Anamitra said:
Now let us look into the fact that an ordinary sphere is commonly used to illustrate the concept of parallel transport in the texts. You did the same thing the following thread:
https://www.physicsforums.com/showpost.php?p=2758350&postcount=25
If you write the three equations [for A(r),A(phi) and A(theta)] for a line of latitude and solve them you get functions like Sin(phi) and cos(phi) in the solutions which are periodic functions of 2pi .
So the vector returns to its original orientation in this case!
Does it?
My daughter was constantly pestering me while I was working on this, so what follows could be full of mistakes.
Tangent spaces of an "ordinary sphere'' are 2-dimensional, and vectors in these tangent spaces don't have r components. Therefore, set r=1 and A^{r}=0. Your three equations then become two equations,
<br />
\begin{equation*}<br />
\begin{split}<br />
\frac{dA^{\phi }}{d\phi }+\cot \theta A^{\theta } &=0 \\<br />
\frac{dA^{\theta }}{d\phi }-\sin \theta \cos \theta A^{\phi } &=0.<br />
\end{split}<br />
\end{equation*}<br />
This set of equations can be solved elegantly using 2x2 matrices, but I'll solve it using a different method. Differentiating the first equation gives
\frac{d^{2}A^{\phi }}{d\phi ^{2}}+\cot \theta \frac{dA^{\theta }}{d\phi }=0,
and using the second equation in this gives
\frac{d^{2}A^{\phi }}{d\phi ^{2}}+\cos ^{2}\theta A^{\phi }=0.
Since \theta is constant along a line of longitude, this is just a harmonic oscillator equation with solution
A^{\phi }\left( \phi \right) =c_{1}\sin \left( \phi \cos \theta \right) +c_{2}\cos \left( \phi \cos \theta \right)
Then,
<br />
\begin{equation*}<br />
\begin{split}<br />
\frac{dA^{\phi }}{d\phi }\left( \phi \right) &= \cos \theta \left[ c_{1}\cos \left( \phi \cos \theta \right) -c_{2}\sin \left( \phi \cos \theta \right) \right] \\<br />
&=-\cot \theta A^{\phi }<br />
\end{split}<br />
\end{equation*}<br />
and, from the first equation,
A^{\theta }\left( \phi \right) =\sin \theta \left[ c_{1}\cos \left( \phi \cos \theta \right) -c_{2}\sin \left( \phi \cos \theta \right) \right].
Consequently,
<br />
\begin{equation*}<br />
\begin{split}<br />
A^{\phi }\left( 0\right) &= c_{2}\cos \left( \phi \cos \theta \right) \\<br />
A^{\theta }\left( 0\right) &= c_{1}\sin \theta \cos \left( \phi \cos \theta \right) <br />
\end{split}<br />
\end{equation*}<br />
and
<br />
\begin{equation*}<br />
\begin{split}<br />
A^{\phi }\left( 2\pi \right) &= c_{1}\sin \left( 2\pi \cos \theta \right) +c_{2}\cos \left( 2\pi \cos \theta \right) \\<br />
A^{\theta }\left( 2\pi \right) &= \sin \theta \left[ c_{1}\cos \left( 2\pi \cos \theta \right) -c_{2}\sin \left( 2\pi \cos \theta \right) \right].<br />
\end{split}<br />
\end{equation*}<br />
As a simple example, take c_{1}=1/\sin \theta and c_{2}=0, so that
<br />
\begin{equation*}<br />
\begin{split}<br />
A^{\phi }\left( 0\right) &= 0 \\<br />
A^{\theta }\left( 0\right) &= 1<br />
\end{split}<br />
\end{equation*}<br />
and
<br />
\begin{equation*}<br />
\begin{split}<br />
A^{\phi }\left( 2\pi \right) &= \sin \left( 2\pi \cos \theta \right) /\sin \theta \\<br />
A^{\theta }\left( 2\pi \right) &= \cos \left( 2\pi \cos \theta \right).<br />
\end{split}<br />
\end{equation*}<br />
What does this mean?
