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Lagrange multipliers for finding geodesics on a sphere

  1. Jan 29, 2013 #1

    Leb

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    1. The problem statement, all variables and given/known data
    Find the geodesics on a sphere [itex]g(x,y,z)=x^{2}+y^{2}+z^{2}-1=0[/itex]
    arclength element [itex]ds=\sqrt{dx^{2}+dy^{2}+dz^{2}}[/itex]




    2. Relevant equations

    [itex]f(x,y,z)=\sqrt{x'^{2}+y'^{2}+z'^{2}}[/itex] where [itex]x'^{2} \text{means} \frac{dx^{2}}{ds^{2}}[/itex] and not [itex]d^{2}x/ds^{2}[/itex]

    3. The attempt at a solution

    Using the fact that [itex]x^{2}+y^{2}+z^{2}=1[/itex] I get three equations if of the form
    [itex]d^{2}x/ds^{2}=2\lambda x[/itex] i.e. the double derivative of x w.r.t. s
    [itex]d^{2}y/ds^{2}=2\lambda y[/itex]
    [itex]d^{2}z/ds^{2}=2\lambda z[/itex]

    My lecturer now says, that we have to differentiate the constraint (g) twice w.r.t. s to get
    [itex](dx/ds)^{2}+(dy/ds)^{2}+(dz/ds)^{2} = -2 \lambda (x^{2}+y^{2}+z^{2})[/itex]
    since the LHS = 1 and the brackets in the RHS = 1 , the lecturer concludes, that [itex]\lambda = -0.5[/itex]. Now I am most confused here, as I do not see where the double differentiation happened. Nor do I see how this helps to determine that xA+By+z=0 (A,B = const.) which defines the great circle.

    Thanks.
     
  2. jcsd
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