Foundations — Christoffel symbols — intro

This page assumes you have seen none of the notation in the parent note. We build every symbol from the ground up, in the order that each one needs the previous.

0. What a "vector on a surface" even means

Before any Greek letters, look at the picture below. An arrow lives at a point. To describe it with numbers, we need reference arrows at that point — a little "grid" telling us "this way is direction 1, that way is direction 2." The vector is then a recipe: "so much of direction 1, so much of direction 2."

In flat Cartesian coordinates the two reference arrows are ${\mathbf{e}}_x$ (points East, always) and ${\mathbf{e}}_y$ (points North, always). In Polar Coordinates they are ${\mathbf{e}}_r$ (points outward from the origin) and ${\mathbf{e}}_{\theta }$ (points "counterclockwise around"). The polar ones change direction from point to point — that is the whole story.

1. Coordinates and the index notation $x^i$

2. The partial derivative ${\partial }_j$

Applied to any labelled quantity $Q$, the symbol ${\partial }_{j}Q$ just means "the rate of change of $Q$ as you step along direction $x^j$." We will apply it to the metric $g_{ij}$ once that is defined in Section 4.

3. The dot product $\mathbf{a}\cdot \mathbf{b}$ and lengths

A special case: $\mathbf{a}\cdot \mathbf{a}$ is the arrow's length squared. That is why the dot product is the machine that measures distance.

4. The metric tensor $g_{ij}$

Let us compute the polar metric directly here, straight from the arrows, so nothing is borrowed. The reference arrow ${\mathbf{e}}_r$ has length $1$, so ${\mathbf{e}}_r\cdot {\mathbf{e}}_r=1$. A one-radian nudge in angle moves you a real distance $r$ (arc length = radius × angle), so the arrow ${\mathbf{e}}_{\theta }$ has length $r$, giving ${\mathbf{e}}_{\theta }\cdot {\mathbf{e}}_{\theta }=r^2$. The two arrows are perpendicular, so their dot product is $0$. Hence $g_{rr}=1,g_{\theta \theta }=r^2,g_{r\theta }=g_{\theta r}=0.$

Now that $g_{ij}$ exists, the object ${\partial }_r{g}_{\theta \theta }={\partial }_r(r^2)=2r$ from Section 2 has a concrete meaning: "how the length-squared of the angular arrow grows as you move outward."

5. The inverse metric $g^{km}$ (upper indices)

6. Upper vs lower indices, and "raising/lowering"

7. The Einstein summation convention (the invisible $\sum$)

8. Basis vectors that CHANGE: ${\partial }_j{\mathbf{e}}_i$

This is the object everything is really about. In Cartesian coordinates ${\mathbf{e}}_x,{\mathbf{e}}_y$ never move, so ${\partial }_j{\mathbf{e}}_i=0$. In polar coordinates they rotate, so ${\partial }_j{\mathbf{e}}_i\ne 0$. The figure shows the twist.

Why the two lower indices are symmetric

9. The formula that builds $\Gamma$ from the metric

We now have every symbol needed to state — and see why — the central formula. The point of the whole topic is that $\Gamma$ can be computed without ever knowing the arrows, using only the metric table $g_{ij}$ and its derivatives.

Why it is even possible. Start from $g_{ij}={\mathbf{e}}_i\cdot {\mathbf{e}}_j$ and differentiate with the product rule: ${\partial }_k{g}_{ij}=({\partial }_k{\mathbf{e}}_i)\cdot {\mathbf{e}}_j+{\mathbf{e}}_i\cdot ({\partial }_k{\mathbf{e}}_j).$ Each $\partial \mathbf{e}$ is a Christoffel combination, so derivatives of the metric carry the same information as the $\Gamma$'s. Writing this in three cyclic versions (rotating $i,j,k$), then adding two and subtracting the third, cancels the unwanted pieces using the symmetry from Section 8 and isolates one $\Gamma$. Raising the leftover index with $g^{km}$ gives:

10. Vector components $V^k$ and the covariant derivative ${\nabla }_j$

Prerequisite map

Read top-to-bottom: reference arrows plus the dot product give the metric; the derivative of the changing arrows plus the inverse metric and summation give the Christoffel symbol; that feeds the covariant derivative — and later the Geodesic Equation and Riemann Curvature Tensor.

Equipment checklist