Timelike v. spacelike, is it arbitrary?

[PeterDonis]

In that sense, the choice of "North" is *not* arbitrary; if by "North" I mean "the direction pointing at a place on the Earth's surface which is on its axis of rotation", then I can't arbitrarily choose which direction that is.

This also helps clarify, by the way, why the "isotropy" question is not relevant. "Isotropy" in the analogy I drew corresponds to the fact that there are *two* directions, at any point on the Earth's surface, that point towards where the Earth's rotation axis intersects the surface: North and South. I can indeed make an arbitrary choice of which one I label "North" and which one I label "South"; the physics doesn't pick out either one as "preferred". But that arbitrary choice doesn't change the fact that only two particular points on the Earth's surface, the two Poles, are on the Earth's axis; I can't change which points they are by making an arbitrary choice.

Similarly, in a stationary spacetime like Minkowski spacetime, the choice of which timelike direction to label "future" and which timelike direction to label "past" is arbitrary. The spacetime geometry itself doesn't pick out either direction as preferred (unlike the FLRW case, where the two timelike directions *are* different, because the change in the scale factor picks out one direction as "expanding" and the other as "contracting"). But that doesn't mean I can arbitrarily choose which dimension is timelike, any more than I can arbitrarily choose where the Earth's axis of rotation is.

 

[Dale]

Isotropy is irrelevant, as are the coordinates used to represent the metric. A Lorentzian metric will have 4 eigenvalues, three will have one sign and one will have the other sign. The eigenvector corresponding to the other sign is timelike. Eigenvectors are independent of coordinates.

 

[PeterDonis]

"isotropy" in the standard usage means "isotropy in the 3 spatial dimensions" in your usage.

Perhaps it's also worth expanding a bit on why this is standard usage. It's because "rotation" in a plane that includes the timelike dimension works differently than rotation in a plane that only includes spacelike dimensions. Invariance under the latter type of rotation is what motivates the term "isotropy".

Briefly, rotation in a plane that includes a timelike dimension (say the t-x plane in ordinary Minkowski coordinates) induces extra phenomena like time dilation and length contraction; rotation in a purely spacelike plane (say the x-y plane in ordinary Minkowski coordinates) does not. This gives another way to pick out the timelike dimension: you can induce a boost (a rotation that includes the timelike dimension and includes the extra phenomena mentioned above) in any of three orthogonal directions, meaning that there are three mutually orthogonal planes in spacetime that include the timelike dimension. The three mutually orthogonal directions in which the boost can be induced pick out the three spacelike dimensions; the fourth one, common to all three boost planes in spacetime, is the timelike dimension.

 

[BruceW]

Minkowski spacetime is stationary, because it has a timelike Killing vector field. What that means is that we can find timelike curves in Minkowski spacetime along which the metric remains unchanged. I think that stationary + isotropic in the standard usage is what corresponds to "isotropic in all 4 dimensions" in your usage.

FLRW spacetime is *not* stationary: there is no family of timelike curves along which the metric remains unchanged. So non-stationary + isotropic in the standard usage would correspond to "isotropic in the 3 spatial dimensions, but not in the time dimension" in your usage.

ah right, sorry for using incorrect terminology again. thanks for explaining the proper way to say it. I'll remember that for next time.

 

[BruceW]

But there is no arbitrary choice, of coordinates or anything else, I can make that will change the distance from Quito to Nairobi, or the distance between either of them and the North Pole...

true. Maybe this is a good way to 'mathematically' explain what I mean. yes, in relativity, our symmetry operations preserve the norm of vectors. But I don't see how this is justified. How about if I add another symmetry operation that does not preserve the norm of vectors? for example if we denote vectors by (V0,V1,V2,V3) then the operation of swapping V1 with V0 (for all vectors). But, we leave the metric as it was. clearly this does not preserve the norm. But doesn't it result in the same system? (Apart from the fact that some spacelike vectors will now become timelike vectors).

As said earlier, if our metric is isotropic, then doing the V1,V0 swap will change the system. for example, the vector might be the 4-momentum of matter in an FLRW universe. Once we do the swap, our system is no longer isotropic. But what about for a general metric that is not isotropic? (or, is isotropic, but also stationary?)

 

[BruceW]

as an example, for the Minkowski spacetime. Suppose we have metric (-1,1,1,1) and I am traveling along vector (1,0,0,0) and you are traveling along vector (0,1,0,0). So we would say I am traveling along a timelike curve and you are traveling along a spacelike curve. Now if we do the V1,V0 swap, I am now traveling along vector (0,1,0,0) and you are traveling along vector (1,0,0,0). So we would now say that I am traveling along a spacelike curve and you are traveling along a timelike curve.

