Timelike v. spacelike, is it arbitrary?

[Dale]

OK, you can call it 'geometry', but it is still arbitrarily defined geometry. It does not correspond to anything physical.

It corresponds to physical light cones.

 

[BruceW]

We can measure time-like spacetime intervals using a single inertial clock that passes through both events. This does not require any synchronization convention since there is only one clock and it does not require any ruler.

However, for space-like spacetime intervals, even though we are measuring it with a single ruler, it is not sufficient that two points on that ruler pass through both events, they must pass through at the same time as determined by two synchronized clocks located at the two points of measurement.

this is a good example of what I am talking about. right, so let's say we have defined what are the time-like intervals. And 'normal' matter all moves along these time-like intervals. So now, we can use some normal matter to measure any time-like spacetime interval. (for example, a muon beam acts a very good clock, since the muon's mean lifetime is accurately known). i.e. we send a muon beam along a time-like interval, and the fraction remaining tells us the arc length along that path.

But now, suppose we come across some not-normal matter that moves along a space-like interval. Again, suppose it is like a muon, i.e. the amount of muons that decay depends only on the arc length that the beam has traveled. Now, we could use this weird matter to measure the arc length along a space-like interval, simply by observing the fraction of muons that remain, since this tells us the arc length along that path.

So again, the point I am trying to get across is that if we assume (as part of the theory of relativity) that all matter travels along the timelike curves, then yes there is a physical difference between which group of curves we label as 'timelike' and which group of curves we label as 'spacelike'. But if we do not make such an assumption, then the choice is arbitrary.

 

[Dale]

well, we both agree that when there is no matter around, such a model works fine, right?

I disagree.

 

[BruceW]

Not without changing the physics. Once you have specified if the curve lies inside or outside the light cone then the physics is set and the timelike or spacelike character of the line is fixed.

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.

 

[Dale]

specifying the curve to lie inside or outside the light cone is an arbitrary choice that we can make.

Not without changing the physics.

 

[BruceW]

Not without changing the physics.

when you say 'the physics', if you mean an arbitrarily defined non-physical convention, then I agree. Else, I disagree.

Edit: ah, or if you want to say that we define all matter to move along the timelike curves, then I would agree that in a universe with matter, we have a non-arbitrary way to assign timelike and spacelike curves.

 

[Dale]

when you say 'the physics', if you mean an arbitrarily defined non-physical convention, then I agree. Else, I disagree.

Edit: ah, or if you want to say that we define all matter to move along the timelike curves, then I would agree that in a universe with matter, we have a non-arbitrary way to assign timelike and spacelike curves.

In a universe with only radiation and no matter we can physically distinguish a line which lies within a light cone from a line which lies outside a light cone. There is nothing arbitrary about it. A line inside is timelike; a line outside is spacelike.

 

[BruceW]

we seem unable to understand each other's line of reasoning. I am trying though :) OK, so let's say we have components a,b,c,d (and the order I have listed them does not imply anything about the metric). (And I have avoided the usual t,x,y,z because that would imply which component should be the timelike component). Now, let's say I look at a path that goes completely along the c component. Is this path timelike or spacelike?

There is no way to tell, because I have not told you which component is the timelike component. And further, it doesn't matter which component I choose as the timelike component. The only way it would be important is if we define matter to move along timelike curves, since I would then have to define the timelike component so that matter travels along timelike curves.

 

[WannabeNewton]

There is no need to look at components in a frame. Forget components, because that is not geometry! You don't need to tell me anything about that. All I need to check is if the squared norm of the tangent vector is everywhere negative.

 

[Dale]

The only way it would be important is if we define matter to move along timelike curves, since I would then have to define the timelike component so that matter travels along timelike curves.

No. It would also be important if you have radiation. You would have to define the timelike compnent to be the one inside light cones.

The reason that I cannot tell which is which is because you eliminated both matter and radiation (i.e. your scenario involves no physics). As soon as there is any physics, whether it is matter or radiation, the choice is constrained by the physics. A choice which is constrained by the physics is not arbitrary.

 

[BruceW]

No. It would also be important if you have radiation. You would have to define the timelike compnent to be the one inside light cones.

The reason that I cannot tell which is which is because you eliminated both matter and radiation (i.e. your scenario involves no physics). As soon as there is any physics, whether it is matter or radiation, the choice is constrained by the physics. That makes it not arbitrary.

alright, say there is a beam of light with tangent vector (0,1,1,0) Now is the path with tangent (0,0,1,0) timelike or spacelike?

 

[Dale]

alright, say there is a beam of light with tangent vector (0,1,1,0)

There is no such thing. It doesn't satisfy Maxwell's equations.

Give me any EM radiation field which actually satisfies Maxwell's vacuum equations (since you want to eliminate matter) with physically possible boundary conditions in terms of your a,b,c,d coordinates and I can tell you which is the timelike and which is the spacelike.

 

[BruceW]

what do you mean? I haven't said which component is the timelike component.

 

[ghwellsjr]

We can measure time-like spacetime intervals using a single inertial clock that passes through both events. This does not require any synchronization convention since there is only one clock and it does not require any ruler.

However, for space-like spacetime intervals, even though we are measuring it with a single ruler, it is not sufficient that two points on that ruler pass through both events, they must pass through at the same time as determined by two synchronized clocks located at the two points of measurement.

this is a good example of what I am talking about. right, so let's say we have defined what are the time-like intervals. And 'normal' matter all moves along these time-like intervals. So now, we can use some normal matter to measure any time-like spacetime interval. (for example, a muon beam acts a very good clock, since the muon's mean lifetime is accurately known). i.e. we send a muon beam along a time-like interval, and the fraction remaining tells us the arc length along that path.

