# Distance Function vs. Metric

• pmb_phy
They do mention the triangle inequality and show how it can be violated, but they don't go into quite as much detail as in MTW.In summary, Schut'z Geometrical methods of mathematical physics and Walds General Relativity use a distance function which is not the same as the metric. Schutz refers to a "distance function" between two points on page 1 and as one exampe this function is the usual Euclidean distance function, and other examples are similar. In Wald he speaks of an open ball in Rn in which the Euclidean function is used as the "open ball" function about a particular point. The purpose of the distance function

#### pmb_phy

I almost every single case that I've seen the distance function is identical to the metric. However there are two important exceptions which are found in Schut'z Geometrical methods of mathematical physics and in Walds General Relativity. Schutz refers to a "distance function" between two points on page 1 and as one exampe this function is the usual Euclidean distance function, and other examples are similar. In Wald he speaks of an open ball in Rn in which the Euclidean function is used as the "open ball" function about a particular point. The purpose of the distance function is to defined open and closed sets. The metric doesn't neccesarily have this ability.

Other texts use the term "distance function" and "metric" to mean the same thing. Has anyone every heard of this or seen this in those texts? Has anyone seen it as a problem in any case either for or against the distance function being the same as the metric or not being the same as the metric?

Thank you

Pete

Say you have a non-empty set X.
pmb_phy said:
I almost every single case that I've seen the distance function is identical to the metric.
To me "distance function" or "metric" are just two different names for one and the same thing, namely a function $d:X\times X\rightarrow \mathbb{R}$ satisfying some requirements I'm sure you're familiar with.

pmb_phy said:
In Wald he speaks of an open ball in Rn in which the Euclidean function is used as the "open ball" function about a particular point. The purpose of the distance function is to defined open and closed sets. The metric doesn't neccesarily have this ability.

I wouldn't express it that way. See, if you're given this set X you can try to give it some structure. One such thing would be a topology, i.e. you specify a certain subset of the power set of X, which you call the open (or closed, for that matter) sets. Now if you already have a metric on your set, you might want to have a very special type of open-set-definition: the so-called induced topology. Notice, that this is a completely free choice, you could also have different topologies. In the induced topology you define the open balls using the metric. Then you go on and call a subset of X open, if... (I'm sure you know the definitions).
So it is not the purpose of the metric, it's just leads to a certain, comfortable choice.

The only difference I have ever seen or heard of between distance functions and metric, is that you could define a function for the distance between a subset of X and a point: $d(z,A):=inf_{y\in X}(d(z,y))$.

I've seen problems where the metric isn't simply the distance function. Specifically, if I remember correctly, in the Lagrangian formulation of Fermets principle of least time, you defined the metric as m_ij = n(x)*g_ij, where g_ij is the "usual" metric, n is the refractive index. Instead of giving you the true distance when integrated, m_ij gives you the "time" taken for light to travel between two points. You now seek a path of least "time distance" between two points in the usual way you would for regular distance.

m_ij is a metric, but it's not measuring true distance here.

pmb_phy said:
The purpose of the distance function is to defined open and closed sets. The metric doesn't neccesarily have this ability.
Distance function and metric are the same thing.

These kind of "explanations" make no mathematical sense at all, and are simply designed to either ignore the difficulties with manifolds that have no valid metric or, and this is worse, to impress the reader with the idea that there really is no issue at all.

In most relativity textbooks it is simply assumed that there is a valid metric or it is simply mentioned that it is a pseudo-metric.

Not all manifolds have a metric. A metric for instance must be Hausdorff and the triangle inequality must hold. We can prove that, for instance, a Minkowski metric is not a metric since it violates the triangle inequality and is non-Hausdorff.

In the case of relativity, there is no complete mathematical model that can capture all its complexities. Things like singularities, geodesic incompleteness, non-Hausdorff conditions, imprisoned incompleteness remain problematic in mathematical modeling.

