Simple question on curvature

MeJennifer

Consider a photon emitted at space-time event E₁ and absorbed at space-time event E₂ in curved space-time.

Since the arc length of the worldline between both events is 0 how can we, with validity, claim that such a path is curved in space-time?

Does it not seem to be more correct to claim that in curved space-time all geodesic worldlines are curved except for null geodesics?

Were is the error in my reasoning?

cesiumfrog

A geodesic is the closest concept (in general, differential geometry) to a "straight line". When paths are described as "curved", this is purely an illustrative term, attempting to convey an appropriate image to people with less experience of non-Euclidean geometry.

I think you've just (mis)taken something too seriously/literally, but if you need to discuss it further then I think you ought to state who makes the claim you mention, what context they wrote in, what definition they chose for "curved", and why you think null geodesics should be called less "curved" than other geodesics.

MeJennifer

A nul geodesic has an arc length of zero, how do you propose to define curvature for something that has zero length?

cesiumfrog

I believe this is covered somewhere in the first couple chapters of MWT.. though out of curiousity, how would you categorise a path that has null length (over any segment) and is not a geodesic?

Mentz114

MJ, the proper length is 0, but the spatial separation and temporal separation are not both 0, just equal in magnitude and opposite in sign.

MeJennifer

Right, but here we are dealing with events in curved space-time not some Lorentz variant slicing of space and time for a particular observer. This seems like simply avoiding rather than adressing the issue. Clearly only space-time vectors are invariant under Lorentz transformations.

Good question, if I am not mistaken even Hawking and Ellis in "The Large Scale Structure of Space-Time" seem to avoid this particular, perhaps non-physical, case.

Mentz114

Why not ? If it's a path it can be curved.

According to the adopted criterion for curvature, null geodesics can be curved.
You probably need to redefine curvature to satisfy this aesthetic urge.
I don't think it's an issue. Just nomeclature.

pmb_phy

Since the arc length of the worldline between both events is 0 how can we, with validity, claim that such a path is curved in space-time?[/quote]Always with the great questions MJ.

The length between the two events is a bad word to use in GR. Its the spacetime interval that is zero. In differential geometry one uses a distance function, which is a different function from the metric tensor. One uses this distance function to define closed sets, whether two points are "nearby" etc. For this reason also one can tell whether two points are close or far apart.

Another point is that curvature cannot be determined by a single geodesic. One has to have more than one geodesic to determine if spacetime curvature is lurking somewhere nearby. That's about the best I can explain it at this time (i.e. off the top of my head).

Good luck

Pete

MeJennifer

I am happy to use any term that you like pmb_phy, but to me it seems that that does not explain anything.
So if the interval is zero, then how can this "path" be curved?

Yes, and so in my example of two space-time events of an emitted and absorbed photon, how far are those events apart in your opinion?

The distance between two events in space-time is Lorentz invariant right?
The arc length of the space-time path of an observer in space-time represents his proper time right?
The arc length of the space-time path between an emited and absorbed photon in space-time is zero right?

So how could some possibly claim that in curved space-time such a null geodesic is curved?

Obviously timelike geodesics are curved in curved spacetime since they have an arc length that is non zero.

nrqed

One cannot say anything about the curvature of a manifold using only a single geodesic. This is clear even if you look at something much simpler, let's say the surface S_2 of a sphere embedded in ordinary 3-dimensional space (no problem with negative terms in the metric there!!) .

what you are asking is the same as if someone would say :look, a geodesic from one pole to the other pole has a length of [itex] 2 \pi R [/itex]. How can one conclude that this space is curved?

WHat would you answer to that question? Don't you agree that the geodesics are clearly curved in that case? Do you see that you are asking exactly the same question (it does not matter that in your question the spacetime interval is zero as opposed to a finite value in my example, the point is the same in both cases: giving a number for a single geodesic tells us nothing about curvature).

The answer is of course that to know about the curvature of a manifold, one must have more information than the "lenght" (or spacetime interval) of a single geodesic or curve. One possible way to learn about the curvature is to go through a closed path following 3 geodesics or more.

nrqed

So answer my even simpler question: on S_2, how can *you* say that a geodesic is curved? Can you answer that question? If not, maybe we should start by discussing *that* much simpler example before tackling a non-positive definite metric!

MeJennifer

Simple, geodesics follow the curvature of space-time so if space-time is curved then obviously the geodesics follow that curvature as well. But a path that has an arc length of zero, what is it supposed to follow that is curved?

I think you know what I am talking about, but it seems you rather want to avoid this discussion. Perhaps because you are confused by it too?

Note: something when wrong in posting my prior message, but nrqed quoted the gist of it.

pmb_phy

A path being curved doesn't have a lot of meaning to me. If the path is a geodesic in embedded in a 2-D surface then the geodesics (of which there are many) will form a helix. This curve is the "straightest possible" curve. In fact this surface has no intrinsic curvature so the path can't be said to be curved in the Riemann sense (as in embedded).
Sorry but when speaking in terms of GR the term "far apart" has no meaning to me.
Yes.
If you mean "spacetime interval" instead of "arc length" then you're right. Otherwise you'll have to define "arc length" for me.
What is arc-length?
[quote[
So how could some possibly claim that in curved space-time such a null geodesic is curved?
[/quote]Got me. I never said nor thought of them as being of such.
Perhaps its obvious to you, but not to me.

Pete

robphy

As suggested by some of the earlier posts,...

there are two different notions of curvature implied by your question:
worldline-curvature (geodesic curvature, 4-acceleration) and spacetime-curvature (Riemann curvature).

The Riemann curvature of a spacetime determines the deviation from flatness, which also determines which spacetime paths are geodesic and which are not.

The geodesic curvature of a spacetime path determines the deviation of that path from being geodesic. [A noninertial observer [with nonzero 4-acceleration] has a nongeodesic timelike path.] (This should not be confused with "geodesic deviation", which describes the behavior of a family of nearby geodesics.)

So, given a spacetime [with or without spacetime-curvature], its geodesics are those paths with zero worldline-curvature. What is important for a geodesic is that its tangent-vector is parallel-transported... When there is a metric available, the norm of this tangent-vector is used to classify the type of geodesic (spacelike, timelike, or null)... and can be used to characterize geodesics in terms of extremal (or stationary) values of arc-lengths.

possibly useful:
http://ocw.mit.edu/NR/rdonlyres/Phys...72/0/gr2_2.pdf

pmb_phy

So non-geodesics have curvature? That makes sense now that I think of it. I didn't know how to phrase that. Thanks Rob.

Pete

robphy

Non-geodesic paths have nonzero worldline-curvature (somewhere along the path).

masudr

We use an affine parameter.