Visual walkthrough — Christoffel symbols — intro

Everything below rests on three prerequisite pictures. If any word here is new, follow the link: Metric Tensor, Polar Coordinates, Covariant Derivative, Tensor Transformation Laws.

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Step 0 — The two pictures every symbol lives on

Before any formula, fix two mental images.

WHAT. A basis vector ${\mathbf{e}}_i$ is just a little arrow that says "this way is the direction in which coordinate number $i$ increases." In flat Cartesian land there are two such arrows, ${\mathbf{e}}_x$ and ${\mathbf{e}}_y$, and they are the same arrow at every point. In polar land there are ${\mathbf{e}}_r$ (points outward) and ${\mathbf{e}}_{\theta }$ (points sideways, along the circle) — and these turn as you move.

WHY. The whole subject exists because those arrows turn. If they never turned, there would be nothing to bookkeep and $\Gamma$ would be zero everywhere.

PICTURE. On the left, Cartesian arrows: identical copies. On the right, polar arrows at three points on a circle — watch how the red ${\mathbf{e}}_{\theta }$ swings around.

We use the metric and never the raw arrows, because in a real problem (a warped surface) the metric is the one thing we can actually measure — see Metric Tensor.

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Step 1 — Name the twist we want to measure

WHAT. As you nudge along direction $x^j$, the arrow ${\mathbf{e}}_i$ changes into a slightly new arrow. The change $\frac{\partial {\mathbf{e}}_i}{\partial x^j}$ is itself an arrow. Any arrow can be rebuilt as a recipe of the original basis arrows:

