In my lecture notes, normal coordinates are defined as the following:

Def.: Let $\left(\mathrm{e}_\mu\right)$ be a basis of $\mathcal{T}_p\left(\mathcal{M}\right)$. Normal coordinates in a neighborhood of $p\in\mathcal{M}$ are defined as the coordinate chart that assigns to $q=e\left(\mathbf{X}_p\right)\in\mathcal{M}$ the coordinates of the vector $X^\mu$.

and the lecturer proceeds to mathematically proof that in normal coordinates together with a Levi-Civita connection, the metric reduces to the Minkowski metric $g_{\mu\nu} = \eta_{\mu\nu} = \mathrm{diag}(-1, +1, +1, +1)$. The lecturer proceeds to define such a frame as a local inertial frame.

However, I felt that the proof was too abstract as it is an entirely based on Mathematical arguments alone. I am wondering if there is a more intuitive way to understand the equivalence of normal coordinates and local inertial frames, or a physical interpretation of this mathematical result.

  • $\begingroup$ When you go to a 'locally inertial frame' you don't experience any gravitational forces. In other words, the potential is zero; so the first derivative of $g_{\mu \nu}$ at that point is zero. A manifold is locally $\mathbb{R}^4$ and curvature makes it first appearance at second order in $h$. $\endgroup$ – Avantgarde Apr 8 at 16:47

Consider how you would actually construct a local inertial coordinate system as a freely falling observer. You would take a set of rigid rods (speed of sound within them almost equal to the speed of light) and extend them in your vicinity. The rods would also have labels written on them with a marker, so that you can check at which point of the rod events happen. Finally, you would also keep a clock on yourself that accurately measures time.

Now, you would observe events and assume that special relativity holds, ignorant of curved space and time, and label events according to your local time $t$ and positions $x^i$ on the rod where you see events happen. (You are aware that the signal gets to you at the speed of light and you correct for that in your coordinates.) Now ask yourself, what is the meaning of the coordinates $t,x^i$ in terms of a curved space-time viewpoint?

So, first of all, you are a freely falling observer moving along a geodesic. So the time $t$ you measure with your clock is the proper time along a geodesic. So if you take your four-velocity $u^\mu$ and exponentiate it from the moment $p$, $Exp_p(u^\mu)$, the exponential will take you by unit time along your trajectory from $p$. You can then see that the moment you would label with $t$ is obviously at the point $Exp_p((t-t_p) u^\mu)$. In other words, $t$ is one of the normal coordinates! Now for the rods, these extend roughly along space-like geodesics, at least if the curvature and time-variability of the field in your frame are small. So the labels on them are affine parameters along spatial geodesics. So you see by the same argument as for $t$ that $x^i$ represent the spatial part of the normal coordinates. In other words, freely falling observers naturally describe their environments in Riemann normal coordinates!


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