These two systems are the same, so in this particular case, the V1,V0 swap is a symmetry operation of the system. It is not a symmetry operation of the Poincare group, but it is a symmetry operation of the system (which is the important thing in the first place). I know the issue here is that this is just a particular example. So it doesn't prove anything about other systems.

 

[BruceW]

As said earlier, if our metric is isotropic, then doing the V1,V0 swap will change the system. for example, the vector might be the 4-momentum of matter in an FLRW universe. Once we do the swap, our system is no longer isotropic. But what about for a general metric that is not isotropic? (or, is isotropic, but also stationary?)

even in this case, maybe the system remains the same after the swap. Before the swap, all the matter was moving in the V0 direction, and the motion in the other directions was zero. Then after the swap, all the matter is moving in the V1 direction and the motion in the other directions is zero. So now, the system is 'isotropic' in the 3 directions V0,V2,V3. (and sorry, I know that the correct terminology for isotropic means a symmetry specifically in the directions V1,V2,V3. But I hope you'll excuse this abuse of notation, since I have explained what I mean by it).

So again, the only meaning I can see for the assignment of 'timelike' curves is that matter is defined to move along the timelike curves. (and if we forced this definition, then the above 'symmetry operation' would not be allowed).

 

[Dale]

These two systems are the same, so in this particular case, the V1,V0 swap is a symmetry operation of the system.

No, it isn't. In one case I can send signals to you but not vice versa, and in the other you can send signals to me but not vice versa. This is an easily distinguishable asymmetry.

 

[Dale]

So again, the only meaning I can see for the assignment of 'timelike' curves is that matter is defined to move along the timelike curves.

I don't know why you keep making this absurd statement when we have already discussed the fact that radiation defines the timelike direction too. Radiation and the metric identify timelike curves every bit as well as matter.

 

[PeterDonis]

How about if I add another symmetry operation that does not preserve the norm of vectors? for example if we denote vectors by (V0,V1,V2,V3) then the operation of swapping V1 with V0 (for all vectors). But, we leave the metric as it was.

This is a logical contradiction. The metric determines the norms of all vectors; if you change the norm of any vector, you are by definition changing the metric.

 

[BruceW]

I don't know why you keep making this absurd statement when we have already discussed the fact that radiation defines the timelike direction too. Radiation and the metric identify timelike curves every bit as well as matter.

But as I said in the post before that, when we do the symmetry operation, we change all vectors. So the tangent vector of the path of the beam of light would also change.

 

[BruceW]

This is a logical contradiction. The metric determines the norms of all vectors; if you change the norm of any vector, you are by definition changing the metric.

no, I'm saying that I change the vectors, but leave the metric alone.

 

[Dale]

But as I said in the post before that, when we do the symmetry operation, we change all vectors. So the tangent vector of the path of the beam of light would also change.

Then that will change the metric. You cannot change the path of a pulse of light this way without changing the metric.

 

[BruceW]

ah wait, you're right. the arc length of a beam of light could become non-zero under my 'symmetry operation'. So the system definitely doesn't stay the same... wow, that was a fairly simple way to show that the timelike curves are not arbitrary. Thanks man. yeah, I guess I was wrong about that.

Now I'm trying to think of what that means / absorb the message. The null curves tell us which dimension is the timelike one?

we already briefly mentioned a beam of light. But because I only talked about one beam of light, that would not specify the timelike dimension. As long as we have two beams going in different directions, then that tells us the timelike dimension.

So as long as we have a way to identify null curves, then we know the timelike dimension. And I think it is OK to say that generally, we always can identify the null curves.

edit: and thanks to PeterDonis for continuing to try to get through to me. It did help that you kept asking me to think more mathematically / be less hand-wavey about what I was saying.

 

[PeterDonis]

As long as we have two beams going in different directions, then that tells us the timelike dimension.

Exactly.

I think it is OK to say that generally, we always can identify the null curves.

Yes. That's why you often see people talking about the light cones (which are just the sets of null vectors at each event) as the things to focus on when you're trying to analyze the structure of a spacetime.

thanks to PeterDonis for continuing to try to get through to me. It did help that you kept asking me to think more mathematically / be less hand-wavey about what I was saying.

You're welcome! I'm glad it's clear now.

 

[robphy]

wow, that was a fairly simple way to show that the timelike curves are not arbitrary. Thanks man. yeah, I guess I was wrong about that.

Now I'm trying to think of what that means / absorb the message. The null curves tell us which dimension is the timelike one?

we already briefly mentioned a beam of light. But because I only talked about one beam of light, that would not specify the timelike dimension. As long as we have two beams going in different directions, then that tells us the timelike dimension.

