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In general relativity we introduce local inertial frames to be such frames where the laws of special relativity holds. Let $\xi^{\alpha}$ the coordinates in the local inertial frame, so we get $$ds^2=\eta_{\alpha \beta}d \xi^{\alpha} d \xi^{\beta}.$$ If we switch the frame of reference to coordinates $x^{\mu}$ : $\xi^{\alpha}= \xi^{\alpha}(x^0, x^1, x^2, x^3)$ and with $$g_{\mu \nu} (x)= \eta_{\alpha \beta} \frac{\partial \xi^{\alpha}}{\partial x^{\mu}} \frac{\partial \xi^{\beta}}{\partial x^{\nu}}$$ we get:

$$ds^2=g_{\mu \nu}d x^{\mu}(x) d x^{\nu}.$$

I don't understand why it isn't possible to find a transformation to get $$ds^2=\eta_{\alpha \beta}d \xi^{\alpha} d \xi^{\beta}$$ on the whole or almost the whole manifold? Because $g_{\mu \nu}(x)$ is still the same on the whole manifold?

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If $ds^2=\eta_{\alpha \beta}d \xi^{\alpha} d \xi^{\beta}$ were true for all points of space, we would have no curvature, hence no gravity!

Take for example a sphere (the Earth), locally we can measure distances by $ds^2=dx^2+dy^2$, but this can't hold for two arbitrary points on the sphere. In fact, this coordinate system changes from point to point (think of a tangent plane on the sphere).

We would have to replace the local coordinates, which you called $\xi^\alpha$ (the cartesian coordinates $x$ and $y$ in this case) and replace them by some other global coordinates, such as the angles $\theta$ and $\phi$. (Note that we would still need to patches to cover the total sphere). Then, the distance between two arbitrary points would be calculated using $$ds^2=r^2 \sin^2 \theta d\phi^2 + r^2 d\theta^2$$

So curvature is what makes us introduce $g_{\mu\nu}$ and the global coordinates $x^\mu$.

A local inertial frame would see no gravity and would be able to do Special Relativity, for a small region there is no significant curvature. To continue the analogy of the Earth, you wouldn't appreciate curvature in many kilometers, but the local region would be much smaller than the whole patch. Note that any world map (a whole patch) will present distorsion because of curvature, but a small road map won't have any distorsion.

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  • $\begingroup$ Ok, but the biggest problem I had, is how large can the local inertial frame be? In case of the S² sphere we only need two charts to define an atlas. Must the local inertial frame be located in a small open subset around a point x of the manifold or could it be the whole chart? If the last one ist correct how it makes sense to the meaning of local inertial frame, where the gravity should more or less the same in the whole inertial frame? $\endgroup$ – Alpha001 Nov 20 '15 at 23:07
  • $\begingroup$ @Alpha001, the local inertial frame IS a chart. So all of the equations one writes for a chart, like the metric expression, remain valid in the whole chart, but not in the whole manifold. You picked out a very specific case in which the manifold happens to be a surface embedded in the Euclidean space. To describe the majority of these surfaces one chart is enough (besides the sphere which asks for two). But for general manifolds, including spacetime, this is not the case. $\endgroup$ – Mr. K Nov 21 '15 at 9:03
  • $\begingroup$ In many books i read that the local inertial frames must small enough (so that the gravity is almost constant) like a free falling elevator or a labratory in a satellite. For me the expression "small enough" and a whole chart dosen't match in general. Or is it the case that alle charts of every possible spacetimes are very "small" (not like S²)? $\endgroup$ – Alpha001 Nov 21 '15 at 9:32
  • $\begingroup$ @Alpha001 I've edited the question. As for the "size" of the charts I think that in most cases you can have four charts to cover the whole manifold (math.stackexchange.com/questions/75594/…). After all, in most cases the manifold is $\mathbb{R}^4$, unless we are using a more complex topology (like in a black hole). $\endgroup$ – jinawee Nov 21 '15 at 10:56
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In Riemannian geometry there is a beautiful theorem which states that a manifold with a symmetric connection is locally flat everywhere if and only if the curvature tensor vanishes. Therefore, in a locally flat coordinates such that $\Gamma_{jk}^i=0$, $g_{ij}$ is constant throughout the chart and a linear transformation can be used to diagonalize the metric into flat metric $\eta_{ij}$. In this case, and in this case only, it would be possible to use the flat metric in the whole chart.

This is not the case in general though, because usually the curvature tensor does not vanish. But still is possible to find a coordinate in a point $p$ of the manifold such that $g_{ij}(p)=\eta_{ij}(p)$ as long as the torsion tensor vanishes (which is the case in GR). This is called geodesic coordinates or normal coordinates. But this is done differently for each point $p$ and it does not mean that second derivatives of the metric, and therefore the curvature, is zero and that is the why you cannot extend the flat metric for the whole manifold (unless the curvature vanishes). Remember also that the metric, as a tensor, is independent on the coordinate frame. Although its coordinates $g_{ij}$ change from one frame to another, the abstract object $g=g_{ij}dx^i\otimes dx^j$ remains the same.

