Parallel transport of a vector.

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I am having trouble understanding the concept of parallel transport of a vector along a closed curve. It is said that if the space where the curve resides has a curvature the orientation of the vector will change when it comes back to its original position. Can you help me in visualizing this process?
 

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  • #2
Spinnor
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Stand at the North Pole of the Earth and then head south to the equator while pointing your arm south. At the equator turn 90 degrees to the left and travel the same distance along the equator (1/4 the circumference of the Earth) while continuing to point your arm to the south. After traveling this distance turn 90 degrees to the left and return to the North Pole while pointing your arm to the south. When you return to the North Pole your arm will point in a direction 90 degrees rotated from its starting direction. You have "parallel transported" your arm around the surface of the Earth.

See:

http://sidk.info/intro_manifolds/node16.html [Broken]

or see:

http://www.google.com/search?tbm=isch&hl=en&source=hp&biw=1000&bih=521&q=parallel+transport&btnG=Search+Images&gbv=2&oq=parallel+transport&aq=f&aqi=g1&aql=&gs_sm=s&gs_upl=2613l2613l0l4205l1l1l0l0l0l0l257l257l2-1l1

Hope this helps.
 
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  • #3
tiny-tim
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Hi Avijeet! :smile:
I am having trouble understanding the concept of parallel transport of a vector along a closed curve. It is said that if the space where the curve resides has a curvature the orientation of the vector will change when it comes back to its original position.
If you mean a smooth closed curve, that's wrong

parallel transport around a smooth curve by definition keeps the arrow (or whatever) at the same angle, so it doesn't change when it returns to the start.

The change comes when we parallel transport along straight lines (geodesics) …

if there is, say, a triangle of straight lines, then the orientation changes on going all the way round the triangle …

see http://en.wikipedia.org/wiki/File:Parallel_transport.png" [Broken] :wink:
 
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  • #4
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See:
http://sidk.info/intro_manifolds/node16.html [Broken]
Hope this helps.
Thanks. Please help me out with this. In the path 1 it seems arrows are drawn tangential to the surface. So, initially what was pointing right points downwards by the time it reaches end of segment 1. During the segment 2 the arrow doesn't change direction and reaches starting point of 3 still pointing downwards. From then on it again moves up tangentially along the path.

Should the vector move parallel to itself, in segments 1 and 3, it doesn't look parallel to itself? The angle between successive vectors is changing.
 
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  • #5
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see http://en.wikipedia.org/wiki/File:Parallel_transport.png" [Broken] :wink:
Hi,
Thanks. Here's my doubt. In the above figure the arrows are definitely parallel in the segments NB and BA. I am not able to understand why the same is not true for the segment AN?
 
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  • #6
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I am having trouble understanding the concept of parallel transport of a vector along a closed curve. It is said that if the space where the curve resides has a curvature the orientation of the vector will change when it comes back to its original position. Can you help me in visualizing this process?
For me, the easiest way to visualize this is with a cone. A cone is locally flat everywhere except at the point, and so you can physically construct a cone from a flat sheet of paper simply by cutting out a wedge from the paper and bringing the two edges of the wedge together.

So, take your cone and draw 2 arbitrary closed paths on it, one that goes across the cut twice and one that goes across only once. Now, on each path pick some point as your starting point and draw a vector. Now, you parallel transport that vector by making small steps along your path and drawing new vectors which are parallel to the old vectors. This is easy to do when you flatten out the cone, but of course you have to close the cone and do it pretty carefully across the cut.

Once you do that you should be able to see a clear difference between the two paths. The path that crossed twice will be parallel-transported back to the same direction, it did not include any curvature. The path that crossed once will be parallel-transported back to a different direction, it did include curvature (the point of the cone).
 
  • #7
tiny-tim
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Hi Avijeet! :smile:
Here's my doubt. In the above figure the arrows are definitely parallel in the segments NB and BA.
That's because you're comparing them "across the join" at B.

Similarly, if you compare the arrows in AN and NB "across the join" at N, they are definitely parallel.
I am not able to understand why the same is not true for the segment AN?
You're comparing them "across the join" at A

but A is where the journey starts (and ends), while N and B were in the middle …

in the middle, the arrows must be parallel, because that's how we decide how to move from one straight side to the next …

follow the arrow from A to N to B and back to A. :wink:
 
  • #8
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You're comparing them "across the join" at A
Hi, Thanks for the reply.
Can you please explain how you can tell that the vector is parallel to itself during the transport through the segment AN (I am not able to figure that out :smile:)
 
  • #9
tiny-tim
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Yup :smile:

it starts at A, pointing along the curve

(I know it looks as if it's at about 10°, but I'm sure it's intended to be along the curve)

as it goes along AN, it stays pointing along the curve

when it reaches N, it's still pointing in the direction in which segment AN would continue, ie about 90° to segment NB

At that point, however, we start moving the arrow along the new segemnt …

but it's the same arrow, so it starts at 90° to the new segment (NM), and carries on at 90° to NB until it gets to B

(and at that point it's at about -20° to segment BA, and it carries on at -20° to BA until it gets to A)
 
  • #10
Bill_K
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If you mean a smooth closed curve, that's wrong … parallel transport around a smooth curve by definition keeps the arrow (or whatever) at the same angle, so it doesn't change when it returns to the start.
Sorry tiny-tim, but that is wrong. The statement of the original poster most certainly does apply for smooth curves as well as broken ones. Triangles and such are usually offered as examples only because the result is easier to calculate and easier to visualize.
 
