Timelike v. spacelike, is it arbitrary?

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  • #1
BruceW
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EDIT: Split from https://www.physicsforums.com/showthread.php?t=704345

after a bit of searching, I found "An introduction to general relativity and cosmology" jerzy plebanski, andrzej Krasinski, where they don't use different terms for proper time and proper length. They just use the word 'arc length' to mean either. And they use the equation:
[tex]l = \int \left( | g_{\alpha \beta}( \gamma ) \frac{d x^{\alpha}}{d \gamma} \frac{d x^{\beta}}{d \gamma} | \right)^{1/2} \ d \gamma[/tex]
Where the line | means 'take absolute value'. So maybe I should use the phrase 'arc length' instead. But then it still implies a length, when I might be measuring proper time. I guess at least they are using a different phrase, instead of what I was doing, which was using the old phrase 'proper time' to mean a new thing.

The main reason to use a definition that does not discern between proper time and proper length is that in general relativity, we often don't define a coordinate time, and we can do lots of useful calculations, without ever caring about whether we have a proper length or a proper time. In this sense, the difference between proper time and proper length in general relativity is artificial. (i.e. sure we can add in a coordinate time, but it is not necessary for using relativistic equations). Further, I thought it would be OK to say that the 'arc length' along a null geodesic is just zero. But as wannabenewton said, null worldlines can't be parameterised by proper time. So I think I was wrong to say that the 'arc length' is zero along a null worldline, since it is not possible to define it along a null worldline?

Also, several of our lecturers at undergraduate would use non-standard definitions, to force us to get used to the fact that people don't always play by the same definitions. This is why I have a spirit of using whatever definitions, and making sure I don't assume that someone I am talking to is using a certain definition that I am used to. But I suppose that since I am not an expert in any field of physics, maybe I should not be so carefree with using non-standard definitions myself. And about being homework helper here, if the other helpers think that I should stick to the standard definitions, then I will do that.
 
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  • #2
PeterDonis
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The main reason to use a definition that does not discern between proper time and proper length is that in general relativity, we often don't define a coordinate time, and we can do lots of useful calculations, without ever caring about whether we have a proper length or a proper time. In this sense, the difference between proper time and proper length in general relativity is artificial.
I disagree with this as you state it, because the difference between proper time and proper length is independent of coordinates, since the difference between timelike and spacelike curves is independent of coordinates. There may be cases where you don't need to care about the difference, but that doesn't make the difference artificial; it's there whether or not you define coordinates that reflect it, and whether or not you actually make use of it in a particular calculation.

I think I was wrong to say that the 'arc length' is zero along a null worldline, since it is not possible to define it along a null worldline?
It is possible to define arc length along a null curve; you just can't use proper time as the affine parameter to do so (the ##\gamma## in the equation you gave is an affine parameter, and it can be proper time or proper length, but it doesn't have to be). The arc length is indeed zero along a null curve, so you weren't wrong to say that.

Also, several of our lecturers at undergraduate would use non-standard definitions, to force us to get used to the fact that people don't always play by the same definitions. This is why I have a spirit of using whatever definitions, and making sure I don't assume that someone I am talking to is using a certain definition that I am used to.
If you're going to do that, you at least need to be explicit about what definition you're using up front, if it's a non-standard one. Otherwise other people are likely to assume you're using the standard definition, which can cause confusion, as in this thread.

But I suppose that since I am not an expert in any field of physics, maybe I should not be so carefree with using non-standard definitions myself.
It's not so much a matter of being an expert vs. not being an expert, as of trying to facilitate communication in general. Standard definitions of terms exist to help with that. That doesn't mean you *have* to use them, but it means it's easier to communicate if you do.

And about being homework helper here, if the other helpers think that I should stick to the standard definitions, then I will do that.
A more precise way of stating the point I was trying to make here is that, unless you have evidence to the contrary, it's safest to assume that textbooks in the field are using the standard definitions, so if a person who is studying from a textbook asks a question, they're probably used to seeing the standard definitions. But if you see evidence to the contrary, then, as you said, you should make sure to find out what definition they are actually using, to avoid confusion.
 
