Minkowski space metric space?

  • Thread starter damnedcat
  • Start date
  • #1
14
0

Main Question or Discussion Point

is minkowski space a metric space. As best as i can remember a metric space is a set with a metric that defines the open sets. With this intuition is Minkowski space a metric space. I mean i think it should be, but according to one of the requirements for a metric:

d(x,y)=0 iff x=y
triangle inequality is also broken.

but this is not true for points along the worldline with v=c in minkowski space.

Or is the requirement above just for euclidean spaces?
 

Answers and Replies

  • #2
14
0
I guess my question is is the mikowski metric really a metric since it violates:

d(x,y)=0 iff x=y
triangle inequality
 
  • #3
George Jones
Staff Emeritus
Science Advisor
Gold Member
7,370
975
No, the Minkowski "metric" is not a metric in the sense that you mean.
 
  • #4
29,788
6,127
You are correct, the Minkowski metric is not technically a metric for exactly the reason you mention. What it is, is a pseudometric, which is a metric with that specific requirement removed. So the Minkowski metric is a pseudo-Euclidean metric and the Schwarzschild metric is a pseudo-Riemannian metric.

A lot of times the "pseudo" is dropped in the GR and SR literature as it is cumbersome and can be assumed as a given, but occasionally the distinction is important as you have discovered.
 
  • #5
Fredrik
Staff Emeritus
Science Advisor
Gold Member
10,851
408
On the other hand, the set needs a topology (a specification of which sets to call "open"), because that's needed to define what it means for a function to be "continuous", among other things, and the standard way to define a topology is to just use the Euclidean metric* and say that a set is open if and only if it's a union of open balls.

*) There are actually two functions that we could call the "Euclidean metric". One of them is a tensor field, and the other is a distance function. I meant the distance function. Edit: I actually couldn't have meant the tensor field, since the topology needs to be defined before the manifold structure that's needed to define tensor fields.

You are correct, the Minkowski metric is not technically a metric for exactly the reason you mention. What it is, is a pseudometric, which is a metric with that specific requirement removed. So the Minkowski metric is a pseudo-Euclidean metric and the Schwarzschild metric is a pseudo-Riemannian metric.
I think it's more common to say that the Minkowski metric is a metric, but not a Riemannian metric. It can be described as pseudo-Riemannian, or Lorentzian.
 
Last edited:
  • #6
George Jones
Staff Emeritus
Science Advisor
Gold Member
7,370
975
I think it's more common to say that the Minkowski metric is a metric, but not a Riemannian metric. It can be described as pseudo-Riemannian, or Lorentzian.
But even a Riemannian "metric" is not a metric in the sense that damnedcat means in the original post. damnedcat means the metric for what Bishop and Goldberg calls a topological metric space. It is true that a Riemannian metric can be used to generate naturally a topological metric (by looking at greatest lower bounds of lengths of curves joining points), while a pseudo-Riemannian metric (like used for general relativity) cannot.
 
  • #7
Fredrik
Staff Emeritus
Science Advisor
Gold Member
10,851
408
But even a Riemannian "metric" is not a metric in the sense that damnedcat means in the original post. damnedcat means the metric for what Bishop and Goldberg calls a topological metric space.
I understood that. (Assuming that a "topological metric space" is a metric space with the the metric topology).

It is true that a Riemannian metric can be used to generate naturally a topological metric (by looking at greatest lower bounds of lengths of curves joining points), while a pseudo-Riemannian metric (like used for general relativity) cannot.
This is what I'm thinking: Minkowski spacetime is the set [itex]\mathbb R^4[/itex], with the topology defined by the Euclidean metric (distance function), the smooth manifold structure defined by all the smooth functions from open subsets of [itex]\mathbb R^4[/itex] into [itex]\mathbb R^4[/itex], and the specific Lorentzian metric tensor field that we're all familiar with.

Since a manifold is "a topological space such that blah-blah", it has to have a topology before we define the manifold structure, and it has to have a manifold structure before we define a metric tensor field.
 
  • #9
14
0
But even a Riemannian "metric" is not a metric in the sense that damnedcat means in the original post. damnedcat means the metric for what Bishop and Goldberg calls a topological metric space. It is true that a Riemannian metric can be used to generate naturally a topological metric (by looking at greatest lower bounds of lengths of curves joining points), while a pseudo-Riemannian metric (like used for general relativity) cannot.
Let me see if i get what ur saying. So there are 2 types/classes of metrics involved in my question: a topological metric d(x,y) and a Riemannian/Pseudo-Riemannian metric. The Riemannian metric generates the distance function metric d(x,y). However a pseudo riemannian metric cannot generate such a function/metric.

(d(x,y) = 0 iff x=y)
 
  • #10
andrewkirk
Science Advisor
Homework Helper
Insights Author
Gold Member
3,836
1,418
Interesting! The physics texts I've seen (plus Wikipedia!) describe Minkowski space as a manifold but don't state what topology is assumed, even though a topology is needed in order for it to be a manifold.
Fredrik suggests the implied topology is that arising from the Euclidean distance metric d(x,y)=(x0-y0)^2+(x1-y1)^2+(x2-y2)^2+(x3-y3)^2. That makes sense although it's a little troubling that this is not stated anywhere in the texts I've seen. Are there any other candidates?

