Geodesics subject to a restriction

  • #1
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Hi, I'm trying to solve a differential geometry problem, and maybe someone can give me a hand, at least with the set up of it.

There is a particle in a 3-dimensional manifold, and the problem is to find the trajectory with the smallest distance for a time interval ##\Delta t=t_{1}-t_{0}##, subject to the restriction that the velocity module is a constant ##k##.

The problem without the restriction is well known, and can be solved by the following lagrangian:
##S=\int_{t_{0}}^{t_{1}}\sqrt{g_{ij}v^{i}v^{j}}dt##
Where ##g_{ij}## is the metric tensor, and ##v^{i}## is the velocity vector. By using the Euler-Lagrange equation:
##\frac{\partial \sqrt{g_{ij}v^{i}v^{j}}}{\partial x^{j}}=\frac{d}{dt}\left(\frac{\partial \sqrt{g_{ij}v^{i}v^{j}}}{\partial v^{j}}\right)##
However, the restriction being precisely ##\sqrt{g_{ij}v^{i}v^{j}}=k##, doesn't allow to set up the problem as before (or that's what I think). Would you know the correct way to solve it?

Thanks in advance.
 

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  • #2
andrewkirk
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It's not clear what the problem is asking. If the additional constraint has an effect (and I am not yet convinced that it does, given that a geodesic is sometimes described as a 'constant speed trajectory') then the result will not be a geodesic, and so we cannot use the geodesic equation. But then the only constraint we have is the constant speed one, and there are infinitely many curves that have constant speed. So there is no unique solution. For all those curves the particle will travel distance ##k(t_1-t_0)## in the allotted time.

Also, I am not sure what is meant above by 'with the smallest distance'. From the context it appears to mean 'smallest distance travelled' which, again, will be ##k(t_1-t_0)##. But it could also mean the smallest (geodesic) distance from start to end point. Is that what was intended? If so, that problem could have a unique solution, but I expect it will be possible for the distance to be zero - ie that it finishes where it starts, and that there will be multiple ways of achieving that result.
 
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  • #3
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It's not clear what the problem is asking. If the additional constraint has an effect (and I am not yet convinced that it does, given that a geodesic is sometimes described as a 'constant speed trajectory') then the result will not be a geodesic, and so we cannot use the geodesic equation. But then the only constraint we have is the constant speed one, and there are infinitely many curves that have constant speed. So there is no unique solution. For all those curves the particle will travel distance ##k(t_1-t_0)## in the allotted time.
I think geodesics are described as constant speed trajectories due to the use of a natural parametrization, as Wikipedia says:
https://en.wikipedia.org/wiki/Geodesic
At least, that's what I understood. What I'm intending to find is not a geodesic, because there is a restriction that prevents you to take the shortest path. As if you are using GPS to find a direction, but the street recommended by the app is closed to traffic, so you need to take another one, not as good as the original.

Also, I am not sure what is meant above by 'with the smallest distance'. From the context it appears to mean 'smallest distance travelled' which, again, will be ##k(t_1-t_0)##. But it could also mean the smallest (geodesic) distance from start to end point. Is that what was intended? If so, that problem could have a unique solution, but I expect it will be possible for the distance to be zero - ie that it finishes where it starts, and that there will be multiple ways of achieving that result.
I mean the smallest geodesic distance (subject to the restriction), yes.
 

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