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Parallel transportation of a vector along a closed triangle

  • Thread starter Manu_
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  • #1
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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
 

Answers and Replies

  • #2
18,079
7,499
Thanks for the post! This is an automated courtesy bump. Sorry you aren't generating responses at the moment. Do you have any further information, come to any new conclusions or is it possible to reword the post?
 
  • #3
TSny
Homework Helper
Gold Member
12,404
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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.
 

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