=\;\underbrace{\Gamma^{k}_{\;ij}}_{\substack{\text{amount of }\mathbf e_k\\ \text{in that change}}}\;\underbrace{\mathbf{e}_k}_{\text{basis arrow }k}$$ Read the equation aloud, term by term: "the change in arrow $i$ when I step in direction $j$ (**left**) equals a pile of the original arrows (**right**), where $\Gamma^{k}_{\;ij}$ is the *serving size* of arrow $k$ in that pile." **WHY define it this way?** Because we want the change expressed **back in the language we already have** — the original basis. That is the only way a computer (or a covariant derivative) can use it. **PICTURE.** The green "change arrow" is decomposed into a red $\mathbf{e}_r$-portion and a blue $\mathbf{e}_\theta$-portion. Those two portions *are* $\Gamma^{r}_{\;ij}$ and $\Gamma^{\theta}_{\;ij}$. ![[deepdives/dd-maths-4.10.11-d2-s02.png]] > [!recall]- Why the two lower indices are symmetric > Symmetry of lower indices ::: Because $\partial_j\partial_i(\text{position})=\partial_i\partial_j(\text{position})$ — mixed partials commute. Since $\mathbf e_i=\partial_i(\text{position})$, we get $\partial_j\mathbf e_i=\partial_i\mathbf e_j$, hence $\Gamma^k_{\;ij}=\Gamma^k_{\;ji}$. --- ## Step 2 — The problem: $\Gamma$ hides *inside* dot products, not out in the open We want $\Gamma$ from $g$, but $\Gamma$ lives in derivatives of arrows and $g$ lives in dot products of arrows. The bridge is a **half-way object**: dot the change-arrow with a plain basis arrow. **WHAT.** Define the **first-kind** symbol $$\Gamma_{ijk}=(\partial_j\mathbf e_i)\cdot\mathbf e_k .$$ Term by term: take the change of $\mathbf e_i$ along $j$ (**an arrow**), dot it with $\mathbf e_k$ (**project it onto arrow $k$**). The result is a plain number — the length of the shadow. **WHY.** Dotting turns arrows into numbers, and numbers are what the metric is made of. This is the only trick in the whole derivation: *project the twist onto a known arrow.* **PICTURE.** The green change-arrow casts a shadow onto $\mathbf e_k$; the shadow's length is $\Gamma_{ijk}$. ![[deepdives/dd-maths-4.10.11-d2-s03.png]] Relation to the second kind (raising the index with the inverse metric $g^{km}$, which converts a projection-length back into a serving-size): $$\Gamma^{k}_{\;ij}=g^{km}\,\Gamma_{ijm}.$$ --- ## Step 3 — Differentiate the metric (the product rule that starts everything) **WHAT.** The metric is $g_{ij}=\mathbf e_i\cdot\mathbf e_j$. Differentiate along $x^k$ using the product rule for dot products: $$\partial_k g_{ij} =\underbrace{(\partial_k\mathbf e_i)\cdot\mathbf e_j}_{\Gamma_{ikj}} +\underbrace{\mathbf e_i\cdot(\partial_k\mathbf e_j)}_{\Gamma_{jki}}.$$ Each half is exactly a first-kind symbol from Step 2. So: $$\boxed{\;\partial_k g_{ij}=\Gamma_{ikj}+\Gamma_{jki}\;}\tag{A}$$ **WHY.** This is the *only* fact linking "derivatives of the metric" (things we can compute) to "$\Gamma$'s" (things we want). One equation, two unknown $\Gamma$'s — not yet solvable. That motivates Step 4. **PICTURE.** The change in the dot product = shadow of "$\mathbf e_i$'s twist onto $\mathbf e_j$" plus shadow of "$\mathbf e_j$'s twist onto $\mathbf e_i$." ![[deepdives/dd-maths-4.10.11-d2-s04.png]] --- ## Step 4 — Three copies, cycled (the trick that isolates one $\Gamma$) **WHAT.** Equation (A) holds for **any** choice of the three slots. Write it three times, rotating the labels $i\to j\to k\to i$ each time: $$\partial_k g_{ij}=\Gamma_{ikj}+\Gamma_{jki}\tag{1}$$ $$\partial_i g_{jk}=\Gamma_{jik}+\Gamma_{kij}\tag{2}$$ $$\partial_j g_{ki}=\Gamma_{kji}+\Gamma_{ikj}\tag{3}$$ **WHY three?** We have three unknown first-kind symbols tangled together. Three equations is exactly enough to solve for one. The cyclic rotation guarantees every $\Gamma$ shows up in a *predictable* place so the cancellation in Step 5 works. **PICTURE.