5
$\begingroup$

I am reading Roger Penrose's The Road to Reality. In section 17.3, I encounter the following passage. To give a context, Penrose was explaining that even though an Aristotelian spacetime can be defined as a Cartesian product of Euclidean spaces $\mathbb{E}^1\times\mathbb{E}^3$, Galilean spacetime, however, has a bundle structure with $\mathbb{E}^3$ (set of all simultaneous events) being fibers over $\mathbb{E}^1$ (time), and there is no natural identification between the points on one fiber to the next (no absolute space). Then he asks

What kind of structure must we assign to our ‘Galilean spacetime’ $\mathcal{G}$? It would certainly be far too strong to endow our fibre bundle $\mathcal{G}$ with a bundle connection (§15.7).${}^{[17.1]}$ What we must do, instead, is to provide it with something that is in accordance with Newton’s first law. This law states that the motion of a particle, upon which no forces act, must be uniform and in a straight line. This is called an inertial motion. In spacetime terms, the motion (i.e. ‘history’) of any particle, whether in inertial motion or not, is represented by a curve, called the world line of the particle. In fact, in our Galilean spacetime, world lines must always be cross-sections of the Galilean bundle; see §15.3.${}^{[17.2]}$ and Fig. 17.3.) The notion of ‘uniform and in a straight line’, in ordinary spatial terms (an inertial motion), is interpreted simply as ‘straight’, in spacetime terms. Thus, the Galilean bundle $\mathcal{G}$ must have a structure that encodes the notion of ‘straightness’ of world lines. One way of saying this is to assert that $\mathcal{G}$ is an affine space (§14.1) in which the affine structure, when restricted to individual $\mathbb{E}^3$ fibres, agrees with the Euclidean affine structure of each $\mathbb{E}^3$.

I would like to think that affine structure is being invoked because in $4$-dimensions (in $\mathcal{G}$) one can then define straight lines that would then correspond to uniform motion. All it does is makes inertial motion possible. However, I am having difficulty understanding that. Because in the Aristotelian case, there was an identification between the points of the hyperplanes $\mathbb{E}^3$ at one time to the next, that is the spatial points remained at the same location (absolute space) ["We take the attitude that a particular spatial point, at one moment of time, is the same spatial point, at a later moment of time" (sec. 17.1)]. In such a setting it makes sense to talk about the world line of a particle in uniform motion — a straight line with a tilt. But if we do not have a natural identification of points on one fiber to the next how we can still say that straight tilted lines correspond to uniform/inertial motion? If we choose to identify the points on the hyperplanes by means of a family of parabolas, then straight lines would not represent uniform motion — parabolas will?

The passage that follows the previous one goes over my head completely. Penrose continues

Another way is simply to specify the $\infty^6$ family of straight lines that naturally resides in $\mathbb{E}^1\times\mathbb{E}^3$ (the ‘Aristotelian’ uniform motions) and to take these over to provide the ‘straight-line’ structure of the Galilean bundle, while ‘forgetting’ the actual product structure of the Aristotelian spacetime A. (Recall that $\infty^6$ means a 6-dimensional family; see §16.7.)

  • What does this even mean?
  • What are these $6$ parameters?
  • What does it mean to say "straight line structure of Galilean bundle" that too forgetting about the product structure?
  • What does any of these have to do with our standard Galilean transformation between observers we all love and cherish?
  • And finally, how are points on the hyperplanes identified?
$\endgroup$
1
  • 1
    $\begingroup$ don’t have the time to fully elaborate now (maybe I’ll answer later…), but the lines he’s referring to are the straight lines in $\Bbb{R}^4$ which do not lie inside of any of the hyperplanes $\{t\}\times \Bbb{R}^3$; these are characterized by 6-parameters. By transport of structure, one can provide a formal definition. But note that this is extra structure. You have to impose it by-hand. Galilean transformations are essentially the “structure preserving maps” (preserve the bundle structure, the fiber-wise distance, lines, …). $\endgroup$
    – peek-a-boo
    Jun 27 at 5:42

