Question #110779
1. Find the smooth curve (C1) realizing the shortest distance between two given points in a plan (usual Euclidean R2).
1
Expert's answer
2020-04-20T15:28:11-0400

d2xλdt2+Γμνλdxμdtdxνdt=0\frac{d^2x^{\lambda}}{dt^2}+\Gamma^{\lambda}_{\mu\nu}\frac{dx^{\mu}}{dt}\frac{dx^{\nu}}{dt}=0 is geodesic equation.

For R2\mathbb R^2 we have g11=g22=1g_{11}=g_{22}=1, g12=g21=0g_{12}=g_{21}=0. So since Γμνλ=12gλm(gmμxν+gmνxμgμνxm)\Gamma^{\lambda}_{\mu\nu}=\frac{1}{2}g^{\lambda m}\left(\frac{\partial g_{m\mu}}{\partial x^\nu}+\frac{\partial g_{m\nu}}{\partial x^\mu}-\frac{\partial g_{\mu\nu}}{\partial x^m}\right), we have Γμνλ=0\Gamma^{\lambda}_{\mu\nu}=0 for every (λ,μ,ν){1,2}3(\lambda,\mu,\nu)\in\{1,2\}^3.

Then geodesic equation for R2\mathbb R^2 is d2x1dt2=0,d2x2dt2=0\frac{d^2x^1}{dt^2}=0, \frac{d^2x^2}{dt^2}=0. We have x1=At+B,x2=Ct+Dx^1=At+B, x^2=Ct+D, that is ACt=C(x1B)=A(x2D)ACt=C(x^1-B)=A(x^2-D), so Ax2+Ex1+F=0Ax^2+Ex^1+F=0, where E=C,F=AD+BCE=-C, F=-AD+BC.

It is equation of straight line.

Answer: straight line between two given points.


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