But now, suppose we come across some not-normal matter that moves along a space-like interval. Again, suppose it is like a muon, i.e. the amount of muons that decay depends only on the arc length that the beam has traveled. Now, we could use this weird matter to measure the arc length along a space-like interval, simply by observing the fraction of muons that remain, since this tells us the arc length along that path.

Your supposition is speculation, isn't it? It has nothing to do with the universe we live in, the physics we use to describe it, or teaching relativity which is the purpose of this forum.

So again, the point I am trying to get across is that if we assume (as part of the theory of relativity) that all matter travels along the timelike curves, then yes there is a physical difference between which group of curves we label as 'timelike' and which group of curves we label as 'spacelike'. But if we do not make such an assumption, then the choice is arbitrary.

There is no assumption in the theory of Special Relativity that matter travels along timelike curves. There are two assumptions (postulates) and from them we get the three distinctly different categories of spacetime intervals and the concept of Proper Time. Your misunderstanding that Proper Time or a time-like spacetime intervals are no different than Proper Length or space-like spacetime intervals or null spacetime intervals is not part of teaching Special Relativity. You are attempting to promote something different.

 

[Dale]

what do you mean?

I mean that, strictly speaking, a beam of light is a fiction. It does not satisfy Maxwell's equations. There is always some divergence of the beam, and that divergence (required to satisfy Maxwell's equations) identifies the other two spatial dimensions.

Radiation satisfies $\Box A^{\mu} = 0$, and if you provide any physically possible A which satisfies that equation then timelike and spacelike are determined.

 

[BruceW]

Your supposition is speculation, isn't it? It has nothing to do with the universe we live in, the physics we use to describe it, or teaching relativity which is the purpose of this forum.

I was just saying that if we use the definition that all matter travels along timelike curves, then yes, we can say which curves are timelike and which are spacelike. But If we do not use this definition, then we cannot.

Your misunderstanding that Proper Time or a time-like spacetime intervals are no different than Proper Length or space-like spacetime intervals or null spacetime intervals is not part of teaching Special Relativity. You are attempting to promote something different.

I agree that we can choose a set of time-like curves and a set of space-like curves. But I am saying that for a given physical situation, it is our choice for which ones are time-like and which ones are space-like. (Unless we define all matter to travel along timelike curves, in which case the choice is made for us by this definition).

 

[Dale]

BruceW, besides the physical point which I am making there is the geometrical point which WBN and PD are making. The geometry is determined by the metric. Any Lorentzian metric will have three eigenvalues with one sign and one eigenvalue with the opposite sign. The eigenvector corresponding to the one with the opposite sign is timelike. No matter, no physics, pure relativity, pure geometry. The only way that you can make it ambiguous is by NOT specifying the geometry (i.e. by not specifying the metric).

Simply put, there is no way in which the choice is arbitrary. It is constrained by the geometry, it is constrained by the physics, that is as far from arbitrary as is possible. The only possible ambiguity/arbitraryness comes from not specifying either the geometry or the physics.

 

[PeterDonis]

But now, suppose we come across some not-normal matter that moves along a space-like interval. Again, suppose it is like a muon, i.e. the amount of muons that decay depends only on the arc length that the beam has traveled. Now, we could use this weird matter to measure the arc length along a space-like interval, simply by observing the fraction of muons that remain, since this tells us the arc length along that path.

Have you actually tried to construct such a model? If so, please show your work, as I asked before. If not, you are just waving your hands and assuming that such a model would work the way you say and still be consistent. That's why I asked you before to actually do the math, instead of just speculating. (And you should really look into the literature on tachyon models; as I've said a couple of times already, there are subtleties lurking there.)

 

[ghwellsjr]

Have you actually tried to construct such a model? If so, please show your work, as I asked before. If not, you are just waving your hands and assuming that such a model would work the way you say and still be consistent. That's why I asked you before to actually do the math, instead of just speculating. (And you should really look into the literature on tachyon models; as I've said a couple of times already, there are subtleties lurking there.)

Isn't it still speculation, even if he does come up with consistent math?

 

[PeterDonis]

OK, so let's say we have components a,b,c,d (and the order I have listed them does not imply anything about the metric). (And I have avoided the usual t,x,y,z because that would imply which component should be the timelike component). Now, let's say I look at a path that goes completely along the c component. Is this path timelike or spacelike?

There is no way to tell, because I have not told you which component is the timelike component.

No, there's no way to tell because you haven't told us what the metric is. Which means, as DaleSpam said, that you aren't doing physics; you're just throwing letters and numbers around. As soon as you define a metric, you have defined which paths are timelike and which are spacelike. You don't have to make any assumptions about what kinds of objects travel on what kinds of paths.

 

[PeterDonis]

Isn't it still speculation, even if he does come up with consistent math?

"Speculation" is a broad word. PF's prohibition is on "personal" speculation. There is, as I mentioned, plenty of literature on tachyon models, so talking about those models and what kinds of subtleties arise in them would not be personal speculation, even though no tachyons have ever actually been observed.

 

[PeterDonis]

for a given physical situation, it is our choice for which ones are time-like and which ones are space-like.

Incorrect; the "physical situation" includes the metric, which, as I said, determines which curves are timelike and which are spacelike, regardless of what kinds of objects travel on what kinds of paths.

 

[PeterDonis]

we both agree that when there is no matter around, such a model works fine, right?

No, I don't, because you haven't shown me such a model. I can't say anything about such a model until you actually show me one.

 

[BruceW]

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? (apart from the fact that matter only travels along timelike curves).

I would (intuitively) say that our new physical description would also work. As long as our equations are manifestly covariant, then I would be surprised that a (-1,1,1,1) metric would work but a (1,-1,1,1) metric would not. (again, ignoring the fact that matter travels along timelike curves).

edit: 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).

 

[WannabeNewton]

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

 

[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.