Parallel transport of tangent vector....(geodesic)

[Apashanka]

https://www.google.com/url?sa=t&sou...Vaw3UvOQyTwkcG7c7yKkYbjSp&cshid=1551081845109
Here in page 55 it is written that geodesic is a curve whose tangent vector is parallely transported along the curve.
Now if there is a curve in 2-D which is determined by λ(length along the curve) .In 2-D cartesian coordinate system the tangent vector at every λ will point along the x(unit) and y(unit) direction that means they are parallely transported along the curve that means any curve in 2-D cartesian coordinate system is a geodesic.
But the same curve in 2-D polar coordinate (r,θ) ,at each λ the tangent vector points along the r(unit) and θ(unit) which differs along the curve ,that is the same curve is not a geodesic in 2-D polar coordinate system ,is it so??

 

[Ibix]

Parallel transport means, loosely, that you move a vector along some path while keeping it pointing in the same direction. In Euclidean space, only straight lines have a tangent vector that, moved along the line and kept pointing in the same direction, remains a tangent vector. As a counter example, imagine a circular path. At some point the tangent points in the +x direction. Parallel transport it a quarter of the way round the circle. The vector still points in the +x direction, but the tangent vector here is in the +y direction.

Coordinates are irrelevant to this fact. The additional complexity in curved spacetime is in the definition of "pointing in the same direction". That's handled by the connection coefficients.

 

[Dale]

In 2-D cartesian coordinate system the tangent vector at every λ will point along the x(unit) and y(unit) direction that means they are parallely transported along the curve that means any curve in 2-D cartesian coordinate system is a geodesic.

This is not correct. In flat space only straight lines parallel transport their tangent vector. I am not sure exactly why you would think that curved paths parallel transport their tangent vector. Could you explain your reasoning?

 

[Apashanka]

This is not correct. In flat space only straight lines parallel transport their tangent vector. I am not sure exactly why you would think that curved paths parallel transport their tangent vector. Could you explain your reasoning?

Here V¹ and V² are the tangent vectors at every λ which point along x(unit) and y(unit) throughout the curve which means they are parallel transported... that's why I am saying.

 

[PeterDonis]

Here V1 and V2 are the tangent vectors at every λ which point along x(unit) and y(unit)

You are mistaken about what tangent vectors are. Tangent vectors point along the curve. They don't point along the coordinate axes.

 

[Dale]

Here V1 and V2 are the tangent vectors

Oh, there is the mistake. Neither V1 nor V2 are tangent vectors to this curve. The tangent vectors would point diagonally along the curve.

A tangent vector can be thought of as being formed by taking two points on the curve, say $(x(\lambda),y(\lambda))$ and $(x(\lambda+\Delta \lambda),y(\lambda+\Delta \lambda))$, drawing the line connecting those two points, and taking the limit as $\Delta \lambda$ goes to 0. V1 and V2 do not fit that description.

 

[Apashanka]

Oh, there is the mistake. Neither V1 nor V2 are tangent vectors to this curve.

Yes sir I got it now they are the components...

 

[Apashanka]

You are mistaken about what tangent vectors are. Tangent vectors point along the curve. They don't point along the coordinate axes.

Yes sir I got it now they are only the components...

 

[Apashanka]

Yes sir I got it now they are the components...

For this case at any λ the tangent vector points in dx(λ)/dλ$\hat x$+dy(λ)/dλ$\hat y$

 

[Dale]

For this case at any λ the tangent vector points in dx(λ)/dλ$\hat x$+dy(λ)/dλ$\hat y$

Yes, that is correct

 

[Apashanka]

You are mistaken about what tangent vectors are. Tangent vectors point along the curve. They don't point along the coordinate axes.

Here T (tangent vector) is not pointing along the curve??

 

[stevendaryl]

View attachment 239314
Here T (tangent vector) is not pointing along the curve??

No, T is pointing perpendicular to the curve. Here's an illustration of what is meant by parallel transport.

We have a curve drawn in blue. The tangent vectors at two points, A and B, are drawn in black. The vector drawn in gray is the parallel-transport of the tangent vector at A to the point B. Since the parallel-transport of the tangent vector at A is not equal to the tangent vector at B, that means the curve is not a geodesic.

 

[Apashanka]

No, T is pointing perpendicular to the curve. Here's an illustration of what is meant by parallel transport.

We have a curve drawn in blue. The tangent vectors at two points, A and B, are drawn in black. The vector drawn in gray is the parallel-transport of the tangent vector at A to the point B. Since the parallel-transport of the tangent vector at A is not equal to the tangent vector at B, that means the curve is not a geodesic.

View attachment 239320

Yes sir got it now...thank you