  • #11
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Sorry tiny-tim, but that is wrong. The statement of the original poster most certainly does apply for smooth curves as well as broken ones. Triangles and such are usually offered as examples only because the result is easier to calculate and easier to visualize.
I am bit confused now. The explanation given by tiny-tim will hold only if the curve changes discontinuously. Otherwise how do you explain the change of behavior from moving along the segment during AN to keeping retaining the old direction NB in the figure?
http://en.wikipedia.org/wiki/File:Parallel_transport.png
 
  • #12
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Sorry tiny-tim, but that is wrong. The statement of the original poster most certainly does apply for smooth curves as well as broken ones. Triangles and such are usually offered as examples only because the result is easier to calculate and easier to visualize.
What about the case when you circumnavigate the Earth around the Equator. Would not the arrow return to the starting point pointing in the same direction it started in?
 
  • #13
tiny-tim
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Hi Bill! :smile:

It depends what we mean by "parallel transport of a vector along a (closed) curve" …

if we mean parallel transport along a general curve but relative to a coordinate frame, then yes of course it applies anyway …

I was dealing with the simpler case of parallel transport relative to the curve itself (and with the curve being a straight line-segment).
 
  • #14
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Hi,
Thanks. Here's my doubt. In the above figure the arrows are definitely parallel in the segments NB and BA. I am not able to understand why the same is not true for the segment AN?
In the linked figure,

Parallel_transport.png


the arrow is approximately parallel to segments AN and BA and roughly at right angles to path NB. The idea is to keep the arrow parallel to itself rather than parallel to the path. Imagine having two arrows and an assistant. The assistant goes ahead and his places his arrow parallel to the first. You then leap frog your assistant and go further ahead and then place your arrow parallel to that of your assistant and so on. The trick is that the arrow is not exactly parallel with each transition, but as near to parallel as possible while remaining tangential to the surface. For example if you started at A on the equator with your arrow tangential to the surface and pointing North, if you exactly parallel transported the arrow it would be pointing directly upwards at the North pole and no longer tangential to the surface. So parallel transporting a vector is actually doing a really bad job of keeping the vector parallel :tongue:. Sack the transporters!
 
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  • #15
jambaugh
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Try it this way...

Imagine you're working on a bumpy golf course. You and your buddy are transporting a telescoping pole (your vector). I paint a path in chalk you are to follow.
You and your buddy grab the pole as it lies at a position where you are both one arms length from the point where the chalk line crosses the middle of the pole and walk on each side of the chalk line. You each take one equal length step at a time in unison. Note the pole may start at any angle so one of you may start behind the other.

The metric determines distance each of you travels which this procedure is designed to keep equal. As you both walk and the curve bends your relative attitudes will change, one of you may fall further behind the other or begin to catch up. Note that the pole is telescoping so both its length and direction can change.

Now take a magic wand and make both you and your buddy shrink (but stay strong enough to carry the pole) and repeat. In the limit as you two become infinitesimally small and infinitesimally close together then the ground becomes effectively flat and keeping "parallel" to the path becomes easier.

This is parallel transport.

[edit] Hmm... Note that each step should be parallel to the path where it is crossing the pole, not relative to where you are closest to the path. This means you can either or both end up on the path and indeed cross the path several times. If the path is a circle on flat ground you will each end up following two equivalent circles which intersect twice (though you won't ever bump into each other as you reach the crossing points at different times.)
 
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  • #16
tiny-tim
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I am bit confused now.
Sorry. :redface:

Bill_K is correct that if we round off the corners of triangle ANB with very tiny arcs,

but keep the arrow at the same angle to the coordinate axes while we do the turn (instead of at the same angle to the arc itself), then the result will still look the same as in the diagram.

I was sticking to the case of the segments being along the coordinate axes anyway (as shown in the diagram) … so that it makes no difference … since I gathered that that was the situation you needed help with.
 
  • #17
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Sorry. :redface:

Bill_K is correct that if we round off the corners of triangle ANB with very tiny arcs,

but keep the arrow at the same angle to the coordinate axes while we do the turn (instead of at the same angle to the arc itself), then the result will still look the same as in the diagram.

I was sticking to the case of the segments being along the coordinate axes anyway (as shown in the diagram) … so that it makes no difference … since I gathered that that was the situation you needed help with.
This gets a little awkward when we get to a coordinate singularity such as the North Pole. For example if we start at A and move North towards N with the vector parallel to a line on longitude, when we get to the North pole and start moving South again along path NB we are now keeping the vector parallel to lines of latitude and at right angles to the lines of longitude and no longer maintaining the same angle to the coordinate axes if we count the lines of longitude and latitude as coordinate axes. Also the observers are supposed to be so small that the surface looks flat and they have no idea of the global coordinate system and they are in effect carrying out a survey of the surface with no initial coordinate system in place.
 
