2$ term as we did in the diagonal cases.

2. Every entry is a constant, so every derivative is $0$. ${\Gamma }_{22}^1=\frac{1}{2}[g^{11}\cdot 0+g^{12}\cdot 0]=\boxed{0}.$ Why this step? The trap this cell teaches: off-diagonal metric does not by itself create Christoffel symbols. Only coordinate-dependence does. A constant metric — even a skewed one — is still flat with all $\Gamma =0$.

Verify: a constant-coefficient linear coordinate change from Cartesian keeps basis vectors fixed in direction, so no twisting, so $\Gamma =0$. Matches Ex 1's logic. ✔ (We really needed the $g^{12}$ term to be present in the setup to make the point that it vanishes only because the derivatives vanish, not the metric.)

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Ex 5 — Cell E: the sphere — many symbols, a sign that flips with latitude

Forecast: on a sphere, moving East gets "cheaper" near the poles. Which of these two symbols should blow up as $\theta \to 0$?

The figure shows lines of longitude squeezing together toward the pole. The metric coefficient ${\sin }^2\theta$ (plum curve) measures how much real distance one unit of $\varphi$ buys — it is largest at the equator ($\theta =\pi /2$) and zero at the poles.

1. ${\Gamma }_{\varphi \varphi }^{\theta }$: set $k=\theta ,i=j=\varphi$, diagonal $\Rightarrow g^{\theta \theta }=1$. ${\Gamma }_{\varphi \varphi }^{\theta }=\frac{1}{2}(1)(0+0-{\partial }_{\theta }{g}_{\varphi \varphi })=-\frac{1}{2}{\partial }_{\theta }({\sin }^2\theta )=-\frac{1}{2}\cdot 2\sin \theta \cos \theta =\boxed{-\sin \theta \cos \theta }.$ Why this step? Same shape as the polar Ex 2: the only surviving derivative is ${\partial }_{\theta }{g}_{\varphi \varphi }$, sitting in the minus slot.

2. Read its sign across the whole range. For $0<\theta <\frac{\pi }{2}$ (northern hemisphere) $\cos \theta >0$ so this is negative; at the equator $\theta =\frac{\pi }{2}$ it is $\boxed{0}$; for $\frac{\pi }{2}<\theta <\pi$ (southern) $\cos \theta <0$ so it flips positive. Why this step? Cell E demands we cover all cases of the coordinate — the sign genuinely changes at the equator, and vanishes exactly there.

3. ${\Gamma }_{\theta \varphi }^{\varphi }$: set $k=\varphi ,i=\theta ,j=\varphi$, diagonal $\Rightarrow g^{\varphi \varphi }=1/{\sin }^2\theta$. ${\Gamma }_{\theta \varphi }^{\varphi }=\frac{1}{2}\cdot \frac{1}{{\sin }^2\theta }{\partial }_{\theta }({\sin }^2\theta )=\frac{1}{2}\cdot \frac{2\sin \theta \cos \theta }{{\sin }^2\theta }=\boxed{\cot \theta }.$ Why this step? We now target the "East-tracking" symbol: upper index $k=\varphi$, direction/vector pair $\theta ,\varphi$. As in Ex 3 the diagonal inverse metric leaves only $m=\varphi$, and the single surviving derivative ${\partial }_{\theta }{g}_{\varphi \varphi }$ sits in a plus slot, so this symbol is positive where $\cos \theta >0$.

4. Limiting behaviour. As $\theta \to 0^{+}$ (approaching the north pole), $\cot \theta \to +\infty$: this symbol blows up. Why this step? That is the forecast answer — the East-tracking symbol diverges at the pole because "East" spins infinitely fast there (a coordinate singularity, not a real one).

Verify: at the equator $\theta =\pi /2$: $\cos (\pi /2)=0$ so ${\Gamma }_{\varphi \varphi }^{\theta }=0$ and $\cot (\pi /2)=0$ — both symbols vanish, matching the fact that near the equator the sphere looks locally like flat Cartesian paper. ✔

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Ex 6 — Cell F: the degenerate limit $r\to 0$

Forecast: what physical quantity is $1/r$ near $r=0$, and is the space actually singular there?

1. Take the limit. $\underset{r\to 0^{+}}{\lim }{\Gamma }_{r\theta }^{\theta }=\underset{r\to 0^{+}}{\lim }\frac{1}{r}=+\infty .$ Why this step? We must check limiting inputs — the origin is exactly where polar coordinates are degenerate ($\theta$ is undefined there).

2. Diagnose: is this the geometry or the coordinates? Compute ${\Gamma }_{\theta \theta }^r=-r$ at the same point: ${\lim }_{r\to 0}(-r)=0$, perfectly finite. Why this step? If the space were singular, no coordinate choice would give finite geometric invariants. Here the plane is manifestly flat and smooth at the origin — Cartesian coordinates see nothing special.

3. Conclude. The divergence of $1/r$ is a coordinate artefact: at $r=0$ the direction "$\theta$-increasing" is undefined (all angles meet), so its turn-rate card reads $\infty$. This is exactly like the sphere's pole in Ex 5.

Verify: switch to Cartesian at the origin — all $\Gamma =0$ there (Ex 1). A quantity that is $\infty$ in one chart and $0$ in another cannot be geometrically real; it is coordinate-borne. This confirms $\Gamma$ is not a tensor (a tensor that is zero in one basis is zero in all). ✔

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Ex 7 — Cell G: word problem — reading a fictitious force

Forecast: which two Christoffel numbers from Ex 2 and Ex 3 should show up, and how many times each?