So as long as we have a way to identify null curves, then we know the timelike dimension. And I think it is OK to say that generally, we always can identify the null curves.

In a (n+1)-dimensional spacetime,
the thing that distinguishes "timelike" is
the causality [i.e. the causal-future ordering, a partial-ordering] defined by the light-cone structure (and a conventional choice of which direction is "future").
No analogous ordering exists for the spacelike-directions [unless n=1].
[This is similar to my earlier post about no-closed-timelike curves vs closed-spacelike curves in (3+1)-Minkowski.]
One could argue that there is a symmetry between timelike and spacelike for (1+1)-Minkowski.

In (3+1)-spacetimes, the sum of two non-parallel future-directed null vectors is a future-timelike vector.

However, the null-curves may not be enough in more exotic examples like (2+2) or (n+2) or even degenerate cases (n+m+p) for (-,+,0). The light-cone structure [the ordering] is lost. One would still have different types of tangent-vectors... but a label of "time" to one of them would no longer be appropriate.

 

[Naty1]

edit: I see Robphy posted while I was writing...

from BruceW:

The null curves tell us which dimension is the timelike one?

What does this mean?

Shouldn't this read "The null curve tell us which dimension is the LIGHT-LIKE one.."??

 

[Dale]

edit: I see Robphy posted while I was writing...

from BruceW:

What does this mean?

Shouldn't this read "The null curve tell us which dimension is the LIGHT-LIKE one.."??

The null curves are light like, but they also form the boundary between timelike and spacelike. So curves on one side are timelike and curves on the other side are lightlike.

 

[Naty1]

they also form the boundary between timelike and spacelike. So curves on one side are timelike and curves on the other side are lightlike.

yes, I get that, but can you tell from null curve alone which is which?? That's what BruceW seems to have posted.

 

[PeterDonis]

yes, I get that, but can you tell from null curve alone which is which?? That's what BruceW seems to have posted.

Yes; as BruceW said, two null vectors pointing in different directions define a timelike vector. (Just take the vector sum of the two null vectors; it will be timelike. For example, the two null vectors $(1, 1, 0, 0)$ and $(1, 0, 1, 0)$ add up to $(2, 1, 1, 0)$, which is timelike.)

 

[WannabeNewton]

yes, I get that, but can you tell from null curve alone which is which?? That's what BruceW seems to have posted.

If you have the set of all null geodesics passing through an event in space-time, you can pass to the tangent space at that point using the exponential map to get a cone (null cone) and the interior of the cone will let you determine the set of all time-like geodesics through that point and the exterior of the cone will let you determine the set of all space-like geodesics through that point because the null cone partitions the tangent space at that point into space-like vectors (exterior), null vectors (cone itself), and time-like vectors (interior).

 

[robphy]

Yes; as BruceW said, two null vectors pointing in different directions define a timelike vector. (Just take the vector sum of the two null vectors; it will be timelike. For example, the two null vectors $(1, 1, 0, 0)$ and $(1, 0, 1, 0)$ add up to $(2, 1, 1, 0)$, which is timelike.)

The null-vectors need to be both future-pointing or both past-pointing for their sum to be timelike... as noted in my earlier post.
Future-pointing-null plus past-pointing-null can be spacelike.

 

[PeterDonis]

The null-vectors need to be both future-pointing or both past-pointing for their sum to be timelike... as noted in my earlier post.
Future-pointing-null plus past-pointing-null can be spacelike.

Ah, yes, good point. At least I chose an example that met this requirement.

 

[yuiop]

Below the event horizon, the radial spatial dimension is said to become timelike. It has the special property that objects can can only travel in one direction (towards the singularity). Outside the event horizon, the time dimension is of course timelike and it also has the special quality that objects can only travel in one direction (towards the future). Doesn't anyone else find that curious? It suggests that the timelike dimension is the only dimension that has this unique one way property.

Conversely, the time dimension below the event horizon becomes spacelike and it is OK for objects (and light) to go backwards or forwards in coordinate time in that location.

 

[WannabeNewton]

Is it? Aren't we ourselves placing the physical restriction that material particles must travel on future-directed time-like curves?

 

[PeterDonis]

Below the event horizon, the radial spatial dimension is said to become timelike. It has the special property that objects can can only travel in one direction (towards the singularity).

But there is also the "white hole" solution, the time reverse of the black hole, which inside its horizon has the property that objects can only travel along timelike curves in one direction, *away* from the singularity. So the time symmetry is still there; it's just that you have to look at the full set of solutions to see it. What picks out the "black hole" solution as the one we actually use is experimental observation: we observe plenty of objects that are good candidates to be black holes, but we've never observed any object that's a good candidate to be a white hole.