Most textbooks simply avoid talking about these things, which I think is a bad thing. Take for instance MTW "Gravitation", if you read this book you would get the impression that there are zero problems in modeling GR, not one single problematic topic is discussed, the book is conformal to the core. And the authors can hardly argue there was not enough space in the book to at least mention some of them. Contrast that book to, for instance, Hawking and Ellis "The Large Scale Structure of Space-Time", not a textbook, but here at least attempts are made to address those situations.

Do these problems interfere with run of the mill relativity? No they don't, but still, it ought to be worth mentioning them in a textbook. Last edited:
MeJennifer said:
Distance function and metric are the same thing.
I disagree. The distance function defines the struction of a manifold. The Minkowski metric is such that two events in spacetime in, SR, can have a large distance between then even if the spacetime interval is zero. The distance function can thus tell us which points are in an epsilon neighborhood while the later cannot. This allows one to make sense out of the difference between the two. By reading the definition of a spacetime interval and the distance function as defined in those two texts then by their definition they are very much not the same thing. Notice that the distance function is not a tensor whereas the metric is.
These kind of "explanations" make no mathematical sense at all, and are simply designed to either ignore the difficulties with manifolds that have no valid metric or, and this is worse, to impress the reader with the idea that there really is no issue at all.
I disagree. As Schutz states it in his math book; from page 3
Notice that we have used the distance function d(x, y) to define neighborhoodsand thereby open sets. We say that d(x, y) has induced a topology on Rn. By this we mean that it has enabled two us to define open sets in Rn which have the properties

Ti) if O1 are O2 are open so is there intersection

Tii) the union of any collection (possibly infinite in number) of open sets is open

Please note however that my disagreement with you is not 100% solid and baked into concrete. But this question has arisen from contrary concepts in different textbooks, each of which makes perfectly good sense.

However I know an expert on these things and will e-mail him and see what he says. I'll aks if I can place his response here in a quote with nothing changed.

Best Regards MJ

Pete

pmb_phy said:
The distance function defines the struction of a manifold. The Minkowski metric is such that two events in spacetime in, SR, can have a large distance between then even if the spacetime interval is zero. The distance function can thus tell us which points are in an epsilon neighborhood while the later cannot. This allows one to make sense out of the difference between the two. By reading the definition of a spacetime interval and the distance function as defined in those two texts then by their definition they are very much not the same thing. Notice that the distance function is not a tensor whereas the metric is.
In differential geometry the metric is the distance function, by definition!

By introducing alternative terminology the issue is simply circumvented, not resolved.

By the way, Finsler geometry, resolves some of those issues, while a Riemannian geometry operates on a positive definite metric, Finsler geometry operates on metrics without such a quadratic restriction. Finsler geometry enjoys two curvature tensors, and is Riemannian only if the Cartan tensor vanishes.

But, perhaps I am wrong, even on Finsler geometries there seem to be unresolved geometric-topological relationships. Anyone could comment on this?

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MeJennifer said:
In differential geometry the metric is the distance function, by definition!

By introducing alternative terminology the issue is simply circumvented, not resolved.

As I said, I'll get back to you. It will be a message from a well-known GRist. You might have even read his text. He may even side with you. But it appears to me that you're ignoring what I'm saying. I said that some texts define them to be identically the samel, as you said. But some, like Wald and Schutz, don't. Do you have one of these texts? Wald uses the term "Open ball" instead of the term "distance function" so that might be the resolution there. I.e. perhaps Wald didn't want to use the term "distance" anywhere near the term "open ball" and thus I foolled myself intop believing something else. But there is no uncertaintly in what Schutz wrote. Do ypu really believe that I'm so naive to jump to such conclusions without putting a lot of thought into it??

I'll remain silent (won't read more of this thread) until/if I get a response and he allows me to post in open forum.

All my best MJ

Pete

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Hi Pete, I am not 100% sure what the question is here but I'm going to make a few very elementary observations in the hope that some of them hit on target.

