Timelike v. spacelike, is it arbitrary?

[PAllen]

PAllen:

I don't doubt it if you say so, but wonder about the implications.

What does this mean? Can we attribute this to some physical characteristic? Also you mention 'independence'...and I wonder if in our FLRW cosmological model, and SR and GR, that condition is satisfied. Does 'independence' relate to isotropy and homogeneaty of
the FLRW model...

On orthogonality, I don't know if there is some cute analogy for the properties of Lorentzian space. Mathematically, a Lorentzian metric has the feature that for any timelike direction from a point, there is an orthogonal spatial 3-plane; and for any spacelike direction, there is an orthogonal 2+1 plane. But for light-like, by definition, this is a vector whose norm = dot product with itself = zero. Thus it is 'orthogonal' in some sense to itself and to not to any independent 3-surface . There is an independent spacelike 2-plane orthogonal to a light like direction; there is also an orthogonal 3 surface that can be formed from the light like vector cross the independent orthogonal spatial two plane. This surface is not independent and is neither spatial nor a 2+1 surface.[edit: corrected; see Robphy's post below, and a message] ).

The comment about independence simply means that the most common coordinates are based on orthonormal basis vectors, but this is not a requirement for coordinates. In GR, many common coordinates (e.g. Gullestrand-Panlieve) do not have an orthogonall basis (Schwarzschild coordinates are orthogonal in the region they cover; not orthonormal - this canpt be achieved except locally in GR) . If you see non-diagonal elements in the metric, you know that the basis there is non-orthogonal.

So, while orthogonality of basis is not required for a coordinate system, independence is what is required - if one basis vector is a linear combination of others, you will not be able to reach points not already reachable by the others. Thus, 4 alleged basis vectors, where one is not independent of the others, only covers a 3-space. This has nothing at all to do with isotropy, homogeneity, cosmology etc. - it is just a requirement for constructing coordinates.

 

[robphy]

But for light-like, by definition, this is a vector whose norm = dot product with itself = zero. Thus it is 'orthogonal' in some sense to itself and to no other direction (timelike, lightlike, or spacelike).

Given null vector (t,x,y,z)=(1,1,0,0), the spacelike vector (0,0,1,1) has zero dot-product with it.
http://books.google.com/books?id=wp2A7ZBUwDgC&pg=PA84 (4.)

 

[PAllen]

Given null vector (t,x,y,z)=(1,1,0,0), the spacelike vector (0,0,1,1) has zero dot-product with it.
http://books.google.com/books?id=wp2A7ZBUwDgC&pg=PA84 (4.)

Oops, you're right. There is a whole 2-plane orthogonal to a light like direction. Prior post corrected.

 

[WannabeNewton]

Given a null vector $\lambda^a$, if by light-like "direction" you mean $\text{span}\{\lambda^a \}$ then yes any null vector $\gamma^a$ such that $\gamma^a \lambda_a = 0$ implies $\gamma^a \in \text{span}\{\lambda^a \}$ and the converse holds as well. However I don't know what is actually being connoted by the term light-like "direction".

 

[PeterDonis]

There is a whole 2-plane orthogonal to a light like direction.

But only a 2-plane, as opposed to a 3-plane or a 2+1-plane. The third "orthogonal" direction for a null vector is *parallel* to the vector; that's the difference between null vectors and timelike/spacelike ones.

 

[PAllen]

But only a 2-plane, as opposed to a 3-plane or a 2+1-plane. The third "orthogonal" direction for a null vector is *parallel* to the vector; that's the difference between null vectors and timelike/spacelike ones.

Right, this was immediately obvious to me.

 

[PAllen]

Given a null vector $\lambda^a$, if by light-like "direction" you mean $\text{span}\{\lambda^a \}$ then yes any null vector $\gamma^a$ such that $\gamma^a \lambda_a = 0$ implies $\gamma^a \in \text{span}\{\lambda^a \}$ and the converse holds as well. However I don't know what is actually being connoted by the term light-like "direction".