  • #3
BruceW
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I disagree with this as you state it, because the difference between proper time and proper length is independent of coordinates, since the difference between timelike and spacelike curves is independent of coordinates.
True, but only because we define matter to travel along timelike curves. For example, in a universe with radiation but no matter, then surely there is no way to discern which curves are 'timelike' and which curves are 'spacelike'. So then, I would say it is not a relativity concept, but a physics of matter concept. (Since really, it hinges on the assumption that all matter travels along timelike curves, which is experimentally true, but not required by relativity, as far as I know). But then maybe you would group that under a concept of relativity. I would prefer not to.

PeterDonis said:
It is possible to define arc length along a null curve; you just can't use proper time as the affine parameter to do so...
Oh, right. Thank you for explaining that. I wasn't sure about it. So in my preferred language: it is possible to define the arc length along a null curve, but you can't use the arc length as the affine parameter to do so.

PeterDonis said:
If you're going to do that, you at least need to be explicit about what definition you're using up front, if it's a non-standard one. Otherwise other people are likely to assume you're using the standard definition, which can cause confusion, as in this thread.
Yeah, that's what I mean. Some of our lecturers would not explicitly tell us which definition they are using, or change notation without telling us (during the lecture), to keep us on our toes / make sure we were don't assume that the standard definition is being used as default, but to think about what the lecturer means. But I appreciate that on a forum, it is probably best to keep to the rule 'assume standard definition unless explicitly stated otherwise'. Also, sorry for derailing this thread a bit.
 
  • #4
PeterDonis
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we define matter to travel along timelike curves.
Not really; we discover that there are two fundamentally different kinds of objects, those with nonzero rest mass (that travel on timelike curves) and those with zero rest mass (that travel on null curves). Ordinarily we use the term "matter" for the first kind of object and "radiation" for the second. I suppose that counts as a "definition" of the term matter, but I don't think that's the kind of definition you meant.

For example, in a universe with radiation but no matter, then surely there is no way to discern which curves are 'timelike' and which curves are 'spacelike'.
Sure there is. You can use the null curves, i.e., curves with zero squared length, to define a coordinate grid and construct a metric, and you will find that there are other curves with positive or negative squared length, and the two kinds of curves are clearly distinguished from each other. Or, you could use the null curves to define light cones at each event, and you would find that there were other curves lying inside (timelike) or outside (spacelike) the light cones, with the light cones themselves providing a clear boundary between the two.

the assumption that all matter travels along timelike curves, which is experimentally true, but not required by relativity, as far as I know).
What's required by relativity is that objects with nonzero rest mass travel on timelike curves, and objects with zero rest mass travel on null curves. But we don't "assume" that "matter" is the first kind of object; we discover, experimentally, which objects fall into each category, and we adopt the term "matter" to refer to the objects that fall into the first category (nonzero rest mass).

So in my preferred language: it is possible to define the arc length along a null curve, but you can't use the arc length as the affine parameter to do so.
Yes, you could put it that way: since arc length is zero along any null curve, obviously you can't use it to parameterize the curve.
 
  • #5
BruceW
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Sure there is. You can use the null curves, i.e., curves with zero squared length, to define a coordinate grid and construct a metric, and you will find that there are other curves with positive or negative squared length, and the two kinds of curves are clearly distinguished from each other. Or, you could use the null curves to define light cones at each event, and you would find that there were other curves lying inside (timelike) or outside (spacelike) the light cones, with the light cones themselves providing a clear boundary between the two.
Yes, but this is an arbitrary choice. For a specific physical universe, we still have the choice of what curves are spacelike and what curves are timelike. In other words, for a specific physical universe, if we look at a specific non-null geodesic, and ask 'is it time-like or space-like?' then the answer simply depends on our arbitrary choice.

edit: I wish I could explain better, but as I've said, I'm still a beginner to general relativity. Maybe start with the simple case of special relativity. ds^2 = -dt^2+dx^2+dy^2+dz^2 Now why did you choose 't' to get the negative sign? It is an arbitrary choice. You could instead give 'y' the negative sign, it is just as valid to do that. And this affects which curves are time-like and which curves are space-like.

another edit: the clue is in the names really. relativity puts space and time on an equal footing. The component we choose as 'time' (i.e. to get the negative sign) is an arbitrary choice. There is no non-arbitrary choice of 'space' and 'time', there is only spacetime.
 
