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There is an expression indicating the change of the vector parallel translation along a closed infinitesimal curve in curvilinear coordinates (one way of introducing curvature tensor): $$ \Delta A_{k} = \oint \Gamma^{i}_{kl}A_{i}dx^{l}. $$

Then, before using Stokes' theorem, there need to require the uniqueness of the integrand.

So, Christoffel symbol and vector's component expanded to a series near some point ${x^{\alpha}}^{0}$ on a curve:

$$ \Gamma^{i}_{kl} = {\Gamma^{i}_{kl}}_{0} + (\partial_{\alpha}\Gamma^{i}_{kl})_{0}(x^{\alpha} - {x^{\alpha}}^{0}) + \frac{1}{2}(\partial_{p\alpha}\Gamma^{i}_{kl})_{0}(x^{\alpha} - {x^{\alpha}}^{0})(x^{p} - {x^{p}}^{0}) + ..., $$

$$ A_{i} = (A_{i})_{0} + (\partial_{\alpha}A_{i})_{0}(x^{\alpha} - {x^{\alpha}}^{0}) + \frac{1}{2}(\partial_{p \alpha}A_{i})_{0}(x^{\alpha} - {x^{\alpha}}^{0})(x^{p} - {x^{p}}^{0}) + ... $$ After that there are a words about the fact that the ambiguity of the operation of parallel transport is covered in quadratic terms. What are some reasons to "believe" that?

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Are you describing something in a book? What book? –  Ben Crowell May 7 '13 at 12:08
From Fock, "The theory of Space Time & Gravitation", or from Landau, "Field Theory". –  PhysiXxx May 7 '13 at 12:11

1 Answer 1

up vote 2 down vote accepted

Obviously, you are talking about the derivation in Landau-Lifshitz book. I admit that this is not the clearest explanation of what is going on. (On the other hand that is not the first and not the last place like that in Landau-Lifshitz...)

Anyway, I don't really see what is your problem -- just use Stokes' theorem from (6.19): $$\Delta A_i = \frac12\int df^{lm}\left(\partial_l(\Gamma_{km}^iA_i)-\partial_m(\Gamma_{kl}^iA_i)\right)$$ Then substitute your decompositions, account for the fact that everything with index "0" is constant, expand and you'll get something like (but note the absence of a linear term in $x-x^0$): $$\Delta A_i = \frac12 \int df^{lm}\left((R_{klm}^i)_0(A_i)_0+(\text{some other stuff}_{\alpha\beta lm})_0(x^\alpha-x^{\alpha0})(x^\beta-x^{\beta0})+...\right) $$ Finally, you just remember that integration is over infinitesimal volume $\Delta f^{lm}$, therefore $(x^\alpha-x^{\alpha0})(x^\beta-x^{\beta0})\sim\Delta f^{\alpha\beta}$ and this terms are, indeed, of the second order.

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"...just use Stokes' theorem from..." I can't use it if integrand isn't single-valued function. Furthermore, the expression for the derivative $\partial_{l}A_{i} = \Gamma^{n}_{il}A_{n}$ can be used only after the requirements of "uniqueness" of $A_{n}$ along the curve. –  PhysiXxx May 7 '13 at 21:10
@PhysiXxx yor decompositions are single-valued? –  Kostya May 7 '13 at 21:51
Only up to the quadratic terms of the expansion. So, my question is about a my question about why it is considered. –  PhysiXxx May 7 '13 at 21:55
@PhysiXxx Have you noticed that quadratic terms (whatever they are) are, indeed, of the second order in $\Delta f^{lm}$? –  Kostya May 8 '13 at 8:46

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