# Riemann normal coordinates and inertial frames in general relativity

The weak equivalence principle states that for infinitesimal patches of space and time, the laws of nature are identical to that of special relativity. This is mathematically reflected in the existence of Riemann normal coordinates at every point on the manifold. Geodesics are straight lines in these coordinates as expected since particles are expected to move in straight lines in special relativity. However, I don't completely understand the relationship between these coordinates and inertial frames.

Consider the case of a particle moving along a geodesic; we can construct normal coordinates at every point on its trajectory. However, depending on how I define my tetrad basis, the coordinates of each of the points close to the particle will be different. It appears to me that there is a special orientation to the tetrad basis we can choose. We can define the "time" basis vector to be tangent to the trajectory of the particle and the other "space" vectors to be orthogonal to this vector. Then, as the particle moves along its path in an infinitesimal time, the magnitude of the "space" coordinates will not vary with only the "time" coordinate changing. This is an inertial frame in which geodesics are straight lines and the particle does not move in its own reference frame (for an infinitesimal time at least before we switch to a different set of normal coordinates adapted to a different point). Is this what we mean by an inertial frame? If so, what is the interpretation of the normal coordinates that can be constructed such that the "time" basis is not tangent to the trajectory of the particle, and where all components change as the particle moves?

Now on the other hand, say a particle is moving along a curve in the manifold with a non-zero four acceleration. At every point in time on its trajectory, it's still possible to construct normal coordinates at the position of the particle. If we follow the same procedure as before and define the "time" vector to be tangent to the trajectory of the particle at a point p, the geodesic along that tangent starting at p will appear to be parameterised by coordinates given by: $$(t,0,0,0)$$ (where t is the parameter). However, since the particle is not moving along a geodesic, the particles position cannot simply be parameterised by $$(t,0,0,0)$$ and there must be a change in position as well. This means that there are no normal coordinates in which a an accelerating particle can see geodesics as straight lines and simultaneously be at "rest" (in the normal coordinates). Is this a reflection of the fact that accelerating observers can't be in inertial reference frames? In that case, can the frame we constructed be considered a co-moving reference frame?

I'm not sure if this the right way to think about normal coordinates, so any help is appreciated.

• The "frame" concept is sometimes used for special coordinate systems in flat spacetime (or in local regions where spacetime is approximately flat), but the concept of a coordinate system is more general. In general, coordinate systems are not tied to observers. For more about this theme, see How do frames of reference work in general relativity, and are they described by coordinate systems?, and probably other posts in this search result. – Chiral Anomaly May 4 at 0:59
• Yes, I've seen them being used for observers a few times. But as far as I understand (which is not very much, I haven't studied them), aren't they the same as defining a non-holonomic basis? – Chandrahas May 4 at 2:08
• A coordinate system doesn't need a basis, and a basis doesn't need a coordinate system. They are two different concepts. A coordinate system simply labels the points of $N$-dimensional spacetime using an $N$-tuple of real numbers that vary smoothly from one point to the next. A basis is a set of $N$ linearly independent vectors (or one-forms) that varies smoothly from one point to the next. Given a coordinate system, we can use it to construct a basis, but not all bases can be consructed that way. – Chiral Anomaly May 4 at 2:21

Yes, this is a (local) inertial frame. If the time direction is not aligned with the tangent vector to the geodesic, then the 4-velocity of the particle will not be $$\mathbf u = \frac{\partial}{\partial t}$$; in other words, you have constructed an inertial frame in which the particle in question is moving. This frame is related to the rest frame of the particle by a Lorentz boost.