Local inertial frame

Question

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?

Answer 1

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.

Answer 2

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.

Answer 3

$g_{\mu \nu}(x)$ 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.

Answer 4

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.

Answer 5

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.