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Answers for “如何从欧拉-拉格朗日方程得到测地线方程?”

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The geodesic equation is obtained by applying the Euler-Lagrange equation to the Lagrangian L=gijx˙ix˙j\mathscr{L} = g_{ij}\dot{x}^i\dot{x}^j. The resulting differential equation for the curve components is d2xkdλ2+Γijkdxidλdxjdλ=0\frac{d^2x^k}{d\lambda^2} + \Gamma^k_{ij}\frac{dx^i}{d\lambda}\frac{dx^j}{d\lambda} = 0, where Γijk\Gamma^k_{ij} are the connection coefficients.

Conditions: λ\lambda is an affine parameter.; Γijk\Gamma^k_{ij} are connection coefficients.; The equation is presented as the general solution of the Euler-Lagrange step for F.

Understand why

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Minimizing the length functional LL is described as tricky. When λ\lambda is an affine parameter, minimizing the simpler functional F=∫ABgijx˙ix˙j dλF = \int_A^B g_{ij}\dot{x}^i\dot{x}^j\,d\lambda is equivalent to minimizing LL and is much easier.

Conditions: λ\lambda must be an affine parameter.; FF is used as an easier surrogate for LL under that condition.

Meet the concept

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An affine parameter is the special choice of the curve parameter λ\lambda for which the scalar proportionality function f(λ)f(\lambda) in the tangent evolution equation vanishes identically. Specifically, the tangent vector T~\tilde{T} satisfies dT~dλ=f(λ)T~\frac{d\tilde{T}}{d\lambda} = f(\lambda)\tilde{T}, and the parameter is affine if f(λ)=0f(\lambda)=0 for all λ\lambda.

Conditions: T~\tilde{T} is the tangent to the geodesic.; f(λ)f(\lambda) is a scalar function of the parameter.; The affine case is the subcase f(λ)=0f(\lambda)=0 for all λ\lambda.

Meet the concept

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The Lagrangian L\mathscr{L} used to minimize the integral FF via the Euler-Lagrange equation is L=gijx˙ix˙j\mathscr{L} = g_{ij}\dot{x}^i\dot{x}^j. This is the integrand of the functional FF.

Conditions: Applied to the integral FF.; x˙i\dot{x}^i and x˙j\dot{x}^j are derivatives with respect to λ\lambda.; gijg_{ij} are metric components.

Find a method

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The length functional LL between endpoints AA and BB is defined as the integral of the square root of the squared norm of the tangent vector with respect to the parameter λ\lambda: L=∫ABgijx˙ix˙j dλL = \int_A^B \sqrt{g_{ij}\dot{x}^i\dot{x}^j}\,d\lambda. Minimizing this functional yields the geodesic curves.

Conditions: Endpoints are AA and BB.; x˙i\dot{x}^i and x˙j\dot{x}^j denote derivatives with respect to λ\lambda.; gijg_{ij} are metric components.