Timelike v. spacelike, is it arbitrary?

[PeterDonis]

I'm trying to think of different ways to explain the point I'm trying to get across... OK, suppose we have some model for the universe, and we have chosen a metric with signature (-1,1,1,1). OK, now instead let's say we choose the metric with signature (1,-1,1,1). Then is our new physical description going to work as well?

We don't know, because you haven't given us a physical description. You haven't given us an actual metric; you've just played around with which order you write the coordinates. Show us an actual metric--an actual expression for $ds^2$ in terms of your coordinates--and we'll be able to tell you which curves are timelike and which are spacelike. But if all you've written down is a 4-tuple of signs, you haven't written down any physics; you've just juggled coordinate labels, and they're not really labeling anything anyway since you haven't written down a metric.

to be more specific, when I say "instead choose the metric with signature..." I mean keep things like the stress-energy tensor and the distribution of matter the same. But choose the metric to be different (i.e. choose a different timelike component).

This makes no sense. I really, really think you need to stop waving your hands and try doing this for a specific, explicit model, so you can see what's actually involved. You are relying way too much on your intuition without grounding your intuition in any actual physics. By a specific, explicit model I mean, once again, writing down an explicit expression for $ds^2$ in terms of some set of coordinates, as well as any other physical quantities you want--stress-energy tensor, whatever. Then try to "choose the metric to be different" in that specific case and see what happens.

 

[BruceW]

What you described is not a different signature; the number of $+$ and $-$ signs are the same in both.

That's good. I want it to still have the same number of plus and minus signs. Ah, I see. the term 'signature' just means the number of plus and minus signs. OK, sorry I was not using correct terminology. right, what I meant to say, is to keep the same signature, but change which component is the 'odd one out'.

 

[PeterDonis]

what I meant to say, is to keep the same signature, but change which component is the 'odd one out'.

This has no meaning unless you write down an expression for $ds^2$, as I keep on asking you to do. The ordering of the coordinates means nothing by itself.

 

[BruceW]

We don't know, because you haven't given us a physical description. You haven't given us an actual metric; you've just played around with which order you write the coordinates. Show us an actual metric--an actual expression for $ds^2$ in terms of your coordinates--and we'll be able to tell you which curves are timelike and which are spacelike.

hmm. Now I think about it, any metric which is isotropic in 3 of its dimensions is forced to have the other dimension as the timelike dimension. But this argument doesn't work for a metric which is not isotropic. Alright, I concede a lot of ground here! :) If our metric is isotropic in 3 of its dimensions, then I agree that the timelike curves are automatically defined for us. But I still maintain that for a non-isotropic metric, we have an arbitrary choice of which dimension is the timelike one.

 

[BruceW]

as an example of a metric which is not isotropic in just 3 of its dimensions, we have the simple metric: ds^2 = -dt^2 + dx^2 + dy^2 + dz^2
In this case, our choice of timelike dimension is arbitrary.

 

[PeterDonis]

I still maintain that for a non-isotropic metric, we have an arbitrary choice of which dimension is the timelike one.

This is incorrect; isotropy has nothing to do with it. Perhaps the problem here is that you have a mistaken understanding of what "isotropy" means:

as an example of a metric which is not isotropic in just 3 of its dimensions, we have the simple metric: ds^2 = -dt^2 + dx^2 + dy^2 + dz^2
In this case, our choice of timelike dimension is arbitrary.

This metric *is* isotropic; it's just standard Minkowski spacetime. I think you need to clarify what you think "isotropic" means.

 

[BruceW]

This is incorrect; isotropy has nothing to do with it. Perhaps the problem here is that you have a mistaken understanding of what "isotropy" means:

No, isotropy in 3 of the dimensions (but not the other one) is the entire reason why we can tell which curves are the timelike ones. And thank you very much for helping me to realize this. I think it is quite an important point in general relativity that I will try to remember.

This metric *is* isotropic; it's just standard Minkowski spacetime. I think you need to clarify what you think "isotropic" means.

ah, but this metric is isotropic in all four dimensions. Therefore it does not force us to choose a particular dimension to be the timelike one. And I think I know what isotropy means. In wiki's page on the FLRW metric: "The FLRW metric starts with the assumption of homogeneity and isotropy of space." As they say, it is isotropic in the 3 'space' dimensions but not in the 'time' dimension. And it is the very fact that it is isotropic in 3 of the dimensions that forces us to choose the other dimension as the timelike dimension.

