1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Lagrange multipliers for finding geodesics on a sphere

  1. Jan 29, 2013 #1


    User Avatar

    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.

  2. jcsd
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Can you offer guidance or do you also need help?
Draft saved Draft deleted

Similar Discussions: Lagrange multipliers for finding geodesics on a sphere