2 Answers 2

5
$\begingroup$

As you can see in his very first rhetoric question “What kind of structure must we provide our Galilean spacetime $\mathcal{G}$?” he is already telling you that you need extra structure on top of merely having a fiber bundle (where the fibers are Euclidean spaces, and the base is a 1-dimensional Euclidean space). So, if all you have is a fiber bundle, then you can’t make sense of anything he’s talking about. In those few pages, Penrose gives you 3 different outlines for “implementing”, in mathematical fashion, this additional structure, called straightness, that conveys the the physical idea we have about inertial motion when talking about a Galilean spacetime/universe:

  • the first proposal is that rather than merely requiring the total space of your fiber bundle to be a topological space (homeomorphic to $\Bbb{R}^4$), it must in addition carry the structure of an affine space. Notice that the imposition of an affine structure is extra information. That’s what allows him to talk about straight lines in the total space. Of course, there’s some compatibility condition of how this affine structure must interact with the bundle structure. Notice that if he did not specify an affine structure on the total space, then you cannot talk about straight lines. To reiterate: you impose, by hand, an affine structure on the total space which is ‘compatible’ with that of the fibers.

This should hopefully address your second paragraph’s concerns. For your remaining questions, the answer is already essentially what I said in the comments, but I’ll elaborate. Note that everything I’m about to say is simply an exercise in formal mathematical writing; particularly, in expressing what ‘structure’ we have, and what ‘structure-preserving morphisms’ ought to look like, etc.


Galilean Spacetimes: Penrose’s second approach.

This is mainly going to be definitions

Definition. (Almost-Galilean spacetime)

An $(n+1)$-dimensional almost-Galilean spacetime, $\mathcal{G}$, (a made-up term btw) consists of the following tuple of information: \begin{align} \mathcal{G}=\bigg((\Bbb{E}^1,\text{Or}), \Bbb{E}^n, (\Gamma,\gamma,\Bbb{E}^1),\mathcal{I}\bigg), \end{align} where

  • $(\Bbb{E}^1,\text{Or})$ is an affine space whose underlying vector space is real 1-dimensional inner product space, and carries an orientation.
  • $\Bbb{E}^n$ refers to an affine space whose underlying vector space is a real $n$-dimensional inner product space.
  • $(\Gamma,\gamma,\Bbb{E}^1)$ is a smooth fiber bundle with total space $\Gamma$ that is homeomorphic to $\Bbb{R}^{1+n}$, projection $\gamma$, base space $\Bbb{E}^1$, with typical fiber $\Bbb{E}^n$ ( with metrics and orientations as described in the previous two bullet points).
  • $\mathcal{I}$ is a collection of smooth sections of the fiber-bundle $(\Gamma,\gamma,\Bbb{E}^1)$. We shall refer to elements of $\mathcal{I}$ as the inertial wordlines.

A definition is a definition, and I can make up whatever (logical) definition I like, but let me briefly elaborate on why I defined things this way. Firstly, in our intuitive notion of time, we think of it as one dimensional, and we ought to be able to say how much time passed by between two events so we need the structure of an affine space, and it has to have a distance function (which we take to be Euclidean). Also, we would like to distinguish future and past, so we require an orientation. Next, our models for space $\Bbb{E}^n$ having Euclidean geometry is something we expect; the new feature of Galilean relativity over ‘Aristotlean’ is that this distance is only defined within fibers of the fiber bundle (i.e we can only speak of distance for simultaneous events). Next, we come to the most important part from the notion of dynamics, the $\mathcal{I}$. Note that physically, a worldline means in (pre-Einstein) terms, a curve in the total space $\Gamma$ (i.e the spacetime), which tells us all the events we’ve been at/will ever be at. So at any given time, you must have been at some event corresponding to that time (and you can’t be at two different events at the same time), so this is exactly what a section of a fiber bundle is trying to tell us. Now, it is a definition that we have to make as to which sections we shall call inertial. Some choices of may give us something physically intuitive/reasonable, others not. For example, if I chose $\mathcal{I}$ to be a singleton consisting of only one section, then it means I’m trying to convey that in this model for Galilean spacetime, there is only one-inertial motion, which of course isn’t what we want physically want.