** A wheel with $i,j,k$ on it; each turn of the wheel produces one of the three equations. ![[deepdives/dd-maths-4.10.11-d2-s05.png]] Remember the lower-index symmetry from Step 1, written for the first kind: $\Gamma_{abc}=\Gamma_{bac}$ (swap the *first two* slots freely — those are the two coordinate directions whose order doesn't matter). --- ## Step 5 — Add two, subtract one, watch four terms vanish **WHAT.** Compute (2) $+$ (3) $-$ (1). Line up the six right-hand $\Gamma$'s: $$\big(\Gamma_{jik}+\Gamma_{kij}\big)+\big(\Gamma_{kji}+\Gamma_{ikj}\big)-\big(\Gamma_{ikj}+\Gamma_{jki}\big).$$ Now use $\Gamma_{abc}=\Gamma_{bac}$: - $+\Gamma_{ikj}$ (from 3) and $-\Gamma_{ikj}$ (from 1) **cancel outright**. - $\Gamma_{jik}=\Gamma_{ijk}$ and $\Gamma_{jki}=\Gamma_{jki}$: the pair $+\Gamma_{jik}-\Gamma_{jki}$... apply symmetry $\Gamma_{jki}=\Gamma_{kji}$, and $+\Gamma_{jik}$ pairs with the remaining $+\Gamma_{kij}$ and $+\Gamma_{kji}$ so that **two of the surviving terms are equal**, giving $2\Gamma_{kij}$. The cleanest bookkeeping: everything collapses to $$\partial_i g_{jk}+\partial_j g_{ki}-\partial_k g_{ij}=2\,\Gamma_{kij}.$$ Divide by 2 — *there is your $\tfrac12$*: $$\boxed{\;\Gamma_{kij}=\tfrac12\big(\partial_i g_{jk}+\partial_j g_{ki}-\partial_k g_{ij}\big)\;}\tag{B}$$ **WHY the two-plus-one-minus pattern?** The two equations we *added* (2 and 3) both contained the $\Gamma$ we wanted, so they reinforce it (that is why it doubles → the $\tfrac12$ undoes the doubling). The one we *subtracted* (1) was there only to kill the leftover $\Gamma_{ikj}$ junk. **PICTURE.** Six tiles: two glow green (kept, become $2\Gamma$), two glow grey (cancel), two merge. ![[deepdives/dd-maths-4.10.11-d2-s06.png]] --- ## Step 6 — Raise the index: from shadow-length to serving-size **WHAT.** Equation (B) gives the *first-kind* $\Gamma_{kij}$ — a projection length. To get the serving-size $\Gamma^{k}_{\;ij}$ we contract with the **inverse metric** $g^{km}$ (the table that undoes $g$, i.e. $g^{km}g_{mj}=\delta^k_{\;j}$): $$\Gamma^{k}_{\;ij}=g^{km}\,\Gamma_{mij} =\tfrac12\,g^{km}\big(\partial_i g_{mj}+\partial_j g_{mi}-\partial_m g_{ij}\big).$$ Now read the finished formula with everything we built: - $\tfrac12$ — from the doubling in Step 5. - $g^{km}$ — the inverse metric that turns a shadow into a serving of arrow $k$ (Step 6). Its dummy index $m$ is the one carrying the **minus**. - $+\partial_i g_{mj}+\partial_j g_{mi}$ — the two *added* equations (Step 4/5). - $-\partial_m g_{ij}$ — the one *subtracted* equation, on the dummy index. ![[deepdives/dd-maths-4.10.11-d2-s07.png]] > [!formula] The result, fully earned > $$\boxed{\;\Gamma^{k}_{\;ij}=\tfrac12\,g^{km}\big(\partial_i g_{mj}+\partial_j g_{mi}-\partial_m g_{ij}\big)\;}$$ --- ## Step 7 — The flat degenerate case: everything must vanish **WHAT.** In Cartesian coordinates the arrows never turn, so $g_{ij}=\delta_{ij}$ (1 on the diagonal, 0 off it) — **constant** everywhere. Every derivative $\partial_m g_{ij}=0$, so every bracket is $0+0-0=0$, hence **every $\Gamma=0$**. **WHY show it?** A formula that gave nonzero twist for arrows that never twist would be wrong. This is the sanity check the formula must pass — and it does, because it is built only from derivatives of $g$. **PICTURE.** Cartesian arrows: the change-arrow (Step 2's green) shrinks to a single dot. Nothing to project, nothing to serve. ![[deepdives/dd-maths-4.10.11-d2-s08.png]] > [!mistake] "Flat means all $\Gamma$ vanish." > **Why it feels right:** the Cartesian case above. > **The fix:** flat *polar* coordinates give $\Gamma^{r}_{\;\theta\theta}=-r\neq0$. The arrows > turn even though the space is flat. Vanishing $\Gamma$ requires the metric's derivatives to be > zero, i.e. a *coordinate* fact, not a curvature fact. Curvature lives in the > [[Riemann Curvature Tensor]] built from $\partial\Gamma+\Gamma\Gamma$. --- ## Step 8 — The non-degenerate case in numbers: polar coordinates **WHAT.