# Lagrange multipliers for finding geodesics on a sphere

1. The problem statement, all variables and given/known data

Find the geodesics on a sphere $g(x,y,z)=x^{2}+y^{2}+z^{2}−1=0$ arc-length element $ds=\sqrt{dx^{2}+dy^{2}+dz^{2}}$

1. Relevant equations

$f(x,y,z)=\sqrt{x′^{2}+y′^{2}+z′^{2}}$where $x′^{2}$ means $dx^{2}/ds^{2}$ and not $d^{2}x/ds^{2}$

1. The attempt at a solution

Using the fact that $x^{2}+y^{2}+z^{2}=1$ I get three equations if of the form $d^{2}x/ds^{2}=2λx$ i.e. the double derivative of $x$ w.r.t. $s$ $d^{2}y/ds^{2}=2λy$ $d^{2}z/ds^{2}=2λz$

My lecturer now says, that we have to differentiate the constraint $g$ twice w.r.t. $s$ to get $(dx/ds)^{2}+(dy/ds)^{2}+(dz/ds)^{2}=−2λ(x2+y2+z2)$ since the LHS = 1 and the brackets in the RHS = 1 , the lecturer concludes, that $\lambda =−0.5$. Now I am most confused here, as I do not see where and why 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.

I was trying to compare this to Geodesics of a Sphere in Cartesian Coordinates but am too daft to see a pattern...

Any help appreciated.

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Differentiate g twice is just to calculate the Lagrangian multiplier $\lambda$, it has nothing to do with the great circle.
The great circle was implied in the equation $d^2x/ds^2=2\lambda x$, which is the Euler-Lagrangian equation for calculating the maximal or minimal length.