Understanding parallel transfer

[pervect]

I tried to in the last message. I know this is dragging out and I am sorry about that. I will soon give up.

I don't see why setting the directional derivative of a vector in the tangent direction to zero is equivalent to parallel transport, dragging a vector along a path in such a way as to keep it parallel to itself. I'm not even sure that "to itself" is correct, since I'm not sure how you define parallelism in curved coordinates. It has been stated that this is the definition of parallel transport. But that does not say what its physical significance is or why we do it.

I'd have to agree that the definition of parallel transport based on the vanishing of the directional derivative in the direction of transport is not all that intuitive, though it's mathematically valid and an accepted definition of parallel transport.

Personally I prefer thinking of parallel transport with more intuitive constructs, though these more intuitive constructions are specific to using the Levi-Civita connection. This can be a problem if one wants to move on to theories (other than GR) that use connections other than the Levi-Civita connection.

One example of a more intuitive construction of parallel transport of this that hasn't (as far as I know) been mentioned in this thread is Schild's ladder <<wiki link>>. Though it is more or less equivalent ideas have been presented.

The physical construction of the Schild's ladder is based on the idea that if you make a quadrilateral with equal sides (of course, you need a metric to find the length of the sides), the opposite sides should be "parallel". Or rather, parallel transported.

The subject of proving that these intuitive or semi-intuitive constructions are equivalent to the definition in terms of the vanishing of the directional derivative (which I use routinely, but don't think about much) is one that I'd have to think about more. But I'm mostly interested in trying to motivate the idea of parallel transport rather than provide hard proofs.

There's one other issue that's worth exploring. If you have a vector field $u^b$, and it's gradient $\nabla_a u^b$ is zero at that point, so that the directional derivative vanishes in all directions, I'm reasonably certain that this implies that the space is flat at that point.

Trying to impose the condition that the gradient of the vector vanishes in all direction over-specifies the problem. If you have a surface that's curved at some point, the best you can do is make the gradient vanish in one direction. When you do this, you've parallel transported the vector in the direction that the gradient vanishes.

The argument for saying that the total vanishing of the gradient $\nabla_a u^b$ implies the surface is flat requires knowing that parallel transporting a vector around a loop on a curved surface rotates the vector. I'm not sure we ever got this far, because we've been mostly talking abut the definition of parallel transport, rather than using it and exploring the consequences. Though you might want to read back over stevedaryl's post, I think he does something similar with a sphere.

If the gradient can be made to vanish in all directions, then you can transport the vector "up", "right", "down", "left", around a small closed quadrilaterial, and the vector will never rotate. Thus the vector must come back to its starting point unrotated. Which is only possible if the space is flat at that point.

 

[vanhees71]

You are very careful to keep the mathematical statements correct. Perhaps you could indicate to me a book for learning that myself. I'd be grateful. Is Schutz''s "Geometrical methods of mathematical physics" good, for instance? Or is it not necessary for what I am trying to understand?

I think the modern abstract formalism is not necessary for a first encounter with GR. I'm myself not an expert in GR (rather in theoretical high-energy hadron/nuclear physics), and I've learned GR out of interest for myself, using Landau/Lifshitz vol. II, which uses the Ricci calculus of general-covariant tensor analysis.

