# Solving constrained Euler-Lagrange equations with Lagrange Multipliers (Geodesics)

I'm trying to solve a calculus of variations geodesics problem using Lagrange Multipliers, showing that the geodesics of a sphere are the so-called great circles. I am using a constrained Lagrangian

$$\int_{a}^{b}\dot{x}^2+\dot{y}^2+\dot{z}^2+\lambda(t)G(x(t),y(t),z(t))dt$$

where G(x,y,z) is the sphere $$x^2+y^2+z^2 = 1$$

Computing the Euler-Lagrange equations, I get the three equations:

$$\ddot{\vec{r}}=\frac{\lambda(t)}{2}\nabla G$$

and the constraint remains: $$\sqrt{x^2+y^2+z^2}=1\ \text{ or equally }\ x^2+y^2+z^2=1$$

Applied to the circle, we find:

$$\ddot{\mathbf{r}}=\lambda(t)\mathbf{r}$$

So now we have four equations, an unknown lambda function, and 3 variables. How do I determine lambda and simplify it enough to solve in Mathematica? This is very unfamiliar material for me, so help is appreciated.

Thanks  $\textbf{Edit: expanding on the solution provided by Qmechanic}$  Differentiating the constraint twice yields the familiar

$$\mathbf{\ddot{r}}\cdot\hat{\mathbf{r}}=-\frac{v^2}{r}$$

And taking the dot product of the EL equations with $\mathbf{\hat{r}}$ and substituting yields

$$-\frac{v^2}{r^2}=\lambda(t)$$

Substituting this back into the original equation for $\lambda(t)$, we find that

$$\ddot{\mathbf{r}}=-\frac{v^2}{r}\mathbf{\hat{r}}$$

This allows an infinite number of geodesics, but choosing the parameterization that leaves v constant, this simplifies to

$$\mathbf{\ddot{r}}=-\mathbf{r}$$

which can be solved for the Cartesian coordinates:

$$\begin{matrix} x=C_1 \sin (t)+C_2 \cos (t) \\ y=C_2 \sin (t)+C_4 \cos (t) \\ z=C_3 \sin (t)+C_6 \cos (t) \\ \end{matrix}$$

Which, with appropriate initial conditions, forms a great circle between the two points.

• $\int_{a}^{b} \sqrt {\dot{x}^2+\dot{y}^2+\dot{z}^2}+\lambda(t)G(x(t),y(t),z(t))dt$ gives the same result as above, right ? Commented Apr 29, 2016 at 5:21
• Yes. $\int |\vec{r}|dt$ and $\int |\vec{r}|^2dt$ should give the same result. I felt the former would be easier. Commented Apr 29, 2016 at 15:03
• You mean the latter Commented Apr 29, 2016 at 18:31

Hints:

1. Deduce from the equations of motion (i.e. the EL equations) that the specific angular momentum ${\bf L}:={\bf r}\times\dot{\bf r}$ is a constant of motion.

2. Deduce that position ${\bf r}$ and velocity $\dot{\bf r}$ are both perpendicular to ${\bf L}$.

3. Deduce that the orbit lies in a plane that goes through the origin.

4. The intersection of the orbit plane with the sphere is hence a great circle.

• Excellent. Thanks for the help! Commented Apr 29, 2016 at 15:04