Can I Use E-L Equations When Parametrizing by Arclength on a Sphere?

In summary, the speaker is struggling with finding the shortest path between two points on a sphere using calculus of variations. They have derived correct differential equations for great circles, but are unsure about the validity of their methods. The speaker describes their approach, which involves using arbitrary parameters to minimize the integral of the Lagrangian. They question whether this approach is correct and if they can use Euler-Lagrange equations when parametrizing by arc length. They also inquire about the applicability of squaring the Lagrangian and whether it is equivalent to minimizing the original Lagrangian.
  • #1
Blazejr
23
2
Hello
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.
 
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  • #2
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.
 
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  • #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.
 
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1. What is the shortest path on a sphere?

The shortest path on a sphere is the geodesic, which is the shortest distance between two points on the surface of a sphere. This path is also known as the great circle arc and is the equivalent of a straight line on a flat surface.

2. How is the shortest path on a sphere calculated?

The shortest path on a sphere is calculated using spherical trigonometry, which involves using the law of cosines and the law of sines to determine the angles and distances between points on the sphere.

3. Can the shortest path on a sphere be found using a simple formula?

No, the shortest path on a sphere cannot be found using a simple formula as it depends on the specific points and the radius of the sphere. It requires more complex mathematical calculations to determine the geodesic.

4. What are some real-world applications of finding the shortest path on a sphere?

Some real-world applications of finding the shortest path on a sphere include navigation and mapping systems, satellite and space travel, and determining the most efficient flight or shipping routes.

5. How does the shortest path on a sphere differ from the shortest path on a flat surface?

The shortest path on a sphere differs from the shortest path on a flat surface because of the curved nature of the sphere. On a flat surface, the shortest path is a straight line, but on a sphere, the shortest path is a curved line following the surface of the sphere.

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