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Understand the major arc connecting two points on a sphere

  1. Aug 21, 2014 #1
    I am not sure if this is the right forum for this question, but I arrived at the question while studying the principle of stationary action so here it is:

    Consider the problem of finding the shortest path between two non-antipodal points on a sphere. Usually one solves this by using calculus of variations and finding trajectories where the distance integral is stationary. There are actually two solutions to this - the minor arc and the major arc of the great circle passing through the points with the minor clearly being the shorter one.

    Here is the thing that has been bothering me. I can see that the minor arc in addition to being stationary is the local minimum. However, I am having trouble trying to figure out if the major arc is a local maximum or local minimum or inflexion point. If I imagine adding a tiny bump to the path, I get a slightly longer path suggesting that the major arc must be a local minimum. However, if I imagine the path as a rubber band that I can continuous deform towards the minor arc, the length gets shorter suggesting a local maximum.

    Clearly I have a major gap in my understanding of stationary paths. Any idea about how I should think about this to resolve the confusion?

    Thanks

    P.S: Sorry if this has been asked before, but I searched through the site (and elsewhere) but could not find anything.
     
    Last edited: Aug 21, 2014
  2. jcsd
  3. Aug 21, 2014 #2

    Simon Bridge

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    Is the greater arc really one of the solutions to the problem of finding the shortest distance between two points on a sphere? Isn't the minor arc a shorter path?

    The two stationary paths are the "straightest" paths - one is longer than the other.

    As you noticed, small deviation for each produces a longer path.
    But small deviations from the minor arc can get you a shorter path than the undeviated major arc.
    This tells you that both are minima, the minor arc is a global minima while the greater arc is a local minima.
    There is no maxima - you can draw in paths of any length you like in a big scribble all around the sphere.
     
  4. Aug 22, 2014 #3
    Hi Simon,

    Firstly, thanks for the response. Of course the greater arc is not a solution to the shortest path problem. However it is a solution to the stationary path problem.

    I also understand that there is no global maxima - the big scribble you mention takes care of that.

    My dilemma arises from the following

    1. I can imagine small variations around the major arc - bumps - that give slightly longer paths.
    2. I can also imagine small variations around the arc - circular path that is not a great circle but close to it - that give slightly shorter path length.

    Hence my question - "Is the major arc a local maxima or minima?". You seem to be convinced that it is a local minima, but #2 above throws me off. I am starting to convince myself that it is locally neither a minima nor a maxima, but an inflection point.

    Sorry about harping on this, but I have come back to physics after over a decade and this time around I want to make sure I have a "deeper" understanding of the math.

    Thanks again for taking the time to reply.
     
  5. Aug 22, 2014 #4

    A.T.

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  6. Aug 22, 2014 #5
    Hi A.T.

    Thanks. Initially, I did think that it might be a saddle point (and it may still be). However, when I try to visualize a surface with saddle point, I run into another problem

    1. Either the surface has no global minimum - which not the case in our example.
    OR
    2. The surface has at least two other stationary points one on either side of the saddle, making a total of three - again not the case in our example.

    Is trying to visualize this, the wrong way to go about it? How do you guys and gals build intuition around this?
     
  7. Aug 22, 2014 #6

    Orodruin

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    As AT said, it is a saddle point, which is rather straightforward to show.

    Without loss of generality we can pick a coordinate system such that both the start and end points are at the equator ##\theta = \pi/2##. Let the start point be ##\phi = 0## and the end point ##\phi = a\pi## (so we have the major arc if ##a > 1##). If we parametrize the path with ##\phi##, we can write a small variation to the arc as
    $$
    \theta(\phi) = \frac \pi 2 + \eta f(\phi).
    $$
    The length of this path (on the unit sphere) is
    $$
    L[\theta(\phi)] = \int_0^{a\pi} \sqrt{\cos^2(\eta f(\phi)) + \eta^2 f'(\phi)^2} d\phi.
    $$
    Differentiating twice wrt ##\eta## and evaluating at ##\eta = 0## gives
    $$
    \left.\frac{d^2L}{d\eta^2}\right|_{\eta = 0} = \int_0^{a\pi} (f'(\phi)^2 - f(\phi)^2) d\phi.
    $$
    This can be made positive by, for example, picking an ##f## which oscillates fast enough, so the arc cannot be a local maximum. That it can be made negative is not as obvious[1], but can be accomplished by letting
    $$
    f(\phi) = \sin\left(\frac{\phi}{a}\right),
    $$
    which fulfills the boundary conditions ##f(0) = f(a\pi) = 0## and results in
    $$
    \int_0^{a\pi} (\frac{1}{a^2}\cos^2(\phi/a) - \sin^2(\phi/a)) d\phi =
    \frac{a\pi}{2} \left(\frac 1{a^2} - 1\right) < 0
    $$
    if ##a > 1##. Thus, the major arc is a saddle point.


    [1] In fact, it can only hold when ##a > 1##, otherwise the minor arc would also be a saddle point. It is a fun exercise to show that this is the case.
     
  8. Aug 22, 2014 #7
    Thanks for the proof Orodruin!

    After my response to AT, I went back and read some more about saddle points and it seems that by definition if you can find a smaller value and a larger value near a stationary point, it is a saddle point. So, by that definition it is a saddle point.

    I was stuck thinking of the saddle point as the typical picture shown in text books. After that I had trouble understanding how such a surface could be extended to include the minor arc without introducing new stationary points.

    I think I need to study functional analysis, etc, to develop a better feel for these kinds of issues.

    Thanks again for your help
     
    Last edited: Aug 22, 2014
  9. Aug 22, 2014 #8

    A.T.

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    For simplicity consider two exemplary modes to perturb a great circle arc:
    M1 : (1 bump) pushes the whole arc in one direction
    M2 : (2 bumps) pushes half of the arc in one direction, the other in the opposite (S-shape)

    For the major arc, M1 will reduce the path length, while M2 will increase it. So if you plot the path length L as function of M1 and M2 you get a saddle. If you move from the saddle point, along M1 in either direction you will arrive at a minimum representing the minor arc. While deviating along M2 always increases the path length.

    2m7d9xx.png

    The surface repeats along M1. But the saddles / minima always represent the same two arcs.
     
    Last edited: Aug 22, 2014
  10. Aug 22, 2014 #9
    Hi AT,

    Thanks for that explanation. The parameterization of the two curves using M1 and M2 really helps see the repetition along M1.

    I think that resolves my question.

    Once again thanks to Simon, AT and Orodruin. Now I can move on to page 8. of Landau Lifshitz Mechanics.
     
  11. Aug 23, 2014 #10

    A.T.

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    You're welcome. Thanks for the interesting question. You basically made all the relevant observations yourself in post #3, that it's not a extremum despite being a stationary point, which leads to a saddle point.
     
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