The idea is to start with a definition of a covariant derivative of vector fields $V^{\mu}(q)$, which leads to a 2nd-rank tensor field $\nabla_{\nu} V^{\mu}$. We work in holonomous bases and co-bases. Thus one would like to have a transformation behavior under general transformations of the coordinates (diffeomorphism)
$$\nabla_{\nu}' V^{\prime \mu}=\frac{\partial q^{\rho}}{\partial q^{\prime \nu}} \frac{\partial q^{\prime \mu}}{\partial q_{\sigma}} \nabla_{\rho} V^{\sigma}.$$
The partial derivative obviously doesn't fulfill this requirement since
$$\partial_{\nu}' V^{\prime \mu}=\frac{\partial q^{\rho}}{\partial q^{\prime \nu}} \partial_{\rho} \left (\frac{\partial q^{\prime \mu}}{\partial q^{\sigma}} V^{\sigma} \right),$$
i.e., you get an extra term
$$\partial_{\nu}' V^{\prime \mu}=\frac{\partial q^{\rho}}{\partial q^{\prime \nu}} \frac{\partial q^{\prime \mu}}{\partial q^{\sigma}} \partial_{\rho} V^{\sigma} +\frac{\partial q^{\rho}}{\partial q^{\prime \nu}} \frac{\partial^2 q^{\prime \mu}}{\partial q^{\sigma} \partial q^{\rho}} V^{\sigma}.$$
Thus we need a connection, symbolized by Christoffel symbols
$$\nabla_{\nu} V^{\mu}=\partial_{\nu} V^{\mu} + {\Gamma^{\mu}}_{\nu \rho} V^{\rho},$$
where the Christoffel symbol is not a tensor but transforms under general coordinate transformations such as to cancel the "bad term" from the partial derivative:
$${\Gamma^{\prime \alpha}}_{\beta \gamma}=\frac{\partial q^{\prime \alpha}}{\partial q^{\mu}} \frac{\partial q^{\nu}}{\partial q^{\prime \beta}} \frac{\partial q^{\rho}}{\partial q^{\prime \gamma}} {\Gamma^{\mu}}_{\nu \rho} - \frac{\partial q^{\prime \nu}}{\partial q^{\beta}} \frac{\partial q^{\prime \rho}}{\partial q^{\gamma}} \frac{\partial^2 q^{\prime \alpha}}{\partial q^{\prime \nu} q^{\prime \rho}}.$$
Thus the Christoffel symbols are not a tensor but transform such that the "bad term" for the transformation of the partial derivative is cancelled.

The choice of the Christoffel symbols seem to be quite arbitrary, and without further constraints they indeed are. The only thing we can say is that the socalled torsion,
$${\Sigma^{\mu}}_{\nu \rho}={\Gamma^{\mu}}_{\nu \rho}-{\Gamma^{\mu}}_{\rho \nu}$$
is a tensor since then the additional term in the transformation of the Christoffel symbol cancels, because the partial derivatives commute.

You can specify the connection further by assuming that for a scalar field the covariant derivative should be the partial derivative, leading to a vector. Also for the partial derivative the product rule should hold, i.e.,
$$\nabla_{\mu} S=\partial_{\mu} S, \quad \nabla_{\rho}(V^{\mu} W^{\nu})=(\nabla_{\rho} V^{\mu})W^{\nu} + V^{\mu} \nabla_{\rho} W^{\nu}.$$
Applying this to the scalar $V^{\mu} W_{\mu}$ one deduces the rule for covariant vector components
$$\nabla_{\nu} W_{\mu}=\partial_{\nu} W_{\mu} - {\Gamma^{\rho}}_{\nu \mu} W_{\rho}.$$
The parallel transport of a vector along a curve is defined by
$$\mathrm{D}_{\tau} V^{\mu} = \mathrm{d}_{\tau} V^{\mu} + \mathrm{d}_{\tau} q^{\nu} {\Gamma^{\mu}}_{\nu \rho} V^{\rho}=\mathrm{d}_{\tau} q^{\nu} \nabla_{\nu} V^{\mu}.$$
Finally since we have a metric we'd also like to have the scalar products of two vectors invariant under parallel transport, leading to
$$\nabla_{\rho} g_{\mu \nu}=0.$$
This also ensures that indices can be lowered and raised for covariant derivatives as usual, i.e.,
$$\nabla_{\mu} V_{\rho}=g_{\rho \sigma} \nabla_{\mu} V^{\sigma}.$$
If we further assume that the torsion tensor vanishes identically this leads to the unique definition of the then symmetric Christoffel symbols as derivatives of the metric:
$${\Gamma^{\mu}}_{\nu \rho}=\frac{1}{2} g^{\mu \sigma}(\partial_{\nu} g_{\sigma \rho} + \partial_{\rho} g_{\sigma \nu}-\partial_{\sigma} g_{\nu \rho}).$$