I wonder whether it is possible to define a natural topology on Minkowski space using the Minkowski semi-metric in some way (eg by defining the distance between two points as the square root of the absolute value of the Minkowski interval between points and then defining open balls based on that distance measure) and if so whether that topology would turn out to be identical to the one based on the Euclidean metric.
 
  • #11
14
0
Interesting! The physics texts I've seen (plus Wikipedia!) describe Minkowski space as a manifold but don't state what topology is assumed, even though a topology is needed in order for it to be a manifold.
Fredrik suggests the implied topology is that arising from the Euclidean distance metric d(x,y)=(x0-y0)^2+(x1-y1)^2+(x2-y2)^2+(x3-y3)^2. That makes sense although it's a little troubling that this is not stated anywhere in the texts I've seen. Are there any other candidates?
can the euclidean distance metric u used above describe minkowski space? wouldn't that imply diag(1,1,1,1) as opposed to diag(-1,1,1,1)?? Not sure if I am thinking about it right.

I wonder whether it is possible to define a natural topology on Minkowski space using the Minkowski semi-metric in some way (eg by defining the distance between two points as the square root of the absolute value of the Minkowski interval between points and then defining open balls based on that distance measure) and if so whether that topology would turn out to be identical to the one based on the Euclidean metric.
I'm thinking not based on my question above.
 
  • #12
Fredrik
Staff Emeritus
Science Advisor
Gold Member
10,851
408
Let me see if i get what ur saying. So there are 2 types/classes of metrics involved in my question: a topological metric d(x,y) and a Riemannian/Pseudo-Riemannian metric. The Riemannian metric generates the distance function metric d(x,y). However a pseudo riemannian metric cannot generate such a function/metric.

(d(x,y) = 0 iff x=y)
That's essentially correct. I think the term "topological metric" is non-standard, and I wouldn't say that a Riemannian metric "generates the distance function". It can be used to define a distance function, but if a distance function was used to define the topology, it would usually be a different distance function.

Interesting! The physics texts I've seen (plus Wikipedia!) describe Minkowski space as a manifold but don't state what topology is assumed, even though a topology is needed in order for it to be a manifold.
Fredrik suggests the implied topology is that arising from the Euclidean distance metric d(x,y)=(x0-y0)^2+(x1-y1)^2+(x2-y2)^2+(x3-y3)^2. That makes sense although it's a little troubling that this is not stated anywhere in the texts I've seen. Are there any other candidates?
I don't have another candidate, but I should mention that the same topology can be defined without reference to a distance function. Define 4-dimensional open intervals, and then define a set to be open if it's a union of open intervals.

I wonder whether it is possible to define a natural topology on Minkowski space using the Minkowski semi-metric in some way (eg by defining the distance between two points as the square root of the absolute value of the Minkowski interval between points and then defining open balls based on that distance measure) and if so whether that topology would turn out to be identical to the one based on the Euclidean metric.
The manifold structure needs to be defined before we define the tensor field. It sounds like what you want to do is to define the standard vector space structure on [itex]\mathbb R^4[/itex] first, then define the bilinear form [itex](x,y)\mapsto x^T\eta y[/itex], and an associated "not really a norm" N by [itex]N(x)^2=x^T\eta x[/itex], and use that to define "open balls" and so on. I haven't spent any time thinking about what the problems are, but I think I read somewhere that it doesn't work.

can the euclidean distance metric u used above describe minkowski space?
What would it mean for a distance function on the underlying set of a manifold to "describe" the manifold?

wouldn't that imply diag(1,1,1,1) as opposed to diag(-1,1,1,1)?? Not sure if I am thinking about it right.
I don't know what "that" refers to (or rather, I don't know what the statement it refers to means), but a distance function doesn't have a matrix of components. A tensor does.
 
  • #13
PAllen
Science Advisor
2019 Award
8,108
1,382
Looking at two of my (old) relativity texts, I came away with the following understanding:

The point set topology of the 4-manifold is established by building up overlapping one-one mappings to Euclidean R-4. This defines open sets and connectedness, but no geometry (for example, little overlapping pieces of Euclidian flat plane can be used to build a torus, sphere, or any other two-manifold, however connected). One imposes some type of invariance rules that different equivalent ways of doing this should have no effect on the topology (these were physics books, so they were vague on this). At this point, there is *no* metric or affine connection established; no distance, no parallel transport, etc. Then, how one imposes connections and distances builds the geometry; in this case, pseudo-euclidean for Minkowski, pseudo-Riemannian for GR.
 
  • #14
Fredrik
Staff Emeritus
Science Advisor
Gold Member
10,851
408
The point set topology of the 4-manifold is established by building up overlapping one-one mappings to Euclidean R-4. This defines open sets...
Each such map [itex]x:U\rightarrow \mathbb R^4[/itex] is required to be a homeomorphism onto x(U). This requirement doesn't make sense if the topology hasn't been defined in advance.
 