That's a good definition.

 

[BruceW]

Is the issue just between the SR definition of spacetime intervals for timelike versus spacelike or is it more than that?

My issue was not specific to GR or SR. It was relevant to both SR and GR. The issue was that I thought we had to use matter to tell us which intervals are timelike and which are spacelike. So I was thinking that the difference between timelike and spacelike was fairly non-fundamental to GR/SR. But (I didn't realize this to begin with) we can also use a couple of beams of light to tell us what the timelike intervals are. And since pretty much all our experiments in relativity must use radiation or matter anyway, I am now satisfied that the distinction between timelike and spacelike is fairly fundamental to GR/SR.

I guess it was an issue of whether it is always possible to do an experiment to find out if an interval is timelike. If it is always possible to do such an experiment, then I am OK with the timelike concept as being fundamental in GR/SR. And since I now know (thanks to you people) that we can use matter or radiation to be able to do such an experiment, I am happy with saying it is always possible to do such an experiment.

In SR is it also for other curves or worldlines rather than just spacetime intervals?

my issue was just with spacetime intervals. I'm not sure what you mean by other curves or worldlines... I guess it would apply to all curves. And worldlines are a particular case, right?

Is the issue purely a mathematical one that could also apply to broader situations beyond the physics that describe our universe?

no not really. It was more of a "what can we take as fundamental to GR/SR due to our assumptions about the kinds of experiments we are always able to do". So not really a mathematical issue.

Is the issue between establishing whether clocks measure time and rulers measure space or the other way around?

I guess that now I have accepted that we can always find the timelike direction at a given point, we are automatically able to tell if something has a timelike or spacelike tangent vector. (which then tells us if we are measuring space or time, as you say).

 

[ghwellsjr]

I'm going to respond to your answers out of order because it makes more sense to me to do it that way:

In SR is it also for other curves or worldlines rather than just spacetime intervals?

my issue was just with spacetime intervals.

Good. A spacetime interval has a specific meaning and I assume you are referring to that specific meaning. In SR, it refers to an invariant property between a pair of events. Given two events, we can measure or calculate the spacetime interval. In any Inertial Reference Frame, it is a straight line through spacetime connecting the two events of interest.

I'm not sure what you mean by other curves or worldlines... I guess it would apply to all curves. And worldlines are a particular case, right?

"Worldline" is just another word for the path through spacetime that goes between two arbitrary events. For a spacetime interval, it is always straight (meaning the longest possible) but for other situations, it can be curved. I asked because I wasn't sure if you were consistently referring exclusively to the straight spacetime interval or some other more general worldline between two events.

Is the issue purely a mathematical one that could also apply to broader situations beyond the physics that describe our universe?

no not really. It was more of a "what can we take as fundamental to GR/SR due to our assumptions about the kinds of experiments we are always able to do". So not really a mathematical issue.

Good, we can set that aside.

Is the issue between establishing whether clocks measure time and rulers measure space or the other way around?

I guess that now I have accepted that we can always find the timelike direction at a given point, we are automatically able to tell if something has a timelike or spacelike tangent vector. (which then tells us if we are measuring space or time, as you say).

I'm going to assume that you agree that clocks always measure time and rulers always measure space. My question was more basic than what you answered as I wasn't referring specifically to spacetime intervals.

Is the issue just between the SR definition of spacetime intervals for timelike versus spacelike or is it more than that?

My issue was not specific to GR or SR. It was relevant to both SR and GR. The issue was that I thought we had to use matter to tell us which intervals are timelike and which are spacelike. So I was thinking that the difference between timelike and spacelike was fairly non-fundamental to GR/SR. But (I didn't realize this to begin with) we can also use a couple of beams of light to tell us what the timelike intervals are. And since pretty much all our experiments in relativity must use radiation or matter anyway, I am now satisfied that the distinction between timelike and spacelike is fairly fundamental to GR/SR.