Finally, the reason why I used the adjective “almost” in “almost-Galilean spacetime” is because I have not given you any conditions on what $\mathcal{I}$ ought to be. A full definition must connect this mathematical model with physics (ultimately experiments) so it must somehow recover our usual notion of inertial motion. We’ll get there soon, but first we follow our mathematical brains :) So, whenever we have a definition of a type of object in math (here an almost-Galilean spacetime of dimension $(n+1)$), we should mention what are the maps between such objects:

Definition. (Morphisms of almost-Galilean spacetimes)

Let $\mathcal{G}_1,\mathcal{G}_2$ be two almost-Galilean spacetimes (not necessarily same dimensions). A morphism from $\mathcal{G}_1$ into $\mathcal{G}_2$ consists of a morphism of the underlying fiber bundles which is an orientation-preserving isometry on the base, a fiberwise isometry, and maps intertial worldlines in $\mathcal{G}_1$ into inertial worldlines in $\mathcal{G}_2$. In more detail, it is a pair of smooth maps $(\Phi,f)$ such that

  • $\Phi:\Gamma_1\to\Gamma_2$ and $f:\Bbb{E}^1\to\Bbb{E}^1$ are such that $\gamma_2\circ\Phi=f\circ\gamma_1$, i.e the following diagram commutes: $\require{AMScd}$ \begin{CD} \Gamma_1 @>{\Phi}>> \Gamma_2 \\ @V{\gamma_1}VV @VV{\gamma_2}V \\ \Bbb{E}^1 @>>{f}> \Bbb{E}^1. \end{CD} Actually, since $\Phi$ determines $f$, it is not uncommon to refer to simply $\Phi$ instead of the pair $(\Phi,f)$.

  • $f:\Bbb{E}^1\to\Bbb{E}^1$ is an orientation-preserving isometry (which implies it is of the form $f(t)=t-t_0$ for some $t_0\in \Bbb{E}^1$).

  • for each $t\in\Bbb{E}^1$, $\Phi$ must restrict to a distance-preserving map of the fiber $(\Gamma_1)_t$ into $(\Gamma_2)_{f(t)}$.

  • for each inertial worldline $l\in\mathcal{I}_1$, the push-forward $\Phi_*l:=\Phi\circ l\circ f^{-1}$ (which is clearly a section of the fiber bundle $(\Gamma_2,\gamma_2,\Bbb{E}^1)$) must lie in $\mathcal{I}_2$.

We denote the set of all such morphisms by $\text{Mor}(\mathcal{G}_1,\mathcal{G}_2)$. In the case $\mathcal{G}_1=\mathcal{G}_2=\mathcal{G}$, we shall simply write $\text{Mor}(\mathcal{G})$.

Given three almost-Galilean spacetimes $\mathcal{G}_1,\mathcal{G}_2,\mathcal{G}_3$, and two morphisms $(\Phi,f)$ and $(\Psi,g)$ from $\mathcal{G}_1\to\mathcal{G}_2$ and $\mathcal{G}_2\to\mathcal{G}_3$ respectively, we define the composition of morphisms to mean the obvious thing $(\Psi\circ\Phi,g\circ f)$, and you can easily verify it satisfies the conditions for being a morphism $\mathcal{G}_1\to\mathcal{G_3}$. We thus have a composition operator $\circ:\text{Mor}(\mathcal{G}_2,\mathcal{G}_3)\times \text{Mor}(\mathcal{G}_1,\mathcal{G}_2)\to\text{Mor}(\mathcal{G}_1,\mathcal{G}_3)$.

An isomorphism from an almost-Galilean spacetime $\mathcal{G}_1$ to $\mathcal{G}_2$ means a morphism which has an inverse which is also a morphism (i.e a $(\Phi,f)$ which is a morphism such that $(\Phi^{-1},f^{-1})$ exists and is a morphism). We denote the set of all isomorphisms $\mathcal{G}_1\to\mathcal{G}_2$ as $\text{Iso}(\mathcal{G}_1,\mathcal{G}_2)$. Finally, in the case $\mathcal{G}_1=\mathcal{G}_2=\mathcal{G}$, we refer to the isomorphisms as automorphisms, and we write $\text{Aut}(\mathcal{G}):=\text{Iso}(\mathcal{G},\mathcal{G})$, and you can easily check that this forms a group under composition. We shall call $\text{Aut}(\mathcal{G})$ the Galilean group of $\mathcal{G}$.

