Timelike v. spacelike, is it arbitrary?

[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)

 

[TrickyDicky]

So it seems it is hard to get rid of the arbitrarines: deciding whether a vector is past or future directed is fully arbitrary for either timelike or null vectors, which makes the distinction timelike vs spacelike ultimately arbitrary or conventional since one has first to decide whether the two null vectors used to decide it are future or past directed. Of course once a convention is chosen one can keep it consistently and in this sense it can't be changed arbitrarily.

 

[PeterDonis]

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 $M$.

Here we are defining timelike vectors in terms of null vectors, so we can't then turn around and define properties of null vectors in terms of timelike vectors. I'm looking for a way to distinguish the two halves of the light cone purely in terms of properties of the null vectors themselves.

 

[robphy]

So it seems it is hard to get rid of the arbitrarines: deciding whether a vector is past or future directed is fully arbitrary for either timelike or null vectors, which makes the distinction timelike vs spacelike ultimately arbitrary or conventional since one has first to decide whether the two null vectors used to decide it are future or past directed. Of course once a convention is chosen one can keep it consistently and in this sense it can't be changed arbitrarily.

Once you pick which is future... do so consistently, everywhere and everywhen.

Then...
future-timelike is completely determined, and thus past-timelike is completely determined.
That a vector is spacelike (which involves the sum of a future-null vector and a past-null vector ) is unchanged by the initial choice of which is future.

 

[PeterDonis]

I'm looking for a way to distinguish the two halves of the light cone purely in terms of properties of the null vectors themselves.

Perhaps the key here is simply that there are two subsets of the null vectors distinguished by their inner products: each subset has positive inner product with other (non-parallel) members of the same subset, but negative inner product with the members of the other subset. That distinguishes the two halves of the light cone; then we just make a choice about which half we label the "future" half.

 

[robphy]

Perhaps the key here is simply that there are two subsets of the null vectors distinguished by their inner products: each subset has positive inner product with other (non-parallel) members of the same subset, but negative inner product with the members of the other subset. That distinguishes the two halves of the light cone; then we just make a choice about which half we label the "future" half.

That's what was implied.
Details (e.g.) in the Geroch links above.

 

[PeterDonis]

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.

Ah, this answers the question I posed in post #98, how we tell the "interior" from the "exterior" of the light cone.

 

[WannabeNewton]

That's called a co-orientation yes. Co-orientation is certainly an equivalence relation on the subset of causal vectors in $T_p M$ and you can arbitrarily choose to label one equivalence class as the future half of the lightcone in $T_p M$ but you won't be able to make a continuous designation of future-directed without a time-orientation (and not all space-times have a time-orientation).

 

[PeterDonis]

you won't be able to make a continuous designation of future-directed without a time-orientation (and not all space-times have a time-orientation).

Yes, understood; I was just trying to get clear about how everything is constructed locally, at a given event. Obviously the local construction of itself does not guarantee anything about global properties like whether the spacetime is time orientable.

 

[WannabeNewton]

Yeah at a given event the method you described is certainly the standard one. In addition to robphy's links, section 2.2 (p. 128) of the following might be of interest: http://www.socsci.uci.edu/~dmalamen/bio/GR.pdf

It's essentially just what you said.

 

[TrickyDicky]

That a vector is spacelike (which involves the sum of a future-null vector and a past-null vector ) is unchanged by the initial choice of which is future.

Oh, ok right, timelike vs spacelike is not affected by the future-past arbitrarines.

 

[PAllen]

Don't know if this will help or confuse anyone, but I thought it would be fun to post a metric for flat Minkowski space for coordinates built from 4 independent lightlike coordinates (can't be orthogonal, because light like directions cannot be 4-orthogonal; independence, however, is all you need for generalized coordinates). At first glance, you could never tell this a metric for Minkowski space with signature (+,-,-,-) but it is (I use a,b,c,e for the coordinates):


d\tau^2 = 4 da db + 2 da dc + 2 da de + 2 db dc + 2 db de + 2 dc de

It is obvious that these coordinates are all light like: taking any 3 constant, the other varying gives you an interval of zero.

Now I will exhibit 4 lines, one timelike, 3 spacelike, then show they are orthonormal, thus showing what I claimed.

1) Consider the line a=b, c=e=0. The line element along this line becomes 4 db^2 , thus timelike.
2) Consider the line a=-b,c=e=0. The line element along this line is -4db^2, thus spacelike.
3) Consider the line a=-c/2, b=-c/2,e=0. The line element along this line is -dc^2, spacelike.
4) Consider the line a=-e/2,b=-e/2,c=0. The line element along this line is -de^2, spacelike.

Conveniently scaled tangent vectors of these lines are:

(1,1,0,0), (1,-1,0,0), (-1,-1,2,0), and (-1,-1,0,2)

Taking the dot product between any pair of these using the given metric yields zero, showing they form an orthonormal set of 1 timelike and 3 spacelike vectors.

Finally, I'll give the transform to Minkowski coordinates:

t = a+b+c+e
x = a-b
y=c
z=e

and the other way:

a = (x+t-y-z)/2
b = (t-x-y-z)/2
c=y
e=z

 

[PAllen]

After my prior post, I was curious to see what was implied by the even more opaque, symmetrical line element in terms of arbitrary coordinates (a,b,c,e):

dadb + dadc + dade + dbdc + dbde + dcde

Again, it is obvious that a,b,c,e are all light like coordinates (see prior post for explanation). I was able to quickly guess a triple of mutually orthogonal vectors per this metric:

(1,1,0,0) , (1,-1,0,0),(-1,-1,1,0)

of which the first has positive norm, and the the other two have negative norm. Thus, already we can guess that this line element is Minkowski space with (+,-,-,-) signature. To establish this we need a 4th mutually orthogonal vector with negative norm. For some reason, I couldn't guess this and had to algebraically solve for it to find that (-1,-1,-1,2) is such a vector.

If we divide each of these vectors by its norm per this metric, we have an orthonormal basis for Minkowski coordinates expressed in terms of these. If we call the directions represented by these vectors, in the order given, as t,x,y,z, and solve for the transform given the known basis vectors (rather laborious, at least the way I did it), we arrive at:

t = a/2 + b/2 + c + e
x = a/2 - b/2
y = c + e/2
z = (e/2)√3

It can be verified that using these to transform from the standard Minkowski metric indeed produces the metric we started with. The reverse transform is:

a = t + x -y - z/√3
b = t - x -y - z/√3
c = y - z/√3
e = 2z/√3

 

[BruceW]

Don't know if this will help or confuse anyone, but I thought it would be fun to post a metric for flat Minkowski space for coordinates built from 4 independent lightlike coordinates (can't be orthogonal, because light like directions cannot be 4-orthogonal; independence, however, is all you need for generalized coordinates).

just blew my mind a little bit. But yeah, I guess it is a sensible 'result', just a bit weird.

 

[ghwellsjr]

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).

I did not respond previously to your post because I thought you had come to terms with this issue in post #84 but since you have posted again, I'm not so sure so I'd like to ask you some questions:

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

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

If the issue is resolved in SR, does that make it also resolved for GR or is GR a separate bigger issue?

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

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

 

[Naty1]

PAllen:

...light like directions cannot be 4-orthogonal...

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...