2

When in a completely flat spacetime, a metric $\eta_{\mu\nu} $ implies that in a stationary reference frame, you are dealing with three cartesian space coordinates, and one time coordinate. On a curved manifold, however, the GR book I'm reading seems to imply that if you can find coordinates such that the metric looks like $\eta_{\mu\nu} $, then at least locally, the coordinates look again like 3 cartesian coordinates and one time coordinate.

This doesn't sit well with me, because it seems on a manifold with some curvature, the coordinates necessary to make the metric look flat at a certain point may be different than coordinates in flat space that make a metric look like $\eta_{\mu\nu} $. So then, regardless of curvature, what about a flat metric at a certain point implies that the coordinates locally look like three cartesian space coordinates and one time coordinates?

I don't believe this is a duplicate of the other post because I understand that any curved manifold looks locally flat, and that coordinates can be chosen such that the metric looks like $\eta_{\mu\nu} $. I am asking about the physical interpretation of those locally flat coordinates.

Emilio Pisanty
  • 132,859
  • 33
  • 351
  • 666
Phantom101
  • 427

2 Answers2

1

Specification of the metric components as a matrix inevitably lands you to a certain coordinate set-up. Look at it this way: You have a pair of events separated infinitesimally so that the spacetime interval is $ds^2$ between them. In a generic coordinate system, this would mean that $g_{\mu\nu}dx^{\mu}dx^\nu=ds^2$. Now, if I go to a reference frame where the metric components look Minkowskian (i.e., my metric components are $\eta_{\alpha\beta}$) then what it means to say so is that the spacetime interval between the two events will be described by $(ds^2=)\eta_{\alpha\beta}dx^\alpha dx^\beta$ where ${x^\alpha}$ are my coordinates. In other words, the spacetime interval, in the reference frame in which the metric looks Minkowskian is given by $-dt^2+dx_1^2+dx_2^2+dx_3^2$ where the $t$ and the $x$s are the coordinates. And this is the very definition of Cartesian coordinates that they add in this way to give you the spacetime interval. So, the fact that when your metric is Minkowskian, your coordinates are Cartesian is not so much of a miracle but sort of a consistency requirement of the definitions involved. But the amazing (in the sense of non-trivial) fact is that at each point (no matter the curvature of the global spacetime) you can always find a frame in which the metric looks Minkowskian locally.

Edit I should add that a space being flat doesn't pin down your metric to Minkowskian nor does the fact that you can always (locally) choose a Minkowskian metric make you able to state that you can always make your space-time locally flat. Flatness is about the triviality of the Riemann tensor, which being a tensorial fact, remains independent of your choice of frame.

0

The physics comes from the Equivalence principle where gravity is equivalent to being in an accelerated frame (and various other ways of stating the equivalence of inertial and gravitational mass). So if free floating in a gravitational field you are in an approximately inertial frame - feels like no forces locally.

So you can use those local coordinates to feel locally no gravity (i.e., no forces). That means the metric locally can be said to be Minkowski, and the connection or derivatives of the metric are locally zero. The second derivatives won't be all zero, meaning that there is some covariant curvature. It's a tidal force, you only feel if you're too big, and therefore no longer local enough. Because of the equivalence principle all parts of your body, if local enough, feel the same force, so you feel no forces exerting any relative acceleration on your body parts (if local enough, if not you'll start feeling the tidal forces due to second derivatives of the metric, i.e., curvature)

This also says that locally special relativity holds, as one would want with no gravitational field.

Bob Bee
  • 13,996