Covariant derivative...

  • I
  • Thread starter Apashanka
  • Start date
392
13

Main Question or Discussion Point

In Carrol's gr notes the covariant derivative of a vector is given as ∇μAϑ=∂μAϑϑμλAλ.....(1)

For a geodesic in 2-D cartesian coordinates the tangent vector is V=##a\hat x+b\hat y##(a and b are constt.)where the tangent vector direction along the curve is ##\hat n=\frac{a\hat x+b\hat y}{\sqrt{a^2+b^2}}##
Now , covariant derivative of a tangent vector along the tangent vector direction for a geodesic is 0.
e.g ##\nabla_\hat nV=0##......(2)
Now how to relate (1) and (2) can anyone please suggest
 

Attachments

Answers and Replies

392
13
For a geodesic curve in 2-D polar coordinates(r,θ) the tangent vector is ##V=\hat r## ,therefore the tangent vector direction is ##\hat r## which we call say 1.
and ##\hat \theta## as 2.
The covariant derivative is therefore ##\nabla_1V=0=∂_r (V)+( )=∂_r(\hat r)+( )=0+() ##
I am in trouble writing out the bracketed term which is ##\Gamma_{1?}^?##
Can anyone please help in sort out this
 
392
13
For a geodesic curve in 2-D polar coordinates(r,θ) the tangent vector is ##V=\hat r## ,therefore the tangent vector direction is ##\hat r## which we call say 1.
and ##\hat \theta## as 2.
The covariant derivative is therefore ##\nabla_1V=0=∂_r (V)+( )=∂_r(\hat r)+( )=0+() ##
I am in trouble writing out the bracketed term which is ##\Gamma_{1?}^?##
Can anyone please help in sort out this
Could you please help in sort out .......@PeterDonis
 
Ibix
Science Advisor
Insights Author
5,472
3,932
For a geodesic curve in 2-D polar coordinates(r,θ) the tangent vector is ##V=\hat r##
This is not correct in general. The only curves whose tangent vectors are ##\hat r## are straight radial lines through the origin.

I do not think you know what a tangent vector is. The tangent vector to a curve is a vector pointing in the instantaneous direction you would face if you were walking along that curve. This does not have anything to do with your coordinates or your basis vectors.
 
392
13
This is not correct in general. The only curves whose tangent vectors are ##\hat r## are straight radial lines through the origin.

I do not think you know what a tangent vector is. The tangent vector to a curve is a vector pointing in the instantaneous direction you would face if you were walking along that curve. This does not have anything to do with your coordinates or your basis vectors.
I have mentioned geodesic curve
 
Ibix
Science Advisor
Insights Author
5,472
3,932
I have mentioned geodesic curve
So? Radial lines through the origin are not the only geodesics of a 2d Euclidean plane. All straight lines are geodesics.
 
802
36
I think you need to learn this stuff from another source. Carrol's notes are not intuitive most of the time.
 
392
13
So? Radial lines through the origin are not the only geodesics of a 2d Euclidean plane. All straight lines are geodesics.
Here I have taken a curve in 2-D(r,θ) with it's one end at the origin parametrised by λ(length along that curve) where at origin λ=0.
For this curve to be geodesic the components of tangent vector are ##V^1##=dr/dλ= 1 along ##\hat r## and ##V^2##=dθ/dλ=0 along ##\hat \theta##,so that the tangent vector is V=##\hat r##.
So what's the problem here.....
And from the geodesic definition ##\nabla_\hat r V=\nabla_\hat r (V^ie_i)=\nabla_\hat r (e_1)=0## that's what I need to prove
My motto is to find how this covariant derivative for tangent vectors work ,,that's why I have taken the simplest example..
The problem here to me is how to decompose the term ##\nabla_\hat r##??(e.g covariant derivative along ##\hat r##)
Can anyone please suggest
 
Last edited:
164
34
So what's the problem here.....
@Ibix is not saying that straight lines through the origin are not geodesics— he’s simply stating that those geodesics form a small subset of all geodesics on a flat plane. So in general the tangent vector of a geodesic using polar coordinates does not equal ##\hat r##.
 
27,062
7,296
The problem here to me is how to decompose the term ##\nabla_\hat r##??(e.g covariant derivative along ##\hat r##)
The covariant derivative along a curve with tangent vector ##V^\mu## is ##V^\mu \nabla_\mu##.
 
pervect
Staff Emeritus
Science Advisor
Insights Author
9,581
851
For a geodesic curve in 2-D polar coordinates(r,θ) the tangent vector is ##V=\hat r##
There is an ambiguity in your statement. You talk about "a geodesic curve in 2-D polar coordinates". You haven't specified a connection, and I'm not sure if you are familiar with the concept of a connection. This is unfortunate, because the covariant derivative isn't uniquely specified unless one defines a connection.

Now, in GR, it is implied that when we talk about a "geodesic curve", we use the Levi-Civita connection. Often, we get lazy about stating this explicitly. If we assume that's what you meant, then there is a subset of geodesic curves in the plane, namely those geodesic curves that pass through the origin, that have a tangent vector of ##\hat{r}##.

However, there are also a large set of geodesic curves with the Levi-Civita connection that don't pass through the origin, and whose tangent vector is something different.

What I think is probably happening is that when you are talking about 'using 2-D polar coordinates', you really mean that you are using a connection other than the Levi-Civiti connection. But I could be wrong about what you're trying to say.

The most striking thing about your posts is that you seem to think that specifying the coordinates matters. It doesn't. Once we specify that we are using the Levi-Civiti connection on the plane, geodesic curves are always the curve of shortest distance between two points. This is hopefully familiar an intuitive.

Then the point is that he curve that is the shortest distance between two points on the plane exists, and is independent of the coordinates used. We don't need to talk about whether we are using polar coordinates or cartesian coordinates to talk about the curve that is the shortest distance between two points.

What I suspect is happening is that you are wandering into the realms of connections other than the Levi-Civita connection without realizing it. Circles about the origin, curves with a constant value for the r coordinate, are not geodesics in the Levi-Civita connection, but may be geodesics with some other connection. But when one opens up the possibility of using connections other than the Levi-Civita connection, one needs to realize that covariant derivatve operators are not unique.

I suspect that it's least confusing at this point to restrict oneself to the Levi-Civiti connection on the plane (or other flat manifolds), as a steppingstone to talking about geodesics with the Levi-Civita connection on manifolds that are not necessiarly flat, and then moving onto using connections that aren't the Levi-Civita connection. But this is not the only way to do things, it's just the way that I'd recommend.
 

Related Threads for: Covariant derivative...

  • Last Post
Replies
10
Views
634
  • Last Post
Replies
13
Views
3K
  • Last Post
Replies
1
Views
2K
  • Last Post
Replies
5
Views
3K
  • Last Post
Replies
10
Views
4K
  • Last Post
Replies
17
Views
670
  • Last Post
Replies
4
Views
359
  • Last Post
Replies
3
Views
3K
Top