Christoffel symbols — intro

WHY do we even need them?

In flat space with Cartesian coordinates $(x,y)$, the basis vectors $\hat{e}_x, \hat{e}_y$ point the same way everywhere. So the derivative of a vector field is just the derivative of its components.

But the moment you use polar coordinates $(r,\theta)$ — or live on a sphere — the basis vectors $\hat{e}_r, \hat{e}_\theta$ rotate as you move. Now if you differentiate a vector, you must account for two things:

1. the change in the components, AND
2. the change in the basis vectors.

WHAT exactly is a Christoffel symbol?

Key facts:

• The two lower indices ($i,j$) are symmetric in a coordinate basis: $\Gamma^{k}_{\;ij}=\Gamma^{k}_{\;ji}$.
• They are NOT tensor components (they don't transform like a tensor) — they encode the coordinate system, not just geometry.

HOW do we derive the formula from the metric?

We want $\Gamma$ purely from the metric $g_{ij} = \mathbf{e}_i\cdot\mathbf{e}_j$, because the metric is the one object we always know.

Step 1 — Differentiate the metric. $\partial_k g_{ij} = (\partial_k \mathbf{e}_i)\cdot \mathbf{e}_j + \mathbf{e}_i\cdot(\partial_k \mathbf{e}_j)$ Why this step? Product rule on $g_{ij}=\mathbf{e}_i\cdot\mathbf{e}_j$.

Define the first-kind symbol $\Gamma_{ijk} = (\partial_j \mathbf{e}_i)\cdot\mathbf{e}_k$. Then: $\partial_k g_{ij} = \Gamma_{ikj} + \Gamma_{jki}$

Step 2 — Write three cyclic permutations. $\partial_k g_{ij} = \Gamma_{ikj}+\Gamma_{jki}$ $\partial_i g_{jk} = \Gamma_{jik}+\Gamma_{kij}$ $\partial_j g_{ki} = \Gamma_{kji}+\Gamma_{ikj}$ Why this step? Three equations let us isolate one $\Gamma$ using the symmetry $\Gamma_{ab c}=\Gamma_{bac}$ in the first two slots... actually symmetry in lower coordinate indices: $\partial_j\mathbf e_i=\partial_i\mathbf e_j$, so $\Gamma_{ikj}=\Gamma_{kij}$ etc.

Step 3 — Add first two, subtract third. After cancellation (using symmetry): $\Gamma_{kij} = \tfrac12\left(\partial_i g_{jk} + \partial_j g_{ki} - \partial_k g_{ij}\right)$

Step 4 — Raise the index with the inverse metric $g^{km}$:

Worked Example 1 — Polar coordinates (the classic)

Metric in 2D polar coordinates: $ds^2 = dr^2 + r^2 d\theta^2$, so $g_{rr}=1,\quad g_{\theta\theta}=r^2,\quad g_{r\theta}=0,\qquad g^{rr}=1,\ g^{\theta\theta}=\tfrac{1}{r^2}.$

Find $\Gamma^{r}_{\;\theta\theta}$. $\Gamma^{r}_{\;\theta\theta}=\tfrac12 g^{rr}\big(\partial_\theta g_{r\theta}+\partial_\theta g_{r\theta}-\partial_r g_{\theta\theta}\big)$ Why? Set $k=r,\,i=j=\theta$ in the boxed formula; only $g^{rr}$ survives the inverse metric (diagonal). $=\tfrac12 (1)\big(0+0-\partial_r r^2\big)=\tfrac12(-2r)=\boxed{-r}$

Find $\Gamma^{\theta}_{\;r\theta}$. $\Gamma^{\theta}_{\;r\theta}=\tfrac12 g^{\theta\theta}\big(\partial_r g_{\theta\theta}+\partial_\theta g_{\theta r}-\partial_\theta g_{r\theta}\big)=\tfrac12\cdot\tfrac{1}{r^2}\cdot 2r=\boxed{\tfrac1r}$ Why? Only $\partial_r g_{\theta\theta}=2r$ is nonzero. All other Christoffel symbols vanish.

Worked Example 2 — Verify a flat case gives zero

In Cartesian coordinates $g_{ij}=\delta_{ij}$ (constant). Every $\partial_m g_{ij}=0$, so every $\Gamma=0$. Why this matters: Christoffel symbols can be nonzero even in flat space (polar!), so $\Gamma\neq0$ does not mean "curved." Curvature lives in derivatives of $\Gamma$ (the Riemann tensor), not in $\Gamma$ itself.

The covariant derivative (where they're used)

Common Mistakes (Steel-manned)

Active-Recall Flashcards

Connections

• Metric Tensor — the source from which all $\Gamma$ are computed.
• Covariant Derivative — where Christoffel symbols are used.
• Riemann Curvature Tensor — built from $\partial\Gamma+\Gamma\Gamma$; true curvature.
• Geodesic Equation — $\ddot x^k+\Gamma^k_{\;ij}\dot x^i\dot x^j=0$.
• Polar Coordinates — the cleanest non-trivial example.
• Tensor Transformation Laws — why $\Gamma$ is not a tensor.

Concept Map

Hinglish (regional understanding)