  • #18
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Here is a suggestion for a parallel transport device. Imagine a two wheeled trolley that you push along with the wheels either side of the path. The two wheels are connected by a differential gear device such that when the wheels are travelling at the same speed the arrow remains in the same direction relative to the trolley. When you go around a corner, one wheel travels faster then the other and the differential device rotates the arrow in the opposite direction (relative to the trolley) to the direction which the trolley is turning in, keeping the arrow tangential to the surface and as nearly parallel to itself as possible as the trolley turns. Hope you can picture that!

On a flat surface the arrow on the trolley will always be pointing in the direction it started in when it return to the start (if it designed correctly) no matter how convoluted the path and so any deviation in the angle of the arrow when you return to the start is a measure of the curvature of the surface.
 
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  • #20
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The figure on that page is as clear as mud (to me anyway) and obviously not intended to give laypersons a better idea of what is going on. Maybe this Java applet might be more intuitive: http://torus.math.uiuc.edu/jms/java/dragsphere/
 
  • #21
Bill_K
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What about the case when you circumnavigate the Earth around the Equator. Would not the arrow return to the starting point pointing in the same direction it started in?
Here's the simplest case I can think of to show how parallel transport works. As you say, consider circumnavigating the Earth. And you're right that if you go around the Equator (or any geodesic) that a parallel-transported vector will come back to its original value, but now consider a non-geodesic path, along a line of constant latitude θ. As you know, a pilot following such a curve will have to keep his rudder turned constantly to one side, to stay on course. And I say that in this case a parallel-transported vector will not come back.

The metric for a sphere is ds2 = R2(dθ2 + sin2θ dφ2). Let v be the unit velocity vector along the circle. It will have components (0, (R sinθ)-1). The parallel transport equation for any vector w is wμvν = 0, or wμ + Γμνσwνvσ = 0. The only nonzero Christoffel symbols are Γθφφ = - sinθ cosθ and Γφθφ = + sinθ cosθ. If we take wμ = (a, b), the equations for a(φ) and b(φ) are da/dφ - sinθ cosθ b = 0, db/dφ + sinθ cosθ a = 0 with solutions a(φ) = A sin Ωφ, b(φ) = A cos Ωφ, where Ω = sinθ cosθ.

What does this show? It shows that a parallel-transported vector will precess with constant angular velocity Ω = sinθ cosθ, where θ is the latitude. By the time we have gone completely around the Earth, φ has increased by 2π and w has precessed by 2πΩ. This is zero at the equator and zero at the North Pole but nonzero in between. So it will not return to its initial direction except at the equator and the Pole.
 
  • #22
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The figure on that page is as clear as mud (to me anyway) and obviously not intended to give laypersons a better idea of what is going on. Maybe this Java applet might be more intuitive: http://torus.math.uiuc.edu/jms/java/dragsphere/
Cannot run the applet : missing plug in :cry:
In the above figure (previous post) the vector neither seems parallel to itself nor is it keeping a constant angle with the curve. Then how is this parallel transport?
 
  • #23
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In the above figure (previous post) the vector neither seems parallel to itself nor is it keeping a constant angle with the curve. Then how is this parallel transport?
If the curve in that figure was on a flat surface then it would be wrong because the vector should always remain parallel to itself in that case, so the curve must be on a 3D surface that curves in a way that is not visible to us. The authors do not make this clear. Some authors set out to make clear complicated concepts to readers new to the subject and other authors set to show how clever they are and without any regard to the reader's better understanding. I suspect the authors of that book are the latter type or they simply assume you already understand the subject before reading it.

A better illustration of what is happening in that figure is this:

Parallel.gif


In the above figure, the vector is neither parallel to itself nor keeping a constant angle relative to the path, but it is a bit more obvious that this is due to the 3D curvature of the surface the curve is on. Remember that "parallel" transport of a vector is not really exactly parallel but as nearly parallel to itself as possible while remaining tangential to the surface as I mentioned earlier. Now if you imagine the above figure with the pretty coloured 3D surface removed so that just the curve is visible then you would have an illustration similar to the figure in the book you referenced.

The same is true if you look at the path in the image in post #14 without the 3D sphere in the background. The vector would just appear to be pointing in random directions when the 3D background surface is not visible.

P.S. You should try and get the required plugin for that applet. It would probably make things a lot clearer to you ;)
 
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  • #24
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Here's the simplest case I can think of to show how parallel transport works. As you say, consider circumnavigating the Earth. And you're right that if you go around the Equator (or any geodesic) that a parallel-transported vector will come back to its original value, but now consider a non-geodesic path, along a line of constant latitude θ. As you know, a pilot following such a curve will have to keep his rudder turned constantly to one side, to stay on course. And I say that in this case a parallel-transported vector will not come back.
Agree.
 
  • #25
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P.S. You should try and get the required plugin for that applet. It would probably make things a lot clearer to you ;)
Hi, Thanks
I got the applet working. Works great. Now I need to spend some time with it and think it over. Thanks again.
 

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