# Flat / curved space metric component dependence on spacetime different locations?

1. Apr 16, 2015

### binbagsss

I'm reading An Intro to GR, Hughston and Tod, it says that in GR the idea is that the geometry of st varies from point to point, represented by allowing the metric to vary over space-time.

It uses a (+,-,-,-) signature and so $proper time=ds^{2}$.

It then makes the comment that proper time depends on the the observers location (compared to SR where it doesn't).

My question:

I believe it is always the case that the comtponents of a metric in GR are functions of space-time, and so I understand that proper time will vary for observers dependent upon location, but,

In flat space am I correct in thinking that the components also can depend upon space-time, e.g use of polar coordinates, and the only case where it doesnt is cartesian coordinates?

So from the above statement I would conclude that proper time also varies from location to location in SR. So shouldnt the statement instead be that if there does not exist coordinate system in which the components of the metric are not functions of space-time, the space-time is curved from which it follows that proper time is location dependent.

Thanks

Last edited: Apr 16, 2015
2. Apr 16, 2015

### Orodruin

Staff Emeritus
Yes, if you have curvilinear coordinates, your metric will in general depend on position. In SR, the coordinates where this is not happening are Minkowski coordinates. Cartesian coordinates are for Euclidean geometry.

3. Apr 16, 2015

### binbagsss

Oh right, thanks. And what's the difference? They are both to describe flat space-time right?

I know that there's a metric with (+,+,+,+) signture where you can't categorize the vectors into tl,null,sl.
And that the Minkowki metric, signature (+,-,-,-) or (-,+,+,+), allows categorization,

But I thought that the difference lies in the metric signature, not the coordinate system.

Thanks.

4. Apr 16, 2015

### Orodruin

Staff Emeritus
Minkowski coordinates are the Minkowski space equivalent of Cartesian coordinates in Euclidean spaces. The transformations between different sets of Minkowski coordinates (ie, Lorentz transformations) are different than the relations between different Cartesian coordinate systems (rotations).

5. Apr 16, 2015

### binbagsss

Am I correct when I say that if there does not exist coordinate system in which the components of the metric are not functions of space-time, the space-time is curved?

6. Apr 16, 2015

### Staff: Mentor

Can you give an exact quote? As you state this, it doesn't seem right, because proper time elapsed for a given observer always depends on that observer's path through spacetime. But I suspect the book actually meant something else (possibly they meant what you suggest as an alternate interpretation, about whether or not the metric coefficients vary from event to event).

7. Apr 16, 2015

### Staff: Mentor

The converse is certainly true: if there does exist a coordinate chart in which the metric coefficients are constant, then spacetime is flat (since obviously all derivatives of the metric coefficients are zero, and therefore the Riemann tensor vanishes identically).

Knowing that, we can easily see that your proposition is true as well, as follows: for the converse to be true, as above, but your proposition to be false, there would have to be at least two flat spacetimes, because we know Minkowski spacetime does have a coordinate chart with constant metric coefficients, and is therefore flat by the converse proposition above (as of course we know it is). In other words, there would have to be some other spacetime that does not admit a coordinate chart in which the metric coefficients are constant, but still has a vanishing Riemann tensor (which is what "flat spacetime" means). But this is not possible; the easiest way to see this is to observe that all the invariants that determine the spacetime geometry are functions of the Riemann tensor, so if the Riemann tensor vanishes identically, all the invariants must also vanish. So any spacetime with vanishing Riemann tensor must have the same geometry as Minkowski spacetime, and must therefore admit a coordinate chart with constant metric coefficients. So any spacetime that does not have a chart in which the metric coefficients are constant must be curved.

8. Apr 17, 2015

### Orodruin

Staff Emeritus
I'm just going to add that this is a local statement. There are flat space-times which are not globally equivalent to Minkowski space.

9. Apr 17, 2015

### Staff: Mentor

Yes, this is a good point, saying that a spacetime is "flat" does not tell you the global topology. However, any flat spacetime, regardless of global topology, must have the same local geometry as Minkowski spacetime, and therefore any open subregion of such a spacetime must admit a chart with constant metric coefficients. The only difference with a global topology other than that of Minkowski spacetime is that you may not be able to cover the entire spacetime with one such chart.