A few remarks here (pretty obvious once you do a bit of math).

  • Notice that almost-Galilean spacetime has a few structures: time, spatial distances, fiber bundle structure, and the inertial structure. So, a morphism (i.e a map which respects the structure) has to respect these four things, hence the four bullet points above.

  • Next, I want to emphasize that the more structure we add, the harder it is for something to be an isomorphism, so if you merely have an isomorphism $(\Phi,f)$ at the level of fiber bundles, then this is not guaranteed to be an isomorphism in the sense of almost-Galilean spacetimes. For example, an arbitrary fiber-bundle isomorphism can have very crazy $f$, but as I mentioned above, time-interval and orientation preserving already force $f$ to be very simply $f(t)=t-t_0$. Another remark: I could have $(\Phi,f)$ which is a morphism $\mathcal{G}_1\to\mathcal{G}_2$, and $(\Phi,f)$ an isomorphism in the sense of fiber bundles, but yet it can still fail to be an isomorphism in the sense of almost-Galilean spacetime. For example, suppose the underlying fiber bundles are the same, so we have the identity fiber-bundle isomorphism $\text{id}$. If I choose $\mathcal{I}_1$ to only consist of one section, but $\mathcal{I}_2$ to consist of all sections, then clearly $\text{id}$ in the forward direction is a morphism, but the inverse fiber-bundle map is not a morphism of almost-Galilean spacetimes (kind of like in topology, the identity map can fail to be a homeomorphism if you choose different topologies on the same space).

  • Some obvious group theory here: the Galilean group $\text{Aut}(\mathcal{G})$ measures the amount of “non-uniqueness in isomorphisms” in the sense that if $\Phi,\Psi\in\text{Iso}(\mathcal{G}_1,\mathcal{G}_2)$ (I omit the base maps for simplicity in notation), then there are unique $\Theta_1\in\text{Aut}(\mathcal{G}_1)$ and $\Theta_2\in\text{Aut}(\mathcal{G}_2)$ such that $\Psi=\Phi\circ\Theta_1=\Theta_2\circ\Phi$ (namely $\Theta_1=\Phi^{-1}\circ\Psi$ and $\Theta_2=\Psi\circ\Phi^{-1}$).

We now come to the simplest and most important example. We consider the trivial fiber bundle $(\Bbb{E}^1\times\Bbb{E}^n,\text{pr}_1,\Bbb{E}^1)$ with projection to the first factor, where we have all the usual Euclidean metrics on the base and on each fiber. Then, for each choice of a collection of sections $\mathcal{I}$ of this fiber bundle, we obtain a corresponding almost Galilean spacetime, $\mathcal{G}_{\text{trivial},\mathcal{I}}$. We shall now finally eliminate the adjective almost, and draw some connections with our usual physical interpretation of what inertial worldlines ought to look like.