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  • $\begingroup$ Thank you. But now I'm a bit confused because your explanation is quite different to the others but my ideas gone in the same direction. So what is true? To your explanation: How matches the free falling elevators or the satellit labratory frame to this? The satellit or the elevator are not only one point of the manifold? $\endgroup$ – Alpha001 Nov 21 '15 at 15:38
  • $\begingroup$ I am not the one who decides what is true here. Keep questioning with critical thinking and you will find the truth by yourself. The principle of equivalence (concerning free falling elevators/laboratory frame) is exactly what I described in the second paragraph and as so, it is only valid in the vanishing neighborhood of $p$. As a exercise you can transform the metric on Minkowski space to a non-inertial frame and see that the free falling equation (in the inertial frame) results in the geodesic equation in the non-inertial frame. $\endgroup$ – Mr. K Nov 21 '15 at 16:02
  • $\begingroup$ All of this you can find in the Szekeres' book on mathematical physics, chapter 18. $\endgroup$ – Mr. K Nov 21 '15 at 16:04
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$g_{\mu \nu}(x)$ means that $g$ is a function of location ($x$) --- so it varies across the manifold, which is the problem.

I think that if $g \ne g(x)$, then necessarily $g = \eta$ ... Hopefully someone else can chime in on that.

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  • $\begingroup$ But shouldn't $g_{\mu \nu}(x)$ be defined on an open subset around x? And since it's still the same expression for $g_{\mu \nu}(x)$ on the manifold why isn't it possible to extend that transformaion to the whole chart or to the whole manifold (not just locally around x)? $\endgroup$ – Alpha001 Nov 20 '15 at 18:51
  • $\begingroup$ @Alpha001 The metric tensor $g$ should be defined on any open subset of the manifold, where it defines a smoothly varying metric at each point --- which, in general, changes depending on the location of the point ($x$). A local inertial frame, by definition, cannot have a varying metric. Perhaps I'm not understanding your question $\endgroup$ – DilithiumMatrix Nov 20 '15 at 19:41
  • $\begingroup$ So the subset around a point where the metric is more or less constant is called the local inertial frame? $\endgroup$ – Alpha001 Nov 20 '15 at 23:03
  • $\begingroup$ @Alpha001 Local inertial frame means the reference frame where the particle at hand is at rest. $\endgroup$ – gented Nov 20 '15 at 23:57
  • $\begingroup$ @Alpha001 yes, exactly! And in a Minkowski metric, that's everywhere. $\endgroup$ – DilithiumMatrix Nov 21 '15 at 15:59
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Let $\mathcal{M}$ be the space time manifold, whose local charts (open sets) are described by $U_i$.

A local coordinate frame $S_i$ is a map $\xi\colon U_i\mapsto \mathbb{R}^N$ such that $\xi(m) = (x_1,\ldots,x_N) \in \mathbb{R}^N, m\,\in U_i$. Let, moreover, $g$ be a $(0,2)$ rank tensor (the metric).

A change of coordinates is any smooth invertible map $f\colon U_i\mapsto U_j$. Under such map the element $\alpha$ of the cotangent bundle transform as $\alpha'(x') = f_*(x)\alpha(x)$, with $f_*$ being the pullback of the map $f$.

It might be possible, given a pair of charts $(U_i,U_j)$, to find $f$ such that the new metric as calculated in $U_j$ will result proportional to the old one in $U_i$; however, since the form $f$ is strongly dependent on the charts and on the point, applying the same $f_*$ to another chart $U_k$ might not do the job (actually, the map $f$ might not even be defined on other charts).

Linear transformations are a very special case because the Jacobian matrix does not depend on the point, since after taking derivatives the dependence on $x$ disappears. This allows to easily extend them to the entire manifold, whereas that may not be possible for any other (non-linear) change of coordinates.

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  • $\begingroup$ But for example if an atlas of the manifold is described by only two charts the would a local inertrial frame be definied on a whole chart of the manifold? It isn't really local in this case? $\endgroup$ – Alpha001 Nov 20 '15 at 22:09
  • $\begingroup$ If you only have two charts and if you can reduce the form of $f_*$ for every point in those two charts, then yes. $\endgroup$ – gented Nov 20 '15 at 22:27
  • $\begingroup$ What are the conditions for this? $\endgroup$ – Alpha001 Nov 21 '15 at 17:30
  • $\begingroup$ This depends on what your manifold is and what the charts are, there is no general rule. The standard example is to that the sphere and see that the coordinates can be flattened everywhere only if you remove of the poles (and glue the two emispheres in a suitable way). $\endgroup$ – gented Nov 22 '15 at 19:02
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Manifolds are defined such that locally they look like Euclidean space; this is why we call them smooth manifolds.

A riemannian manifold is a manifold that locally has some inner product structure, ie a way of measuring length and angles.

Lengths and angles are invariants, hence will have an invariant expression in terms of a local coordinate basis; and hence also a transformation law.

Essentially, locally all you're doing is doing linear algebra.

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  • $\begingroup$ @ibere kuntz: well, looking at the question again, I'm not sure what he's asking: his final two sentences suggests he's asking why transformation laws are always local; but this seems already quite strange, since manifold are constructed locally. $\endgroup$ – Mozibur Ullah Nov 21 '15 at 16:34

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