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  • #6
DrGreg
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Maybe start with the simple case of special relativity. ds^2 = -dt^2+dx^2+dy^2+dz^2 Now why did you choose 't' to get the negative sign? It is an arbitrary choice. You could instead give 'y' the negative sign, it is just as valid to do that. And this affects which curves are time-like and which curves are space-like
But however you label your coordinates and whatever metric signature convention you follow, you'll find 3 coordinates with the same sign and one with the opposite sign. The one on its own is timelike and the other 3 are spacelike.
 
  • #7
PeterDonis
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Yes, but this is an arbitrary choice. For a specific physical universe, we still have the choice of what curves are spacelike and what curves are timelike. In other words, for a specific physical universe, if we look at a specific non-null geodesic, and ask 'is it time-like or space-like?' then the answer simply depends on our arbitrary choice.
No, it doesn't. You can choose which sign convention you want to use for the metric, which determines whether the squared length of (timelike, spacelike) vectors is (negative, positive) or (positive, negative). But you can't choose whether a given curve is timelike or spacelike, because the metric only has one timelike dimension and three spacelike ones. The two are not symmetric.

ds^2 = -dt^2+dx^2+dy^2+dz^2 Now why did you choose 't' to get the negative sign? It is an arbitrary choice. You could instead give 'y' the negative sign, it is just as valid to do that.
That's just changing the names of the coordinates. It's not changing the physics.

And this affects which curves are time-like and which curves are space-like.
No, it doesn't. It just changes which coordinate labels you put on particular curves. It doesn't change the physical nature of the curves.

relativity puts space and time on an equal footing. The component we choose as 'time' (i.e. to get the negative sign) is an arbitrary choice. There is no non-arbitrary choice of 'space' and 'time', there is only spacetime.
Incorrect. See above.
 
  • #8
WannabeNewton
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Given a space-time ##(M,g_{ab})##, the class of space-like and time-like curves do not depend on any choice of coordinates or frames. Taking the ##(-,+,+,+)## sign convention, we say a curve ##\gamma## is time-like (resp. space-like) if ##\xi^a \xi_a < 0## (resp. ##\xi^a \xi_a > 0##) everywhere on ##\gamma## where ##\xi^a## is the tangent vector field to ##\gamma## (in fact for geodesics it is enough to know the sign of ##\xi^a \xi_a## at a single point on the geodesic because ##\xi^b \nabla_b (\xi_a \xi^a) = 2 \xi_a \xi^b \nabla_b \xi^a = 0##). This is purely geometric.
 
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  • #9
BruceW
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No, it doesn't. You can choose which sign convention you want to use for the metric, which determines whether the squared length of (timelike, spacelike) vectors is (negative, positive) or (positive, negative). But you can't choose whether a given curve is timelike or spacelike, because the metric only has one timelike dimension and three spacelike ones. The two are not symmetric.
I don't see why not. Let's say you have some curve in spacetime. You could define the timelike dimension to lie along that curve (in which case you'd label the curve as 'timelike'). Or you could define one of the spacelike dimensions to lie along the curve, in which case you'd label that curve as spacelike. the spacelike and timelike labels are arbitrary.
 
  • #10
WannabeNewton
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No you can't do that. If the tangent vector to a curve satisfies ##\xi^a \xi_a < 0## everywhere on the curve then by definition it is time-like. You cannot change this, it's a geometric property of the curve and the only thing it depends on is the metric tensor.
 
  • #11
BruceW
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when you say "no you can't do that" I'm guessing you mean "we have already defined the timelike direction". But it is the definition of the timelike direction that I am saying is arbitrary.
 
  • #12
WannabeNewton
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But it isn't arbitrary. The tangent vector lies along the "time-like direction" if ##\xi^a \xi_a < 0## and along a "space-like direction" if ##\xi^a \xi_a > 0##. What's arbitrary is the choice of sign convention ##(-,+,+,+)## or ##(+,-,-,-)## in which case the above signs flip but it won't change which curves are time-like and which are space-like.
 