Given a set A, we can define a function d:A x A --> R called by some authors a "distance function" and by others a "metric" as having the properties

i) d(x,y) $\geq$ 0 and =0 iff x=y.
ii) d(x,y)=d(y,x)
iii) d(x,y)+d(y,z) $\geq$ d(x,z) [triangle inequality]

On the other hand, given a differentiabe manifold M, a point p in M, and $T_p(M)$, the tangeant space at p, a metric tensor is (as far as I know), a function $\mathbf{g}^*_p:T_p(M)\times T_p(M)\rightarrow \mathbb{R}$ that is symetrical and non-degenerate. It is easy to show that $\mathbf{g}^*$ is actually an inner product on the real vector space $T_p(M)$.

And an inner product does not satisfy the axioms of a distance function [for starters, the norm can be negative, which violates i)]. So, in conclusion, it seems that the metric tensor at a point p of a diff. manifold is not a distance function.

However, following notational convention for innerproducts, we shall write $\mathbf{g}^*_p(u,v)=<u,v>$. It is a known result from linear algebra that every inner product can be used to define a norm by $||u||:=\sqrt{|<u,u>|}$. Recall that a 'norm' on a real vector space is a function $||\cdot||:V \times V\rightarrow \mathbb{R}$ such that

i) $||u||\geq =0$ with =0 iff u=0
ii) $||av||=|a|||v||$
iii) $||u\pm v||\leq ||u||+||v||$ [triangle inequality]

Thus, every norm can be used to define a distance by d(u,v)=||u-v||.

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People! Instead of stating the obvious and the trival here, has anyone actually taken a first hand look at those texts I was referring to, i.e. Schutz and Wald. If not then please reread the question I posted. Other than these sources then of course I'd say that the metric and the distance function are two different names for the same thing. However the question pertained to two distict places where the authors seem not to use "distance function" and "metrtic" as meaning the same thing.

Folks - Please stick to the original question which was
Other texts use the term "distance function" and "metric" to mean the same thing. Has anyone every heard of this or seen this in those texts? Has anyone seen it as a problem in any case either for or against the distance function being the same as the metric or not being the same as the metric?
In other words, if you don't have both Schutz's math text and Wald's GR text then this question was not directed to you.

Pete

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pmb_phy said:
In other words, if you don't have both Schutz's math text and Wald's GR text then this question was not directed to you.
I have both I have Schutz's text here right in front of me.
The term 'distance function' as used in chapter 1.1 is simply the metric. I do not see anything that could be considered an interpretation that 'distance function' is anything different from a metric here. Later in the text he also uses it for vectors.

By the way he makes one comment which is interesting:

Schutz - Geometrical methods of mathematical physics said:
The idea that a line joining any two points of Rn can be infinitely subdivided can be made more precise by saying that any two points in Rn have neighborhoods which do not intersect. (They will also have some neighborhoods which do intersect, but if we choose small enough neighborhoods we can make them disjoint.) This is called the Hausdorff property of Rn. It is possible to construct non-Hausdorff spaces, but for our purposes they are artificial and we shall ignore them.
I am all but convinced of the validity of the remark marked red.
Is there perhaps some theorem that proves that that is the case? With regards to Wald, he simply provides a short appendix regarding differential geometry in the back of the book. The 'open ball' discussed in chapter 2.1, the term used to describe the immediate region of a given point, is not further worked out and becomes a moot point after Wald writes:

Wald General Relativity said:
Viewed as topological spaces, we shall consider in this book only manifolds which are Hausdorff and paracompact

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To answer directly your question: I have never (in my limited experience) encountered a situation in which 'metric' and 'distance function' meant different things.

I have Wald. Can you give the page in which he asserts that the metric, as opposed to the distance function, cannot be used to define an open-ball topology? I'd be very surprised to actually read that.