I guess it was an issue of whether it is always possible to do an experiment to find out if an interval is timelike. If it is always possible to do such an experiment, then I am OK with the timelike concept as being fundamental in GR/SR. And since I now know (thanks to you people) that we can use matter or radiation to be able to do such an experiment, I am happy with saying it is always possible to do such an experiment.

I'm not sure what experiment you are talking about.

Here's an experiment I would use (assuming, of course, that we are enabled to do anything we want within the bounds of what is physically possible and that we can repeat the experiment over and over again for the same two events until we get it right.)

For any two events, see if it is possible for a single inertial clock to be present at both events. If so, the time interval measured by the clock between the two events is the timelike spacetime interval. This is even more fundamental than SR or GR because it doesn't require any convention. It doesn't even require any theory. It only requires knowing when a clock is inertial.

Can you similarly describe the experiment that you mentioned involving two light beams to tell us what the timelike intervals are?

 

[BruceW]

I'm going to respond to your answers out of order because it makes more sense to me to do it that way:

thanks man, I appreciate it.

"Worldline" is just another word for the path through spacetime that goes between two arbitrary events. For a spacetime interval, it is always straight (meaning the longest possible) but for other situations, it can be curved. I asked because I wasn't sure if you were consistently referring exclusively to the straight spacetime interval or some other more general worldline between two events.

you can measure the spacetime interval along 'not-straight' paths too. I was talking about the spacetime interval in the most general way possible. (i.e. along any general curve). Also, I thought 'worldline' meant timelike curve.

edit: well not a completely general curve. It needs to be smooth I guess.

Can you similarly describe the experiment that you mentioned involving two light beams to tell us what the timelike intervals are?

What I had in mind is pretty much what DaleSpam was saying about lightcones. I just did not realize when he first said it. OK, say you send off two non-collinear beams of light, and say that you can track them over some small distance after they leave your position. The vectors you will measure will be something like (1,1,0,0) and (1,0,1,0) and therefore, we now know that the timelike direction is (1,0,0,0) since this is the component they both have in common.

In SR, this is fairly simple. And in GR, we can use normal coordinates, since then geodesics that go through our origin are approximately just like normal SR vectors. (we just need to make sure that we track the light beams over a short distance after they leave).

edit: and if the light beams have a vector like $( \sqrt{3} ,1,1,1)$ then we can also tell which is the timelike direction, because of the fact that the vector must be a null vector (since it is a beam of light). And in any case, we can use an entire 'light cone' of different light beams by just sending beams of light in lots of directions.

 

[WannabeNewton]

Indeed worldline refers only to time-like curves in space-time. And indeed space-time intervals and worldlines are two different things and are related in exactly the way you have stated.

 

[BruceW]

cool, glad I've got it all sorted out in my head now. thanks newtwon!

 

[ghwellsjr]

thanks man, I appreciate it.

you can measure the spacetime interval along 'not-straight' paths too. I was talking about the spacetime interval in the most general way possible. (i.e. along any general curve). Also, I thought 'worldline' meant timelike curve.

edit: well not a completely general curve. It needs to be smooth I guess.

Worldline is one of those words that is defined differently by different sources. Wikipedia, for example, agrees with you and states that my usage causes confusion. However, Taylor and Wheeler also include lightlike paths. So I'm not sure that there is a commonly accepted standard definition for worldline.

However, there is a commonly accepted standard definition for spacetime interval in SR and it excludes curved paths. It is always the longest worldline between any two arbitrary events. Now see, I can't even say it that way, I have to say it's the longest path between any two arbitrary events.

So now that I realize that you not using the standard definition of "spacetime interval", your answer to the first question in post #129 has to be that your issue is not just with spacetime intervals, correct? And that explains why my posts haven't connected with you, correct?

What I had in mind is pretty much what DaleSpam was saying about lightcones. I just did not realize when he first said it. OK, say you send off two non-collinear beams of light, and say that you can track them over some small distance after they leave your position. The vectors you will measure will be something like (1,1,0,0) and (1,0,1,0) and therefore, we now know that the timelike direction is (1,0,0,0) since this is the component they both have in common.

