Intuitive explanation of parallel transport and geodesics

In summary, the author wonders why parallel transport should be related to the normal and tangent vectors, and why the velocity vector should be parallel to itself. He asks his Diff Geom TA to explain the concept, but the TA cannot answer the author's questions. The author concludes that his life has no meaning and that he can't rest until he knows the answer to his questions.
  • #1
teodron
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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
 
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  • #2
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
 
  • #4
parallel transport is a way to measure the angle sum of a triangle from within the surface itself. try some examples.
 
  • #5
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?
Nope. The amusingly-named Hairy Ball Theorem says that the tangent bundle of the 2-sphere has no nonvanishing sections. In other words, it's impossible to assign a nonzero vector (say, the one you want to associate with "up") to every point on the sphere in a smooth (or even continuous) way. For example, the direction we call "north" doesn't exist at the north pole, and likewise for south. Come to think of it, east and west don't exist at the poles, either. That's because the latitude/longitude system degenerates at the poles. The HBT says that no matter what clever coordinates you come up with, the best you can do is reduce the number of "bad points" to one.

Some keywords for parallel transport: Riemannian metric, affine connection.
 
  • #6
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 non-trivial 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
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.
 
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  • #8
Thank you all for your replies!
The amusingly-named Hairy Ball Theorem says that the tangent bundle of the 2-sphere has no nonvanishing sections. In other words, it's impossible to assign a nonzero vector (say, the one you want to associate with "up") to every point on the sphere in a smooth (or even continuous) way.
I didn't want it to sound as if I am ignoring the fact that the sphere is not parallelizable ( aka the HBT). My intuition was to start with a pseudo "up" direction and try to keep your "eyes" in that direction while moving along the curve from two points in such a way that we don't cross any "bad points" (where the parallel vector field frame vanishes - I forgot the fancy and yet semantically incorrect names of the tools that Diff Geom operates with: section, bundle, fibre, sheaf, etc. and I must re-read a Riemannian Geometry book). As I remember, the thing with parallelizable manifolds is that you can define an universal (kind of moving) frame for the reunion of the tangent spaces (may this be the tangent bundle or whatever the correct name for this object is)..
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 non-trivial 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?
Nice explanation, very natural. It makes sense because if the acceleration were to have any "parallel" components, it would "shift" the advancing direction, hence rendering that curve not "the straightest" path of them all, i.e. no longer a geodesic. As for cutting the surface with a perpendicular knife, it's dubbed the "greedy algorithm" to approximate geodesics on manifolds (probably not producing genuine geodesics..) - how can I prove or disprove this?
Again, thank you..
 
  • #9
teodron said:
Thank you all for your replies!


As for cutting the surface with a perpendicular knife, it's dubbed the "greedy algorithm" to approximate geodesics on manifolds (probably not producing genuine geodesics..) - how can I prove or disprove this?
Again, thank you..

What do you mean by the greedy algorithm?

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
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
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 non-Euclidean 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
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/summary?doi=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
teodron said:
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/summary?doi=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.

try it.
 
  • #14
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?
attachment.php?attachmentid=45273&stc=1&d=1332160692.jpg

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!
 

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  • #15
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.
 
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  • #16
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.
Yes, I was trying to answer to lavinia's suggestion of actually cutting a manifold (e.g. sphere) with a knife that contains the normal vector and this way describing a geodesic. I exemplified by cutting along the 45 deg N lat circle. What I did was to squeeze a cone that contained this circle and had its vertex in the center of the sphere. If you take small Δl elements of that circle and join them with the cone's vertex, you can get lavinia's knife sticking deep inside the sphere. If you move along the circle, you can cut the sphere and see the knife always "containing" the normal.
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 re-read 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
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
Dear teodron,
I've just read your problem here and I'm amazed at how no-one here can give a simple explanation to your problem!
Any closed loop on a surface is just an n-polygon with n -> infinity just like on the Euclidean plane. So think of a latitude circle on the sphere as a regular n-polygon. 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, no-one here can answer at my questions, so i am leaving this forum.
 
  • #19
This image is from Wikipedia:
Parallel_transport.png
 
  • #20
Dom Claude said:
Dear teodron,
I've just read your problem here and I'm amazed at how no-one here can give a simple explanation to your problem!

Unfortunately, no-one here can answer at my questions, so i am leaving this forum.

Your explanation of parallel transport is incomplete. this is because you are assuming that you are transporting around a geodesic triangle on a 2 dimensional surface. But what about an arbitrary curve? Parallel transport around geodesics is a special case and was explained in one of the answers. I guess you missed it.

Before you leave us, be kind enough to give us the full explanation.
 
