# Geometric shape of Minkowski space

1. Jul 7, 2011

### thehangedman

So, suppose for visualization there are only two dimensions: ct and x. Now if the metric where Euclidean, we could visualize this space is a simple plane.

What would be the shape of the "plane" when the metric is +1, -1 (Minkowski)?

Is it somehow hyperbolic?

2. Jul 7, 2011

### bcrowell

Staff Emeritus
One way to think about it is that (using intimidating terminology for a moment) you have topological structure plus metric structure.

Topological structure means whatever is preserved under stretching a rubber sheet. Topologically, Minkowski space is the same as the Euclidean plane, and a coffee cup is the same as a doughnut http://en.wikipedia.org/wiki/Topology#Elementary_introduction , but the plane isn't the same as a doughnut.

Metric structure means how you assign distances, which is different for the Euclidean plane than for Minkowski space. In Euclid's axioms, there is only one that refers to distance measure: given any point to use as a center, and any line segment with that point at an end, you can make a circle with that segment as its radius. The hyperbola plays the same role in Minkowski space as the circle does in the Euclidean plane.

Another way to think about it is that the role played by rotations in the Euclidean plane is played by a certain kind of distortion in Minkowski space: http://www.lightandmatter.com/html_books/0sn/ch07/ch07.html [Broken] In both cases, the transformations preserve "distances" (Pythagorean or Minkowskian), as well as areas and parallelism.

Last edited by a moderator: May 5, 2017
3. Jul 7, 2011

### Ben Niehoff

Minkowski space is flat like a plane.

The metric signature (-+) does not make the plane "somehow hyperbolic". It is still a plane, its curvature vanishes, its geodesics are straight lines, etc.

The metric of a hyperbolic plane looks like this:

$$ds^2 = \frac{dx^2 + dy^2}{y^2}$$

Notice that this metric still has signature (++).

In general, changing the signature from (++++) to (-+++) is not the same thing as introducing curvature. It's actually something quite a bit more radical. Not every manifold (as a topological space) can be given a metric of indefinite signature in such a way that can be extended over the whole manifold. This is called putting a Lorentzian structure on a manifold. It turns out to be related to whether the manifold can have a continuous, non-vanishing vector field.

4. Jul 7, 2011

### pervect

Staff Emeritus
The topology also gives you a sense of neighborhood, that the geometry doesn't.

Anything separated by a light cone from some event P has a "distance" i.e. a Lorentz interval, of zero, but only some of those events are in the neihborhood of P, those are the events that have nearly identical coordinates.

So the "open balls" of topology determine the neighborhood. Projected onto some basis, the Lorentz interval gives you proper times and proper distances. In the light beam example, we can tell that the absolute value of the distance equals the absolute value of the time in some local coordinate system, as it must to make the Lorentz interval zero (dx^t = dx^2), but we can't tell how long each is (what is dx, what is dt), until we specify more about the reference frame - basically, we need the velocity of the observer.

5. Jul 12, 2011

### TrickyDicky

It is worth remembering though that given that the hyperbolic plane can't be embedded in Euclidean space, it must be represented by some model, and among those models there is the hyperboloid model that is a 3-dim Minkowski spacetime. The Minkowski plane can be thought of as a flat model of hyperbolic curve represented as distance.

6. Jul 12, 2011

### Ben Niehoff

This is a bit misleading, which is why I avoided the issue (also the way you've stated it is flat wrong).

The hyperbolic plane (of signature ++) can be isometrically embedded in Minkowski 3-space (of signature -++). That is, the hyperbolic plane can be thought of as a particular 2-dimensional curved surface within Minkowski 3-space. But this does not imply that there is anything "hyperbolic" about Minkowski space itself. Likewise, there is nothing "spherical" about Euclidean space, simply by virtue of the fact that we can isometrically embed spheres in Euclidean space.

1. One doesn't need to have an external space in which to embed the hyperbolic plane in order to have a "model" of it. In fact, the metric I gave above is the Poincare metric for the "upper half plane model", which is a model of hyperbolic space drawn on the upper half of an ordinary Euclidean plane, where the geodesics are drawn as circles centered on the x axis. There are additional models, including the hyperboloid model, but also including two different models drawn within the open disc of the unit circle. One model has straight lines as geodesics; in the other model, geodesics are circles that meet the boundary at 90 degrees. In fact, by conformal mapping, you can make any sort of model you want.

2. There is no such thing as a "hyperbolic curve". All 1-dimensional curves are flat.

3. Minkowski space is not a model of hyperbolic space. Every hyperbolic space can be isometrically embedded in Minkowski space of one dimension higher. This is a different statement.

Some other interesting things can be isometrically embedded in Minkowski space as well. For example, we generally think of a cone as something with a deficit angle at a given point. However, one can have cones with an excess angle as well, and these cones can be isometrically embedded in Minkowski space in such a way that the symmetry group of the cone is a subgroup of the symmetries of Minkowski space. (The light cone, incidentally, has infinite excess angle).

Excess-angle cones can also be isometrically embedded in Euclidean space, but they will have to be "wrinkled up", and hence their symmetries will not coincide with symmetries of Euclidean space.