For each $\tau\in\Bbb{E}^1,p\in\Bbb{E}^n,v\in\Bbb{R}^n$ (the underlying vector space of the affine space), let us define a section $l_{\tau,p,v}:\Bbb{E}^1\to\Bbb{E}^n$ as $l_{\tau,p,v}(t):=(t,p+ (t-\tau)v)$. Note that if you sketch this, it is exactly a “slanted line in $\Bbb{E}^1\times\Bbb{E}^n$”. Now, we consider the collection of all such sections, i.e $\mathcal{I}_{\text{physical}}=\{l_{\tau,p,v}\,:\tau\in\Bbb{E}^1,p\in\Bbb{E}^n,v\in\Bbb{R}^n\}$. By counting parameters, it may seem like there is a total of $1+n+n=2n+1$ of them, but in fact the $\tau$ parameter is redundant, because if I change $\tau$ to some $\tau’$ it will change $p$ to $p+(\tau’-\tau)v$ and keep $v$ as it is, i.e $l_{\tau,p,v}=l_{\tau’,p+(\tau’-\tau)v,v}$. So really, this collection depends only on $2n$ parameters (if $n=3$, this is the 6-parameters Penrose mentions). Note that another way of describing the collection $\mathcal{I}_{\text{physical}}$ is that it is the set of sections $l$ for which there exists a $v\in\Bbb{R}^n$ such that for all $t_1,t_2\in\Bbb{E}^1$, $l(t_1)-l(t_2)=(t_1-t_2,(t_1-t_2)v)$, i.e you’re lying on a straight line in $\Bbb{E}^1\times\Bbb{E}^n$ (with its naturally induced affine structure from each of the factors) which has a direction vector $(1,v)\in\Bbb{R}\times\Bbb{R}^n$ (i.e you “move uniformly in time and space”). The reason I didn’t want to say it this way originally is because you seemed a little uneasy with introducing the affine structure, so I gave a definition which does not in any way use an affine structure on the total space $\Bbb{E}^1\times\Bbb{E}^n$ (but clearly these are describing the same collection of sections).

Let us adopt the notation $\mathcal{G}_{\text{physical},n+1}$ to mean this trivial fiber bundle with this specific choice $\mathcal{I}_{\text{physical}}$ of physically-intuitive inertial worldlines. This is just a definition that I make, and the reason I make it is because it conforms to our (pre-Einstein) view of the world (in a given ‘inertial frame of reference’). Finally, we come to the definition, getting rid of “almost”:

Definition. (Galilean spacetime).

By a Galilean spacetime of dimension $n+1$, we shall mean an almost-Galilean spacetime $\mathcal{G}$ of dimension $n+1$ which is isomorphic to the just-introduced $\mathcal{G}_{\text{physical},n+1}$. A choice of an isomorphism $\mathcal{G}\to\mathcal{G}_{\text{physical}, n+1}$ is called a Galilean frame of reference (or an inertial frame of reference in the sense of Galileo).

So, a Galilean spacetime is one that after “stripping away the disguise” (i.e isomorphism) looks exactly like we expect. The reason we got rid of “almost” is because we have now singled out what corresponds physically to inertial worldlines. Note that even in Penrose’s third description, he doesn’t just say “Galilean spacetime is a fiber bundle with a connection on the total space”. He goes further to specify that the connection has to be torsion-free and have zero curvature; the extra conditions on the connection affect what geodesics (inertial motion) can look like, so he’s trying to connect the math to the physics here. It’s the same thing with the $\mathcal{I}_{\text{physical}}$ above.

Just so we don’t get lost in all the formal mathematical writing (which tbh becomes second nature with some practice) consider the following quote from Landau-Lifshitz:

“It is found, however, a frame of reference can always be chosen in which space is homogeneous and isotropic and time is homogeneous. This is called an inertial frame. In particular, in such a frame a free body which is at rest at some instant remains always at rest.”

So, we see that our definition perfectly embodies what is said here as well.

Coming to your bullet points, we’ve taken care of the first and second. For the third, hopefully it’s clear: it just means we look at a $\mathcal{G}$ which is isomorphic to the standard $\mathcal{G}_{\text{physical},n+1}$, which has a trivial underlying fiber bundle. I guess if we wanted to be more formal, one could define Galilean spacetime to be the isomorphism class of $\mathcal{G}_{\text{physical},n+1}$, but let’s not worry about it. Lastly, what do these have to do with the Galilean transformations we know and love? Well, that’s precisely the automorphism group (which as I mentioned above measures the non-uniqueness in isomorphisms), so it tells us how to go from one Galilean frame of reference to another.


Calculating the Galilean group $\text{Aut}(\mathcal{G}_{\text{physical},n+1})$.

Let us now calculate the automorphism group of the Galilean spacetime $\mathcal{G}_{\text{physical},n+1}$. Let $(\Phi,f)$ be any automorphism. Purely for ease of notation, let us fix an origin $\tau_0\in\Bbb{E}^1$ and $x_0\in\Bbb{E}^n$, so that after making this choice, we can simply work with $\Bbb{R}$ and $\Bbb{R}^n$ (if you don’t like the notational change to $\Bbb{R}$ and $\Bbb{R}^n$, then I invite you to repeat everything below and carry around the $\tau_0$ and $x_0$).

