Intuitive explanation of parallel transport and geodesicsby teodron Tags: explanation, geodesics, intuitive, parallel, transport 

#1
Feb1612, 07:31 AM

P: 7

Hello,
First of all, please excuse me if I posted in the inappropriate place.. While a student few years ago, I used to work a lot with advanced differential geometry concepts, but never got an intuitive view of HOW humanity got to think about parallel transport, why it contained two words that describe an action and a condition and why these words, in reality, do not suffice to carry out this process. Coming back to this issue, I always started with a sphere. I used to move a cigar along an orange, while trying to keep it tangent to a latitude circle. If that was not the equator, I was supposed to get a different position for that vector/cigar when I came back and crossed the start point. It didn't quite happen, so I didn't understand HOW to carry that cigar along the surface of the orange while that same latitude circle and not end up with the same orientation for the cigar. I asked my Diff Geom TA to explain the concept to me, apart from the covariant derivative definition and other unintuitive devices.. He didn't shed any light on the issue. Now I'm at work and can't find peace of mind. I think of this concept in two ways, both of which can be dangerously incorrect: 1. Parallel transport means: move with a particle across a surface, along a predescribed path. You want to keep on track, thus you must shift your velocity vector. If you can do that without pushing or lifting from the surface (no normal velocity), then the path is a geodesic. If the path is a geodesic, then the velocity vector should be parallel to itself at any point on that path embedded in the surface. Now, if you view that vector from the outside space (sphere to 3d space analogy, the vector is in no way parallel to whatever it was looking like at the past instances). Then I think and say to myself: the parallel concept must be related to how the normal looks like and how the tangent space looks like, and thus how they change.. Maybe I'm on the right track: anywhere on the sphere, I know the "up" direction and the "left" direction.. they can be given by the way the sphere/surface is parametrized, right? I think Gauss said something important about this aspect, but I'm interested in analyzing the problem from above, so I need the normal and left vectors.. If I move a bit forward, I have to take a tiny turn to stay on the parallel circle/latitude circle. The turning amount is a vector in the tangent space in that point to the sphere. Then by this same transformation I have to "shift" the cigar/the vector I'm trying to parallel transport. This can be a rotation about the UP/normal vector, that is given by the surface's geometry. Now I end up with another vector that is not parallel, of course, to its previous counterpart. Or is it? I can see that if I reach the start point by encircling the sphere in such a manner, I might end up with a vector pointing in a different direction, thus, being back to the same tangent plane, and having the vectors in this plane, they are not parallel, hence the latitude circles are not geodesics (apart from equators). What I can't understand is why the velocity vector of this latitude circle isn't parallel to itself. It just doesn't add up.. when I return to the start point, it has to point in the same direction, right? 2. Smaller explanation following: a surface, an embedded curve/path/trajectory, a vector field/vector that is parallel transported along the curve. The result? some kind of a ribbon given by the point c(t) on the curve and another point d(t) given by the endpoint of the X(c(t)) tangent vector.. Now the resulting swept/pseudo swept surface is a ribbon like object, at least for a sphere. If that were to be a ribbon resulting from the movement across a geodesic, then this ribbon should have no torsion? Or, at least, the measure that gives the intuitive torsion factor for a curve (torsion free curves are planar, right?) should tell something about that curve as to how close to being a geodesic it actually is. Given such a ribbon, I used to imagine two infinitely close line segments or lines, like the planes of a capacitor (the electrical symbol) and if I try to insert that ribbon through this vent, I will not be able to do it unless it comes from a paralle transport process.. These were the two questions I asked my TA. he couldn't answer, and I'm sincerely haunted by this years old question.. I feel like my life has no meaning, I can't rest not being able to understand this rather essential concept. So, I kindly ask of you to help me out. I should state that I've did a bit of forum research and the subject I've found don't help that much. I don't want to go back to Christoffel symbols, covariant and contravariant fields, tensor derivatives, etc.. I first want to understand how the human mind stumbled upon this concept, then I will be glad to understand how to generalize it. Best wishes! Theodore 



#2
Feb1912, 03:29 PM

P: 134

I don't have any answer, but since nobody said anything yet, I'll say at least this: This semestr I'm (hopefuly) taking a course in Riemannian geometry, and I'm fortunate enough to have researcher in this field as a tutor. So if you are willing to wait some weeks for answer, I'll ask him since I want to understand this myself.
Cheers 



#3
Feb1912, 04:14 PM

P: 2,472

there's some images on google that might help with your understanding, here's one for a shpere where a vector is transported across 3 geodesics and back to where it started and the direction is changed.
http://www.google.com/imgres?hl=en&s...dvaGnCA&zoom=1 



#4
Feb1912, 10:32 PM

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Intuitive explanation of parallel transport and geodesics
parallel transport is a way to measure the angle sum of a triangle from within the surface itself. try some examples.




