# Geodesics doubts

1. Aug 25, 2010

### TrickyDicky

Let's imagine a test particle in outer space not being subjected to any significant force, gravitational(far enough from any massive object) or any other. Its path would be describing a geodesic that follows the universe curvature, right? Would that be an euclidean straight path, or would it follow a curved path, like an ellipse or a hyperbola?

2. Aug 25, 2010

### zhermes

In space it would be euclidean straight, but not in space-time -- due to cosmological expansion.

3. Aug 25, 2010

### TrickyDicky

Aha, in spacetime would be ,then?
Or there is no easy way to answer this?

4. Aug 25, 2010

### zhermes

It would depend sensitively on the particular cosmological model. I'm not sure what kinds of paths are characteristic or common.

5. Aug 25, 2010

### bcrowell

Staff Emeritus
I don't think this is right.

In spacetime, the definition of straightness is a geodesic. So if the OP's question is interpreted as referring to spacetime, then the answer is definitely that it's straight.

In space, there is no way to answer the question, because you can't separate out spatial curvature from spacetime curvature in any unique way; it depends on your coordinates. In rotating coordinates, flat spacetime is spatially curved. To describe the path of a particle over cosmological distances you need to construct a coordinate system that covers cosmological distance scales, and such a coordinate system is highly arbitrary. An observer moving with the particle will describe the particle as being at rest, so its path is a point. If the particle is in motion relative to the Hubble flow, and it passes through galaxy G, which is at rest relative to the Hubble flow, then by symmetry an observer in G always sees the particle is moving in a straight line. A distant observer in galaxy Y, which is moving with its own local Hubble flow, will see both G and the particle as accelerating, and in the geometrically general case the will see the particle's path as some kind of curve (to the extent that it even makes sense to talk about a cosmologically distant observer's view of an object's motion -- in general it really doesn't make sense to do so).

Last edited: Aug 25, 2010
6. Aug 25, 2010

### Hurkyl

Staff Emeritus
If you really mean "Euclidean", then you really haven't understood the idea of "space-time" at all -- you are still thinking in terms of "space" and "time" being separate notions.

Assuming that you do understand the idea of space-time and just lack the words -- you probably meant "Minkowski straight".

Of course, space-time being curved, Minkowski only applies on small scales.

The moon orbits the Earth in a, more or less, straight path through space-time. (Assuming you are thinking in terms of space-time) the only reason you think otherwise is because you are measuring using the wrong metric.

An analogous error is thinking that latitude lines are straight lines on the Earth's surface.

7. Aug 25, 2010

### zhermes

Very astute, and true. But if this was the case for the OP, he would be (effectively) asking "would a straight thing be straight," so i think its fair to assume he means solely in space (see below)

Again, this is very accurate and very true; but (because of the above) a comoving frame is uninteresting, and the G perspective is shortlived; thus the 'Y' case is the most informative to OP's question--in my opinion.

Space and time are separate notions, that's why they're different words. Space and time are intrinsically inseparable, but are none-the-less distinct (hence the somewhat common 3+1 terminology). We, as observers in reference frames non-relativistic to one-another, perceive time and space differently, and thus a discussing one while not the other can be informative and elucidating.

This is a good point to note, but consider the above.

8. Aug 25, 2010

### TrickyDicky

I should have seen that one coming. My question is a bit naive and too sloppy in the choice of words. So yes, by definition a geodesic is the straightest path in a curved spacetime so just by talking about them I'm implying a curved spacetime and a straight path. I guess what I really meant was what kind of curvature does our spacetime have.

My question was exactly that,what kind of curve would that be? But I see that this doesn't seem to have a straightforward answer, according to zhermes due to the fact of cosmological expansion. In a nonexpanding spacetime I guess the curvature would be that of the spatial part of the line element, right?

Another example of sloppiness on my part, I probably should have said Minkowskian, still when I've read descriptions of the Einsten model of the universe, IIRC they talk about a hypersphere embedded in Euclidean ambient space. Perhaps someone can clarify this for me.

9. Aug 25, 2010

### atyy

Spacetime is not expanding.

In general relativity, a test particle falling under the influence of only gravity traverses a spacetime geodesic.

The spacetime curvature of our universe is well approximated by the FRW solution on large scales, and by the Schwarzschild solution near our solar system.

There are many ways to divide the FRW spacetime into "space" and "time". In one of them, "space" is expanding. In another "space" is not expanding. Regardless of how the FRW spacetime is divided into "space" and "time", a test particle falling under the influence of only gravity traverses a geodesic of the FRW spacetime.

10. Aug 25, 2010

### Hurkyl

Staff Emeritus
As I'm familiar with the phrase, "space is expanding" is not a coordinate-dependent phenomenon. If you compute, making use one coordinate chart, that space is expanding near a point, then you will get the same result if you repeat the calculation using any other coordinate chart.

