Parallel transportation of a vector along a closed triangle

In summary, Manu found that the vector orthogonal to the initial vector is the result of parallel transport around the closed triangle.
  • #1
12
1
Hello everyone,

I am trying to solve exercise 7.21 in the "Hobson, Efstathiou, Lasenby, General Relativity. An introduction for physicists."
What is asked is to show that the parallel transportation of a vector, along a closed triangle on a 2-sphere, results in an vector orthogonal to the initial one. The triangle has 3 right angles:

images?q=tbn:ANd9GcRxE_a5WPYsrBZ4sZ_UiBWK-GIk0Ti2eQ3u7ksmu7Hh-5-kIzox2Lo8rQzi_g.jpg


Here is what I tried so far: I have started with
[tex] \Delta v^{a} = -\frac{1}{2} R^{a}{}_{bcd} v^{b}_{P} \oint \! x^{c} \, d x^{d} [/tex]

where [tex] \Delta v^{a} = v^{a}-v^{a}_{P} [/tex] is the change underwent by the vector (the initial one being evaluated at point P), and [tex] R^{a}{}_{bcd} [/tex] is the Riemann tensor.

The metric being [tex] ds^{2} = a^{2} \left( d \theta^{2} +\sin^{2}{\theta} d\phi^{2} \right) [/tex] the only independant component of the Riemann tensor is [tex] R^{\theta}{}_{\phi \theta \phi} = -R^{\theta}{}_{\phi \phi \theta} = -R^{\phi}{}_{ \theta \theta \phi} = R^{\phi}{}_{ \theta \phi \theta} = \sin^{2}{\theta}[/tex]

I then find:
[tex] \begin{split}
\Delta v^{\theta} &= -\frac{1}{2} R^{\theta}{}_{\phi \theta \phi} v^{\phi}_{P} \oint \! \theta \, d \phi -\frac{1}{2} R^{\theta}{}_{\phi \phi \theta} v^{\phi}_{P} \oint \! \phi \, d \theta \\
&= -\frac{1}{2} R^{\theta}{}_{\phi \theta \phi} v^{\phi}_{P} \left( \oint \! \theta \, d \phi - \oint \! \phi \, d \theta \right) \\
\end{split} [/tex]

And the closed path integrals are (A is the pole, B is the left vertex and B the right one):

[tex] \oint \! \theta \, d \phi = \int_{A}^{B} \theta \, d\phi + \int_{B}^{C} \theta \, d\phi + \int_{C}^{A} \theta \, d\phi [/tex]

I then say that [itex]\phi[/itex] does not vary from B to C, since we stay on the equator, so:

[tex] \oint \! \theta \, d \phi = \int_{A}^{B} \theta \, d\phi + \int_{C}^{A} \theta \, d\phi = \theta \left( -\frac{\pi}{2} +\frac{\pi}{2} \right) =0[/tex]

And the other closed integral gives:

[tex] \begin{split}
\oint \phi \! \, d \theta &= \int_{A}^{B} \phi \, d \theta + \int_{B}^{C} \phi \, d \theta + \int_{C}^{A} \phi \, d \theta \\
&= \int_{B}^{C} \phi \, d \theta \\
&= \phi \times \frac{\pi}{2}
\end{split}
[/tex]

In the end, I obtain:

[tex] \Delta v^{\theta} = \frac{\pi}{4} \phi v^{\phi}_{P} \sin^{2}{\theta} [/tex]

And by similar calculations, I get:

[tex] \Delta v^{\phi} = \frac{\pi}{4} \phi v^{\theta}_{P} \sin^{2}{\theta} [/tex]

And from this, I was tempted to use the scalar product between the new vector (denoted with a prime) and the old one to see if they cancel:

[tex] \begin{split} v'^{a}v_{a} &= v'^{\theta}v_{\theta} + v'^{\phi}v_{\phi} \\
&= \left( v^{\theta}+\frac{\pi}{4} \phi v^{\phi}\sin^{2}{\theta} \right) v_{\theta} + \left( v^{\phi}+\frac{\pi}{4} \phi v^{\theta} \sin^{2}{\theta} \right) v_{\phi} \\
&= v^{\theta}v_{\theta} + v^{\phi}v_{\phi} + +\frac{\pi}{4} \phi \sin^{2}{\theta} \left(v^{\phi}v_{\theta} + v^{\theta}v_{\phi} \right)
\end{split}
[/tex]

which don't seem to cancel at all. Could anyone possibly point out my mistake(s)?

Thank you!
Manu
 
  • #3
Manu_ said:
Here is what I tried so far: I have started with
[tex] \Delta v^{a} = -\frac{1}{2} R^{a}{}_{bcd} v^{b}_{P} \oint \! x^{c} \, d x^{d} [/tex]
This equation is only valid for a very small loop. I don't see any way to apply it directly to the large triangle that you are working with. I would suggest using parallel transport around the triangle.
 

What is parallel transportation of a vector along a closed triangle?

Parallel transportation of a vector along a closed triangle is a geometric operation that involves moving a vector from one point to another while keeping it parallel to its original direction. This operation is commonly used in mathematics and physics to study the properties of curved surfaces.

What is the purpose of parallel transportation of a vector along a closed triangle?

The purpose of parallel transportation of a vector along a closed triangle is to understand how vectors behave on curved surfaces. By studying how a vector changes as it is transported along a closed triangle, we can gain insights into the curvature and geometry of the surface.

How is parallel transportation of a vector along a closed triangle performed?

To perform parallel transportation of a vector along a closed triangle, we first choose a starting point and a direction for the vector. Then, we move the vector along the edges of the triangle while keeping it parallel to its original direction. The process is repeated until the vector returns to its starting point.

What are some applications of parallel transportation of a vector along a closed triangle?

Parallel transportation of a vector along a closed triangle has many applications in diverse fields such as differential geometry, physics, and computer graphics. It is used to study the properties of curved surfaces, calculate geodesic paths, and simulate motion on curved surfaces.

What are the limitations of parallel transportation of a vector along a closed triangle?

Parallel transportation of a vector along a closed triangle is limited to closed surfaces, meaning that the starting and ending points must be connected by a continuous path. It also assumes that the surface is smooth and has no abrupt changes in curvature. Additionally, the process can be computationally intensive, making it challenging to perform on complex surfaces.

Suggested for: Parallel transportation of a vector along a closed triangle

Replies
4
Views
587
Replies
1
Views
772
Replies
0
Views
407
Replies
1
Views
648
Replies
7
Views
1K
Back
Top