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In my textbook, it states that a nonholonomic basis is the basis used by a physical observer, representing a basis w.r.t a local Lorentz frame.

Conversely, a coordinate basis represents the global spacetime.

Can someone explain why this should be so?

My current thoughts are that for a physical observer, locally their spacetime is flat and so we can just set up an orthonormal basis, whereas globally spacetime is curved and so any basis would not remain orthonormal. Is this along the right lines, or am I completely missing the mark?

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  • $\begingroup$ Comments: 1) only for a freely falling physical observer does the spacetime look locally Minkowski. 2) it is possible to set up an orthonormal basis throughout a region of spacetime even if the spacetime is curved (the basis won't be holonomic, however). $\endgroup$
    – gj255
    Apr 7 '17 at 11:20
  • $\begingroup$ Which textbook? $\endgroup$
    – Qmechanic
    Jul 30 '17 at 15:07
  • $\begingroup$ Btw, space-time is NOT locally flat. $\endgroup$
    – Bellem
    Mar 25 '19 at 14:11
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First of all, all bases are local.

There is a slight attendum to this though.

Definition: A (smooth) manifold $M$ is said to be parallelizable, if it admits a global basis field. Eg, if $M$ is $n$ dimensional, then there exists $n$ everywhere linearly independent, smooth, global vector fields $\{E_1,...,E_n\}$.

Not every manifold is parallelizable. Now, we say that a (local) frame, $\{E_1,...,E_n\}$ is holonomic, if there exists a coordinate system $(x^1,...,x^n)$ such that $E_i=\partial/\partial x^i$.

Theorem: A local frame $\{E_1,...,E_n\}$ is holonomic if and only if $[E_i,E_j]=0$, eg the frame vector fields commute.

The conclusion to take home is that not every basis field belongs to a coordinate system. In fact, even if a manifold is parallelizable, it doesn't mean it admits a global coordinate frame!

Basically, if a manifold admits a global coordinate frame, it implies that the manifold is homeomorphic to euclidean space, because the exsitence a global coordinate frame means there is a global homeomorphism $\psi:M\rightarrow\mathbb{R}^n$.

Because anholonomic frames are not tied to coordinates, a global anholonomic frame does not imply the existence of such a homeomorphism.

A good example would be the 2-torus. It is parallelizable, but it is clear that it is very different topologically from $\mathbb{R}^2$, for example it is cyclic and has a "handle" (different homotopy classes and all). Another example would be the 3-sphere $S^3$. It is parallelizable (it is diffeomorphic to $\text{SU}(2)$, which is a Lie group and all Lie groups are parallelizable), but it is topologically very different from $\mathbb{R}^3$.

If you are curious about how it is possible for a frame to NOT form a coordinate system, check out my answer here.


Now, onto the actual question.

My current thoughts are that for a physical observer, locally their spacetime is flat and so we can just set up an orthonormal basis, whereas globally spacetime is curved and so any basis would not remain orthonormal. Is this along the right lines, or am I completely missing the mark?

It is possible to set up so-called Riemannian normal coordinates about a point $p$, in which the metric is $\eta_{\mu\nu}$ at $p$, and the Christoffels vanish at $p$. This is a coordinate system, though, with a coordinate (holonomic) frame, but this frame only knows this orthonormality relation at $p$ and not in points away from $p$.

It is however absolutely possible to set up an anholonomic frame in an extended region (though not globally, unless spacetime is parallelizable), which is orthonormal everywhere. The price to pay is that this frame will not belong to a coordinate system, if spacetime is curved.

If you are interested in creating a frame at a point $p$ for an observer, then it will be just an orthonormal frame at that point. If you are willing, you can either extend it to a neighborhood of that point to be a coordinate frame (Riemannian normal coordinates) or an anholonomic frame (orthonormal frame), its your choice.

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