  • As I mentioned above, since $f:\Bbb{R}\to\Bbb{R}$ is an orientation-preserving isometry, there exists $t_0\in\Bbb{R}$ such that for all $t\in\Bbb{R}$, $f(t)=t-t_0$.
  • Since $\Phi$ is a fiberwise isometry, a standard geometry theorem tells us that on each fiber, it is a composition of a translation and an orthogonal map.

Hence, for each $t\in\Bbb{R}$, there exist points $p_t\in\Bbb{R}^n$ and orthogonal linear map $R_t\in\text{O}(n,\Bbb{R})$, such that for all $(t,x)\in\Bbb{R}\times\Bbb{R}^n$, \begin{align} \Phi(t,x)&=(t-t_0,p_t + R_t(x)). \end{align} So far we’ve only used fiberwise information, which is why the $c_t$ and $R_t$ depend on $t$, a-priori in a very complicated fashion. But, by now using the fact that it maps inertial worldlines to inertial worldlines, we can now simplify greatly.

  • First, consider the inertial wordline $l=l_{0,0,0}$ (the first $0$ is the origin in time, the next is the origin in space, and last is the $0$ ‘boost velocity’ $v$), i.e $l(t)=(t,0)$, i.e simply the “straight vertical line through the origin in space”. Since $\Phi$ preserves inertial worldlines, there must exist an inertial worldline $l_{0,b_0,v_0}$ such that $\Phi_*l=l_{0,b_0,v_0}$, or composing by $f$ on both sides, $\Phi\circ l=l_{0,b_0,v_0}\circ f$. Writing this out in full, for all $t\in \Bbb{E}^1$, \begin{align} \left(t-t_0, p_t+0\right)=(t-t_0, b_0+(t-t_0)v_0), \end{align} where I used the fact that $R_t(0)=0$. So, we have figured out that $p_t=b_0+(t-t_0)v_0$, which is a nice simple form.

So, $\Phi(t,x)=\left(t-t_0, b_0+(t-t_0)v_0 + R_t(x-x_0)\right)$.

  • Next, consider the inertial worldlines $l=l_{0,p,0}$, i.e $l(t)=(t,p)$, which is a vertical line through an arbitrary point $p$ in space, not necessarily our choice of origin. Then, there exists some $l_{0,b_1,v_1}$ such that $\Phi\circ l=l_{0,b_1,v_1}\circ f$, i.e for all $t\in\Bbb{E}^1$, \begin{align} \left(t-t_0,b_0+(t-t_0)v_0 + R_t(p)\right)&=\left(t-t_0, b_1+(t-t_0)v_1\right) \end{align} First, plug in $t=t_0$ to get $b_0+R_{t_0}(p)=b_1$, and next plug in $t=t_0+1$ to get $b_0+v_0+R_{t_0+1}(p)=b_1+v_1$. With these two equations, we can rearrange to figure out that for all $t$, \begin{align} R_t(p)&=(b_1-b_0)+(t-t_0)(v_1-v_0)=R_{t_0}(p)+(t-t_0)[R_{t_0+1}(p)-R_{t_0}(p)]. \end{align}

Since $p$ was arbitrary here, we now find that $\Phi$ can be written as \begin{align} \Phi(t,x)&=\bigg(t-t_0,b_0+(t-t_0)v_0+ R_{t_0}(x)+(t-t_0)[R_{t_0+1}(x)-R_{t_0}(x)]\bigg).\tag{$*$} \end{align}

