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advanced28 min studyLesson 804 in Physics

Two geometric conditions determine one connection

Derive the Levi-Civita coefficients and use flat polar coordinates to separate connection values from curvature.

Before you begin
On this page

The question for today

What you’ll understand

  • Explain why ordinary component derivatives require a connection correction.
  • Derive the unique torsion-free metric-compatible connection.
  • Check metric compatibility and component dimensions in polar coordinates.

A vector’s components can vary because the geometric vector varies or because its coordinate basis varies. A connection defines a covariant derivative that handles both effects. A general manifold admits many connections, but a nondegenerate metric selects a unique one once torsion is required to vanish. We will derive that Levi-Civita connection and verify it in a flat coordinate system where the coefficients are nonzero despite zero curvature.

Covariant differentiation compensates changing bases

A partial derivative of vector components does not transform as a tensor under a nonlinear coordinate change. Differentiating the Jacobian introduces an extra term. Connection coefficients supply precisely the compensating transformation. For each upper tensor index the covariant derivative adds a connection term, while each lower index receives a minus term. A connection itself is not a tensor, although the difference between two connections is one because their inhomogeneous pieces cancel. Its coefficients compare nearby tangent spaces according to a specified transport rule.

Metric compatibility means covariant differentiation preserves the metric, so parallel transport preserves inner products. Vanishing torsion means the connection is symmetric in its two lower indices in a coordinate basis. These are independent assumptions for a general affine connection. In conventional metric general relativity both are imposed. In noncoordinate orthonormal frames, the relation between torsion and coefficient antisymmetry also contains the frame commutator; one must not copy the coordinate-basis symmetry rule into every tetrad calculation.

∇μVρ=∂μVρ+ΓρμνVν,∇ρgμν=0,Γρμν=Γρνμ\nabla_\mu V^\rho=\partial_\mu V^\rho+\Gamma^\rho{}_{\mu\nu}V^\nu,\qquad \nabla_\rho g_{\mu\nu}=0,\qquad \Gamma^\rho{}_{\mu\nu}=\Gamma^\rho{}_{\nu\mu}
The last equality expresses zero torsion in a coordinate basis. The metric has Lorentzian signature −+++ in spacetime, although the positive-metric polar example uses the same construction. A Γ component has units [xρ]/([xμ][xν]), so mixed angular coordinates produce different component dimensions.↔ Scroll wide equations horizontally.

Permuting metric compatibility solves for the coefficients

Expand metric compatibility for three cyclic permutations of the metric indices. Add two equations and subtract the third. The lower-index symmetry lets most connection terms cancel, leaving twice a metric-lowered connection coefficient. Multiplying by the inverse metric gives the Christoffel formula. Nondegeneracy of the metric makes that last step possible. Substitution verifies both required properties, establishing existence as well as uniqueness. Neither flatness nor Einstein’s field equation is an assumption of this construction.

For a plane in polar coordinates the metric is diag(1,r²). Its only varying component has radial derivative 2r, giving Γʳφφ=−r and Γφrφ=Γφφr=1/r. These values describe the turning and changing scale of the polar coordinate basis. They do not represent a gravitational field in an invariant sense. Cartesian coordinates make the same connection coefficients vanish everywhere on a flat patch; transforming to polar coordinates reintroduces them without changing the geometry. At r=0 this polar chart itself is invalid.

Γρμν=12gρσ(∂μgσν+∂νgσμ−∂σgμν),Γrφφ=−r,Γφrφ=r−1\Gamma^\rho{}_{\mu\nu}=\frac12g^{\rho\sigma}\left(\partial_\mu g_{\sigma\nu}+\partial_\nu g_{\sigma\mu}-\partial_\sigma g_{\mu\nu}\right),\qquad \Gamma^r{}_{\varphi\varphi}=-r,\quad \Gamma^\varphi{}_{r\varphi}=r^{-1}
The final two expressions use dℓ²=dr²+r²dφ² with r in metres and φ dimensionless on r>0. Thus Γrφφ is in metres and Γφrφ in m⁻¹. The inverse metric is a tensor inverse, not a component-by-component reciprocal when off-diagonal entries are present.↔ Scroll wide equations horizontally.

