Distance Function vs. Metric

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  • #26
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Surely it is that the distance function is a positive definite function defined on the [itex]\mathbb{R}^4[/itex] charts of a manifold. The metric is a tensor (the distinction as made by quasar987), which is not necessarily positive definite (when we feed it two vectors).

Also, is this the source of MeJennifer's complaints about lack of positive-definiteness in standard GR?
 
  • #27
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Surely it is that the distance function is a positive definite function defined on the [itex]\mathbb{R}^4[/itex] charts of a manifold. The metric is a tensor (the distinction as made by quasar987), which is not necessarily positive definite (when we feed it two vectors).

Also, is this the source of MeJennifer's complaints about lack of positive-definiteness in standard GR?
So you are saying that the distance function in GR is Euclidean while the metric is not?

Could you please explain how such a model makes any mathematical sense?
 
  • #28
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When the distance function, say

[tex]d(X_1, X_2) = \sqrt{(t_1-t_2)^2 + (x_1-x_2)^2},[/tex]

is different from the metric, which is

[tex]g(X_1, X_2) = t_1 t_2 - x_1 x_2.[/tex]

The metric tensor was never meant to give distances in the same sense the Euclidean metric does.
 
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  • #29
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When the distance function, say

[tex]d(X_1, X_2) = \sqrt{(t_1-t_2)^2 + (x_1-x_2)^2},[/tex]

is different from the metric, which is

[tex]g(X_1, X_2) = t_1 t_2 - x_1 x_2.[/tex]

The metric tensor was never meant to give distances in the same sense the Euclidean metric does.
You're using the term "metric" as if it meant something different than the Euclidean metric. The Euclidean metric is just one of many metrics.

Pete
 
  • #30
quasar987
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Since confusion can easily arise btw metric and metric tensor, let's refer to the distance function as "distance function" and never as "metric" and let's refer to the metric tensor as "metric tensor" or just "metric". Alright?

This is the convention masudr adopted in his last message. His distance function d is the euclidian distance function, while his metric g is the Minkowski metric (http://en.wikipedia.org/wiki/Minkowski_metric#Standard_basis)
 
  • #31
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When the distance function, say

[tex]d(X_1, X_2) = \sqrt{(t_1-t_2)^2 + (x_1-x_2)^2},[/tex]

is different from the metric, which is

[tex]g(X_1, X_2) = t_1 t_2 - x_1 x_2.[/tex]

The metric tensor was never meant to give distances in the same sense the Euclidean metric does.
What do you mean "never meant". :confused:

Is this mathematics or religion we are discussing?

So masudr, do you claim that the distance between two points in Minkowski space-time is:

[tex]d(X_1, X_2) = \sqrt{(t_1-t_2)^2 + (x_1-x_2)^2 + (y_1-y_2)^2 + (z_1-z_2)^2}[/tex]

:confused:
 
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  • #32
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What do you mean "never meant". :confused:

Is this mathematics or religion we are discussing?

So masudr, do you claim that the distance between two points in Minkowski space-time is:

[tex]d(X_1, X_2) = \sqrt{(t_1-t_2)^2 + (x_1-x_2)^2 + (y_1-y_2)^2 + (z_1-z_2)^2}[/tex]

:confused:
Isn't it amazing how threads obtain a life of their own even when the OP has his answer? You go girl! :biggrin:

Pete
 
  • #33
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What do you mean "never meant". :confused:

Is this mathematics or religion we are discussing?

Hehe, OK I take that statement back.