In SR, this is fairly simple. And in GR, we can use normal coordinates, since then geodesics that go through our origin are approximately just like normal SR vectors. (we just need to make sure that we track the light beams over a short distance after they leave).

edit: and if the light beams have a vector like $( \sqrt{3} ,1,1,1)$ then we can also tell which is the timelike direction, because of the fact that the vector must be a null vector (since it is a beam of light). And in any case, we can use an entire 'light cone' of different light beams by just sending beams of light in lots of directions.

So why not just set off a flash bulb and emit beams of light in all directions? Isn't that just what you are describing in your last edit? But however you are emitting the light, how do you track the beams?

And I still fail to see how this has anything to do with determining the type of the spacetime interval between any two arbitrary events, although I have always been assuming the standard definition and you may be thinking of something entirely different.

 

[WannabeNewton]

A space-time interval is not a worldline. A worldline is the image $\gamma(I)$ of a special type of curve $\gamma:I \rightarrow M$ where $M$ is a space-time and $I \subset \mathbb{R}$ is the unit interval. The space-time interval between two events is a certain real number. Between two locally separated events connected by time-like curves, it is a geodesic that maximizes the length. You are mixing up different concepts.

 

[BruceW]

... So I'm not sure that there is a commonly accepted standard definition for worldline.

I see. yeah, ok.

However, there is a commonly accepted standard definition for spacetime interval in SR and it excludes curved paths. It is always the longest worldline between any two arbitrary events. Now see, I can't even say it that way, I have to say it's the longest path between any two arbitrary events.

Where do you get that definition of spacetime interval from? I'm pretty sure that is not the standard definition. For example, in the twin paradox, we talk about the spacetime interval along the path taken by the accelerating twin. (and since the twin is accelerating, this is not the longest worldline between events).

So now that I realize that you not using the standard definition of "spacetime interval", your answer to the first question in post #129 has to be that your issue is not just with spacetime intervals, correct? And that explains why my posts haven't connected with you, correct?

I don't think it makes a difference either way. My issue is the same whether we are talking about timelike paths or general paths.

So why not just set off a flash bulb and emit beams of light in all directions? Isn't that just what you are describing in your last edit? But however you are emitting the light, how do you track the beams?

exactly. very good question. Once I got to here, I assumed we are always able to do this, but I haven't explicitly shown it... Um I dunno, maybe by placing detectors all around our origin? And then I guess another problem is whether we are always able to place detectors around our origin according to normal coordinates. Since we can put these detectors very close to our origin, I am fairly sure this is always possible, but I'm not certain. Hopefully someone else knows?

And I still fail to see how this has anything to do with determining the type of the spacetime interval between any two arbitrary events, although I have always been assuming the standard definition and you may be thinking of something entirely different.

well, once we have established the timelike direction at our origin, then we can say if any path going through our origin is timelike or spacelike.

edit: replace 'origin' with 'coordinates representing me,here,now' if you prefer.

 

[BruceW]

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).

going back to this, what if we want to know whether a (non-geodesic) smooth curve through the point is timelike or spacelike? I would guess that this is possible. But on the wiki page for normal coordinates, it only mentions that we can write geodesics which go through the point as $(tV^1,...,tV^n)$ and it does not seem to explicitly mention that we can do this for non-geodesic curves through the point.

 

[PeterDonis]

going back to this, what if we want to know whether a (non-geodesic) smooth curve through the point is timelike or spacelike?

First, a clarification: strictly speaking, the terms timelike, spacelike, and null apply to *vectors*, not curves--more precisely, they apply to tangent vectors to curves at particular points.

You can always tell whether a curve's tangent vector at a given point is timelike, null, or spacelike, because the tangent space at a given point looks just like Minkowski spacetime, and we've established here how to tell whether a vector in Minkowski spacetime is timelike, null, or spacelike.