  • #21
I was just trying to make a point.
I am aware that my answer is incomplete.
When you parallel transport a vector along any infinitesimal polygon line on that surface, it is by a geodesic line that you approximate that line on that surface, you cannot differentiate otherwise! Then you project your vector to the plane where the next infinitesimal line lies and so on and so forth...
 
  • #22
Dom Claude said:
I was just trying to make a point.
I am aware that my answer is incomplete.
When you parallel transport a vector along any infinitesimal polygon line on that surface, it is by a geodesic line that you approximate that line on that surface, you cannot differentiate otherwise! Then you project your vector to the plane where the next infinitesimal line lies and so on and so forth...

Thanks but I am confused.

First the pictures are of large triangles not infinitesimal triangles - whatever they are.

Second, defining parallel transport in terms of geodesics assumes that you already know what parallel transport is. Hmmm...

Also it is absolutely false that you need an idea of geodesic to take a covariant derivative of a vector field.

So maybe, you can elaborate your simple true answer so that we can understand.
 
  • #23
After all, any polygon on the surface can be approximated by geodesic line segments!
Actually, it is defined by geodesic line segments, hence any curve. Just think of the intersection of the two infinitesimal line segments, forming the angle of the aforenentioned triangle if you extend them on the surface long enough and resetting your north pole anew each time you pass from one to another.
 
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  • #24
Dom Claude said:
After all, any polygon on the surface can be approximated by geodesic line segments!
Actually, it is defined by geodesic line segments, hence any curve.
parallel transport does not require the idea of a geodesic. You can parallel translate along any smooth curve.

the idea of geodesic already assume that you know what parallel transport is.

parallel transport does not require a closed curve.

So an intuitive description of parallel transport should be based on an intrinsic notion of how a vector field moves along a curve. A geodesic is then a unit speed curve whose tangent is self parallel.

Along a geodesic on a surface, parallel transport just means sliding the vector along the curve keeping its length and angle to the tangent constant. It does not mean this on an arbitrary curve. Also if the intuitive idea is going to help us imagine higher dimensional examples - e.g. space-time - it should not, in my opinion, depend completely on 2 dimensional geometry but should indicate a general principle.
 
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  • #25
Of course! I rephrased my previous answer...!
But the way you do parallel transport in the limit, is along geodesic lines. Please reread my previous answer...
How can you do parallel transport in the limit, if you do not a have basic idea to fall back upon?
And the basic idea is that parallel transport of a vector along a geodesic line, leaves the angle between the vector and the tangent to this line unchanged! Am I wrong? Please correct me!
It is by passing from one geodesic line to another that this angle changes!
Do not be confused by the triangle. It is just an extreme case to elucidate this point.
 
  • #26
lavinia said:
Thanks but I am confused.

So maybe, you can elaborate your simple true answer so that we can understand.

You see my friend that a latitude line is not a geodesic, and you MUST approximate this line by geodesic line segments. The new line differs significally from the original latitude line though in the limit it looks the same! Now move your vector from one geodesic segment to another and watch how it jumps as it moves from one segment to another in the tangent plane like in the triangle case! Write a computer program if you wish... and verify the answers!
That's what teodron tried to understand and nobody told him...
Best regards :approve:
 
  • #27
Dom Claude said:
You see my friend that a latitude line is not a geodesic, and you MUST approximate this line by geodesic line segments. The new line differs significally from the original latitude line though in the limit it looks the same! Now move your vector from one geodesic segment to another and watch how it jumps as it moves from one segment to another in the tangent plane like in the triangle case! Write a computer program if you wish... and verify the answers!
That's what teodron tried to understand and nobody told him...
Best regards :approve:

So I guess by some other method you know what geodesics are. I always thought they were curves whose tangent was parallel along the curve. But your method requires knowing that they are geodesics already.

So I guess if you already know what the geodesics are you can do some approximation. BTW: in higher dimensions do you know a priori just because you have a geodesics, how to parallel translate any vector? Is the approximation method the same in higher dimensions?

It seems completely false that you MUST approximate the line by geodesics. Can you prove this?

But i guess if you do approximate by geodesic polygons you get an easy picture of how the parallel translate moves. Good point. This begs the question of how you know what a geodesic is, I think.
 
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  • #28
Dom Claude said:
Dear teodron,

Unfortunately, no-one here can answer at my questions, so i am leaving this forum.

Good luck.
 
