# Worldline's arc-length

1. Jul 26, 2011

### TrickyDicky

The proper time elapsed between two points in a timelike geodesic in GR is the arc-length of this curve segment.
Does this mean that we have an arc-length parameterization of the geodesic curve, with $\tau$ (proper time) as the parameter?

2. Jul 26, 2011

### bcrowell

Staff Emeritus
This seems to be a question about terminology. I don't think it's common to hear people refer to this as "arc-length parametrization," but it is certainly closely analogous. I think it's more common for people to use phrases like "parametrized by proper time."

Note that this also works for spacelike curves, but in the case of lightlike curves it doesn't, so you have to use some other affine parameter rather than the one defined by the metric.

3. Jul 27, 2011

### TrickyDicky

I'm not sure if it is very commonly used, but it is in any (or most) calculus book(s).
What I wanted to stress is that timelike geodesics, if we want to abstract from an ambient higher dimension, can be parametrized by arc-length, wich is a natural, intrinsic parametrization that all smooth curves admit. In this specific case (timelike geodesics) this parametrization by proper time, also reflects the intrinsic curvature of the manifold, right?
I should have specified I'm referring to physical particles paths, so it doesn't work for spacelike geodesics either.

4. Jul 27, 2011

### Ben Niehoff

Careful here. As bcrowell mentioned, not all smooth curves admit an arc-length parametrization. Lightlike curves have zero arc-length, and so they must be parametrized by something else. This is a funny consequence of Lorentzian signature; in Euclidean signature, your statement is correct.

The proper time elapsed along a single curve is not enough information to give you the curvature, if that's what you mean.

If you know the proper time elapsed along the collection of all curves in some open region U, then this is equivalent to knowing the metric tensor in U. However, in order to get the curvature, you must also know the torsion...that is, you need to know about parallel transport; not just proper time. (If we assume the torsion is zero, then the metric is sufficient to give us the curvature).

Are you interested in geodesics specifically, or general curves? Anyway, a spacelike curve doesn't have a proper time, but it does have a proper length, which is a perfectly fine parameter.

5. Jul 27, 2011

### bcrowell

Staff Emeritus
I meant it's not common to hear people refer to it in relativity as "arc-length parametrization."

6. Jul 27, 2011

### TrickyDicky

I'm trying to be as careful as I can, that is why I constrained my statement to material particle's paths.
Since I'm trying to understand the GR scenario, torsion is assumed to be vanishing when I refer to the specific case of a timelike geodesic. In this case, being the single curve a geodesic, isn't it enough to have the geodesic parametric equations to know its metric and thus the curvature, given the fact that precisely the metric defines the geodesic?
For now, I'm interested just in understanding the timelike geodesics case.

7. Jul 27, 2011

### TrickyDicky

Agreed.

8. Jul 27, 2011

### Ben Niehoff

OK, so assuming zero torsion: If you know the proper time elapsed along all timelike geodesics, I'm not sure if this is actually enough information to reconstruct the full metric tensor, since you don't know anything about spacelike or lightlike geodesics.

The answer depends on whether you can use your knowledge of timelike geodesics to work out the spacelike geodesics using some kind of clever construction (Schild's ladder, maybe?). I couldn't say off the top of my head whether it works.

9. Jul 27, 2011

### TrickyDicky

Do they need to be worked out? What particle's path corresponds to spacelike geodesics?

10. Jul 27, 2011

### TrickyDicky

11. Jul 27, 2011

### bcrowell

Staff Emeritus
Working in the other direction, knowing the lightlike geodesics isn't enough to determine the metric, because that just fixes everything up to a conformal factor. But if you know the lightlike geodesics and also some local proper times, then the proper times can be used to fix the conformal factor and you can find the metric. In other words, you need both light-cones and a clock in order to fix the metric.

So it seems to me that knowing all proper times along all timelike geodesics certainly suffices. If you know the set of all timelike geodesics, then you certainly know the set of lightlike geodesics, since those are in some sense the boundary of the set of timelike ones. Therefore you know the lightlike geodesics and you have information that fixes the conformal factor, and we know that's sufficient to fix the metric.

12. Jul 27, 2011

### TrickyDicky

This is pretty much along the lines of what I had in mind when I started this thread. I must say though, that I was not sure at all this is right (I'm not yet) and I wanted to make sure.

13. Jul 28, 2011

### TrickyDicky

But they are a boundary only if you consider an asymptotically flat universe, that's certainly not the case in FRW cosmologies for instance.

14. Jul 29, 2011

### TrickyDicky

If we know the proper times, how many coordinates are needed to locate a point in a timelike geodesic (the paths of free massive particles)?

15. Jul 29, 2011

### TrickyDicky

anybody?
..I thought this was an easy one.