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Calculus of Variation - Shortest path on the surface of a sphere

  1. Nov 12, 2013 #1
    Refer to "2.jpg", it said that the shortest path on the surface of a sphere is Ay-Bx=z , which is a plane passing through the center of the sphere. I cannot really understand about this. Does it mean that the shortest path is a ring that connects two points with its center at the center of the sphere?
     

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    Last edited: Nov 12, 2013
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  3. Nov 12, 2013 #2

    PAllen

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    It actually says the geodesic is the intersection of a plane through the center with the spherical surface. This is a great circle, as it says.
     
  4. Nov 12, 2013 #3
    Am I right to say that the intersection of a plane through the center of a sphere is a ring(circle), and the short arc of this ring(circle) which pass through the two points is the shortest path?
     
  5. Nov 12, 2013 #4

    PAllen

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    The intersection of a plane through the center of a sphere with its surface, embedded in a higher dimensional space (as the reference you cite presents it), is a ring - a great circle. So yes, you are essentially right. Also, you are correct that the shorter arc is the shortest possible distance.
     
  6. Nov 12, 2013 #5
    Thanks PAllen.
     
  7. Nov 12, 2013 #6
    I have another question, why the solution comes out is an equation of a plane and not an equation of a ring(circle)?
     
  8. Nov 12, 2013 #7

    PAllen

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    They first get the great circle solution purely in terms of two angular coordinates on the sphere's surface. Then they derive that it is the intersection of plane through the center with the sphere.
     
  9. Nov 12, 2013 #8
    Okie, I get it, many thanks for explaining.
     
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