Answer to Question #110779 in Differential Geometry | Topology for brighton

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

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

For "\\mathbb R^2" we have "g_{11}=g_{22}=1", "g_{12}=g_{21}=0". So since "\\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 "\\Gamma^{\\lambda}_{\\mu\\nu}=0" for every "(\\lambda,\\mu,\\nu)\\in\\{1,2\\}^3".

Then geodesic equation for "\\mathbb R^2" is "\\frac{d^2x^1}{dt^2}=0, \\frac{d^2x^2}{dt^2}=0". We have "x^1=At+B, x^2=Ct+D", that is "ACt=C(x^1-B)=A(x^2-D)", so "Ax^2+Ex^1+F=0", where "E=-C, F=-AD+BC".

It is equation of straight line.

Answer: straight line between two given points.


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