  • #29
I tried to give a clue in my two posts as to how to see this in an elementary way, but it seems to have not been helpful. One really does need to carry out the experiments oneself to see it. At least I did. The one I described with a triangle at the corner of a cube helped me but here is another try. Notice that the simplest case of parallel transport of a vector, means to move a tangent vector along a geodesic, keeping it always tangent to that geodesic. (More generally to parallel transport a vector along a geodesic just means to keep it always making the same angle with the oriented tangent of that geodesic. For non geodesic curves one compensates by the angle between the tangent to that curve and the tangent to the geodesic through each point.) For example, a vector moving along a side of a triangle in the Euclidean plane is parallel transported along that side if it remains parallel to the side as it moves, or if it always keeps the same angle with the side.

Parallel transport around a triangle measures exactly the same thing as the angle sum, or angle sum exess, of a geodesic triangle, or geodesic polygons (yes one can manage without geodesics, but it is easier to see what happens with them). Namely what is being measured is average curvature within the polygon. In Riemann’s habilitationshrift he remarks that curvature at a point is angle sum excess or defect, compared to the area, of an infinitesimal triangle. This version of curvature is apparently due earlier to Gauss. The later more global version, of parallel transport, may be due later to Levi - Civita. In fact there is a much earlier Chinese mechanical device for producing it called the south pointing chariot.

http://en.wikipedia.org/wiki/South-pointing_chariot

Have you ever tried to prove that the angle sum of a triangle in the euclidean plane is π by trandslating a vector around the triangle, and noting that if you rotate it through the interior angle at each vertex so as to remain parallel to the new side, then when it returns to its original position, it has rotated exactly through π?

Notice that this process also gives π when performed on a sphere. But here it equals the (negative, i.e. clockwise) sum of the three interior angles through which one has translated, plus the contribution from parallel transporting the vectors around the three sides of the triangle, keeping them parallel to the geodesic sides all the way. Thus the discrepancy after parallel transport (called “holonomy”) = angle sum – π. Hence the angle sum of the geodesic polynomial measures essentially the same phenomenon as does parallel transport around the geodesic polygon, i.e. average curvature of the interior of the polygon.

Thus the angle sum of polygons, which Gauss and Riemann knew reflected curvature, is equivalent to the total holonomy obtained by parallel transport. On a sphere where curvature is constant there is thus also a connection with area. You may see that if the radius, hence also curvature, of the sphere is 1, then the area of a geodesic triangle is equal to the angle sum –π. E.g. in post #19, if all three angles are right, this angle excess equals π/2, exactly the area of the geodesic triangle which in that case covers 1/8 of the sphere. If you expand the triangle to cover the entire sphere, the area equals 4π = 2π.2 = 2π.(euler characteristic of the sphere).

In general you can triangulate a compact surface of genus g > 0 by 4g geodesic triangles, with 6g edges, 4g faces, and 2 vertices. Since there are only two vertices, the total angle sum of all the triangles is 4π. (E.g. a torus is triangulated by covering it by a rectangle with opposite edges identified, and then adding one vertex to the center of that rectangle and joining it to each vertex of the rectangle, obtaining 4 triangles, with 6 edges and 2 (distinct) vertices.)) If you then give the surface a metric of constant curvature -1, the area and angle sum formulas for the triangles adds up to give the formula area.(curvature) = 4g(total angle sum –π) = 4π – 4gπ = 2π(2-2g) = 2π.(euler characteristic of surface). This is called the Gauss - Bonnet formula.
 

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  • #30
By the way teodron, I know it is hard to ask a question about something one is puzzled by, but, I think one reason you got no clear answer to your questions, is that they are fuzzy and hard to pin down, not to mention inaccurate in many points.

e.g."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. "

I for one can hardly understand this. So please for give me for largely ignoring your question and simply trying to explain how elementary a concept parallel transport seems to be, to me at least. I readily admit I am not an expert. I just came up with these views while trying to design a geometry course for 9 year olds that uses no calculus, just elementary geometry.

Oh by the way, the reason the parallel transport around the triangle in post #19 ends up with an angle that differs from its starting position exactly by the angle at the north pole, is that in that very special triangle the angles add up to π plus that north pole angle. This concept of discrepancy of total parallel transport over a triangle or closed curve from zero, is apparently called "holonomy". For a geodesic triangle this is apparently just another name for angle sum excess or defect.
 
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  • #31
Thanks Mathwonk. This is a nice way to look at curvature ( I guess sectional curvature in higher dimensions)I guess one can't really visualize parallel translation unless one knows what the geodesics are and what the metric is. This seems strange to me. I keep thinking there must be a more fundamental intuition. Maybe there is a physical intuition.
 
  • #32
lavinia said:
Thanks Mathwonk. This is a nice way to look at curvature ( I guess sectional curvature in higher dimensions)


I guess one can't really visualize parallel translation unless one knows what the geodesics are and what the metric is. This seems strange to me. I keep thinking there must be a more fundamental intuition.

