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I Geodesics on surfaces

  1. Nov 1, 2016 #1
    Consider the metric of ##S^{2}##: $$ds^{2}=d\theta^{2}+\sin^{2}(\theta)d\phi^{2}$$ Then in order to determine the geodesics on this surface one can minimise the integral $$s=\int_{l_{1}}^{l_{2}}\sqrt{\left(\frac{d\theta}{dl}\right)^{2}+\sin^{2}(\theta)\left(\frac{d\phi}{dl}\right)^{2}}dl$$ where ##l## parametrises a path connecting two points on the surface. We can identify the Lagrangian as ##L=\sqrt{\left(\frac{d\theta}{dl}\right)^{2}+\sin^{2}(\theta)\left(\frac{d\phi}{dl}\right)^{2}}##.

    If one parametrises the path by its arc-length ##s## then ##L=1##. Now, in the case of space-time the arc-length of a time-like path is equal to the proper time between the two endpoints of the path, and hence an arc-length parametrisation corresponds in this case to choosing proper time to parametrise the path.

    My question is, is the analogue of this for a curved surface (such as ##S^{2}##) the proper distance between two points on the surface, defined as ##dl=\sqrt{ds^{2}}##, such that, in the case of ##S^{2}##, the equations of motion are: $$\frac{d^{2}\theta}{dl^{2}}-\sin(\theta)\cos(\theta)\left(\frac{d\phi}{dl}\right)^{2}=0\\ \frac{d^{2}\phi}{dl^{2}}+2\cot(\theta)\frac{d\theta}{dl}\frac{d\phi}{dl}=0$$
     
  2. jcsd
  3. Nov 6, 2016 #2
    Thanks for the thread! This is an automated courtesy bump. Sorry you aren't generating responses at the moment. Do you have any further information, come to any new conclusions or is it possible to reword the post? The more details the better.
     
  4. Nov 7, 2016 #3

    vanhees71

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    We had a discussion about this in the context of GR. Despite the fact that in GR you have a pseudo-Riemannian manifold, everything said there is valid here. Instead of using the "square-root form" of the action, use
    $$L_2=\frac{1}{2} g_{ab} \dot{x}^a \dot{x}^b.$$
    Then automatically the parameter of the geodesic becomes an affine parameter. See the mentioned thread:

    https://www.physicsforums.com/threads/geodesics-and-affine-parameterisation.891679/

    In #17 I give a proof that you can derive any equation of motion (including also external forces other than gravity) as well from this "squared form" of the Lagrangian that you can derive from the "square-root form" with the advantage that the world-line parameter is automatically an affine parameter along the trajectories. Of course this also holds for the special case of no additional forces, i.e., for the geodesics of the manifold.

    That's why in the square-root form the equation for the geodesic gets automatically in the preferred form for using an addin paramater, i.e.,
    $$\mathrm{D}_{l}^2 x^{a}=\ddot{x}^a + {\Gamma^{a}}_{cd} \dot{x}^c \dot{x}^d=0.$$
    You can of course scale your parameter ##l## such that ##\mathrm{d} s^2=g_{ab} \dot{x}^{a} \dot{x}^B \mathrm{d} l^2=\mathrm{d} l^2##, i.e., you can always impose the constraint
    $$g_{ab} \dot{x}^a \dot{x}^b=1.$$
     
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