So masudr, do you claim that the distance between two points in Minkowski space-time is:

[tex]d(X_1, X_2) = \sqrt{(t_1-t_2)^2 + (x_1-x_2)^2 + (y_1-y_2)^2 + (z_1-z_2)^2}[/tex]

:confused:

What I mean is that locally all manifolds look like [itex]\mathbb{R}^n,[/itex] and as far as I know we are all happy with that. If we want to talk about how far away points are locally, it makes most sense to use the Euclidean distance function. Don't you think? The way I see it, is that in terms of nearness, using the Euclidean metric locally makes most sense, since locally a manifold is [itex]\mathbb{R}^n.[/itex]

The usual metric tensor used in GR is a completely different beast: it is physically motivated. I understand that a real physical spacetime is best modelled by a manifold, with this additional structure, the metric tensor. Why is this (indefinite) object motivated? Well, obviously all that stuff about measuring time/distances/angles as experienced by observers in this physical manifold. As I'm sure we all know, it says the "distance" (if we use it to define distance) between two points on a null path is 0. But topologically, the two points aren't necessarily neighbouring, in any way whatsoever. Furthermore, as you have pointed out, the triange inequality is disobeyed if we use the usual GR metric (which is Minkowski in a certain choice of coordinates).

In answer to your question, note that I'm quite happy to define my distance function howsoever I wish, as long as it satisfies a few simple criteria (again, I'm sure you know them; for those that don't, I think the Wikipedia page on distance functions lists them). So yes, if I so felt that day, I could quite easily choose that as my distance function.

Let me ask you a similar question, though: Do you think that two events are separated by the distance function that is induced by our friend, the (0,2) GR metric tensor?
 
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  • #34
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Okay. I mentioned before in this thread that I would e-mail an expert in this area and get back with his response. The expert I asked this question to is Hans C. Ohanian. His response was
Your question about different ways of defining distances in spacetime, for metric vs. topological purposes: The reason why two different distances need to be used in spacetime can best be understood by the example of the light cone. Consider the light cone erected at some point (in flat spacetime or in curved spacetime, it makes no difference). According the spacetime metric, all points on the cone at are zero spacetime distance, and that is fine in regard to the geometry. But it makes no sense in regard to the topology--the light cone is a CONE, and some of its points are widely separated, not nearby or on top of one another. So to define the topology of the light cone (that is, to decide which points are near which, and which points are distinct), we can't use the spacetime metric as a criterion. We need some other criterion that, in a qualitative way, tell us which points are near which. The way this is handled, is by introducing coordinates in spacetime, by means of a nonsingular coordinate system, or overlapping nonsingular coordinate patches. Topologically, points are then considered nearby if their coordinates are nearly equal. (In Euclidean space, we do not need to worry about the distinction between metric separation distances and topological separation--in Euclidean space, we can use the metric distance to determine the topology. In spacetime, things are more tricky.)
Makes sense to me. I do admit that I'm surprised by his answer. But then again I'd be suprised by any answer on this subject so..... .Anybody disagree with the above assertion by Ohanian? If so please post your thoughts.

Best wishes

Pete
 
  • #35
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Indeed, I was trying to hint at this [Ohanian's understanding] above.

I still have one slight unresolved issue. To me, the most obvious distance function in terms of exploring the topology is the Euclidean one

[tex]d(X_1, X_2) = \sqrt{(t_1-t_2)^2 + (x_1-x_2)^2 + (y_1-y_2)^2 + (z_1-z_2)^2}[/tex]

However, as I have noted above, any function which satisfies the axioms of a distance function is valid. Furthermore, the value this function gives is, of course, highly dependent upon the choice of coordinates. So the distance function is not unique in at least 2 different ways.

My issue is, basically, in terms of topology, can it make any difference by choosing a different distance function or by choosing some exotic coordinates? I anticipate that the answer is, of course, no, but would prefer some reassurance by a more knowledgable PF member.
 
  • #36
Haelfix
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"Anybody disagree with the above assertion by Ohanian? If so please post your thoughts"

The source of the confusion it seems to me is the observation that locally spacetime is not R^4 in the most naive way. The isometry group of a local neighberhood in spacetime is not SO(4), but rather SO(3,1)! So we require an additional physical map from R^4 --> R^3 * R^1, and this induces a new distance function.
 

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