#5
Feb2012, 01:27 PM

P: 234

Some keywords for parallel transport: Riemannian metric, affine connection. 



#6
Feb2112, 05:56 AM

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P: 1,716

In Euclidean space a vector field along a curve is parallel if it is constant. A curve is self parallel if its tangent is constant. This is true for a straight line with constant speed parameter.
On a surface, a vector field is parallel if its length is constant and the rate of change of its angle to the tangent is proportional to the geodesic curvature of the curve. If the geodesic curvature is zero, the angle remains constant. So on a surface one just slides a vector along a geodesic keeping its angle constant to the tangent. In higher dimensions it might be possible for the vector to rotate. i am not sure. Check it out and let me know. For a manifold in Euclidean space, a constant speed curve is a geodesic if its acceleration is normal to the manifold. Intuitively, the geodesic does not wiggle along the surface. A vector field is parallel if its change in direction is normal to the surface. I'd like to see some examples for nontrivial surfaces, say a surface of revolution of constant negative curvature. One way to find geodesics is to slice the surface with a knife that is perpendicular to the surface. On a sphere this produces a great circle. What about on a symmetrically shaped torus? What about on a cylinder? 



#7
Feb2112, 01:36 PM

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draw a triangle going around the corner of a cube, with one edge in each face. Try to translate a vector around that triangle, keeping it parallel all the way around. It may help to cut the cube along one edge and flatten it out. Then see where the vector ends up after translation all the way around. Remember how you have to re identify the edges if you cut it. I claim the vector will end up perpendicular to where it started.
Moreover the angel sum of your triangle, measure on the surface of the cube, is 270 degrees, hence off by exactly the amount of the total parallel transport. By the way if this is right, it checks with your sphere example. I.e. a "triangle" made on a sphere by subdividing a circumference into three equal parts, has angle sum off by 360 degree, so is undetectable in this simple way under parallel transport. And the rule for parallel transport is different for curves that are not geodesics. 



#8
Mar1412, 11:16 AM

P: 7

Thank you all for your replies!
Again, thank you.. 



#9
Mar1612, 05:45 AM

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P: 1,716

The perpendicular knife determines a plane that contains the normal to the surface. So the acceleration of a unit speed curve that lies on the surface must be normal to the surface. 



#10
Mar1612, 05:59 AM

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P: 1,716

Physically, one can think of a geodesic on a surface in 3 space, as a line of least constraint. One imagines stretching a rubber band between two points on the surface, pinning it down at the ends, then letting go of it and allowing it to relax. As it releases tension it will change shape and finally come to rest when it has minimized its tension.At this equlibrium it will lie on a geodesic.
This idea of geodesic is formalized as a variational problem where geodesics are curves that locally minimize distance among all possible "near by" curves. 



#11
Mar1612, 06:29 AM

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I would guess  though I haven't read it myself  that Riemann's Habilitation Thesis is one of the first places that parallel transport is described. I would be willing to read through it with you.
It seems reasonable to me that early differential geometers were struggling with the realization that geometry is not conceptually intrinsic to the idea of space. When they saw that the parallel postulate could not be proved from primitive ideas of line and plane they saw that the actual geometry of the universe had to be determined from measurement. I think that gauss attempted to determine whether Euclidean or nonEuclidean plane geometry was true by measuring the sum of the angles of large triangles on the earth. I suppose then if you have two rulers at different points in space, you could ask the question of how you would know if they have the same length. One way might be to parallel translate one of the rods along a curve to the second rod. 



#12
Mar1912, 06:03 AM

P: 7

Let me chime in with what I mean by the "greedy" algorithm. I do believe it's the same concept as proposed here:
http://citeseerx.ist.psu.edu/viewdoc...10.1.1.97.5808 What happens when you cut a sphere with a knife along the 45 deg N parallel circle? I bet I can make the knife contain the normal in each point to the sphere, and still be able to cut the sphere along that exact curve. And that's no geodesic. 



