Difference (if any) between E^1 x E^3 and E^4

[cianfa72]

Hi,

reading this link galilean spacetime a doubt raised to me: which is the difference (if any) between $\mathbf E^1\times\mathbf E^3$ and $\mathbf E^4$ ?
I believe each space there (spaces involved in the cartesian product too) has to be regarded as equipped with standard Euclidean structure

 

[WWGD]

In what "Category" or structure tupe are you comparing? As topological spaces, vector spaces (over the Reals), Manifolds, etc? Euclidean spaces come equipped with a lot of different types of structure.

 

[Periwinkle]

There is no difference between $E^3 \oplus E = E^3 \times E$ and $E^4$, which are Euclidean spaces. However, based on the link you give, you ask why Galilean space-time is not $E^4$?

The definition of Galilean space-time is shown here. The Galilean transformation taught in classical physics in the Galilean space does not change the distance between the two points. This is a vector space with a particular distance concept. When we think of the endpoints of the vectors, we get a homogeneous set of points with an individual distance concept, which is somewhat in line with our general ideas of geometry. It can be seen that Galilean space-time is not $E^4$, because the distance concept is different.

The reason for this difference, in my opinion, is that time does not behave like spatial coordinates. Time is flowing, and there is a specific direction of its flow. Therefore, we cannot expect the combination of space and time to form $E^4$.

In this way, space and time can be incorporated into a geometric unity to reflect the specific role of time as reflected in the Galileo principle of relativity. (So that the structure of space is not changed by Galilean transformation.)

 

[cianfa72]

In what "Category" or structure type are you comparing?

I think affine structure with the underlying (real) vector space. The point is related to the last statement at that link:

"Clearly $E^1×E^3$ fails in the same way, but much more obviously. In this case you can clearly associate the event $(t1,r)$ with the event $(t2,r)$, and so you clearly have a notion of events at different times occurring at the same point in space"

Thus it seems $E^4$ is fundamentally different from it (see for instance Roger Penrose - The Road to Reality p. 385 about Aristotelian spacetime)

 

[WWGD]

Some else to think about, the standard metric in Euclidean $\mathbb R^4$ is $dx_1^2+dx_2^2+ dx_3^2+ dx_4^2$ , not $dx_1^2+dx_2^2+dx_3^2 -t^2$. These are not equivalent.

 

[cianfa72]

Some else to think about, the standard metric in Euclidean $\mathbb R^4$ is $dx_1^2+dx_2^2+ dx_3^2+ dx_4^2$ , not $dx_1^2+dx_2^2+dx_3^2 -t^2$. These are not equivalent.

In this thread we are not talking about Minkowski spacetime. Instead I believe the difference between them is that in $E^4$ does exist a concept of metric for each point belonging to it whereas in $E^1\times E^3$ does not. Here there exist just only a concept of metric inside each space separately so it does not make sense to ask for the distance between two point $(t_1, r_1)$ and $(t_2,r_2)$

 

[WWGD]

So you are dealing with them as metric spaces, meaning having a distance function?

 

[cianfa72]

So you are dealing with them as metric spaces, meaning having a distance function?

Actually I'm not sure about it. Starting from last sentence at the link and reading the section of the aforementioned book, I came out with that "interpretation": there exist a metric/distance function coming from euclidean inner product defined separately in each euclidean space entering in the cartesian product; an inner product or a distance function for the cartesian product itself is not defined though. This way the spaces are really different

 

[Periwinkle]

Actually I'm not sure about it. Starting from last sentence at the link and reading the section of the aforementioned book, I came out with that "interpretation": there exist a metric/distance function coming from euclidean inner product defined separately in each euclidean space entering in the cartesian product; an inner product or a distance function for the cartesian product itself is not defined though. This way the spaces are really different

There is no way to define a standard metric in Galilean space-time. Because the essence of such a metric is that it is invariant against Galilean transformation. However, it can be achieved with the Galilean transformation - thinking about trains running side by side -, that when $t_1$ and $t_2$ are different times, events $t_1, x_1$ and $t_2, x_2$ occur at the same $x$ location. Therefore, the metric cannot depend on $x$, it is $|t_1-t_2|$.

However, events at the same time are separated by the distance defined in $R^3$. The product symbolizes this.