11. Aug 25, 2010

### Hurkyl

Staff Emeritus
The relevant mathematics was originally developed for purposes like studying the properties of Euclidean surfaces that do not depend on how the surface is embedded into Euclidean space. Also, surfaces in 3-space are the most complex examples that can be straightforwardly visualized. I think these descriptions are mainly just to convey the basic ideas of how differential geometry works, and to motivate terms like "curved".

The (IMHO) misleading thing is that they are purely spatial. If you have some surface, you could start drawing a straight line on it, tracing out one centimeter per second. One might get the idea that this is what is meant in general relativity by particles traveling along a geodesic. However, it's not.

12. Aug 25, 2010

### atyy

Really? I just meant da/dt > 0 in the usual FRW coordinates.

13. Aug 25, 2010

### bcrowell

Staff Emeritus
The issue isn't so much coordinate-independence as definition-dependence. Fundamentally, there is no unique definition for the velocity of a distant object in GR, because vectors at distant points can only be compared by parallel transport, which is path-dependent. If you want to describe cosmological redshifts as kinematic Doppler shifts, you can [Bunn and Hogg 2008]. If you want to describe them as gravitational Doppler shifts, you can [Francis 2007].

Bunn and Hogg 2008, http://arxiv.org/abs/0808.1081v2
Francis 2007, http://arxiv.org/abs/0707.0380v1

14. Aug 25, 2010

### George Jones

Staff Emeritus
The expansion scalar (thus coordinate-independent) for the congruence of fundamental FRW observers is positive.

15. Aug 25, 2010

### atyy

Yes, how about observer dependent then, since one presumably need not choose fundamental FRW observers?

16. Aug 25, 2010

### George Jones

Staff Emeritus
Right, but FRW spacetimes arise by demanding spatial isotropy and homogeneity, and the fundamental observers are the observers for which space is homogeneous and isotropic.

17. Aug 25, 2010

### bcrowell

Staff Emeritus
Nobody disputes this, just as nobody disputes the existence of cosmological redshifts. The question is whether to describe these facts as expansion of space. Of the two references I gave in #13, one argues the point of view that they should be described as an expansion space, the other that they shouldn't. The arguments they present are arguments about pedagogy and about which interpretation is more natural. There is no objective basis on which to say that one is correct and one incorrect.

18. Aug 26, 2010

### TrickyDicky

I rephrased it, I swear I meant space, if spacetime expanded we wouldn't notice, would we? but can someone address the quoted question now? Maybe in this less complex scenario I could understand it better.

19. Aug 26, 2010

### TrickyDicky

anybody? to clear up this simple doubt?

20. Aug 26, 2010

### George Jones

Staff Emeritus
I'm away from home for a couple of days, and I have very limited computer access on a very slow connection, but I will make some posts after I get back. If you have access, read section 4.8 from General Relativity: An Introduction for Physicists by Hobson, Efstathiou, and Lasenby. You might be able to read this section from Google Books.

21. Aug 26, 2010

### Ich

No. Expansion of space is purely coordinate dependent. Or better, as George Jones puts it, "expansion" is a property of a congruence, like the one defined by the canonical observers in a FRW metric. It's not a property of spacetime.
Yes, that's why people are talking about "expanding space" at all.
Another more general set of fundamental observers, independent of FRW symmetries, is defined to be at rest in normal coordinates, maybe one could call them "Einstein observers", as they reproduce the inertial frames of SR if curvature is negligible, on which most people base their intuition. Expansion vanishes if you use these observers.
@TrickyDicky:
It don't think you want to see curves in space. You want to see a spacetime diagram of a neighbouring geodesic in some Riemann normal coordinates. They curve away from the origin for accelerated expansion, and curve inwards for decelerated expansion.
However, as bcrowell noted, the defining geodesic at the center is straight by definition.

22. Aug 26, 2010

### bcrowell

Staff Emeritus
This is a very nice way of putting it!

23. Aug 26, 2010

### bcrowell

Staff Emeritus
Amazon gives access to that section through their "Look Inside" feature. I don't see what relevance it has to the present discussion. This is 4.8, "Tensors as geometrical objects?"

24. Aug 26, 2010

### TrickyDicky

Right, but in order to simplify, since I am a novice in this, I was asking for the case of a non-expanding spacetime manifold just to fix concepts before I go into the more geometrically complex FRW metric.

25. Aug 26, 2010

### Ich

The FRW metric is geometrically especially simple.
No matter, in normal coordinates, Newtonian mechanincs works well. Especially if you concentrate on radial motion.
If there is matter between worldlines, there is gravity, and the worldlines will converge.
If there is repulsion, like Dark Energy, worldlines will diverge.
If there is both, well, it depends on which is stronger.

All this works in principle like Newtonian mechanics, no matter which metric you choose to setup spacetime.
In normal coordinates, a FRW spacetime is just another ball of dust.