# Free fall coordinates/Fermi (normal) coordinates

It makes sense intuitively given the equivalent principle, and I've seen many times it stated, that for a free fall (geodesic) path in an arbitrary spacetime, we can choose our coordinate system to look inertial and "flat" locally on the geodesic (meaning $g_{ab}$ is diagonal with $\pm1$'s, Christoffel symbols vanish, but not truly flat because of course the Riemann curvature tensor doesn't vanish unless it just happens to be zero there). So along the geodesic it will look locally like an inertial frame. With some searching, these are called Fermi (normal) coordinates.

While this makes sense intuitively (maybe only because I've heard some variant of that said so many times), I've never seen the details written out. Given an arbitrary spacetime and coordinate system, I see how I can choose new coordinates at some point to make the metric diagonal. I also see there is still some coordinate freedom after doing this. But I'm not sure how to show it is always possible to do this along a path (and a related question seems to indicate that at least this step is possible for even non-geodesics), and I'm not sure how to do the final step and get the Christoffel symbols to vanish as well.

Question:

Please explain/demonstrate to me how for any (psuedo)Riemannian manifold we are free to choose a coordinate system such that the metric is the simple diagonal value and the Christoffel symbols vanish along a geodesic?

Since we use this all the time to apply physical intuition to a problem, it would be nice to see it explained in detail.

• – Qmechanic Dec 5 '14 at 15:03