(Similarly, the expanding FRW solution that we use to describe our universe has a time reverse, the contracting FRW solution. We pick the expanding solution for actual use because we experimentally observe the universe to be expanding.)

 

[Naty1]

yuiop: oh, good point...I forgot about event horizons...'timelike' on both sides...[if that is conventional terminology]

But I think all of you resolved my question about BruceW's post...which was

The null curves tell us which dimension is the timelike one?

and which I did not 'like'...because while the curves do

form the boundary between timelike and spacelike.

null curve ALONE don't distinguish which is which ; one needs an additional piece of information to determine spacelike vs timelike...like two null vectors pointing in the same direction...

And Robphy post...

If you have the set of all null geodesics passing through an event in space-time, you can pass to the tangent space at that point using the exponential map to get a cone (null cone) and the interior of the cone will let you determine the set of all time-like geodesics through that point and the exterior of the cone...

is another 'twist' I hadn't thought about...

thanks...nice insights into spacetime which I found very helpful...

 

[PeterDonis]

the interior of the cone will let you determine the set of all time-like geodesics through that point and the exterior of the cone will let you determine the set of all space-like geodesics through that point

Is there a slick way to capture how the "interior" and "exterior" of the cone are defined, in terms of the null vectors themselves?

 

[robphy]

Causal order.

 

[WannabeNewton]

Is there a slick way to capture how the "interior" and "exterior" of the cone are defined, in terms of the null vectors themselves?

I was thinking of it geometrically but I'm not sure if it's slick in any way. If we take an event $p$, we can find an orthonormal basis $\{e_{\mu}\}$ for $T_p M$ so that $g_{\mu\nu}(p) = \eta_{\mu\nu}$ hence the set of all null vectors at $p$ will be given by $S = \{\lambda = \lambda^{\mu}e_{\mu}:-(\lambda^0)^2 + (\lambda^1)^2 + (\lambda^2)^2 + (\lambda^3)^2 = 0\}$. This is equivalent to a cone in Minkowski space-time with vertex at the origin and in the same way as in Minkowski space-time, the "interior" would just be the set of all points inside of the cone so defined by $S$ and the "exterior" would just be the set of all points outside of the cone so defined by $S$.

 

[robphy]

In (1+1), fix two non-parallel (nonzero) future-null vectors, $\vec u$ and $\vec v.$
When scalars $a$ and $b$ are positive, the vector $a\vec u+b\vec v$ generates all of the future-timelike vectors (inside the future light cone). [These are essentially future-timelike-vectors expressed in "light-cone coordinates"]


added in edit:

My earlier "causal order" comment refers to constructions of the form:
Given a set of events and the set of all ordered-pairs of events which are future-null-related,
you can determine the set of ordered-pairs that are future-timelike-related.
("On the structure of causal spaces" by Kronheimer and Penrose, 1966).

http://www.google.com/search?q=horismos+penrose

 

[PeterDonis]

In (1+1), fix two non-parallel (nonzero) future-null vectors, $\vec u$ and $\vec v.$

But how do I tell that both vectors are future null? Can I tell by their inner product? I can see how I could tell that the vectors were both in the same "direction" (i.e., both future null or both past null) by the sign of their inner product--it should be positive if both are in the same direction. But how do I distinguish one "direction" from the other?

 

[Dale]

yes, I get that, but can you tell from null curve alone which is which?? That's what BruceW seems to have posted.

Yes, the null curves define the asymptote of a set of hyperboloids. On one side the hyperboloid is a hyperboloid of one sheet, on the other it is a hyperboloid of two sheets. The side with one sheet is spacelike, the side with two sheets is timelike, one sheet representing the future and the other representing the past.

 

[WannabeNewton]

But how do I tell that both vectors are future null? Can I tell by their inner product? I can see how I could tell that the vectors were both in the same "direction" (i.e., both future null or both past null) by the sign of their inner product--it should be positive if both are in the same direction. But how do I distinguish one "direction" from the other?

You need a temporal orientation of the space-time first; this is a continuous time-like vector field $t^a$ defined on the space-time $M$. Then at any point $p\in M$, a non-zero causal vector $\lambda^a$ is future directed if $t^a \lambda_a > 0$ and past directed if $t^a \lambda_a < 0$.

 

[robphy]

...in other words, [when time-orientable] pick one case to be future [and propagate that choice consistently].

See this chapter on the Minkowski vector space from Geroch "Mathematical Physics"
http://books.google.com/books?id=wp2A7ZBUwDgC&pg=PA79
http://books.google.com/books?id=wp2A7ZBUwDgC&pg=PA82 (1., 2. and onward)