Galilean spacetime as a fiber bundle

[cianfa72]

TL;DR Summary

Galilean spacetime has been defined as fiber bundle (over absolute time projection).
How to single out physically inertial paths through spacetime

Hi,

reading the book "The Road to Reality" by Roger Penrose I was a bit confused about the notion of Galilean spacetime as fiber bundle (section 17.2).

As explained there, each fiber over absolute time $t$ is a copy of $\mathbf E^3$ (an instance of it over each $t$), there exist no identification between fibers nevertheless the whole bundle (the spacetime) is actually one "thing".

Now, from a physical point of view, I believe the direction of 'inertial motions' can be singled out by zero reading of accelerometers (inertial paths in spacetime are actually those having zero reading of accelerometers following them).

I'm puzzled about how identify them in each fiber (copy of $\mathbf E^3$) without reference to a given inertial reference frame IRF

 

[cianfa72]

I'm puzzled about how identify them in each fiber (copy of $\mathbf E^3$) without reference to a given inertial reference frame IRF

Does really make sense defining them in a frame invariant way ?

 

[robphy]

You might be interested in my comment
https://www.physicsforums.com/threads/affine-space-as-the-spacetime.987440/post-6327867
in the thread about Affine Space.

From http://trautman.fuw.edu.pl/publications/scientific-articles.html ,
this may be of interest:
http://trautman.fuw.edu.pl/publications/Papers-in-pdf/25.pdf
Fibre bundles associated with space-time, Rep. Math. Phys. (Torun) 1, 29–34 (1970)

 

[mma]

Now, from a physical point of view, I believe the direction of 'inertial motions' can be singled out by zero reading of accelerometers (inertial paths in spacetime are actually those having zero reading of accelerometers following them).

I'm puzzled about how identify them in each fiber (copy of $\mathbf E^3$) without reference to a given inertial reference frame IRF

"how identify them in each fiber" - Do I understand that "them" refers here to "the inertial motions"? Then "in each fiber" doesn't make sense. Inertial world lines are sections of the bundle $\mathcal G\to \mathbb E^1$, so they don't run in one fiber, they run across all fibers. As Penrose says, inertial motions are singled out by a connection on $\mathcal G$, such that the inertial motions are the geodesics with respect to this connection. "Connection" means specifying the parallel transport (parallel shift) of tangent vectors of $\mathcal G$. Tangent vectors of $\mathcal G$ are elements of the tangent bundle $T\mathcal G$. (This tangent bundle ($T\mathcal G$) is the bundle that appears under "How is Galilean spacetime represented as a fiber bundle?" in the FAQ above. Not to confuse with the bundle $\mathcal G\to \mathbb E^1$ of Penrose!) "Geodesic" means a curve on $\mathcal G$ so that its tangent vectors are parallel shifts of each other. This is the construction that defines "constant velocity" without reference to any observer. In our model, the equality of two velocities is replaced with being the parallel shifts of each other. Once more again: the trick is that despite the points of different fibers aren't in any connection to each other, the points of the tangent spaces above different points of the base space are identified with each other by the parallel shift defined by a connection. The existence of this connection is the content of Newton's first law in our fiber bundle model.

 

[mma]

I don't see now the FAQ I referred to. Was it deleted, or only hided by some filter?

 

[cianfa72]

Tangent vectors of $\mathcal G$ are elements of the tangent bundle $T\mathcal G$. (This tangent bundle ($T\mathcal G$) is the bundle that appears under "How is Galilean spacetime represented as a fiber bundle?" in the FAQ above. Not to confuse with the bundle $\mathcal G\to \mathbb E^1$ of Penrose!)

Just to be pedantic: an element of $T\mathcal G$ is a tangent vector at a specific point p in $\mathcal G$ since $T\mathcal G$ is defined as the disjoint union of tangent spaces $T_p \mathcal G$ each at different point in $\mathcal G$.

"Geodesic" means a curve on $\mathcal G$ so that its tangent vectors are parallel shifts of each other. This is the construction that defines "constant velocity" without reference to any observer.

In other words a geodesic is a smooth section in $T\mathcal G$ having specific properties (i.e. it is autoparallel according the given connection).

The existence of this connection is the content of Newton's first law in our fiber bundle model.

Yes, recently we had a thread about Newtonian spacetime.

 

[mma]

In other words a geodesic is a smooth section in $T\mathcal G$ having specific properties (i.e. it is autoparallel according the given connection).

I think the name "geodesic" refers to my curve, and your section is a special lift of my curve into the tangent bundle. The specialty of the lift is that $c(t)$ is lifted to $(c(t), \dot c(t))$. In the case of a geodesic, this lift is at the same time, a horizontal lift.

 

[cianfa72]

I think the name "geodesic" refers to my curve, and your section is a special lift of my curve into the tangent bundle. The specialty of the lift is that $c(t)$ is lifted to $(c(t), \dot c(t))$. In the case of a geodesic, this lift is at the same time, a horizontal lift.

