Geodesics on three dimensional graph

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  • Thread starter kairama15
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  • #1
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Summary:
Looking for differential movement of a point along a surface.
Suppose there is a three dimensional graph (such as z=x^2+y^2).

Suppose there is a point on the surface of the 3 dimensional graph, for example at (x,y,z)=(1,1,2).

Suppose the point is moving along the surface (along a geodesic) according to a unit vector, such as <0,1,0>.

Is there a calculation that may be done that determines its next position after a short amount of movement such as ds along its surface according to its unit vector? As well as a calculation that determines the new unit vector after it moves and rotates slightly?

Any help with a general solution to this would be interesting.
 

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  • #3
Office_Shredder
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The answer is yes, you can write down a series of differential equations that the geodesic must satisfy and then you can try to solve them. Are you trying to actually compute exact answers, or just get approximate answers? The exact answer is a bit complicated (I think surprisingly complicated)

https://en.m.wikipedia.org/wiki/Solving_the_geodesic_equations
 
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Great! Could someone help walk me through finding the next point and new directional vector using the geodesic equations you recommended? I'm unfamiliar with determining the metric and Christoffel symbols for this kind of thing. Could you point me in the right direction of how to set it up? :)
 
  • #5
Office_Shredder
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I still think it would help if you could tell us a bit more about what the application is here. Do you need to do this for arbitrary surfaces, or just specific ones? Do you need an exact answer or just an approximation?
 
  • #6
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Thanks for the response. I was looking into general relativity, and was learnong about geodesics. After viewing some videos on youtube showing geodesics on three dimensional graphs I became curious. I don't necessarily want to focus on relativity, but rather, 3d surfaces. If I can figure out how to find the very next point of movement in the above example, as well as the next vector describing its movement Im sure I could figure it out foe other surfaces.

In short, Im interested in how these geodesic equations are used for an approximate geodesic calculation for a 3d graph. Like, how Id set the problem up so i could repeat and iterate it to find how the point moves.
 
  • #7
Office_Shredder
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Ok, I'm far from an expert at this but I guess let's start by talking about metrics. In ##\mathbb{R}^3##, we say the metric is ##ds^2= dx^2+dy^2+dz^2##. ##ds## is how far away a point is if you move by ##(dx,dy,dz)##. There is a far more rigorous definition in terms of inner products on tangent planes of a manifold, but this is a rough description of what a metric means).

Cool. Now let's consider the surface you described. It helps to parameterize the surface with different coordinates so we're going to say the surface has parameters ##(u,v) \to (u,v,u^2+v^2)##.

We want to compute the metric on this surface, i.e. ##ds^2 = Adu^2 + B dv^2 + 2C du dv ##. Here ##A,B,C## are all going to be functions of ##u## and ##v##. Near the origin the parabaloid is basically flat so ##ds^2\approx du^2+dv^2##, but away from the origin a change in ##u## and ##v## causes a big z-direction change also, so ##ds^2 >> du^2+dv^2##. The fact that ##C##exists and is non-zero might not be immediately obvious but turns out to be true. Since we know ##ds^2 > 0## always it turns out you can prove ##AB - C^2 >0##. You can also imagine a "metric" in which this isn't true, e.g the minkowski metric in spacetime.

If we imagine that the coordinate functions such as ##x(u,v)## are sufficiently differentiable, we can get that if we change ##u## and ##v## by a small amount, that
$$dx \approx \frac{\partial x}{\partial u} du + \frac{\partial x}{\partial v} dv$$

And of course as the size of the move becomes arbitrarily small, this approximation is arbitrarily good, to the point of equality. Doing this for ##x=u##, ##y=v##, and ##z=u^2+v^2##, we get
##dx=du##, ##dy=dv##, ##dz= 2udu + 2vdv##.

So
$$ds^2=dx^2+dy^2+dz^2$$
$$= du^2+dv^2 (2udu+2vdv)^2$$
$$= (1+4u^2) du^2 + (1+4v^2) dv^2 + 8uv du dv$$

Alright, I'm going to pause here and let you chew on that for a bit. As a challenge, consider the sphere parametrized by two euler angles. What is the metric of that surface?
 
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  • #8
martinbn
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This has been quite for a while, but just add to the last post of @Office_Shredder , which outlines the general aproach, in the question of the first post one can do it in a different way. Since it is already in the Euclidean space and the induced structure form it. A curve on the surface is a geodesic if it curves only to stay on the surface but not within the surface. More precisely its acceleration vector need to be orthogonal to the surface, which might be shorter and easier to do.
 
  • #9
Office_Shredder
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That's not really sufficient, since there are many curves that lay entirely on a surface but are not geodesics (for example, on a plane you can just draw an arbitrary curve that isn't a straight line).


I suspect, but have never gotten around to checking, that the geodesic satisfies the property that its tangent vector can be projected onto the next tangent plane and that gives you the next tangent vector.
 
  • #10
martinbn
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That's not really sufficient, since there are many curves that lay entirely on a surface but are not geodesics (for example, on a plane you can just draw an arbitrary curve that isn't a straight line).
But their acceleration will not be normal to the surface.
I suspect, but have never gotten around to checking, that the geodesic satisfies the property that its tangent vector can be projected onto the next tangent plane and that gives you the next tangent vector.
If i understand you correctly, yes. The connection is the tangential component of the standart connection of the Euclidean space.
 

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