### Question

Consider a 2-sphere with coordinates ##\left(\theta,\phi\right)## and a metric\begin{align}

{ds}^2={d\theta}^2+\sin^2{\theta}{d\phi}^2&\phantom {10000}(1)\nonumber

\end{align}a) Show that lines of constant latitude (##\phi=\rm{constant}##) are geodesics, and that the only line of constant latitude (##\theta=\rm{constant}##) that is a geodesic is the equator (##\theta=\pi/2##).

b) Take a vector with components ##V^\mu=\left(1,0\right)## and parallel transport it once round a circle of constant latitude. What are the components of the resulting vector, as a function of ## \theta##?

### Answers

Part (a) was quite easy because I have laboured over it before. If I had known that the general solution to $$\frac{d^2V}{d\phi^2}+kV=0

$$was$$

V=A\cos{\left(k\phi\right)}+B\sin{\left(k\phi\right)}

$$then I could have done the second part on my own. I needed some help from Prof Anthony Aguirre at UC Santa Cruz for that differential equation.

It seems that 'parallel transporting' is pretty wonky. Here's how it goes at 70° colatitude (polar angle):

Once the vector has got all round the sphere (at 70°) it's pointing in a new direction, even thought it's length never changes. (It appears to here from perspective effects.) Around equation 3.42 Carroll writes "It follows that the inner product of two parallel-transported vectors is preserved ... This means that parallel transport ... preserves the norm of vectors, the sense of orthogonality and so on.ðŸ’¡ðŸ’¡

Full answer here: Ex 3.05 2-sphere geodesics and parallel transport.pdf (6 pages)

To answer your question "Which are indeed correct. How did he know that?". It is basically the harmonic oscillator. After a few years' study of physics this kind of stuff becomes obvious. I would probably use A exp(-kx) + B exp(+kx) instead, because complex exponentials are in general easier to work with than sines and cosines.

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