mertcan said:
After your posts, I would like to add and ask: In parallel transport, covariant derivative is 0. Also I saw that covariant derivative is the projection of the ordinary derivative to the surface ( it may means that covariant derivative measures how the vector changes on the tangent plane of surface?).
That's a little misleading. An "ordinary" derivative acts on real-valued functions, not vectors. It's just that in flat Euclidean space, using Cartesian coordinates, we can define a covariant derivative so that the basis vectors all have covariant derivative zero. So in that case, the covariant derivative of a vector field becomes just the ordinary derivative applied to each component.
The business about projections is about a curved space that is embedded into flat Euclidean space. In that case, you can relate the covariant derivative in the curved space to the covariant derivative in the Euclidean space. Not every curved space needs to be thought of as embedded in Euclidean space, however. Our universe is curved due to gravity, but it isn't usually thought of as being embedded in any larger space.
Now, If we make a parallel transport, magnitude of vector changes by proportion of christoffel symbol.
No, christoffel symbols are not about the magnitude changing, but about the components changing.
Let's take the simplest example: flat 2-D space using polar coordinates r and \theta. Then:
g_{\theta \theta} = r^2
g_{r r} = 1
(the rest of the components of g are zero).
\Gamma^r_{\theta \theta} = -r
\Gamma^\theta_{r \theta} = \frac{1}{r}
\Gamma^\theta_{\theta r} = \frac{1}{r}
(all the other christoffel symbols are zero). Then let \vec{V} be defined by: \vec{V} = e_\theta, the basis vector in the \theta direction. Now, let's parallel-transport \vec{V} around the path \mathcal{P}(s) defined by letting (\mathcal{P}(s))^r=1, (\mathcal{P}(s))^\theta = s.
Then \frac{d\mathcal{P}^\nu}{ds} \nabla_\nu V^\mu = 0 for parallel transport. So:
\frac{d\mathcal{P}^\nu}{ds} (\partial_\nu V^\mu + \Gamma^\mu_{\nu \lambda} V^\lambda) = 0
Since \mathcal{P}(s) takes us just in the \theta direction, we have \frac{d\mathcal{P}^\theta}{ds} = 1 and \frac{d\mathcal{P}^r}{ds} = 0. So we have:
(\partial_\theta V^\mu + \Gamma^\mu_{\theta \lambda} V^\lambda) = 0
There are only two nonzero values of \Gamma^\mu_{\theta \lambda}: \Gamma^r_{\theta \theta} = -r and \Gamma^\theta_{\theta r} = \frac{1}{r}. So we have:
(\partial_\theta V^r - r V^\theta) = 0
(\partial_\theta V^\theta + \frac{1}{r} V^r) = 0
So V^r = -r \partial_\theta V^\theta. Plugging this into the top equation gives:
\partial_\theta (r \partial_\theta V^\theta) + r V^\theta = 0
So
(\partial_\theta)^2 V^\theta + V^\theta = 0
The general solution is: V^\theta = A cos(\theta) + B sin(\theta)
Since V^\theta = 1 at \theta = 0, we have:
V^\theta = cos(\theta)
Then we can use our equation for V^r:
V^r = -r \partial_\theta V^\theta = r sin(\theta).
The magnitude squared of V is g_{\mu \nu} V^\mu V^\nu = g_{rr} V^r V^r + g_{\theta \theta} V^\theta V^\theta = r^2 sin^2(\theta) + r^2 cos^2(theta) = r^2. So since r=1, the magnitude of V doesn't change.
That is why your vector's magnitude changes due to parallel transport ( embedded space (the surface of the sphere)) in post "6". To sum up: we both say in parallel transport covariant derivative is 0 and covariant derivative is the projection of the ordinary derivative to the surface ( covariant derivative measures how the vector changes on the tangent plane of surface). I think magnitude changes also correspond to vector change. How this is possible??
The magnitude is not (usually) changed by parallel transport, if there is a metric and the covariant derivative is compatible with that metric.