With geodesic curves, once you know its tangent vector at a single point, you know the entire curve, so the above is sufficient to know the causal nature (i.e., timelike, null, or spacelike) of the entire curve--which of course means that nature can't change anywhere along the curve, if it's (timelike, spacelike, null) at one event, it's (timelike, spacelike, null) everywhere.

With non-geodesic curves, however, the causal nature of the curve can *change* from event to event, so in general you need to look at every point on the curve separately to establish whether the curve's tangent vector at that point is timelike, null, or spacelike. In some special cases there's a quicker way to tell: for example, it's easy to show that a curve of constant $r, \theta, \phi$ in Schwarzschild spacetime is timelike for $r > 2M$, null for $r = 2M$, and spacelike for $r < 2M$, just from looking at the Schwarzschild line element, even though none of these curves are geodesics.

 

[BruceW]

right, so it is enough to know the tangent vector of the curve at some point to say if it is timelike at that point. In fact, I see on this page: http://en.wikipedia.org/wiki/Gauss's_lemma_(Riemannian_geometry) some more explanation is given. And it looks like (in a certain sense), any general smooth curve through a point can be identified (locally) with a geodesic through the point.

 

[PeterDonis]

And it looks like (in a certain sense), any general smooth curve through a point can be identified (locally) with a geodesic through the point.

Yes, since a tangent vector at any point defines a geodesic, any non-geodesic curve with the same tangent vector at that point can be locally identified with the geodesic having that tangent vector.

 

[BruceW]

awesome. that's some pretty cool maths. thanks, man!

 

[WannabeNewton]

In fact that is exactly what you are doing when you go to a momentarily comoving locally inertial frame to an arbitrary timelike curve at some event on the curve.

 

[BruceW]

'when you go to a ... to a ...' the grammar does not work ?! But I think I understand what you mean.

Also, I was thinking about it again and I am not sure why they (on wikipedia) go on about the geodesics when they introduce normal coordinates. Any (not necessarily geodesic) smooth curve through the point can be expressed as $(tV^1,...,tV^4)$ so there's nothing interesting about the geodesics through the point is there? Maybe they make such a fuss about the geodesics because they can be uniquely associated with every tangent vector at the point? (there will be many different general curves which have the same tangent vector at the point). And on this wiki page http://en.wikipedia.org/wiki/Gauss's_lemma_(Riemannian_geometry) , they say that "any sufficiently small sphere centered at a point in a Riemannian manifold is perpendicular to every geodesic through the point." But surely it is just as valid to say 'perpendicular to every smooth curve through the point'... it seems weird to me that they would specify geodesic for no particular reason.

 

[PeterDonis]

Maybe they make such a fuss about the geodesics because they can be uniquely associated with every tangent vector at the point?

Yes. Only geodesics have this property.

surely it is just as valid to say 'perpendicular to every smooth curve through the point'

No, because the point of the lemma is that geodesics radiating outward from the chosen point, when they cross a sphere of some small radius centered on the point, are perpendicular to the sphere, and that property is only true for geodesics. Non-geodesic curves will not all be perpendicular to the sphere, because they will "bend" relative to the geodesics. For example, think of curves radiating out from the north pole of a 2-sphere and crossing a curve of constant latitude some small distance south of the pole (the curve of constant latitude is the "sphere centered on the chosen point"--in this case a 1-sphere); the curves will all be perpendicular to the latitude line only if they are all geodesics, i.e., great circles.

Note that this lemma is for Riemannian geometry, not pseudo-Riemannian, so it is not directly applicable to spacetime.

 

[WannabeNewton]

'when you go to a ... to a ...' the grammar does not work ?! But I think I understand what you mean.

What I meant is that if $p$ is an event on some time-like curve $\gamma$ followed by a particle, then we can always find a locally inertial frame momentarily comoving with the particle at $p$ which is basically just the statement made above by Peter that at $p$, we can always find a time-like geodesic which intersects $\gamma$ and has the same tangent vector as $\gamma$.

