In my recent studies of curvature, I worked with the Riemann tensor and the equation:(adsbygoogle = window.adsbygoogle || []).push({});

([itex]\delta[/itex]V)^{a}= A^{[itex]\mu[/itex]}B^{[itex]\nu[/itex]}R^{a}_{b[itex]\mu[/itex][itex]\nu[/itex]}V^{b}

Now previously, I worked in 2D with the 2 sphere. While doing so, I learned that if I set my x^{1}coordinate to be θ and my x^{2}coordinate to be ø, then the vectors that serve to be the legs of the loop that I am transporting around would be as follows:

A^{[itex]\mu[/itex]}= <θ, 0>

B^{[itex]\nu[/itex]}= <0, ø>

and then of course the vector that I parallel transport would be as follows:

V^{b}= <θ, ø>

Now this may work for 2 dimensions, but what if I have 3 or more dimensions? With only 2 vectors being the legs of the loop, there wouldn't be enough vectors for me to give each individual coordinate its own leg with every other component being 0 (as shown above with A^{[itex]\mu[/itex]}and B^{[itex]\nu[/itex]}).

How do I deal with this? Is it even a requirement for every coordinate to have its own leg that is reminiscent of a unit vector? Page 5 on the following PDF gave me the impression that it is a requirement:

http://www.physics.ucc.ie/apeer/PY4112/Curvature.pdf

Is it possible for one of the legs of the loop to have more than one type of coordinate in it (like A^{[itex]\mu[/itex]}= <r , θ, 0>) ?

**Physics Forums | Science Articles, Homework Help, Discussion**

Dismiss Notice

Join Physics Forums Today!

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

# More questions on parallel transport

**Physics Forums | Science Articles, Homework Help, Discussion**