0
$\begingroup$

I know that flat spacetime can be described using a constant spacetime metric. This is the metric of inertial observers who use Cartesian space coordinates.

But consider the case where the metric is not constant anymore (either because the observer uses a different set of coordinates, or beacause the spacetime itself is curved). If we express the metric tensor with the usual $4×4$ matrix $g$, we can have different possibilities. For exampls:

1) the "time-principal-coefficient" of the matrix ($g_{11}$) depends on one of the space coordinates; the other components of $g$ are constant. Example: Rindler metric: uniformly accelerated observer

2) one of the "space-principal-coefficients" ($g_{22}$, $g_{33}$ or $g_{44}$) depend on one of the space coordinates; the other components of $g$ are constant. Example: inertial observer who describes flat spacetime in spherical coordinates

3) both of the above conditions happen simultaneously ($g_{11}$, $g_{22}$, $g_{33}$, $g_{44}$ are space dependent; the other components of $g$ are constant). Example: Schwarzschild metric

It is clear that several other combinations are also possible.

My question is: if I know that one observer is using a specific metric, from this information only (studying carefully the space and time dependency of each component of this metric) am I able to deduce immediatly

1) if he is an inertial or not inertial observer

AND

2) if he is living in a curved or a flat spacetime?

$\endgroup$
  • $\begingroup$ I don't know about the first but you can't tell if spacetime is curved or flat. Take for example the $R^{3}$ Euclidean space. In cartesian co-ordinates the metric is a 3*3 unitary matrix, when in spherical polar co-ordinates the metric looks "funny" and still desribes the same flat space. $\endgroup$ – Thanasis_Kar Dec 2 at 22:40
2
$\begingroup$

Your question is founded on some misconceptions about general relativity.

We don't normally have a spacetime and a coordinate system such that one coordinate is a time coordinate at the others are spatial coordinates.

Coordinate systems don't relate to observers or frames of reference. See How do frames of reference work in general relativity, and are they described by coordinate systems?

If you have a metric and you want to find out if spacetime is curved, you need to compute the Riemann tensor. You can't easily tell from the superficial form of the metric when expressed in some set of coordinates.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.