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Distance on a 2-sphere

  1. Jan 13, 2005 #1
    Okay, say you're given two points on a sphere, using the metric tensor how do you find the distance between the two points? (along the geodesic connecting them)

    By the way, I know how to do it with just plain old vectors in R^3, but I'd like try doing it with the metric tensor.

    Thanks,

    Kevin
     
  2. jcsd
  3. Jan 13, 2005 #2

    pervect

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    Well, if you have the equation of the geodesic, it's fairly easy. Let's suppose the coordinates are x1 and x2, probably lattitude and longitude. Suppose our geodesic is a curve that has coordinates [tex]\mbox{x^1(\lambda)}[/tex] , [tex]\mbox{x^2(\lambda)}[/tex] i.e
    these two functions are the paramaterized geodesic. You can think of [tex]\mbox{\lambda}[/tex] as being a time parameter if you envision an object actually moving along the geodesic. Then to get the distance, you just have to integrate


    [tex]
    \int_{lambda} d \lambda \sqrt{\Sigma_{ij} (g_{ij} {(\frac{d x^i}{d \lambda})(\frac{d x^j}{d \lambda}) )}
    [/tex]


    where [tex]\mbox{\lambda}[/tex] varies from the starting point of the geodesic to the ending point

    Expanding out the sum over i,j in the square root for expositional purposes, we write
    [tex]
    \int_{lambda} d \lambda \sqrt{g_{11} (\frac{dx^1}{d\lambda})^2 + 2 g_{12} \frac{dx^1}{d\lambda} \frac{dx^2} {d\lambda} + g_{22} (\frac{dx^2}{d\lambda})^2}
    [/tex]

    If you don't have the equation for the geodesic, you have to solve the differential equations for geodesic motion

    [tex]
    \frac{d^2 x^u}{d \lambda^2} + \Gamma^u{}_{ab}(\frac{dx^a}{d \lambda})(\frac{dx^b}{d \lambda})
    [/tex]
     
  4. Jan 14, 2005 #3
    well the geodesic is going to be part of a great circle so I should probably just try to parameterize it...Thanks, I'll try it out.
     
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