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This comes from an earlier post. Why is this approximation true?

We can always choose coordinates such that the Christoffel symbols vanish at an arbitrary point $x_0$ of our choosing, but the presence of spacetime curvature means that they will not vanish in a nonzero neighborhood of $x_0$. The best we can do is choose Riemann normal coordinates, in which case $$\Gamma^i_{\ \ jk}(x) \approx \frac{1}{2}R^i_{\ j\ell k}(x_0) (x^\ell-x_0^\ell)$$

-J.Murray. https://physics.stackexchange.com/questions/614243/how-do-you-describe-a-geometry-where-the-christoffel-symbols-vanish

I believe $\Gamma$ and ${R^a}_{bcd}$ should be of the same order, due to how ${R^a}_{bcd}$ is defined in terms of $\Gamma$ by parallel transport. This approximation seems related to Taylor's theorem.

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  • $\begingroup$ It doesn't affect the spirit of your question, but I've fixed my linked answer. As Mike very nicely lays out, my previous expression was for the spin connection rather than the Christoffel symbol. $\endgroup$
    – J. Murray
    Feb 13, 2021 at 21:42

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I think that the correct expressions for Riemann normal coordinates about an arbitrarily chosen origin $x^\mu=0$ are
$$ g_{\mu\nu}(x)= \delta_{\mu\nu}- \frac 13 R_{\mu\sigma \nu\tau}(0) x^\sigma x^\tau + O(|x|^3). $$ and
$$ {\Gamma^{\lambda}}_{\mu\nu}(x)= -\frac 13 (R_{\lambda\nu\mu\tau}+R_{\lambda\mu\nu\tau})x^\tau+ O(|x|^2). $$

Suppose that we have constructed coordinates at O such that $$ g_{\mu\nu}(x)= \delta_{\mu\nu}+\frac 12 A_{\mu \nu\sigma \tau} x^\sigma x^\tau + O(|x|^3), $$ where $A_{\mu\nu\sigma\tau}$ is symmetric under $\mu\leftrightarrow \nu$ and under $\sigma\leftrightarrow \tau$. Then $$ {\Gamma^{\lambda}}_{\mu\nu}(x)=\frac 12(A_{\lambda\mu\nu\tau}+A_{\lambda\nu\mu\tau}-A_{\mu\nu\lambda\tau})x^\tau+O(|x^2|), $$ and $$ R_{\rho\sigma\mu\nu}(0)= \frac 12(A_{\rho\nu\sigma\mu}-A_{\nu\sigma\mu\rho}+A_{\mu\sigma\nu\rho}-A_{\rho\mu\sigma\nu}). $$ We can verify that this curvature tensor satisfies the pair exchange symmetry $R_{\mu\nu\rho\sigma}= R_{\rho\sigma\mu\nu}$, the two antisymmetries $$ R_{\rho\sigma\mu\nu}=- R_{\sigma\rho\mu\nu}= -R_{\rho\sigma\nu\mu} $$ and the first Bianchi identity: $$ R_{\rho\sigma\mu\nu}+R_{\rho\mu\nu\sigma}+R_{\rho\nu\sigma\mu}=0. $$ Now in d=4, for example, the array of numbers $A_{\mu\nu\sigma\tau}$ has 10$\times$10=100 independent entries while its symmetries lead $R_{\rho\sigma\mu\nu}$ to have only 20. We have, however, at our disposal 80 degrees of freedom in the coefficients ${b^\mu}_{\rho\sigma\tau}$
$$ x^\mu \to x^\mu + {b^\mu}_{\rho\sigma\tau} x^\rho x^\sigma x^\tau+\ldots $$ of a local change-of-coordinates expansion that keeps the metric euclidean up to quadratic corrections. We should therefore be able to find a co-ordinate system in which $A_{\nu\sigma\mu\rho}$ is expressed in terms of $R_{\rho\sigma\mu\nu}$. Indeed, counting degrees of freedom shows that in any number of dimensions we can use a cubic change of co-ordinate to reduce $A_{\mu\nu\sigma\tau}$ to the form $$ A_{\mu\nu\sigma\tau}= \frac{c}{2}(R_{\mu\sigma\nu\tau}+R_{\nu\sigma\mu\tau}). $$ Note that this expression has the correct $\mu\leftrightarrow \nu$ and $\sigma\leftrightarrow \tau$ symmetries. If we plug this form for $A_{\mu\nu\sigma\tau}$ into the formula for $R_{\rho\sigma \mu\nu}$ and the use the symmetries (including the first Bianchi identity) we find $$ R_{\rho\sigma \mu\nu} =\frac{c}{4}(-6 R_{\rho\sigma \mu\nu}), $$ and so $c=-2/3$.