MeJennifer said:
I am all but convinced of the validity of the remark marked red.
Is there perhaps some theorem that proves that that is the case? That's one of the most basic things about $\mathbb{R}^n$. On $\mathbb{R}^n$ you have the standard metric (induced by the dot product), which gives you the distance between two points. So if you take the two points with a distance - say - r, then you just look at two open balls around the two points with radius < r/2, for example.

pmb_phy: Having had a look at Schutz Geometrical Methods and his General Relativity I think your confusion arises mainly because of the following two details.

1) Often in the context of special relativity one has two different notions of distance: First, the spacetime interval, which is a measure of distance for two events in spacetime (i.e. maybe at different times). Second, the distance between two poins in space at a fixed point of time. Notice that this difference can also be made in classical mechanics, where instead of the Minkowski $\mathbb{R}^n$ you simply have the Euclidean $\mathbb{R}^n$.

2) In $\mathbb{R}^n$, no matter "which way you go" for no matter how long, you'll always stay in $\mathbb{R}^n$. This is not the case on manifolds! Therefore, you might not be able to define the topology you want using the metric of the manifold itself. Often, one then considers the embedding of the manifold in $\mathbb{R}^n$ for some n and takes the metric on $\mathbb{R}^n$ and the induced topology on the manifold. The important thing is, that those two metrics are distinct. This is quite intuitive: Imagine you're an ant, living on a sphere. Then to you, distance from one point to the other is quite different from distance for the observer, who can pass thru the inner part of the sphere, whereas you have to go all the way round.

pmb_phy said:
In Wald he speaks of an open ball in Rn in which the Euclidean function is used as the "open ball" function about a particular point. The purpose of the distance function is to defined open and closed sets. The metric doesn't neccesarily have this ability.
Precisely in what context can't the metric define the open and closed sets?

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cliowa said:
That's one of the most basic things about $\mathbb{R}^n$. On $\mathbb{R}^n$ you have the standard metric (induced by the dot product), which gives you the distance between two points. So if you take the two points with a distance - say - r, then you just look at two open balls around the two points with radius < r/2, for example.
Shutz is talking about non-Hausdorff manifolds here. It is far from obvious, to me at least, that this is the case for non-Hausdorff manifolds.

cliowa said:
1) Often in the context of special relativity one has two different notions of distance: First, the spacetime interval, which is a measure of distance for two events in spacetime (i.e. maybe at different times). Second, the distance between two poins in space at a fixed point of time.
The second kind of distance has absolutely nothing to do with the distance function or metric of a manifold.

cliowa said:
2) In $\mathbb{R}^n$, no matter "which way you go" for no matter how long, you'll always stay in $\mathbb{R}^n$. This is not the case on manifolds! Therefore, you might not be able to define the topology you want using the metric of the manifold itself. Often, one then considers the embedding of the manifold in $\mathbb{R}^n$ for some n and takes the metric on $\mathbb{R}^n$ and the induced topology on the manifold. The important thing is, that those two metrics are distinct. This is quite intuitive: Imagine you're an ant, living on a sphere. Then to you, distance from one point to the other is quite different from distance for the observer, who can pass thru the inner part of the sphere, whereas you have to go all the way round.
Sorry, but I fail to see what your point is.
In Riemannian geometry we do not even concern ourselves with higher dimensional embeddings and extrinsic curvature.

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MeJennifer said:
Shutz is talking about non-Hausdorff manifolds here. It is far from obvious, to me at least, that this is the case for non-Hausdorff manifolds.
Schutz said:
The idea that a line joining any two points of Rn can be infinitely subdivided can be made more precise by saying that any two points in Rn have neighborhoods which do not intersect. (They will also have some neighborhoods which do intersect, but if we choose small enough neighborhoods we can make them disjoint.)This is called the Hausdorff property of Rn. It is possible to construct non-Hausdorff spaces, but for our purposes they are artificial and we shall ignore them.
And this is talking about non-Hausdorff manifolds? Especially: "they are artificial and we shall ignore them"? I don't follow.
Clearly, for non-Hausdorff spaces the statement that any two distinct points have disjoint neighborhoods is not necessarily true, as implied by the definition of Hausdorff. There are however various other fine distinctions in the same spirit. Have a look at http://en.wikipedia.org/wiki/Separation_axiom#Definitions_of_the_axioms".

