# Local Charts in General Relativity

We may consider a "local" region in curved spacetime (local in respect of the spatial and the temporal coordinates). A "local inertial frame" may be constructed by some transformation that produces flat spacetime locally. This transformation produces the diagonal [1,-1,-1,-1] in an approximate manner.

A physical point:

You are performing some experiment in a small laboratory room (stationary w.r.t. the Earth) for a small period of time. Are you in a local inertial frame? If there was a freely falling lift in front of you, it would have been a better approximation to the inertial frame concept. So your laboratory does not qualify to be an inertial frame when compared with the falling lift. (The difference becomes even more conspicuous if you imagine "gravity" to be a 100 or a thousand times stronger.)

Now, let's move into the problem:

We take a small region of curved spacetime and use some suitable transformation to produce a "local inertial" frame. For this transformation we are disregarding all the anisotropies and inhomogeneities of the surrounding space in the original manifold.Incidentally the objective behind creating local inertial frames is to apply SR. The associated Lorentz transformations are supposed to hold good in the ideal conditions of isotropy and homogeneity of space

Our transformation transforms a small region of curved spacetime to a small region of flat spacetime. To this small/finite flat spacetime, we add the rest of it, ignoring all the anisotropies and inhomogeneities of the surroundings in the original curved spacetime [original manifold]. In effect we are assuming a "Global Transformation" without working it out! (And we expect correct results when we go back to the original space by the reverse transformations) We are in effect assuming a "Global Transformation" from curved spacetime to flat spacetime. Is it not as sacrilegious as "Flattening a Sphere"?

How much unreliability is the aforesaid error going to cast on calculations performed in the local inertial frames?To what extent will this error affect the Principle of Equivalence?

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Forgive my ignorance, but is there a question in there? –  Warrick Nov 14 '11 at 9:09
The question basically boils down to "Can you find a coordinate system in which a curved space becomes flat?" And the answer is "no, because the norm of the Riemann tensor is nonzero in all coordinate systems." –  Ron Maimon Nov 14 '11 at 9:46

You cannot do a coordinate transformation in which a curved space becomes flat, and this is because the nonzero curvature is intrinsically meaningful.

The way you define a meaningful notion of curvature in two dimensions is by drawing triangles. In flat space, the sum of the angles in a triangle is 180 degrees, because when you walk around a triangle, turning to face the new walking direction as you hit each vertex, you make one full turn in orientation when you walk once around the triangle, so that the sum of the exterior angles is 360 degrees.

The curvature of a 2 dimensional plane is defined by the difference between the sum of the angles and 180 (or the sum of the exterior angles and 360). This is proportional to the area for small triangles, as you can see by dividing the triangle in two, and noting that the angle-deficit adds up over the two sub-triangles to the big triangle.

The amount of deficit angle per unit area for an infinitesimal triangle defines the intrinsic Gauss curvature of a 2d surface, and this is the content of the Riemann tensor. If a space is flat, this Riemann tensor vanishes indentically. You can compute the Riemann tensor from the metric tensor, and if it is nonzero in one coordinate system it is nonzero in all coordinate systems.

This proves that you can't choose a global frame where GR turns into SR.

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Tensor equations are supposed to stay invariant under coordinate transformations where the line element is preserved,ie, the value of ds^2 remains unchanged.This is true so long as we are within the same manifold.In General Relativity the tensor equations are considered invariant in all manifolds.The geodesic equation or Maxwell's equations preserve their form[tensor form:covariant form] in all manifolds including flat spacetime –  Anamitra Palit Nov 14 '11 at 9:59
"Our transformation transforms a small region of curved spacetime to a small region of flat spacetime. To this small/finite flat spacetime, we add the rest of it, ignoring all the anisotropies and inhomogenities of the surroundings in the original curved spacetime [original manifold]. In effect we are assuming a "global transformation" without working it out! ".I would like the quoted portion to be considered in view of the answer above[by Ron Miamon] –  Anamitra Palit Nov 14 '11 at 23:23
@Palit: It is not clear what you are doing--- you are doing formal manipulations which seem incoherent. The transformation which flattens out a "small region" of a curved spacetime only works to lowest order in the size of the region, it is really at a single point, not over any finite size region. There is no way to extend it beyond a single point, as pointed out in Qmechanic's answer below. –  Ron Maimon Nov 14 '11 at 23:58
In Response to Ron Miamon:In the transformation process,you are leaving out a huge portion of the manifold which may have inhomogeneities and anisotropies.The basic reason behind the transformation is to obtain a small flat spacetime region is to apply SR.But can you really apply the Lorentz Transformations in the tiny transformed space,given the vast amount of anisotropies and inhomogeneities in the surrounding space? –  Anamitra Palit Nov 15 '11 at 5:10
@Anamitra Palit: You can't. It only works at a point. –  Ron Maimon Nov 15 '11 at 7:38

1) OP's first sentence(v5) is

We may consider a "local" region in curved spacetime (local in respect of the spatial and the temporal coordinates).

It varies from author to author whether locally means

1. in a point,
2. or in a sufficiently small open neighborbood.

OP adapts the latter meaning in the first sentence.

2) OP's second sentence(v5) is

A "local inertial frame" may be constructed by some transformation that produces flat spacetime locally.

The second sentence is true, if one means locally in a point (1), but it is in general not true if one means locally in a sufficiently small open neighborbood (2).

The Ricci tensor $R_{\mu\nu}$ is a tensor field, which implies that if it is non-zero at a point $p$ in some local coordinates, it will be non-zero at $p$ in every local coordinate system (that covers $p$).

Concerning interpretation (1), given an arbitrary point $p$ (and some local coordinates $x^{\mu}$ in a neighborhood of $p$), one may prove that there exists a local coordinate transformation $x^{\mu}\to x^{\prime \nu}(x)$, such that the metric in the new primed coordinates at $p$ becomes the flat metric

$$\left.g^{\prime}_{\mu\nu}\right|_p~=~\eta_{\mu\nu},$$

and the corresponding first derivatives of the primed metric at $p$ vanish,

$$\left.\frac{\partial g^{\prime}_{\mu\nu}}{\partial x^{\prime\lambda}}\right|_p~=~0.$$ cf. normal coordinates.

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You have pointed out the local considerations.The metric in the local tangent space is a flat one.But our aim,essentially, is to apply SR in the local context.The exclusion of non-local regions is a serious flaw since the aforesaid non-local region might contain anisotropies and inhomogeneities that come in the way of applying SR in the local context. Nevertheless we apply SR locally "as if" the non-local region was flat space time.A "Global Transformation" is assumed to be valid –  Anamitra Palit Nov 15 '11 at 9:08