 

[BruceW]

What I meant is that if $p$ is an event on some time-like curve $\gamma$ followed by a particle, then we can always find a locally inertial frame momentarily comoving with the particle at $p$ which is basically just the statement made above by Peter that at $p$, we can always find a time-like geodesic which intersects $\gamma$ and has the same tangent vector as $\gamma$.

ah right. yeah, that makes sense to me.

 

[BruceW]

No, because the point of the lemma is that geodesics radiating outward from the chosen point, when they cross a sphere of some small radius centered on the point, are perpendicular to the sphere, and that property is only true for geodesics. Non-geodesic curves will not all be perpendicular to the sphere, because they will "bend" relative to the geodesics.

that doesn't make sense to me. I thought we were saying that any non-geodesic curve close to the point can be locally identified with a geodesic through that point. In other words, that non-geodesic curves which go through the point can also be represented in normal coordinates as $(tV^1,...,tV^4)$. So, I thought that in the normal neighbourhood of the point, all curves are geodesic curves, in effect.

Note that this lemma is for Riemannian geometry, not pseudo-Riemannian, so it is not directly applicable to spacetime.

that's interesting. Is there some reason why it does not work for spacetime? Is it because we don't have the same notion of 'perpendicular' for pseudo-Riemmanian geometry?

 

[PeterDonis]

I thought we were saying that any non-geodesic curve close to the point can be locally identified with a geodesic through that point.

In the sense that they both have the same tangent vector at the point, yes. But only at the point. See below.

In other words, that non-geodesic curves which go through the point can also be represented in normal coordinates as $(tV^1,...,tV^4)$.

No; only geodesics will appear as straight lines in normal coordinates throughout the entire coordinate patch surrounding the point. Non-geodesic curves will appear as curved lines that are tangent to the geodesic straight lines at the chosen point.

Is there some reason why it does not work for spacetime?

Yes; you can't define a "sphere" the same way, because the metric is not positive definite. There might be theorems which are somewhat analogous for spacetime, but I don't know of any off the top of my head.

 

[BruceW]

In the sense that they both have the same tangent vector at the point, yes. But only at the point. See below.
...
No; only geodesics will appear as straight lines in normal coordinates throughout the entire coordinate patch surrounding the point. Non-geodesic curves will appear as curved lines that are tangent to the geodesic straight lines at the chosen point.

aahh, ok, I get it now. I suppose that is the entire point of the exponential map in the first place. Thanks very much for explaining all this to me. I appreciate it. And I have one last question, well, really I'm not totally sure of why they talk about a normal neighbourhood of the point, within which the geodesics are in the form $(tV^1,...,tV^4)$
I mean, what I expect is that in some kind of limit of 'closeness' to the point, the geodesics would tend to the form $(tV^1,...,tV^4)$ But they (wikipedia) explain it in a different way. They use this idea of a normal neighbourhood, and don't mention limits as far as I can tell.

 

[WannabeNewton]

Neighborhoods encode all the information about limits in topological spaces. Keep in mind that an arbitrary space-time manifold has no natural metric so $\epsilon$-$\delta$ definitions of limits don't apply. In topological spaces, neighborhoods fully characterize limits; for example a sequence $(x_i)_{i\in \mathbb{N}}$ in a topological space $X$ converges to a point $p\in X$ iff for any neighborhood $U$ of $p$, there exists an $N\in \mathbb{N}$ such that for all $n > N$, $x_n \in U$. This is quite similar in form to the definition of the convergence of a sequence of points in a metric space except that there is no metric when dealing with arbitrary topological spaces hence no notion of "closeness" in the metric sense (of course for a metric space the two definitions are equivalent); this is for example why you don't naturally see Cauchy sequences defined for arbitrary topological spaces. You would probably gain a better understanding of these things by shifting through a proper Riemannian geometry text e.g. Lee "Riemannian Manifolds: An Introduction to Curvature" or O'Neill "Semi-Riemannian Geometry With Applications to Relativity".

 

[PeterDonis]

I suppose that is the entire point of the exponential map in the first place.

Yes, exactly.