MeJennifer said:
The second kind of distance has absolutely nothing to do with the distance function or metric of a manifold.
...
Sorry, but I fail to see what your point is.
In Riemannian geometry we do not even concern ourselves with higher dimensional embeddings and extrinsic curvature.

I'm sorry for the confusion I may have caused: I was trying to clear up pmb_phy's confusion, not one which you might have.

Best regards...Cliowa

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cliowa said:

And this is talking about non-Hausdorff manifolds? Especially: "they are artificial and we shall ignore them"? I don't follow.
Ops, you are right, Schutz does not mention non-Hausdorff conditions, I misread.

cliowa said:
Clearly, for non-Hausdorff spaces the statement that any two distinct points have disjoint neighborhoods is not necessarily true, as implied by the definition of Hausdorff.
That was indeed my point.

I'm beginning to think I understand where I was getting things mixed up. I was using the distance function in Eq. (1.1) for all spaces just as Wald seemed to be doing. I thought that this was a universal type of metric which allows one to determine which points are outside a neighboorhood and which are inside the neighboorhood. The condition "Ti" on page 3 seemed not to apply to Minkowski spacetime given that it also had a metric, i.e. the minkowski metric. Perhaps I can ask Schutx himself. Can anyone tell me how to determine whether a point is in a neighborhood. E.g. suppose the origin of a circular disk in spacetime is located in the (ct,x) plane. I.e. the disk has its center at (0,0). Suppose we also consider a event which is at (1000, 1000). Using the minkowski metric would seem have this event inside the neighboorhood. What criteria is used to say "inside neighboorhood" etc.?

Pete

pmb_phy said:
What criteria is used to say "inside neighboorhood" etc.?
What do you mean criteria pmb_phy?

You simply apply the distance function, e.g. the metric to determine that.
In fact you gave a perfectly valid example, so I am not sure what you are asking? Furthermore it should be clear from your example that a manifold with a Minkowski metric is not Hausdorff!

MeJennifer said:
What do you mean criteria pmb_phy?
Usually when one speaks of a neighborhood they say something like "The point x is inside the neighboorhoor S". So how do you do this with the Minkowski metric? The minkowski "distance function" will give some negative values of what in Euclidean geometry one would call the "radius of the neighborhood.

Furthermore it should be clear from your example that a manifold with a Minkowski metric is not Hausdorff!
You keeo bringing up the Hausdorff property of a space. Why?

Best regards

Pete

pmb_phy said:
You keeo bringing up the Hausdorff property of a space. Why?

Why indeed...

pmb_phy said:
Usually when one speaks of a neighborhood they say something like "The point x is inside the neighboorhoor S". So how do you do this with the Minkowski metric? The minkowski "distance function" will give some negative values of what in Euclidean geometry one would call the "radius of the neighborhood.

You keeo bringing up the Hausdorff property of a space. Why?
Ok then Pete, never mind, no more Haudorff.

Perhaps someone else can explain your question without mentioning the Hausdorff condition.

Good Luck!

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pmb_phy said:
Usually when one speaks of a neighborhood they say something like "The point x is inside the neighboorhoor S". So how do you do this with the Minkowski metric? The minkowski "distance function" will give some negative values of what in Euclidean geometry one would call the "radius of the neighborhood.

Correct me if I'm wrong, but the so called Minkowski metric is a metric tensor, not a metric as in "a distance function". And as I pointed out in my first post, a metric tensor is not a distance function.

A neighborhood of a point in a topological space X is a subset of an open containing the point. So questions relative to neighborhoods in Minkowski spacetime need the knowledge of what the topology of the spacetime is. And I think that this is an open question. At least, this is the impression I got from the replies I got to my thread https://www.physicsforums.com/showthread.php?t=153887.

