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Shortest path on sphere

  1. Sep 16, 2014 #1
    I'm struggling with well-known problem of finding shortest path between two points on a sphere using calculus of variations. I managed to find correct differential equations of great circles, but I'm not confident about validity of methods I used. Below I describe my approach.

    In solutions of this problem I found online it was assumed that curve we want to find can be parametrized by angle [itex]\theta[/itex] (of spherical coordinates, which I'm going to use from now on). This is correct - geodesics on sphere are great circles and we can always choose coordinates on sphere such that both points lie on one meridian, say [itex]\phi=0[/itex]. However, I do not want to use my knowledge about those geodesics before I actually find them. Therefore, I have no way of knowing that [itex]\theta[/itex] is correct global parametrization of this curve - it could go along some line of latitude (theta=const.) for some time. Therefore I need to assume [itex]\theta , \phi[/itex] to be independent variables parametrized by some arbitrary parameter along the curve (call it time [itex]t[/itex]).

    Problem comes down to finding functions [itex]\theta (t), \phi (t)[/itex] that minimize integral:
    [tex]\int L \mathrm{d}t = \int \sqrt{\dot{\theta}^2+\sin ^2 \theta \dot { \phi } ^2} \mathrm{d}t[/tex]
    It follows imidietely from Euler-Lagrange equation that quantity [itex]\frac{\sin ^2 \theta \dot{ \phi}}{L}[/itex] is conserved. E-L equation for second variable produce complicated and messy equation.

    Here comes my idea. Variable [itex]t[/itex] is arbitrary parametrization of the curve. We can parametrize it by any other parameter that grows monotonically with time. It is easy to check that function under the integral is unaffected by this change of variables. Therefore I can use length of curve [itex]s[/itex] as parameter. If I do that we have that [itex]L(s)[/itex] is actually constant (as it is just rate of change of length w.r.t length) and E-L equations yield:
    [tex]\sin ^2 \theta \dot {\phi}=const., \qquad 2 \ddot {\theta} = \sin ( 2 \theta ) \dot {\phi}^2 [/tex]
    I've read on some website that those are correct equations for great circles. I can't check, because I don't know how to solve them anyways. If anyone can give me a hint on that I would be grateful, but that's not what I wanted to ask.
    Problem is that I am not sure if this approach is correct. Can I really use E-L equations when I parametrize by arclength? This is after all equivalent to imposing (in language of classical mechanics) non-holonomic constraint: [itex]\frac{\mathrm{d}L}{\mathrm{d}s}=0[/itex]. Therefore [itex]\dot{\theta},\dot{\phi}[/itex] are no longer independent.

    Is this approach completely wrong and I got good equations "by accident" or am I missing something? If it is, how can I fix my reasoning to find those equations (without simplifying assumption that the curve can be parametrized by [itex]\theta[/itex]?). Thanks in advance.
  2. jcsd
  3. Sep 16, 2014 #2


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    Using calculus of variations is of course the right path for solving such a problem and because the "Lagrangian" is a function of only the variables and their first derivative, of course you should use Euler-Lagrange equations. But there is something tricky here.
    Making [itex] \int_{t_1}^{t_2} \sqrt{\dot \theta^2+\sin^2{\theta}\dot \varphi^2} dt [/itex] stationary gives rise to a very complicated differential equation.(I actually derived that equation). So one may choose to make [itex] \int_{t_1}^{t_2} (\dot \theta^2+\sin^2{\theta}\dot \varphi^2) dt [/itex] stationary instead, as you did. The guy may say:"no big deal, its just the square of that!" But its actually not!
    I'm not telling its wrong, but it of course is not trivial and needs to be proved that you can make [itex] \int L^2 dt [/itex] stationary instead of [itex] \int L dt [/itex]!
    So that's the only tricky part I see in your calculations and everything else seems fine to me.
    But about your question. Great circles may have complicated parametrizations, because you may have a great circle which is running not parallel to a constant [itex] \varphi [/itex] circle. But there is always a change of coordinates that makes that complicated parametrization to a very simple one because you can always change coordinates to make that great circle a circle of constant [itex] \varphi [/itex]. So if you can find a reparametrization that transforms your differential equations to [itex] \dot \varphi=0 [/itex] and [itex] \ddot \theta =0 [/itex], then your differential equations are describing a great circle. I think that will do in theory, don't know about its applicability.
    Last edited: Sep 16, 2014
  4. Sep 16, 2014 #3
    Squaring Lagrangian is not really what I've done. Here's how I exactly got to my equations.
    I assumed that parametrization is chosen such that [itex]L=\frac{\mathrm{d}s}{\mathrm{d}t}=1[/itex]. We then have [itex]\frac{\mathrm{d}L}{\mathrm{d}t}=0[/itex] and [itex]\frac{\partial L}{\partial \theta}=\frac{\sin 2\theta \dot{\phi}^2 }{2L}[/itex] and [itex]\frac{\mathrm{d} }{\mathrm{d} t}\frac{\partial L}{\partial \dot{ \theta}}=\frac{\mathrm{d} }{\mathrm{d} t}\frac{\dot{\theta}}{L}=\frac{1}{L}\frac{\mathrm{d} }{\mathrm{d} t}\dot{\theta}[/itex].
    Second equation was extracted similarly.

    By choosing parametrization by arc length I make L constant function (when considered as a function of such chosen "time" only). This makes E-L equations much simpler, as shown above. What I am not sure about is whether E-L equations are applicable if I parametrize by arc length.
    I suspect I shouldn't be able to use E-L equation here because that makes derivatives of coordinates [itex]\dot{\theta},\dot{\phi}[/itex] dependent variables.

    Oh and regarding minimizing L^2: I actually checked if curve x(t) minimizing integral of L^2 implies that it minimizes integral of L. That is NOT true, unless [itex]\frac{\mathrm{d}L}{\mathrm{d}t}=0[/itex] which is in this case true only if we choose parametrization by arc length. So again comes the question, are we allowed to do that.
    Last edited: Sep 16, 2014
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