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Lesson 21 of 30

Geodesics as critical points of energy

Prove the first-variation formula with fixed endpoints.

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A curve traversed over time interval [0,T] has energy E=(1/2)∫||γ′||²dt and length L=∫||γ′||dt. If its speed is constant, E=L²/(2T). If the same trace is traversed unevenly, length stays fixed but energy rises.

This makes energy a convenient variational functional for geodesics with a fixed time interval. It avoids the square root in the length integrand while selecting constant-speed parametrizations at critical points.

Predict

Predict whether a stationary energy curve must minimize globally. A long great-circle arc is stationary but not shortest. First variation detects criticality, not the sign of every second variation or the best path among distant competitors.

Predict what happens to boundary terms when endpoints are fixed. The variation field vanishes there. If endpoints are allowed to move, those terms encode additional conditions and cannot be discarded.

Experiment

Compare two speed profiles on [0,1] along the same straight segment of length two.

Speed profileLengthEnergy
Constant speed 222
Speed 1 for half the time, then 325/2

The piecewise profile illustrates the inequality; smooth approximations give the same strict-energy phenomenon. Compute the integral rather than comparing only average speed. The square makes uneven motion more costly.

Understand

Let Γ(t,ε) be a smooth variation, γ(t)=Γ(t,0), and V=∂εΓ at ε=0. Differentiating energy gives E′(0)=∫⟨D_ε∂tΓ,γ′⟩dt. The connection is torsion-free and the parameter vector fields commute, so D_ε∂tΓ=D_tV. Metric compatibility and integration by parts yield E′(0)=[⟨V,γ′⟩]₀ᵀ−∫⟨V,D_tγ′⟩dt. For fixed endpoints the bracket vanishes.

If γ is geodesic, the integral vanishes for every variation. Conversely, stationarity under all fixed-endpoint variations includes all smooth compactly supported tangent fields V along coordinate subintervals. The fundamental lemma forces D_tγ′=0. Thus stationary energy curves are precisely affinely parametrized geodesics under the stated smoothness assumptions.

Cauchy–Schwarz gives L²=(∫speed)²≤T∫speed²=2TE. Equality holds exactly for constant speed. This explains why constant-speed length minimizers are energy minimizers on a fixed interval, while a stationary point need not be a global minimum.

Apply

Repair “The Euler–Lagrange equation proves the shortest path.” It proves a necessary critical-point equation, and in this setting characterizes stationarity. Local minimization will follow in a normal neighborhood from Gauss’s lemma; global minimization depends on cut points and completeness.

A great-circle arc with length between πR and 2πR supplies a concrete counterexample to the global claim. It satisfies the geodesic equation, but the complementary shorter arc connects the same endpoints.

For optional depth, allowing endpoints on submanifolds leads to orthogonality boundary conditions from the retained bracket. Second variation introduces curvature and Jacobi fields. These are additional analyses, not automatic consequences of setting the first derivative to zero.

Continue

State the competitor class, fixed endpoint data and parameter interval when making a variational claim. Continue to the exponential map and Gauss lemma for an actual local minimizing theorem.

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Practice and reasoning

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A curve on [0,2] has constant speed 3. Find its energy (1/2)∫speed² dt.

Integers, fractions and finite decimals are accepted. Expressions such as sqrt(2) are not parsed by this checker.

A fixed-endpoint stationary energy curve is

Your response

For a smooth fixed-endpoint variation with variation field V, derive E′(0)=−∫⟨V,D_tγ′⟩dt.

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Show stationarity under all fixed-endpoint variations implies the geodesic equation.

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For a curve on [0,T], prove L²≤2TE and identify equality.

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Give a geodesic segment on a sphere that is not globally minimizing.

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Reflect and continue

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