#13
Mar1912, 06:48 AM

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#14
Mar1912, 07:38 AM

P: 7

I haven't tried it literally (I suppose a knife and an orange would do..), but I did manage to splash some odd looking depiction of what I'm imaging in a png file. I am cutting along the 45N parallel. My task is: make sure my knife contains the surface normal and that I do not cut something else apart from the curve. I guess that's not impossible, and here's why: imagine a cone with the vertex at the center of the sphere, containing the parallel circle. Cutting that circle means keeping the knife tangent to the cone's surface, fixing the knife's tip at the center of the sphere. That seems to be possible. What am I doing wrong?
Now, see the "blue" band the normal vectors then describe? That band is a cone frustum/ cone section. If that were to a great circle, then that band would've been a simple difference of two discs. My moral is: I can cut the surface and keep the knife along the normal field, and that would not forbid me to "turn" the knife while staying along the path. In that case, the knife is guided by the surface of the cone. The knife constraint, therefore, needs to be restated. I'd say, if correct!, that this "knife mechanism" must produce a "band" (or trail ribbon, as CG guys call it), that can be pulled through some sort of slit without "pushing" on the walls of that slit. In other words, that ribbon surface must be pulled through a a rectangular opening (  < the slit ) just like the one from an ATM credit card machine without having the ribbon ripped, teared or damaged in any way from this process. I tend to think that the cone bit an cutting the parallel works because I drag my knife along the surface and exert pressure (other than advancing pressure), while trying to keep on the trajectory. This concept of a ribbon also appears here > http://en.wikipedia.org/wiki/Torsion_tensor (in the wikipedia picture of a torsion along the geodesic). Is it in any way connected to the manner in which one "cuts a surface with a knife"? Kudos for your replies!! 



#15
Mar1912, 12:29 PM

P: 2,043

The blue vectors are not tangent to the sphere, they are normal. I'm not sure if what you're trying to do is do parallel transport along the sphere, but you have your cone upside down for that. The parallel transport on a sphere is the same as parallel transport along the lip of a tangent cone "hat" so to speak (flip the cone upside down). The difference in parallel transported vector's angle from the original vector is the same as the deficit angle for the cone.




#16
Mar2012, 05:55 AM

P: 7

I understood what you said: the cone's tangent planes coincide with the sphere's tangent planes, and hence the parallel transported vectors lie on a cone. If developed in a plane, the vectors shift in orientation by exactly the angle deficit, as you specified. Although very important as an example, this cannot be generalized, whereas Lavinia's "method" of "knifing" the manifold should work regardless of the surface (we cannot fit a developable surface to contain those vectors for any kind of surface, unfortunately, otherwise I'd have been satisfied with putting "cone hats" on manifolds and detect how much off a curve is from being a true geodesic). Regards! P.S. I am continuing this debate to provide students like I have been with a starting point before they dive into the Del operator, Christoffel symbols and other wonders that efficiently hide any geometric elegance from their users. I am extremely angry at most math professors for writing the same information in their lecture notes, books, etc. and providing students with stupid pictures of obvious things. When they start discussing tensors, curvature, bundles, geodesics, they again conjure up a sphere or a cylinder and draw the obvious. None of them tries to actually digest these things, as to explain and track their origins. I would bet my life that Riemann and Levi Civita didn't submit to swallowing up definitions and formulae, then using them to develop ground breaking mathematical devices. It simply cannot be. Have we advanced so much that we cannot stop any longer and analyze the intuitive meaning of the concepts we even develop our PhD theses on? If so, I am very much disappointed and sad. I don't have the time to reread the whole theory, but I do believe one can understand some concepts without maneuvering all those general aspects (resuming to 3d curvature, torsion of 2d manifolds and curves). I am willing to swear that most of my University Professors were nothing more than skilled users of calculus, many of them lacking more fundamental insights on the objects they were invoking in their papers. Can it be that a TA can't provide you with answers to some questions derived from what your professor writes on the blackboard? I sincerely want people to understand such things without wasting many hours that most do not have.. also, not all students are as bright as Einstein to grasp the notions without much detailing. Kind regards to all of you interested people. 



#17
Mar2012, 05:22 PM

P: 2,043

If you are trying to draw geodesics, that circle you drew out is certainly not a geodesic. Geodesics on a sphere are great circles.




#18
Mar2112, 04:19 AM

P: 10

Dear teodron,
I've just read your problem here and i'm amazed at how noone here can give a simple explanation to your problem! Any closed loop on a surface is just an npolygon with n > infinity just like on the Euclidean plane. So think of a latitude circle on the sphere as a regular npolygon. The simplest polygon is a triangle in the most extreme case. So on the sphere, let us degenerate a latitude circle to a spherical triangle... This triangle has two vertices on the equator say, and one on the north pole. Now start with a vector on the equator parallel to a side of this triangle, a longitudinal line that is, and move your vector up this line ( it does not change its angle with it until it reaches the north pole, identically zero that is! ). But it forms an angle with the other longitudinal line when it reaches the north pole. So when you parallel transport this vector down to the equator again along the other side, and back on the equator to where you started it points to a different direction. The turn in radians of the vector is exactly the angle of the triangle at the north pole and this is parallel transport. Unfortunately, noone here can answer at my questions, so i am leaving this forum. 


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