7. Jul 12, 2011

### thehangedman

Ok. Thank you all for your help. One thing I was thinking about was to take the proper time and use that as a "third" dimension so I could visualize things. If you do that, then eucledian space is just a cone. z^2 = x^2 + y^2. Minkowski space would just be the curve z^2 = x^2 - y^2.

8. Jul 12, 2011

### TrickyDicky

Ah, I see when something can be misleading you rather avoid it than clarify it.

This is quite silly, I didn't imply anything, Minkowski space has no curvature, period.
If what you claim is that Minkowski space is totally unrelated with hyperbolic geometry , you are flat wrong. But I won't bother debating this, just read some geometry, ever heard of hyperbolic numbers?

This is ridiculous, I have never said what you attribute me ,nor suggested it. Why do you recur to a "straw man" trick?

You mean 1-dimensional curves have no intrinsic curvature. Surely you are not denying the existence of hyperbolas or of curves in general just because their curvature is extrinsic, are you?

I wrote "hyperboloid model", not "Minkowki space model", right? Precisely what you call a different statement is what I was trying to transmit, but I guess I'm sloppy.

9. Jul 12, 2011

### thehangedman

Thank you. This makes complete sense. Donuts really just a flat plane like Euclid had, but a different formula for distances. Couldn't you transform one to the other by making time real / vs imaginary?

10. Jul 12, 2011

### thehangedman

Not donuts. Stupid iPhone autocorrect. "it's" ...

11. Jul 12, 2011

### TrickyDicky

Yes, it's called the Minkowski trick (when going from Minkowski to Euclidean), aka Wick rotation.

12. Jul 12, 2011

### Ben Niehoff

My object was to avoid overloading the OP with irrelevant information that only muddles the issue. He was genuinely confused as to whether Minkowski space is or is not flat.

I was responding to what you wrote below, which I feel is highly misleading (emphasis mine):

Perhaps this was merely a typo, but what you wrote appears to claim that Minkowski 3-space (a flat space of 3 dimensions and indefinite signature) is the same thing as the hyperboloid model (a curved space of 2 dimensions and definite signature).

Of course I have, and I've used split-biquaternions to explore representations of the Lorentz algebra.

"Hyperbolic geometry" merely refers to geometry on spaces that are negatively curved. It is neat that some of these spaces can be isometrically embedded in Minkowski space, but this is not an essential feature.

There are other hyperbolic spaces which cannot be isometrically embedded in Minkowski space. Anti-deSitter space, for example, is a space of signature (1, d) and constant negative curvature. It can be isometrically embedded in $\mathbb{R}^{2, d+1}$, but not in Minkowski space.

To me, the phrase "their curvature is extrinsic" is incorrect, from a strict mathematical point of view. All 1-dimensional curves are flat. Their embedding within some other manifold might have extrinsic curvature. But of a 1-dimensional curve, the curvature does not belong to the curve, in the strictest sense; it is a property of the embedding.

Similarly, an ordinary 2-torus can be given a completely flat metric. The 2-torus can also be embedded in $\mathbb{R}^3$. However, despite the fact that the 2-torus is flat, its embeddings in $\mathbb{R}^3$ always have extrinsic curvature. Only in $\mathbb{R}^4$ can one embed the torus isometrically, such that the extrinsic curvature is equal to the intrinsic curvature (i.e., zero).

As a manifold, there is no difference between a hyperbola and the real line.

13. Jul 13, 2011

### TrickyDicky

I think I already clarified that this is not what I was claiming and admitted that my sloppy writing could be misleading. Instead of is should read is in. Thanks for the correction.

What is just neat to you might be significative to others, this is purely subjective. What is objective is that there is many connections between Minkowski space and hyperbolic geometry. Not only the embedding "neatness" : The space of velocities in Minkowski spacetime is hyperbolic, there is hyperbolic orthogonality: hyperbolic right angles, etc...
You argue as if I was claiming that Minkowski geometry is hyperbolic geometry. I could as well have stressed the close relation between the Minkowskian and Euclidean plane (even more close with the complex plane). But the OP referred to the similarities with the hyperbolic plane, and once it is cleared up the obvious fact that the minkowki plane doesn't have negative curvature, it is perfectly licit to talk about the connections with hyperbolic notions.

Totally irrelevant since I never said that all hyperbolic spaces can be embedded in Minkowski space. I rather tried to stress the fact that hyperbolic spaces can't be embedded in n+1 Euclidean spaces.

That is why is called extrinsic. Your nitpicking is quite absurd here. Extrinsic curvature is not a mathematically incorrect term at all. It is implicit that is always referred to the embedding. Following your "strict" policy you shouldn't call them curves either because they are flat.

Sure, but the discussion was not in terms of manifolds and topology but clearly geometrical.

Last edited: Jul 13, 2011
14. Jul 13, 2011

### TrickyDicky

More connections, the same way the Euclidean plane can be described with complex numbers (complex plane), the Minkoskian plane can be described with hyperbolic numbers (split-complex plane). And in the Minkowki plane hyperbolic trigonometry arises naturally from the rectangular hyperbola like the usual circular trig does in the Euclidean plane.
In order not to confuse anybody, when I say hyperbolic I'm not referring to hyperbolic geometry but to anything related to or with the shape of hyperbolas.