 

[WWGD]

In this thread we are not talking about Minkowski spacetime. Instead I believe the difference between them is that in $E^4$ does exist a concept of metric for each point belonging to it whereas in $E^1\times E^3$ does not. Here there exist just only a concept of metric inside each space separately so it does not make sense to ask for the distance between two point $(t_1, r_1)$ and $(t_2,r_2)$

EDIT At any rate, the " Standard" homeomorphism between $E_1 \times E_3$ and $E_4$ ( As product Topological spaces) takes a pair $x_{11}, (x_{21}, x_{22}, x_{23}) \text in E_1 \times E_3$ and sends it to $(x_{11}, x_{21}, x_{22}, x_{23}) \text in E_4$ You may check what other properties this product preserves. Maybe this shows time-space cannot be "decomposed".

 

[WWGD]

This is a problem that interested me a while back in my * pointset Topology days: Under what conditions is a Topological space (homeomorphic to ) a product. But there is more than Topology involved in here, which makes it harder/more interesting.

*somewhat bizarre, given it is largely " tapped out".

 

[cianfa72]

However, events at the same time are separated by the distance defined in $R^3$. The product symbolizes this.

ok here you mean the cartesian product I believe...

another point: consider a three dimensional euclidean affine bundle over one dimensional affine euclidean space: this structure is definitively different from the cartesian product $E_1 \times E_3$ (each endowed with standard euclidean structure), right ?

 

[Periwinkle]

ok here you mean the cartesian product I believe...

Yes, Vladimir Igorevich Arnold is thinking of Cartesian products in his book Mathematical Methods of Classical Mechanics. He calls $\mathbb {R} \times \mathbb {R}^3$ the Galilean coordinate space, which, as he writes, has a natural Galilean structure. As I wrote, $R \times R^3$ refers to this physical structure.

However, there is no coordinate system in the space-time of the physical world. Therefore, Galilean space-time is mathematically a four-dimensional affine space, in which a concept of distance is introduced, in a unique way that was mentioned above.

another point: consider a three dimensional euclidean affine bundle over one dimensional affine euclidean space: this structure is definitively different from the cartesian product $E_1 \times E_3$ (each endowed with standard euclidean structure), right ?

I don't see this before myself. Because I assume that the dimension of the tangent space is the same as the dimension of the embedded space.

 

[cianfa72]

However, there is no coordinate system in the space-time of the physical world. Therefore, Galilean space-time is mathematically a four-dimensional affine space, in which a concept of distance is introduced, in a unique way that was mentioned above.

I read Arnold book about Galilean affine spacetime definition. Could you elaborate a bit about the aforementioned sentence in bold ?

I don't see this before myself. Because I assume that the dimension of the tangent space is the same as the dimension of the embedded space.

What I'm trying to say is that in fiber bundle structure (the fiber here is the three dimensional euclidean affine space $E^3$) we have basically multiple "copies/instances" of the same fiber space $E^3$ over the base space (i.e. $E^1$) but with no identification each other. How we have to understand the product $E^1 \times E^3$ ? As multiple copies/instances of $E^3$ over $E^1$ that however are all identified each other this time ?

To put it in another words, to you $E^1\times E^3$ is a (trivial) example of fiber bundle ?

 

[Periwinkle]

I read Arnold book about Galilean affine spacetime definition. Could you elaborate a bit about the aforementioned sentence in bold ?

This statement was: There is no coordinate system in the space-time of the physical world.

I mean precisely what Newton writes about the absolute space, time and the relative position in Principia (explanatory note for definitions):

Paragraph II, the first sentence, [5]​

Translated text: The absolute space by its own nature, without reference to any kind of exterior, always remains similar and immobile.​

As Newton writes, places in the absolute space are not visible - as I wrote, in the physical space-time there is no coordinate system - so we use tangible quantities instead.

Paragraph IV, fourth subparagraph, the first sentence [5]​

Translated text: It is true, however, that these places of space cannot be seen and cannot be differentiated from one another by our senses. Therefore, the location and distance of things are compared to something we consider immobile.​

Paragraph I, the first sentence [7]​

Translated text: Absolute exact and mathematical time, by its own nature, without any reference to any external thing, proceeds smoothly.​

What I'm trying to say is that in fiber bundle structure (the fiber here is the three dimensional euclidean affine space $E^3$) we have basically multiple "copies/instances" of the same fiber space $E^3$ over the base space (i.e. $E^1$) but with no identification each other. How we have to understand the product $E^1 \times E^3$ ? As multiple copies/instances of $E^3$ over $E^1$ that however are all identified each other this time ?

To put it in another words, to you $E^1\times E^3$ is a (trivial) example of fiber bundle ?

These statements should be carefully considered.