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I have a doubt regarding the Riemann tensor in a LIF. The general expression of the Riemann tensor is:

$R^{\alpha}_{\beta \mu \nu} = \Gamma ^{\alpha}_{\beta \nu, \mu} - \Gamma ^{\alpha}_{\beta \mu, \nu} -\Gamma ^{\alpha}_{\kappa \nu} \Gamma ^{\kappa}_{\beta \mu} +\Gamma ^{\alpha}_{\kappa \mu} \Gamma ^{\kappa}_{\beta \nu} \tag{1}$

where $A_{ \alpha \beta, \mu \nu } = \dfrac{\partial A_{\alpha \beta}}{\partial x^{\mu}\partial x^{\nu}}$

we recognize the first two parts which are linear in the second derivative and the other two parts nonlinear in the first derivatives of metric tensor.

The R. tensor has a very nice form when computed in a Locally Inertial Frame:

$R^{\sigma}_{ \beta \mu \nu} = \dfrac{1}{2}g^{\sigma \alpha}[g_{\alpha \nu, \beta \mu} - g_{\alpha \mu, \beta \nu} +g_{\beta \mu, \alpha \nu} -g_{\beta\mu, \alpha \nu} ] \tag{2} $

We know that in flat spacetime and consequently in a LIF Christoffel symbols vanish. The nonlinear part of $(1)$ is zero, thus we only have the second derivatives of metric tensor i.e. $(2)$ which are related to the derivatives of Christoffel symbols in $(1)$.

The WELL known definition of Local Inertial Frame (or LIF) is a local flat space which is the mathematical counterpart of the general equivalence principle. If we know $g_{\mu\nu}$ and their first derivatives (i.e. $\Gamma^{\alpha}_{\mu \nu}$) in the point $X$, in a general spacetime we can always determine a locally (inertial) frame $\xi^{\alpha}(x)$ in the neighborhood of $X$. From (e.g. {1}) the following expression:

$ \dfrac{\partial ^2 \xi ^{\beta}}{\partial x^{\mu} \partial x^{\nu} } = \dfrac{\partial \xi^{\beta}}{\partial x^{\lambda}} \Gamma^{\lambda}_{\mu \nu} \tag{3}$

we are able to write the series expansion near $X$ up to the second order:

$ \underset{x \approx X }{ \xi^{\beta}(x)} = \xi^{\beta}(X) + [\dfrac{\partial \xi^{\beta}(x)}{\partial x^{\lambda}}]_{x=X} (x^{\lambda}- X^{\lambda}) + \dfrac{1}{2}[\dfrac{\partial \xi^{\beta}(x)}{\partial x^{\lambda}}\Gamma^{\lambda}_{\mu\nu}]_{x=X} (x^{\mu}- X^{\mu}) (x^{\nu}- X^{\nu}) + \text{higher orders} := \\ := a^{\beta} + b^{\beta}_{\lambda}(x^{\lambda}- X^{\lambda}) + \dfrac{1}{2} b^{\beta}_{\lambda} \Gamma^{\lambda}_{\mu\nu}(x^{\mu}- X^{\mu}) (x^{\nu}- X^{\nu}) + \text{higher orders} $

in addition, since it must be a locally flat space we have to relate the old frame to the new via $\eta_{\mu \nu}$ :

$g_{\mu \nu}(X)= \eta_{\alpha \beta} \dfrac{\partial \xi(x) ^{\alpha}}{\partial x^{\mu}}|_{x=X}\dfrac{\partial \xi (x)^{\beta}}{\partial x^{\nu}}|_{x=X} \equiv \eta_{\alpha \beta} b^{\alpha}_{\mu} b^{\beta}_{\nu}$

From the previous equation we find $b^{\beta}_{\mu}$. As regards $a^{\beta}$ there is an ambiguity but we have still the freedom to make a Lorentz transformation and the new frame is still locally inertial.

My question is: in a LIF why are Christoffel symbols equal to zero but their derivatives not?

My possible answer:

If we differentiate the rhs of $(3)$ with respect $x^{\sigma}$ we can use again $(3)$ with with other indices:

$ \dfrac{\partial ^3 \xi ^{\beta}}{\partial x^{\sigma} \partial x^{\mu} \partial x^{\nu} } = \dfrac{\partial \xi^{\beta}}{\partial x^l} \Gamma ^l _{\sigma \lambda} \Gamma^{\lambda}_{\mu \nu} + \dfrac{\partial \xi^{\beta}}{\partial x^s} \Gamma ^{s}_{\mu \nu , \sigma} \tag{4} $

After moving in a LIF (i.e. connections vanishes), from $(4)$ we can isolate $ \Gamma ^{s}_{\mu \nu , \sigma}$ .

What do you think?

{1} Carroll, S. M. (2019). Spacetime and geometry. Cambridge University Press.