Zero coefficients at one event do not erase curvature

Normal coordinates can set the Levi-Civita coefficients to zero at one event, together with first metric derivatives there. Their derivatives need not vanish, and curvature generally survives. Over an open neighborhood, vanishing curvature is the relevant local condition for making the metric constant in suitable coordinates; global topology can still obstruct one global inertial chart. A divergence of selected coefficients is likewise inconclusive until the chart domain and curvature or extension are checked. The geometric connection and its coordinate representation must remain conceptually separate.

The function lab exposes the derivative used in the polar example. Reset and set both controls to a=1,x=1, predicting y=1 and slope two. Reset and set a=1,x=2, predicting y=4 and slope four. Interpreting y as a normalized gφφ=r² identifies the derivative needed in the Christoffel formula. The lab does not evaluate a metric inverse or a connection, so Γʳφφ=−r must still be derived with the correct index positions and factor one half. Its parabola is not a curvature plot.

Work it through

Verify metric compatibility at one polar radius

In the flat polar plane, take r=2 m. Compute Γʳφφ and Γφrφ, then check ∇r gφφ explicitly with gφφ=r².

  1. The Christoffel formula gives Γʳφφ=−r=−2 m. The coefficient with an upper angular index is Γφrφ=1/r=0.50 m⁻¹.

  2. The partial derivative is ∂r gφφ=2r=4 m. It is nonzero even though the plane is flat.

  3. Both lower φ indices contribute a minus connection term: ∇r gφφ=4−2×(0.50)×4=0 m. This verifies compatibility rather than incorrectly requiring every partial metric derivative to vanish.

The result

The two coefficients are −2 m and 0.50 m⁻¹, and their two index corrections cancel the 4 m partial derivative exactly.

The living notebook

Explore the model

Interactive lab

Interactive lab · function

Inspect the varying angular metric coefficient

Reset and set BOTH controls to a=1,x=1, predicting y=1,slope=2. Reset and set a=1,x=2, predicting y=4,slope=4. Assign y to normalized r². This supplies derivative data only, with no metric inverse, connection, or curvature solver.

Inspect the varying angular metric coefficientThis lab uses only y = ax². Playback traces the curve within the fixed integration interval from 0 to x; it does not estimate numerical error or change the chosen point. Current readings: y at chosen x: 0.608; Tangent slope · 2ax: 1.35; Signed area · ax³/3: 0.182. All parameters are adjustable with the labeled controls below.−2.4−1.201.22.4−9−4.504.59xy(x, ax²)
y = ax²TangentSigned integral
y at chosen x
0.608
Tangent slope · 2ax
1.35
Signed area · ax³/3
0.182
t = 0 s

This lab uses only y = ax². Playback traces the curve within the fixed integration interval from 0 to x; it does not estimate numerical error or change the chosen point. Parameter changes pause playback and recompute the model at the current elapsed time. Reset restores both parameters and returns time to zero. Playback starts paused. Adjust either parameter, then choose Play or Step.

Look forA varying metric component can produce nonzero Christoffel coefficients in flat coordinates.

Pause & reason

Check your understanding

Which pair of conditions uniquely selects the Levi-Civita connection of a nondegenerate metric?

Your turn

Try it yourself

On a one-dimensional spatial patch take dℓ²=exp(2x/L)dx² with x,L in metres and L=2 m. Find Γˣxx at x=0. Show how y=Lexp(x/L), with y>0, makes the metric constant and explain the curvature.

Take a moment to make a prediction or work it out on paper. Open a hint when you need one.

Give me a hint

Use Γ=(1/2)g⁻¹∂xg. Differentiate y to transform the line element.

Compare with the solution

The coefficient is Γˣxx=1/L=0.50 m⁻¹, independent of x. Since dy=exp(x/L)dx, the metric becomes dℓ²=dy² and its y-coordinate coefficient vanishes. A one-dimensional metric has identically zero intrinsic Riemann curvature; the nonzero x-coordinate connection encoded the nonlinear labeling on this patch, not a curved one-dimensional geometry.

Keep the idea

Take it with you

  • Connections repair component derivatives but are not tensors themselves.
  • Metric compatibility and zero torsion yield the Christoffel formula.
  • Nonzero or divergent coefficients require chart and curvature checks before a geometric interpretation.

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