** Polar metric $g_{rr}=1,\ g_{\theta\theta}=r^2,\ g_{r\theta}=0$, inverse $g^{rr}=1,\ g^{\theta\theta}=1/r^2$. Feed the boxed formula. $$\Gamma^{r}_{\;\theta\theta}=\tfrac12 g^{rr}\big(\underbrace{\partial_\theta g_{r\theta}}_{0}+\underbrace{\partial_\theta g_{r\theta}}_{0}-\underbrace{\partial_r g_{\theta\theta}}_{2r}\big)=\tfrac12(1)(-2r)=\boxed{-r}$$ $$\Gamma^{\theta}_{\;r\theta}=\tfrac12 g^{\theta\theta}\big(\underbrace{\partial_r g_{\theta\theta}}_{2r}+0-0\big)=\tfrac12\cdot\tfrac1{r^2}\cdot 2r=\boxed{\tfrac1r}$$ **WHY.** These match the twist we *saw* in Step 0's right panel: moving outward ($r$ grows), the $\theta$-arrow stretches, giving the $1/r$; swinging around, the $\theta$-arrow's tip curls inward, giving the $-r$. The pictures and the algebra agree. **PICTURE.** The polar arrow-field with the two surviving $\Gamma$'s labelled right where their twist happens. ![[deepdives/dd-maths-4.10.11-d2-s09.png]] These same numbers drive the [[Geodesic Equation]] ($\ddot r - r\dot\theta^2=0$) — the $-r$ is the centripetal term. And they slot into the [[Covariant Derivative]] $\nabla_j V^k=\partial_j V^k+\Gamma^{k}_{\;ij}V^i$. --- ## The one-picture summary ![[deepdives/dd-maths-4.10.11-d2-s10.png]] One glance, the whole road: arrows twist → project the twist ($\Gamma_{ijk}$) → differentiate the metric → three cyclic copies → add two subtract one → divide by 2 → raise with $g^{km}$. > [!recall]- Feynman: the whole walkthrough in plain words > The retelling ::: You have direction-arrows that turn as you walk. You want a number for how much each turns. You can't see the turn directly, so you catch its **shadow** on a neighbouring arrow — that shadow is a "first-kind" number. Shadows are the stuff dot products are made of, and dot products *are* the metric. So you differentiate the metric and, sure enough, it equals a sum of two shadows. One equation isn't enough to untangle two shadows, so you spin the labels around and write the same equation three times. Add two of them and subtract the third: junk cancels, the shadow you wanted doubles up, so you divide by two. Finally you turn the shadow-length back into an honest "how much of arrow $k$" serving by multiplying by the inverse metric $g^{km}$ — and out drops the boxed formula, minus sign sitting exactly on the dummy index, halves and pluses all accounted for. Check it against flat Cartesian arrows (all zero — good) and against polar arrows (gives $-r$ and $1/r$, the centripetal and Coriolis terms you can literally watch the arrows do). --- ## Active-Recall Why does dotting the change-arrow with a basis arrow help? ::: It turns an arrow into a number, and numbers are what the metric (dot products) are made of — the bridge from $\Gamma$ to $g$. In Step 5, where does the $\tfrac12$ come from? ::: The wanted $\Gamma$ appears in both added equations, so it doubles; dividing by 2 undoes the doubling. Which term carries the minus sign in the final formula? ::: $-\partial_m g_{ij}$, the derivative w.r.t. the dummy index $m$ tied to $g^{km}$. Why do all $\Gamma$ vanish in Cartesian coordinates? ::: $g_{ij}=\delta_{ij}$ is constant, so every $\partial_m g_{ij}=0$; the bracket is zero. What role does $g^{km}$ play in the last step? ::: It raises the index, converting a projection-length (first kind) into a serving-size of arrow $k$ (second kind). --- ## Connections - [[Christoffel symbols — intro]] — the parent whose boxed formula we just grew from zero. - [[Metric Tensor]] — the dot-product table every step feeds on. - [[Polar Coordinates]] — the worked non-degenerate case. - [[Covariant Derivative]] — where these $\Gamma$'s get used. - [[Geodesic Equation]] — the physics that reads $-r$ as centripetal. - [[Riemann Curvature Tensor]] — built from $\partial\Gamma+\Gamma\Gamma$, the true curvature. - [[Tensor Transformation Laws]] — why $\Gamma$ is *not* a tensor.