Geodesic Sphere Homework: Prove Great Circle Path is Shortest

  • Thread starter Thread starter roeb
  • Start date Start date
  • Tags Tags
    Geodesic Sphere
AI Thread Summary
The discussion focuses on proving that the shortest path between two points on a sphere is a great circle using the Euler-Lagrange equation. The integrand's independence from phi simplifies the equation, leading to a constant value for phi prime. The hint suggests that by aligning the z-axis with one of the points, the constant c can be determined to be zero, indicating that phi is constant. This implies that the path traced by varying theta while keeping phi constant corresponds to a great circle. Ultimately, the analysis confirms that the geodesic on a sphere is indeed a great circle.
roeb
Messages
98
Reaction score
1

Homework Statement


L = R \int_{\theta_1}^{\theta_2} \sqrt{1 + sin^2(\theta ) \phi ' ^ 2} d\theta

Use the result to prove that the geodesic (shortest path) between two given points on a sphere is a great circle. [Hint: The integrand f(phi,phi_prime,theta) in the result is independent of phi so the Euler-Lagrange equation reduces to partial_f/partial_phi_prime = c, a constant. This gives you phi_prime as a function of theta. You can avoid doing the final integral by the following trick: There is no loss of generality in choosing your z axis to pass through the point 1. Show that with this choice the constant c is necessarily zero and describe the corresponding geodesics.

Homework Equations



\partial f / \partial x = d/du \partial f / \partial x'

The Attempt at a Solution



I am having a bit of difficulty interpreting this problem. Using the Euler-Lagrange equation I get the following:
\phi ' ^ 2 = C^2 / ( sin^4 (\theta ) - C^2 sin^2(\theta )
The hint kind of confusing me. How can 'c' be zero? If I were to have the z-axis at point 1, I would think that \theta_1 = 0 if anything.

Can anyone help guide me on the right path for this problem?
 
Physics news on Phys.org
roeb said:
I am having a bit of difficulty interpreting this problem. Using the Euler-Lagrange equation I get the following:
\phi ' ^ 2 = C^2 / ( sin^4 (\theta ) - C^2 sin^2(\theta )

It might be more useful to write this as:

\left( sin^4 (\theta ) - C^2 sin^2(\theta ) \right) \phi ' ^ 2 = C^2

I think the point of the hint is that the above relation holds for all allowable values of \theta and that \theta=0 is one such allowable value. What does C^2 equal when \theta=0 assuming that \phi' is bounded (i.e. not infinite)? Since C is a constant, it must have this value for all \theta.
 
Thanks for your reply.

Now I can see why C = 0. If I am doing my math correctly that means \phi (\theta ) = A (another constant).

If \phi (\theta ) is just some constant, does that mean this constant A will be a segment of the great circle between two locations on the sphere?
 
As the choice of axis is arbitrary, you can choose one such that \theta is zero. Then varying \phi (which as you point out should have no \theta dependence) is just tracing out a great circle.
 
Thread 'Help with Time-Independent Perturbation Theory "Good" States Proof'
(Disclaimer: this is not a HW question. I am self-studying, and this felt like the type of question I've seen in this forum. If there is somewhere better for me to share this doubt, please let me know and I'll transfer it right away.) I am currently reviewing Chapter 7 of Introduction to QM by Griffiths. I have been stuck for an hour or so trying to understand the last paragraph of this proof (pls check the attached file). It claims that we can express Ψ_{γ}(0) as a linear combination of...
Back
Top