In general relativity there is always a locally flat system, for which the transformed metric is the Minkowski metric up to second order at one point. But this does not mean, that the system, described by the new coordinates is automatically an inertial frame. Is that right?

If this is true: why can I apply Lorentz transformation of special relativity in that metric? The metric is locally flat so SR applies - but the frame is nevertheless no inertial frame. Isn't that a contradiction, because Lorentz-transformations are valid only in inertial systems?

I'm confused...

  • 2
    $\begingroup$ Though it isn't obvious from the title this is effectively a duplicate of How can locally Euclidean space of zero curvature accumulate to non-zero global curvature? $\endgroup$ Aug 16, 2021 at 17:22
  • $\begingroup$ I can transform locally always to a coordinate system, where the metric is like Minkowski metric? This would be normal coordinates, but not coordinates, which describe a free falling system. The question is: Even when I have a fixed point on earth, I can find a coordinate system which the metric is Minkowski like. But, as acceleration is not zero this is of course not an inertial system. Can I nevertheless apply special relativity in such frame? $\endgroup$
    – MichaelW
    Aug 16, 2021 at 17:46
  • $\begingroup$ Your metric at a fixed point would locally look like a Rindler metric not Minkowski. $\endgroup$ Aug 16, 2021 at 19:11

1 Answer 1


I tis very important to notice that you can do this at a given point, but not at any given curve. In fact, it is true that for geodesics you can define a locally flat coordinate system such that the Christoffel symbols vanish along the curve. This is compatible with the idea of the equivalence principle for $inertial$ observers.

However, for more general trajectories, it is not possible to define a such a coordinate system: the Christoffel symbols get corrections due to the acceleration of the trajectory, and these are compatible with the notion of fictious forces that are seen due to non-inertiality.

The way to analyze the problem above is by means of the Fermi normal coordinates (FNC). These are a coordinate system that is associated with a timelike trajectory and describes "what an observer undergoing this trajectory sees". In case you want learn more about this, I recommend https://arxiv.org/abs/1102.0529. In any case, it is possible to show that the metric in FNC can be written locally around the curve (up to first order in acceleration and curvature) as

$$\begin{aligned} &g_{\tau \tau}=-\left(1+2 a_{\mathrm{i}}(\tau) x^{i}+R_{0 \mathrm{i} 0 \mathrm{j}}(\tau) x^{i} x^{j}\right), \\ &g_{\tau i}=-\frac{2}{3} R_{0 \mathrm{jik}}(\tau) x^{j} x^{k} \\ &g_{i j}=\delta_{\mathrm{ij}}-\frac{1}{3} R_{\mathrm{ikjl}}(\tau) x^{k} x^{l} \end{aligned},$$ where $\tau$ is the proper time of the trajectory (and the timelike coordinate of the FNC) and $x^i$ denote the proper distance from the trajectory, which are the spacelike trajectories of the FNC. $a(\tau)$ is the trajectories proper acceleration and $R(\tau)$ is the curvature tensor along its motion.

The Christoffel symbols to first order in acceleration and curvature then read

$$\begin{aligned} \Gamma_{i j}^{\tau} &=\frac{1}{3}\left(R_{0 \mathrm{ijm}}+R_{0 \mathrm{jim}}\right) x^{m} \\ \Gamma_{\tau i}^{\tau} &=a_{\mathrm{i}}+R_{0 \mathrm{i} 0 \mathrm{~m}} x^{m} \\ \Gamma_{\tau \tau}^{\tau} &=0 \\ \Gamma_{j k}^{i} &=\frac{1}{3}\left(R_{\mathrm{j} \mathrm{km}}^{\mathrm{i}}+R_{\mathrm{k} \mathrm{jm}}^{\mathrm{i}}\right) x^{m} \\ \Gamma_{\tau j}^{i} &=R_{0 \mathrm{mj}}{\mathrm{j}} x^{m} \\ \Gamma_{\tau \tau}^{i} &=a^{\mathrm{i}}+R_{0}^{\mathrm{i}}{ }_{0 \mathrm{~m}} x^{m} \end{aligned},$$ which can be seen to vanish if $a^i = 0$ (inertial motion) and $x^i = 0$ (on top of the curve).


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