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What exactly are great circles, and how does one derive them? Given that the Lagrangian is: $$ L =\frac {1}{2}(\dot\theta^2 + \sin^2\theta\dot\phi^2)$$ it was written that the great circles were $$\theta = \tau$$ and $$\phi = \phi_0$$

I have no idea how this comes about. Is there some decent procedure to calculate the great circles? I don't want the answer nor do I want you to try to solve it, I just want to be guided in a way to understand how to get them alone.

  • great circle = meridian (with respect any, arbitrarily fixed, North pole) – Valter Moretti Sep 27 '14 at 13:49
  • Related question by OP: http://physics.stackexchange.com/q/137422/2451 – Qmechanic Sep 27 '14 at 14:00
  • Excuse me @ValterMoretti, does it mean that θ=τ and ϕ=ϕo are correct great circles given that lagrangian? – PhilosophicalPhysics Sep 27 '14 at 14:03
  • Actually I do not think to understand the question well. However, the motion of a particle with that Lagrangian is always along great circles (geodesics) of the spherical surface. What you can prove here is that the indicated motion, in fact along a meridian, solves EL equations with suitable initial conditions (e.g. initial point at the north pole and initial velocity tangent at the north pole). Since the Lagrangian is rotationally invariant, all motions are of this type via a suitable (obvious) re-definition of the spherical coordinates.. – Valter Moretti Sep 27 '14 at 14:16

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As Wikipedia says, a great circle is a circle formed by the intersection of a sphere and a plane that passes through the center of the sphere.

The great circles parametrized by $\theta =\tau$ and $\phi =\text{const}$ are not all the great circles. They are only the `vertical' great circles, that is, the great circles formed by intersection with a plane that contains the $z$-axis. (You will find that the general solution to the Euler-Lagrange equations of your Lagrangian is $\theta (t)=at+b$, $\phi =\text{const}$, for some constants $a$ and $b$. We an then just define $\tau :=at+b$, so that $\theta =\tau$. This is nothing more than a simple reparametrization.) If you need help visualizing these curves, I suggest you work on your understanding of spherical coordinates.

Now I will sketch how one actually shows that these great circles are the equations of motion. You should find that the Euler-Lagrange equations are \begin{align} \ddot{\theta} & =\dot{\phi}^2\sin (\theta )\cos (\theta ) \\ \frac{\mathrm{d}}{\mathrm{d}\, t}\left( \sin (\theta )^2\dot{\phi}\right) & =0. \end{align} The latter equation implies that $\sin (\theta )^2\dot{\phi}=\text{const}$. As the particle will at some point be located at one of the poles, we find that this constant is $0$ (by plugging in $\theta =0,\pi$). Thus, $\dot{\phi}=0$. The first equation then becomes $\ddot{\theta}=0$.

Jonathan Gleason
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