Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top
  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.

share|cite|improve this question
up vote 1 down vote accepted

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.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.