  • #15
PAllen
Science Advisor
2019 Award
8,108
1,382
Each such map [itex]x:U\rightarrow \mathbb R^4[/itex] is required to be a homeomorphism onto x(U). This requirement doesn't make sense if the topology hasn't been defined in advance.
Do you know why this is true? It seems to me I can impose structure on an amorphous set of points by the mapping. That is, I declare each open set in the R4 patch to define a corresponding open set in the manifold under 'construction'.
 
  • #16
DrGreg
Science Advisor
Gold Member
2,290
895
Each such map [itex]x:U\rightarrow \mathbb R^4[/itex] is required to be a homeomorphism onto x(U). This requirement doesn't make sense if the topology hasn't been defined in advance.
I might be missing something subtle, but I don't see why you can't start with a set (that will become the manifold) without a topology, functions [itex]x:U\rightarrow \mathbb R^4[/itex] which have all the correct properties except those that require a topology, and then define your topology to be the inverse images of the open sets of [itex]\mathbb R^4[/itex].

EDIT: written at the same time as PAllen's reply!
 
Last edited:
  • #17
George Jones
Staff Emeritus
Science Advisor
Gold Member
7,370
975
Do you know why this is true? It seems to me I can impose structure on an amorphous set of points by the mapping. That is, I declare each open set in the R4 patch to define a corresponding open set in the manifold under 'construction'.
I might be missing something subtle, but I don't see why you can't start with a set (that will become the manifold) without a topology, functions [itex]x:U\rightarrow \mathbb R^4[/itex] which have all the correct properties except those that require a topology, and then define your topology to be the inverse images of the open sets of [itex]\mathbb R^4[/itex].

EDIT: written at the same time as PAllen's reply!
With a bi of care, this can be done.
 
  • #18
andrewkirk
Science Advisor
Homework Helper
Insights Author
Gold Member
3,836
1,418
I might be missing something subtle, but I don't see why you can't start with a set (that will become the manifold) without a topology, functions [itex]x:U\rightarrow \mathbb R^4[/itex] which have all the correct properties except those that require a topology, and then define your topology to be the inverse images of the open sets of [itex]\mathbb R^4[/itex].
I think the problem with that is that there are an infinity of different choices of atlas (an atlas being a set of coordinate patches that covers the manifold together with the bijections from the patches to R4) and, without the constraint that the prior continuity requirement provides, different atlases will result in different topologies on the manifold. One atlas might make the manifold a torus and another might make it a sphere. In fact, even the dimension of the manifold will not be well-defined.

So if you don't start out with a topology on the manifold, you can't create a well-defined one via coordinate maps.

If you put some additional constraints on what type of coordinate maps you will allow, you can do it, but I think those constraints will define a topology. For instance you might say the only coordinate maps allowed are those whose domain is an open box in Minkowski space, but if you do so then you are imposing the box topology on the space (which Fredrik indicated above is a way to put a topology on the space without using a distance function).
 
  • #19
PAllen
Science Advisor
2019 Award
8,108
1,382
I think the problem with that is that there are an infinity of different choices of atlas (an atlas being a set of coordinate patches that covers the manifold together with the bijections from the patches to R4) and, without the constraint that the prior continuity requirement provides, different atlases will result in different topologies on the manifold. One atlas might make the manifold a torus and another might make it a sphere. In fact, even the dimension of the manifold will not be well-defined.

So if you don't start out with a topology on the manifold, you can't create a well-defined one via coordinate maps.

If you put some additional constraints on what type of coordinate maps you will allow, you can do it, but I think those constraints will define a topology. For instance you might say the only coordinate maps allowed are those whose domain is an open box in Minkowski space, but if you do so then you are imposing the box topology on the space (which Fredrik indicated above is a way to put a topology on the space without using a distance function).
The point is to define some topology. The dimension ambiguity I'm not sure I understand if all patches are to R4. The difference is between 'finding coordinate systems for some pre-established topology' versus choosing what you call an atlas to establish the topology. Of course, the nature of patches determines everything about the topology (see my connectedness example).

The key, for me, is something has to establish the topology before you can start defining geometry, but that something need not involve any distance function. This avoids having a topologic distance function separate from the metric distance function. This is the way some old relativity books I have seemed to define it: impose open sets and connectedness from the overlapping mappings to R4 patches; then start imposing geometry on top of the topology.
 
  • #20
George Jones
Staff Emeritus
Science Advisor
Gold Member
7,370
975
I have attached part of page 21 from Introduction to Smooth Manifolds by John M. Lee.
 

Attachments

Related Threads on Minkowski space metric space?

  • Last Post
2
Replies
49
Views
4K
  • Last Post
2
Replies
28
Views
4K
Replies
4
Views
1K
  • Last Post
3
Replies
54
Views
22K
  • Last Post
Replies
4
Views
2K
  • Last Post
Replies
4
Views
2K
Replies
4
Views
7K
  • Last Post
Replies
2
Views
1K
  • Last Post
4
Replies
87
Views
15K
  • Last Post
Replies
7
Views
3K
Top