1. Radial ($k=r$) equation. The only non-zero ${\Gamma }^r$ is ${\Gamma }_{\theta \theta }^r=-r$. $\ddot{r}+{\Gamma }_{\theta \theta }^r\dot{\theta }\dot{\theta }=\ddot{r}-r{\dot{\theta }}^2=0.$ Why this step? Plug the geodesic equation with $k=r$; only the $\theta \theta$ term survives from Ex 2.

2. Angular ($k=\theta$) equation. The non-zero ${\Gamma }^{\theta }$ are ${\Gamma }_{r\theta }^{\theta }={\Gamma }_{\theta r}^{\theta }=1/r$ (symmetric lower indices!). The sum ${\Gamma }_{ij}^{\theta }{\dot{x}}^i{\dot{x}}^j$ picks up both orderings: $\ddot{\theta }+\frac{1}{r}\dot{r}\dot{\theta }+\frac{1}{r}\dot{\theta }\dot{r}=\ddot{\theta }+\frac{2}{r}\dot{r}\dot{\theta }=0.$ Why this step? The factor of $2$ is the whole point — because $i,j$ are symmetric, the double sum counts $r\theta$ and $\theta r$ separately, doubling $1/r$.

3. Interpret. $\ddot{r}-r{\dot{\theta }}^2=0$ is Newton with the centripetal term; $\ddot{\theta }+\frac{2}{r}\dot{r}\dot{\theta }=0$ is angular-momentum conservation (the Coriolis term). Both "forces" are pure $\Gamma$ — they are artefacts of the rotating basis, not real pushes.

Verify: multiply the second equation by $r^2$: $r^2\ddot{\theta }+2r\dot{r}\dot{\theta }=\frac{d}{dt}(r^2\dot{\theta })=0$, so $r^2\dot{\theta }=$ const — conserved angular momentum, exactly what a free straight-line puck must have. ✔

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Ex 8 — Cell H: the exam twist (a conformal rescale + tensor trap)

Forecast: $e^{2f}$ is never zero, and its derivative brings down $2f^{\prime }$. Guess the shape of the answer.

1. Inverse metric. Diagonal with entries $e^{2f}$, so $g^{xx}=g^{yy}=e^{-2f}$. Why this step? Inverse of a diagonal matrix is reciprocals; we need $g^{xx}$ for the contraction.

2. Apply the formula, $k=i=j=x$. ${\Gamma }_{xx}^x=\frac{1}{2}{g}^{xx}({\partial }_x{g}_{xx}+{\partial }_x{g}_{xx}-{\partial }_x{g}_{xx})=\frac{1}{2}{g}^{xx}{\partial }_x{g}_{xx}.$ Why this step? When all three indices coincide, "two plus, one minus" collapses to a single derivative.

3. Evaluate. ${\partial }_x{g}_{xx}={\partial }_x{e}^{2f}=2f^{\prime }{e}^{2f}$, and $g^{xx}=e^{-2f}$: ${\Gamma }_{xx}^x=\frac{1}{2}{e}^{-2f}\cdot 2f^{\prime }{e}^{2f}=\boxed{f^{\prime }(x)}.$ Why this step? The exponential cancels cleanly — a signature move for conformal metrics.

4. Rebut the examiner (Cell H trap). Take $f(x)=x$ so ${\Gamma }_{xx}^x=1$, a non-zero constant. The metric $e^{2x}(d{x}^2+d{y}^2)$ is conformally flat, and a non-zero $\Gamma$ does not prove curvature; only the Riemann Curvature Tensor built as $\partial \Gamma +\Gamma \Gamma$ (an invariant) decides. Because $\Gamma$ carries a non-tensorial second-derivative piece under coordinate change (see Tensor Transformation Laws), you can always make it non-zero by a bad coordinate choice. Why this step? This is the recurring exam intent: separate "coordinate twisting" from "genuine curvature."

Verify (limiting check): if $f$ is constant, $f^{\prime }=0$ so ${\Gamma }_{xx}^x=0$ — the metric $e^{2f}{\delta }_{ij}$ with constant $f$ is a mere uniform rescale (still flat Cartesian), giving all $\Gamma =0$, matching Ex 1. ✔

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Recall One-line placement drill

Given a new problem, which cell is it? Constant metric ::: Cell A/D — all $\Gamma =0$, whether diagonal or skewed. Diagonal metric, one entry depends on a coordinate ::: Cell B/C — one derivative survives; sign depends on plus vs minus slot. Metric coefficient vanishes somewhere (pole, origin) ::: Cell E/F — expect a $\Gamma$ to diverge; it is a coordinate artefact. Someone claims curvature from $\Gamma \ne 0$ ::: Cell H — rebut; only Riemann decides.

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Connections

• Parent: Christoffel Symbols — Intro — the formula and derivation this page drills.
• Metric Tensor — the input to every example here.
• Geodesic Equation — where Ex 7's fictitious forces come from.
• Covariant Derivative — the operator these symbols correct.
• Riemann Curvature Tensor — the invariant that (unlike $\Gamma$) truly detects curvature (Ex 8).
• Polar Coordinates · Tensor Transformation Laws — coordinate machinery behind Ex 2–3 and the Ex 8 trap.