See in particulars the posts by Robphy before MJ's Hausdorff obsession took over the thread. I'm purely joking MJ, I enjoyed the discussion a lot and I think it's great that you're not afraid to ask deep challenging questions.

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MeJennifer said:
Ok then Pete, never mind, no more Haudorff.

Perhaps someone else can explain your question without mentioning the Hausdorff condition.

Good Luck!
I did ask you why you kept mentioning the Hausdorff condition, not to stop mentioning it.

In any case by "criteria" I meant above is obvious. For the event E(ct,x) to be inside the disk S of radiius r, which is in the ct-plane centered at (0,0) then the criteria that must be satisfied for E to be within this disk is

sqrt(x2 + (ct)2) < r

The distance function here is d(ct,x) = qrt(x2 + (ct)s2)

S of points will be open if every point x in S has a neighborhood entirely contained in S.

I can't see how to state these criteria without the distance function d(ct,x) = qrt(x2 + (ct)2) beiung defined. Now it may be the case that there are two distance functions defined on the manifold.

Pete

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pmb_phy said:
I did ask you why you kept mentioning the Hausdorff condition, not to stop mentioning it.

In any case by "criteria" I meant above is obvious. For the event E(ct,x) to be inside the disk S of radiius r, which is in the ct-plane centered at (0,0) then the criteria that must be satisfied for E to be within this disk is

sqrt(x2 + (ct)s2) < r

The distance function here is d(ct,x) = qrt(x2 + (ct)s2)

What is s? quasar987 said:
What is s? A mistake (which I have corrected, thanks) Pete

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Surely it is that the distance function is a positive definite function defined on the $\mathbb{R}^4$ 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?

masudr said:
Surely it is that the distance function is a positive definite function defined on the $\mathbb{R}^4$ 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?

When the distance function, say

$$d(X_1, X_2) = \sqrt{(t_1-t_2)^2 + (x_1-x_2)^2},$$

is different from the metric, which is

$$g(X_1, X_2) = t_1 t_2 - x_1 x_2.$$

The metric tensor was never meant to give distances in the same sense the Euclidean metric does.

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masudr said:
When the distance function, say

$$d(X_1, X_2) = \sqrt{(t_1-t_2)^2 + (x_1-x_2)^2},$$

is different from the metric, which is

$$g(X_1, X_2) = t_1 t_2 - x_1 x_2.$$

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

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)

masudr said:
When the distance function, say

$$d(X_1, X_2) = \sqrt{(t_1-t_2)^2 + (x_1-x_2)^2},$$

is different from the metric, which is

$$g(X_1, X_2) = t_1 t_2 - x_1 x_2.$$

The metric tensor was never meant to give distances in the same sense the Euclidean metric does.
What do you mean "never meant". Is this mathematics or religion we are discussing?

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

$$d(X_1, X_2) = \sqrt{(t_1-t_2)^2 + (x_1-x_2)^2 + (y_1-y_2)^2 + (z_1-z_2)^2}$$ Last edited:
MeJennifer said:
What do you mean "never meant". Is this mathematics or religion we are discussing?

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

$$d(X_1, X_2) = \sqrt{(t_1-t_2)^2 + (x_1-x_2)^2 + (y_1-y_2)^2 + (z_1-z_2)^2}$$ Isn't it amazing how threads obtain a life of their own even when the OP has his answer? You go girl! Pete

MeJennifer said:
What do you mean "never meant". 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:

$$d(X_1, X_2) = \sqrt{(t_1-t_2)^2 + (x_1-x_2)^2 + (y_1-y_2)^2 + (z_1-z_2)^2}$$ What I mean is that locally all manifolds look like $\mathbb{R}^n,$ 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 $\mathbb{R}^n.$

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 modeled 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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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

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

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

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 knowledgeable PF member.