For two nearby points in General Theory of Relativity. The change in the vector components when parallel transported is given by

$$\mathrm{d}A^\mu = {\Gamma^\mu}_{\sigma\nu} \mathrm{d}x^\sigma A^\nu$$

Now, since the parallel transport change must depend on the path taken between the two points, what path does the christoffel symbol equate to being taken between the two points for parallel transporting a vector?

Geometric/intuitive pictures/arguments will be appreciated alongside rigour as I am from a physics background.


$dx^\sigma$ in the formula means that the infinitely small transport is considered. For two infinitely close points the path dependence vanishes (unless you add some finitely sized loop to the infinitely small displacement). When you want to make the transort along some finite path, you should integrate that formula and get $$\Delta A^\mu=\int_L \Gamma^\mu_{\sigma\nu}A^\nu(x)\,dx^\sigma.$$ Now the path dependence is clearly apparent in the choice of integration path. For some path $L_1$ you will get one result, and for some other path $L_2$ - some other result. Be careful that $A^\nu(x)$ should be taken at the point where you have transorted it so far, and not be the initial vector.

Geometric intuition for these matters is like this. The geometry is flat with respest to the quantities of first order, and ball-like (/saddle-like) with respest to the quantities of second order. So as long the formula $dA^\mu=\Gamma^\mu_{\sigma\nu}A^\nu\,dx^\sigma$ considers the Chistoffel symbols which are first order, and the first power of $dx^\sigma$ is used, there is no path dependence. $\Gamma^\mu_{\sigma\nu}$ taken in some point describe possible linear distortions of coordinate grid on flat background. That is, unit segments can grow or shrink, and coordinate lines can converge and diverge. You can check that taking two consecutive infinitesimal displacements in different orders ($dx_1^\sigma$ then $dx_2^\sigma$, or $dx_2^\sigma$ then $dx_1^\sigma$) makes the same $dA^\mu$ - use the explicit expression for $\Gamma^\mu_{\sigma\nu}$.

  • $\begingroup$ If path dependence vanishes for an infinitesimal distance, then why is only christoffel symbol used and not some other connection? $\endgroup$ – Isomorphic Aug 24 '14 at 15:41
  • $\begingroup$ The only other connection that can be used is a torsion generalisation of Christoffel symbol. It would not lead to Euclidean geometry in the flat space case. It can be used in some extension of GR, but for GR itself it is unneeded. $\endgroup$ – firtree Aug 24 '14 at 15:50

The christoffel symbol is a function of $(x)$ so different for each paths, so the resulting parallel transport vector for each paths are different as show in this figure enter image description here

If $$\Delta A^\mu=\int_L \Gamma^\mu_{\sigma\nu}A^\nu(x)\,dx^\sigma.$$ is indepedent of paths as you may expected, it will be means that $$0=\oint_C \Gamma^\mu_{\sigma\nu}A^\nu(x)\,dx^\sigma= A^\mu -\parallel_C A^\mu=\frac1 2 R^\mu{}_{\lambda\alpha\beta}A^\lambda\,\iint_\sigma dx^\alpha \wedge dx^\beta $$ Which means the spacetime is flat. To be more clear I will show another more familiar figure but with the same settings

enter image description here

The second and first picture indeed the same situation since parallel transport invariance under reverse the direction so we have the third figure below enter image description here


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.