  • #13
PeterDonis
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Let's say you have some curve in spacetime. You could define the timelike dimension to lie along that curve (in which case you'd label the curve as 'timelike').
You can't do that arbitrarily, because, as I said, the metric has one timelike and three spacelike dimensions.

For example: say I have coordinates ##(t, x, y, z)## set up, with the line element ##ds^2 = - dt^2 + dx^2 + dy^2 + dz^2##. Consider the straight line ##L1## that goes from ##(0, 0, 0, 0)## to ##(0, 1, 0, 0)##. Its squared length is 1.

Suppose I try to "define" the timelike dimension to lie along this curve ##L1## (i.e,. along the ##x## axis). If your claim is true, we should be able to do this. But can we? No; to see why, consider the following other curves all starting at the origin ##(0, 0, 0, 0)##: the straight line ##L2## from the origin to ##(0, 0, 1, 0)##, the straight line ##L3## from the origin to ##(0, 0, 0, 1)##, and the straight line ##L0## from the origin to ##(1, 0, 0, 0)##. The squared lengths of ##L2## and ##L3## are 1; but the squared length of ##L0## is -1.

All four of these lines are orthogonal, as is easily seen by observing that the dot product of any pair of them is zero. But the line ##L1##, the one you are claiming can be "defined" to be the timelike dimension, has the *same* sign of its squared length as two other mutually orthogonal lines, ##L2## and ##L3##. That can only be true of a spacelike line. The only one of the four lines that can possibly be timelike is ##L0##. That's a physical fact, independent of whatever coordinates we choose to adopt.
 
  • #14
BruceW
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@WannabeNewton- I agree that flipping the sign convention is not important. I'm saying that we can choose whichever component we want to be the timelike component, and this does change which curves are timelike and which are spacelike.
 
  • #15
BruceW
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For example: say I have coordinates ##(t, x, y, z)## set up, with the line element ##ds^2 = - dt^2 + dx^2 + dy^2 + dz^2##. Consider the straight line ##L1## that goes from ##(0, 0, 0, 0)## to ##(0, 1, 0, 0)##. Its squared length is 1.

Suppose I try to "define" the timelike dimension to lie along this curve ##L1## (i.e,. along the ##x## axis)...
OK, now the line L1 has squared length -1
edit: and the others will all have squared length +1, under the new definition of the timelike dimension.
 
  • #16
PeterDonis
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independent of whatever coordinates we choose to adopt.
I suppose I should expand on this a bit, in view of the earlier exchange in this thread about deciding to put the minus sign on the ##y## coordinate instead of the ##t## coordinate. Suppose we tried something like that; suppose we said, let's put the minus sign on the ##x## coordinate, i.e., along the direction of the curve ##L1## that we want to define to be the timelike dimension. That is, we want to define the line element to be ##ds^2 = dt^2 - dx^2 + dy^2 + dz^2##.

Can we do this? If we keep the coordinate labels of events constant, the answer is no, because it would change a geometric invariant, the squared length of ##L1##; with this new line element, the squared length of ##L1## would be -1, not 1. Changing a geometric invariant changes the geometry.

We could, however, change coordinate labels: we could say that we are simply relabeling the coordinates so that ##L1## now goes from the origin to ##(t, x, y, z) = (1, 0, 0, 0)##, and ##L0## now goes from the origin to ##(0, 1, 0, 0)##. Then we have the same geometry with different coordinate labels on it; but the curve ##L0## is still the one timelike curve out of the four, so we haven't "defined" a different timelike dimension, we've just changed how we label the dimensions with coordinates.
 
  • #17
PeterDonis
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OK, now the line L1 has squared length -1
edit: and the others will all have squared length +1, under the new definition of the timelike dimension.
See my follow-up post; you can't do that, because it's changing the geometry.
 