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    $\begingroup$ May it be the fact that you can always define a locally flat spacetime, but without being able to eliminate tidal forces? $\endgroup$
    – Rob Tan
    Aug 4, 2020 at 15:12
  • $\begingroup$ @Rob In a local inertial frame you do not "feel" gravity. In fact a free falling body equation of motion can be mapped into another of the kind x'' = 0 i.e. no gravity force acting. If you live in a LIF you perceive a flat space so a null Riemann tensor but if you are from a curved space then you are able to define a LIF in which the derivative of connections are relevant (see eqs (3) and (4)). The point is "how do you match these two facts?". In my opinion, in order to test a non zero R. tensor you have to "use" a scale distance greater than LIF scale i.e. an infinitesimal scale. $\endgroup$
    – Siderius
    Aug 4, 2020 at 15:31
  • $\begingroup$ @Rob In other words Riemann tensor and connections would be zero up to a O(||x-X||) correction which is perceived beyond the scale definition of a LIF (centered in X). $\endgroup$
    – Siderius
    Aug 4, 2020 at 15:32
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    $\begingroup$ I don't know if we are talking of the same thing. If you have a point $\boldsymbol{p}\in\mathcal{M}_n$ on a manifold you are always able to define a local geodesic frame with the coordinates transformation $$\tilde{x}^\alpha =x^\alpha-p^\alpha+\frac{1}{2}\left.\Gamma_{\beta\gamma}^\alpha\right|_{\boldsymbol{p}}\left(x^\beta-p^\beta\right)\left(x^\gamma-p^\gamma\right)$$ such that $\left.\tilde{\Gamma}_{\beta\gamma}^\alpha\right|_{\boldsymbol{p}}=0$. Is that the LIF you are talking about? $\endgroup$
    – Rob Tan
    Aug 4, 2020 at 17:45
  • $\begingroup$ @Rob you have shown something of second order. But the definition I have studied have a linear term as I have just written i the edit. $\endgroup$
    – Siderius
    Aug 4, 2020 at 21:06

2 Answers 2

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I think I have answered this before but one can construct co-ordinates in which $$ g_{\mu\nu}(x)= \delta_{\mu\nu}- \frac 13 R_{\mu\sigma \nu\tau}(0) x^\sigma x^\tau + O(|x|^3),\\ {\Gamma^{\lambda}}_{\mu\nu}(x)= -\frac 13 (R_{\lambda\nu\mu\tau}(0)+R_{\lambda\mu\nu\tau}(0))x^\tau+ O(|x|^2). $$ Similarly we we can construct local vielbein frames in which we have a co-frame and spin connection $$ e^{*a}_\mu(x)= \delta_{a \mu}- \frac 16 R_{a \sigma \mu\tau}(0) x^\sigma x^\tau +O(x^2),\\ {\omega^a}_{b\mu}(x)=- \frac 12 {R^a}_{b\mu\tau}(0)x^\tau+O(|x|^2). $$

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A way of constructing a local and inertial frame is using Riemann normal coordinates. As stated by S. Carroll book:

(Riemann normal coordinates) provide a realization of the locally inertial coordinates (...). They are not unique; there are an infinite number of non-Riemann normal coordinates systems (related to a neighborhood of the point p) in which $g_{\mu \nu }(p)= \eta_{\mu \nu }$ and $\partial_{\sigma} (p)=0$ but in an expansion around $p$ they will differ from Riemann normal coordinates only at third order in $x^{\mu}$.

Riemann normal coordinates lead us to the metric tensor expression:

$g_{\mu \nu }(x)= \eta_{\mu \nu} + C_{\mu \nu, \alpha \beta} x^{\alpha} x^{\beta}+.. $

where $C_{\mu \nu, \alpha \beta}$ is the second order coefficient which depends on the second derivatives of $g_{\mu \nu}$.

This definition could be taken as local if $x \approx 0$.

In this way the derivatives of Christoffel symbols have sense because, roughly speaking, the second derivative of the metric tensor (i.e. what characterizes Ch. symbols) could give a constant. In fact, as e.g. already stated here:

$\partial_{l} \Gamma^{\lambda}_{\rho \nu} = \eta^{\lambda \tau}( C_{\tau \nu, k \rho} + C_{\tau \rho , k \nu} - C_{\rho \nu , k \tau}) \delta_{l}^k + ...$

In this way we may even have:

$lim_{x \rightarrow 0}\Gamma^{\lambda}_{\rho \nu} =0$

with

$lim_{x \rightarrow 0}\Gamma^{\lambda}_{\rho \nu,\alpha} \neq 0$

In conclusion the sentence

After moving in a LIF (i.e. connections vanishes), from (4) we can isolate the derivative of Ch. symbols

isn't right because Ch. symbols are not tensors i.e. they can be zero in a frame and non-zero in another frame; for this reason the general covariance principle cannot be applied (like when we move from an the expression with ordinary derivatives to the same with covariant derivatives).

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