Dear lavinia,
I hope this my last post, before the administrator deletes my account...
May I remind you that one can find a geodesic betweeen two points also by using Euler-Lagrange formula as the shortest distance in the calculus of variations...
I hope this clears all misconceptions about parallel transport in your mind.
 
  • #33
thanks lavinia. you have helped me too here by your example of a paragon of patience.
 
  • #34
for this point of view of curvature lavinia, take a look at page 4.A - 14 of volume II of spivak's differential geometry book, his translation of riemann's "on the hypotheses which lie at the foundation of geometry", section II.3, I believe.
 
  • #35
Dom Claude said:
Dear lavinia,
I hope this my last post, before the administrator deletes my account...
May I remind you that one can find a geodesic betweeen two points also by using Euler-Lagrange formula as the shortest distance in the calculus of variations...
I hope this clears all misconceptions about parallel transport in your mind.

ok so your program is to figure out the geodesics on the surface first - say by measuring shortest paths - then use parallel translation on them to estimate parallel translation on arbitrary curves. That is nice.
 
<h2>1. What is parallel transport?</h2><p>Parallel transport is a concept in differential geometry that describes how a vector or tensor field is transported along a curve or surface in a smooth and consistent manner. It is used to define the notion of a "straight line" on a curved surface, known as a geodesic.</p><h2>2. How does parallel transport relate to geodesics?</h2><p>Parallel transport is closely related to geodesics, which are the shortest paths between two points on a curved surface. Geodesics are defined as the curves along which parallel transport of a vector or tensor field is maintained. In other words, geodesics are the curves that are "straight" according to the concept of parallel transport.</p><h2>3. What is the intuitive explanation of parallel transport?</h2><p>The intuitive explanation of parallel transport is that it describes how a vector or tensor field is "dragged" along a curve or surface without changing its direction. Imagine a ball rolling along a curved surface - as the ball moves, it maintains its orientation in space, even though the surface itself may be curved. This is similar to how parallel transport maintains the direction of a vector or tensor field along a curve or surface.</p><h2>4. How is parallel transport used in real-world applications?</h2><p>Parallel transport is used in many fields, including physics, engineering, and computer graphics. In physics, it is used to describe the motion of particles in curved spacetime, as predicted by Einstein's theory of general relativity. In engineering, it is used to calculate the stress and strain on materials that are curved or deformed. In computer graphics, it is used to create realistic animations of objects moving on curved surfaces.</p><h2>5. What are some common misconceptions about parallel transport and geodesics?</h2><p>One common misconception is that parallel transport always results in a straight line on a curved surface. In reality, the concept of parallel transport only guarantees that the direction of a vector or tensor field remains unchanged, not necessarily its path. Another misconception is that geodesics are always the shortest paths between two points on a curved surface. While this is often true, there are cases where geodesics may not be the shortest path due to the curvature of the surface.</p>

1. What is parallel transport?

Parallel transport is a concept in differential geometry that describes how a vector or tensor field is transported along a curve or surface in a smooth and consistent manner. It is used to define the notion of a "straight line" on a curved surface, known as a geodesic.

2. How does parallel transport relate to geodesics?

Parallel transport is closely related to geodesics, which are the shortest paths between two points on a curved surface. Geodesics are defined as the curves along which parallel transport of a vector or tensor field is maintained. In other words, geodesics are the curves that are "straight" according to the concept of parallel transport.

3. What is the intuitive explanation of parallel transport?

The intuitive explanation of parallel transport is that it describes how a vector or tensor field is "dragged" along a curve or surface without changing its direction. Imagine a ball rolling along a curved surface - as the ball moves, it maintains its orientation in space, even though the surface itself may be curved. This is similar to how parallel transport maintains the direction of a vector or tensor field along a curve or surface.

4. How is parallel transport used in real-world applications?

Parallel transport is used in many fields, including physics, engineering, and computer graphics. In physics, it is used to describe the motion of particles in curved spacetime, as predicted by Einstein's theory of general relativity. In engineering, it is used to calculate the stress and strain on materials that are curved or deformed. In computer graphics, it is used to create realistic animations of objects moving on curved surfaces.

5. What are some common misconceptions about parallel transport and geodesics?

One common misconception is that parallel transport always results in a straight line on a curved surface. In reality, the concept of parallel transport only guarantees that the direction of a vector or tensor field remains unchanged, not necessarily its path. Another misconception is that geodesics are always the shortest paths between two points on a curved surface. While this is often true, there are cases where geodesics may not be the shortest path due to the curvature of the surface.

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