  • #18
BruceW
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I agree that once we have defined a timelike dimension, then it is clear which are the timelike curves and which are the spacelike curves. The point is that our choice of the timelike dimension is arbitrary in the first place. OK, you can call it 'geometry', but it is still arbitrarily defined geometry. It does not correspond to anything physical.
 
  • #19
PeterDonis
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I agree that once we have defined a timelike dimension, then it is clear which are the timelike curves and which are the spacelike curves. The point is that our choice of the timelike dimension is arbitrary in the first place. OK, you can call it 'geometry', but it is still arbitrarily defined geometry. It does not correspond to anything physical.
I don't understand what you mean here. If you mean we can choose which coordinate label to put on the timelike dimension, sure, nobody is questioning that. But if that's all you've been saying, it hasn't been at all clear.

If you mean something else, then you'll need to explain in more detail what you mean, because I'm not getting it.
 
  • #20
PeterDonis
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our choice of the timelike dimension is arbitrary
Perhaps this will help clarify where you're coming from: are you saying that I can make an arbitrary choice of whether to call the dimension that points into my future (for example, from the event of me right now to the event of me a minute from now) timelike or spacelike?
 
  • #21
BruceW
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yeah, if we ignore the fact that matter travels along timelike paths. For example, if some matter traveled along spacelike paths, then who is to say which path is timelike and which is spacelike? As it happens, we have only observed matter that travels along the same kind of path as us. But this is not required by relativity. The point I'm trying to make is that since we observe all matter to travel along the same kind of path as us, we are able to define the timelike paths as paths that matter travel along. But this has nothing to do with relativity. (Unless you include this as part of the theory of relativity, but I would prefer not to).
 
  • #22
PeterDonis
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yeah, if we ignore the fact that matter travels along timelike paths. For example, if some matter traveled along spacelike paths, then who is to say which path is timelike and which is spacelike?
The fact that there are three dimensions with one sign of the squared length, and only one with the other sign of the squared length. The odd one out is the timelike dimension.

As it happens, we have only observed matter that travels along the same kind of path as us. But this is not required by relativity.
If you mean, relativity does not prohibit tachyons, technically that's true, yes, but there are subtleties. The Usenet Physics FAQ has a good, if brief, discussion.

However, even if we did observe tachyons, that would not change the fact that there is one dimension with a squared length of the opposite sign from the other three, as above. That one dimension is the timelike dimension.

The point I'm trying to make is that since we observe all matter to travel along the same kind of path as us, we are able to define the timelike paths as paths that matter travel along.
That's not how we define timelike paths. We define them as I did above. The fact that matter travels along timelike paths is an additional observation; it's not a definition of what "timelike" means.
 
  • #23
BruceW
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The fact that there are three dimensions with one sign of the squared length, and only one with the other sign of the squared length. The odd one out is the timelike dimension.
agreed. And the one we choose to be 'odd one out' is an arbitrary definition. So this means the timelike dimension is an arbitrary definition.

PeterDonis said:
That's not how we define timelike paths. We define them as I did above. The fact that matter travels along timelike paths is an additional observation; it's not a definition of what "timelike" means.
We agree on this, at least.
 
  • #24
PeterDonis
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And the one we choose to be 'odd one out' is an arbitrary definition.
I'm not sure what you mean by this. Going back to my post #56, are you saying that we can make an arbitrary choice about whether the dimension that points from me now to me a minute from now is the "odd one out"? If you are, you're wrong; that dimension has to be the odd one out. Think, once again, about the definition I gave earlier: it's not physically possible to find a dimension orthogonal to the one that points into my future, but which also has the same sign of its squared length as the one that points into my future.
 
  • #25
BruceW
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I'm not totally sure what you mean either. Do you mean "If we define the dimension which points into my future as timelike, then there is no dimension orthogonal to my future that is timelike" ? I agree with that. What I am saying is that in the first place, we can choose the dimension which points into my future as spacelike. And then there will be a dimension orthogonal to my future that is timelike.

This definition of 'which paths are timelike' and 'which paths are spacelike' is arbitrary, but I agree that if I define myself to be on a timelike path, then the orthogonal dimensions are spacelike. ('orthogonal dimensions' is probably incorrect terminology, but oh well).
 

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