 

[robphy]

I agree that once you have specified if the curve is inside or outside the light cone, then the timelike or spacelike character of that line is fixed. What I am saying is that to begin with, specifying the curve to lie inside or outside the light cone is an arbitrary choice that we can make.

(Sorry for jumping back into the thread...)

In (3+1)-dimensional Minkowski spacetime,
the set of curves with [everywhere-]spacelike-tangent-vectors includes closed-spacelike curves
whereas the set with [everywhere-]timelike-tangent-vectors has no closed-timelike curves.

To detach the physical ideas of "space" and "time" from the words used,
it might be helpful to call the timelike curves "curves of the first kind (associated with [with my signature convention] positive square-norm)"
and the spacelike curves "curves of the second kind (associated with negative square-norm)".

 

[PeterDonis]

No, isotropy in 3 of the dimensions (but not the other one)

I think I see what you are saying here, but your terminology is highly nonstandard. See below.

this metric is isotropic in all four dimensions. Therefore it does not force us to choose a particular dimension to be the timelike one. And I think I know what isotropy means. In wiki's page on the FLRW metric: "The FLRW metric starts with the assumption of homogeneity and isotropy of space." As they say, it is isotropic in the 3 'space' dimensions but not in the 'time' dimension. And it is the very fact that it is isotropic in 3 of the dimensions that forces us to choose the other dimension as the timelike dimension.

Ok, here's how to state what you just stated here in standard terminology:

Minkowski spacetime and FLRW spacetime are both isotropic, because "isotropy" in the standard usage means "isotropy in the 3 spatial dimensions" in your usage.

Minkowski spacetime and FLRW spacetime are also both homogeneous; I think your usage of that word is the same as the standard usage.

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.

Having said all that: you still have not shown how we can arbitrarily choose the timelike direction in Minkowski spacetime. A spacetime being stationary does not allow you to do that. If you think it does, then please show how, explicitly. Don't just wave your hands and say you think it can be done. Show how it can be done.

 

[PeterDonis]

Show how it can be done.

Perhaps it will help if I give an example that doesn't involve the timelike/spacelike distinction, to make it clearer why I and others are reacting so strongly to the idea that choosing the timelike direction is arbitrary.

Consider three points on the Earth's surface: the North Pole; Quito, Ecuador; and Nairobi, Kenya. Both of the cities are (to a good enough approximation for here) on the Earth's equator, and Nairobi is due East of Kenya, whereas the North Pole is due North of both.

Now consider a claim analogous to the one you (BruceW) have been making: the choice of which direction is "North" is arbitrary. I can just as easily set up coordinates in which Nairobi is due North of Quito.

Of course I can choose labels for the directions in any way I want; but all that does is change the meanings of the labels. If I pick labels for the directions such that Nairobi is due North of Quito, then what had been called the North Pole is now the West Pole, and is due West of both Quito and Nairobi. No actual geometry has changed; no actual physics has changed. I've just changed labels.

I could even make a more drastic coordinate change: I could pick coordinates such that Nairobi is due East of both Quito and the "North" Pole. But "North" in the name of the Pole now has to be in quotes because it's only North of Quito; it's not North of Nairobi in these new coordinates. Here I have not only changed labels, I've changed the orientation of the coordinate grid on the Earth (basically I've exchanged latitude and longitude).

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, or the angles of the triangle formed by those three points (here "triangle" really means the figure on the Earth's surface formed by the three great circles connecting the pairs of points). There is also no arbitrary choice I can make that will change the fact that the "North Pole" (however it's labeled in my chosen coordinates) is on the Earth's axis of rotation, and Quito and Ecuador are both in a plane perpendicular to that axis. 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.

In the case of spacetime, what I and others have been saying is that your claim that you can arbitrarily choose the timelike direction is *not* like the claim that you can arbitrarily choose coordinates on the Earth, as above; it's like the claim that you can arbitrarily choose whether the North Pole or some other point is on the Earth's axis of rotation, or whether Quito and Nairobi, or some other points, are in the plane perpendicular to that axis. That's why we're so skeptical of your claim.

 

[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.)