Taylor expansion and parallel transport

  • I
  • Thread starter mertcan
  • Start date
  • #1
340
6
hi, first of all in this image there is a fact that we have parallel transported vector, and covariant derivative is zero along the "pr"path as you can see at the top of the image. I consider that p, and r is a point and in the GREEN box we try to make a taylor expansion of the contravariant component of Z vector at the "p" and "r" points. But in the RED box we try to make a taylor expansion of the contravariant component of Z vector at the "u" and "r" points. Here is my confusion: Although there is not a "u" point, instead is just "p" and "r" points, How can we use such a point???? What does "u"point stand for????

I quoted from Differential Geometry Fionn Fitzmaurice page 36
 

Attachments

Answers and Replies

  • #2
340
6
I have seen so many views, but I have not been replied for a long time..... I am waiting your valuable return
 
  • #3
5
0
Hi.

The point ##u = h_{ds}(q)##, i.e. transporting from ##p## to ##q## by ##dt## and then from ##q## to ##u## by ##ds##. It's the same construction as section 6.3.
 

Related Threads on Taylor expansion and parallel transport

  • Last Post
Replies
14
Views
4K
  • Last Post
Replies
10
Views
10K
  • Last Post
Replies
9
Views
2K
  • Last Post
Replies
10
Views
4K
  • Last Post
Replies
1
Views
2K
  • Last Post
Replies
1
Views
3K
  • Last Post
Replies
3
Views
2K
  • Last Post
Replies
2
Views
1K
Replies
44
Views
19K
  • Last Post
Replies
11
Views
498
Top