Thus we can find coordinates in which
$$ g_{\mu\nu}(x)= \delta_{\mu\nu}- \frac 13 R_{\mu\sigma \nu\tau}(0) x^\sigma x^\tau + O(|x|^3). $$ and
$$ {\Gamma^{\lambda}}_{\mu\nu}(x)= -\frac 13 (R_{\lambda\nu\mu\tau}+R_{\lambda\mu\nu\tau})x^\tau+ O(|x|^2). $$ These new coordinates are precisely the geodesic normal coordinates. In other words, straight lines through the origin in ${\mathbb R}^d$ are geodesics to $O(|x|^2)$.

We can do something similar for Vielbeins and the assocated "spin connection". We seek an orthonormal vielbein co-frame ${\bf e}^{*a}$ such that $$ e^{*a}_\mu e^{*a}_\nu=g_{\mu\nu}= \delta_{\mu\nu}- \frac 13 R_{\mu\sigma \nu\tau}(0) x^\sigma x^\tau + O(|x|^3). $$ We can take $$ e^{*a}_\mu= \delta_{a \mu}- \frac 16 R_{a \sigma \mu\tau}(0) x^\sigma x^\tau +\ldots $$ and confirm that $$ g^{\mu\nu} e^{*a}_\mu e^{*b}_\nu= \delta^{ab}+ O(|x|^3). $$

Knowing the ${\bf e}^{*a}$, we compute the spin connection from the definition $$ \nabla_\mu {\bf e}^{*a}=-{\omega^{a}}_{b\mu} {\bf e^{*b}} $$ as $$ {\omega^a}_{b\mu}(x)= -e^\nu_b(\partial_\mu e^{*a}_\nu-{\Gamma^\sigma}_{\nu\mu}e^{*a}_\sigma)\\ = \frac 16 ({R^a}_{\mu b\tau}+{R^a}_{\tau b\mu})x^\tau-\frac 13({R^a}_{\mu b\tau}+{R^a}_{b\mu\tau})x^\tau+\ldots\\ = \left(-\frac 16 {R^a}_{\mu b\tau}-\frac 16 {R^a}_{\tau\mu b}-\frac 13 {R^a}_{b\mu\tau}\right)x^\tau+\ldots\\ =\left(+\frac 16 {R^a}_{b\tau\mu}-\frac 13 {R^a}_{b\mu\tau}\right)x^\tau+\ldots\quad\hbox{(Bianchi)} \\ = - \frac 12 {R^a}_{b\mu\tau}(0)x^\tau+O(|x|^2). $$ This last equation looks like the one you cited, so it is for the orthonormal frame-field "spin" connection and not for the coordinate frame Christofflel symbols

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  • $\begingroup$ +1 Thanks for working out the details on my mix-up. I'll go fix my other answer :) $\endgroup$
    – J. Murray
    Feb 13, 2021 at 21:38
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This is not a general expression, it applies in a very special set of coordinates $x^{\hat{\mu}}$ known as Riemann normal coordinates. These are defined by the derivatives of the metric being zero at the origin of the coordinates $g_{\hat{\mu}\hat{\nu},\hat{\kappa}}|_{x^{\hat{\lambda}} = 0} = 0$. This leads to non-trivial coordinate conditions that generic coordinates simply do not fulfill!

As a result, you have vanishing Christoffel symbols at the origin, and $$g_{\hat{\mu}\hat{\nu}} = g_{\hat{\mu}\hat{\nu}}|_0 + g_{\hat{\mu} \hat{\nu},\hat{\kappa}\hat{\lambda}}|_{0}x^{\hat{\kappa}} x^{\hat{\lambda}} +... $$ From that you can derive that $$g_{\hat{\mu} \hat{\nu},\hat{\kappa}\hat{\lambda}}|_{0} = -\frac{1}{3} R_{\hat{\mu}\hat{\kappa}\hat{\nu}\hat{\lambda}}$$ Then it is easy to derive the identity in the OP.

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I think I made a mistake in that answer - the standard expression for the Christoffel symbols in normal coordinates is

$$\Gamma^i_{\ \ jk} = \frac{1}{3}(R^i_{\ \ j\ell k} + R^i_{\ \ k \ell j})(x^\ell-x_0^\ell)$$

I think my expression is for the spin connection $\omega^i_{\ \ jk}$. I'll fix that, but it doesn't affect the motivation for your question.


In Riemann normal coordinates, the metric goes like

$$g_{\mu\nu} \approx \eta_{\mu\nu} -\frac{1}{3} R_{\mu\alpha \nu\beta} (0) x^\alpha x^\beta +\mathcal O(x^3)$$ Since the $\Gamma$'s are constructed from the derivatives of the metric, you can construct them immediately and see that they are linear in $x$. Note that the $R$ which appears in this expansion is the Riemann-Christoffel tensor evaluated at the coordinate origin.

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