Yes sorry, you're right. The geodesic is the autoparallel curve in $\mathcal G$ (the base space), not the (associated) lifted smooth section in $T\mathcal G$.

Btw, what is an horizontal lift ?

 

[mma]

Btw, what is an horizontal lift ?

If a connection is given on a fiber bundle, then a horizontal lift of a curve $c$ running in the base space is a curve $d$ in the bundle which is a lift of $c$ (that is $\pi(d)=c$), and its tangent vector in every point is horizontal. The connection determines in every point of the bundle the horizontal subspace of the tangent space above this point, and a tangent vector is said to be horizontal, if it is in the horizontal subspace.

 

[cianfa72]

The connection determines in every point of the bundle the horizontal subspace of the tangent space above this point, and a tangent vector is said to be horizontal, if it is in the horizontal subspace.

Ah ok, so for instance in the case of cylinder viewed as fiber bundle over the circle $\mathbb S$, the connection given by its embedding in $\mathbb R^3$ determines the horizontal vector subspace at each point. An horizontal lift of $\mathbb S$ is therefore a particular smooth section on the bundle (or the horizontal lift of a curve in $\mathbb S$).

 

[mma]

Ah ok, so for instance in the case of cylinder viewed as fiber bundle over the circle $\mathbb S$, the connection given by its embedding in $\mathbb R^3$ determines the horizontal vector subspace at each point. An horizontal lift of $\mathbb S$ is therefore a particular smooth section on the bundle (or the horizontal lift of a curve in $\mathbb S$).

Almost, except a curve on a smooth manifold is not a set, but it is a mapping from an interval of $\mathbb R$ to the manifold. That is, beyond the image of this mapping, the parametrization also matters. So it is not a section, but a curve of which image is a section.

In your exanple, if the embedding is the surface $H=\{(x,y,z):x^2+y^2=1\}$, and the connection defines the horizontal subspace of the tangent space at each point as its subspace spanned by $\partial_x$ and $\partial_y$, and if we embed $\mathbb S$ into $\mathbb R^2$, then a horizontal lift of the curve $c:\mathbb R\to \mathbb S: t\mapsto (\cos t,\sin t)$ to the point $(x,y,z)$ is the curve $d:\mathbb R\to H: t\mapsto (\cos t,\sin t, z)$.

 

[mma]

I don't see now the FAQ I referred to. Was it deleted, or only hided by some filter?

It seems that FAQ is visible only when I'm not logged in. But this answer at least misses the point if not misleading there.

How is Galilean spacetime represented as a fiber bundle?

Galilean spacetime is represented as a fiber bundle by considering the 4-dimensional space-time continuum as the base space and the set of all possible velocities as the fiber space. This means that for every point in the base space, there is a corresponding set of velocities in the fiber space.

The problem is with this answer that it describes the tangent bundle $T\mathcal G$ of the Galilean spacetime $\mathcal G$, not $\mathcal G$ itself as a fiber bundle, That is, $\mathcal G$ is the base space of $T\mathcal G$, not the total space of the bundle. But it hides the difference between the Galilean and Einsteinian spacetime, namely, that $\mathcal G$ itself is also a fiber bundle (as being its total space) with a one-dimensional base space (the time), in contrast to the Einsteinian spacetime. where this bundle structure (the projection to the time) is missing.

How can I react to it to make the answer more appropriate? Where is this FAQ actually?

 

[cianfa72]

Almost, except a curve on a smooth manifold is not a set, but it is a mapping from an interval of $\mathbb R$ to the manifold. That is, beyond the image of this mapping, the parametrization also matters. So it is not a section, but a curve of which image is a section.

Sorry, maybe I was sloppy. Ok for the definition of a curve $\gamma$ as a smooth mapping from an $\mathbb R$-interval to the manifold $\mathbb S$ (as in post#11 example). Then in order to lift it to the fiber bundle (i.e. to the cylinder) one needs a section $s$ from $\mathbb S$ to the cylinder. Therefore the lift is basically the composition $s \circ \gamma$, I believe.

 

[mma]

Sorry, maybe I was sloppy. Ok for the definition of a curve $\gamma$ as a smooth mapping from an $\mathbb R$-interval to the manifold $\mathbb S$ (as in post#11 example). Then in order to lift it to the fiber bundle (i.e. to the cylinder) one needs a section $s$ from $\mathbb S$ to the cylinder. Therefore the lift is basically the composition $s \circ \gamma$, I believe.

Exactly.

 

[WWGD]

If a connection is given on a fiber bundle, then a horizontal lift of a curve $c$ running in the base space is a curve $d$ in the bundle which is a lift of $c$ (that is $\pi(d)=c$), and its tangent vector in every point is horizontal. The connection determines in every point of the bundle the horizontal subspace of the tangent space above this point, and a tangent vector is said to be horizontal, if it is in the horizontal subspace.

Is this in reference with Ehresman connections?

 

[mma]

Is this in reference with Ehresman connections?

Yes, of course. What else?

 

[WWGD]

Yes, of course. What else?

Just refreshing the material.