  • Finally, we consider the third main type of inertial worldlines, $l=l_{0,0,v}$, i.e $l(t)=(t,tv)$. Once again, there must exist some $l_{0,b_2,v_2}$ such that $\Phi\circ l=l_{0,b_2,v_2}\circ f$, i.e for all $t\in\Bbb{R}$, \begin{align} \bigg(t-t_0,b_0+(t-t_0)v_0+ tR_{t_0}(v)+t(t-t_0)[R_{t_0+1}(v)-R_{t_0}(v)]\bigg)=(t-t_0, b_2+(t-t_0)v_2). \end{align} Note I used linearity of $R$ to pull out $R(tv)=tR(v)$. Look at the second components, notice we have a quadratic on the LHS, and a degree 1 function of $t$ on the RHS. So, the coefficient of the quadratic term must vanish (or if you want, differentiate both sides twice), i.e $R_{t_0+1}(v)-R_{t_0}(v)=0$. We derived this for arbitrary $v$, therefore, our expression $(*)$ simplifies to $\Phi(t,x)=(t-t_0,b_0+(t-t_0)v_0+R_{t_0}(x))$.

Notice that modulo some notation, this is exactly what the usual Galilean transformations look like. Therefore, every element of our automorphism group $\text{Aut}(\mathcal{G}_{\text{physical},n+1})$ gives rise to a Galilean transformation in the usual sense, and the converse is very easily verified. And it’s pretty easy to see that the mapping $\text{Aut}(\mathcal{G}_{\text{physical},n+1})\to \text{Galilean}_{n+1}$ between our automorphism group and the usual Galilean group is a group isomorphism

Thus, the Galilean group of $\mathcal{G}_{\text{physical},n+1}$, defined here as its automorphism group, coincides exactly with the usual Galilean group as given in Wikipedia. For completeness, let’s count its dimension as a Lie group (since we have a group isomorphism, we can inherit the Lie group structure as well): it is $1+n+n+\frac{n(n-1)}{2}=\frac{(n+1)(n+2)}{2}$. We have $1$ time-translational freedom, $n$ spatial translations, $n$ boost velocities, and $\frac{n(n-1)}{2}$ rotations. So, in the case $n=3$, we get a $\frac{(3+1)(3+2)}{2}=10$-dimensional Lie group, as expected.


Summary, and further remarks.

I should emphasize that the most difficult part of all of this is actually identifying what physical structures we would like in our mathematical model. From here, it’s just “standard mathematics”, in terms of writing definitions, coming up with definitions. Then, quantify the uniqueness/non-uniqueness of certain things (e.g our automorphism groups). Once we have all of this, we just crunch away at our definitions, and out pops the beloved Galilean group :) By the way, you may also want to take a look at Aristotelian vs Galilean relativity in terms of bundles; it’s written at a lower level of detail, and perhaps it might address some issues.

If you now want to talk about ‘Newtonian spacetime’, with Newtonian gravity, then you throw away $\mathcal{I}$ here, and you replace it with a torsion-free connection (which has curvature). That will give rise to a corresponding set $\mathcal{I}$ of sections, the “inertial worldlines in Newtonian gravity”. If you go to SR, then you throw away the bundle structure entirely. If you go to GR, then you make the Lorentzian metric itself dynamical and so on.

$\endgroup$
10
  • $\begingroup$ +1 thanks for the great answer. I have read it once but can't say that I have been able to understand everything. I'll probably read this a couple of more times and ask a few questions before accepting the answer. In the meantime do you know of any references for this type of rigorous mathematical treatment of spacetime structure starting from Galilean relativity up to special relativity? (For GR there seem to be many books that are expressly mathematical like Hawking and Ellis etc. But before I start with those books I feel I must understand pre-GR relativity in mathematical details as well) $\endgroup$ Jun 28 at 15:51
  • $\begingroup$ "We consider the trivial fiber bundle $(\mathbb{E}^1\times \mathbb{E}^n,{\rm pr}_1,\mathbb{E}^1)$" Why though? Didn't we argue that spacetime does not have a Cartesian product structure that lets one identify the same spatial points at different times like $(t_1,\boldsymbol{x})$ and $(t_2,\boldsymbol{x})$ (like what happens in absolute space)? We introduced a bundle structure to avoid this, right? I don't understand this. Maybe I'm missing something. $\endgroup$ Jun 28 at 17:32
  • $\begingroup$ @FaberBosch In the few paragraphs afterwards I tried to explain that. I mean it’s because firstly we can and it makes the presentation simple. Notice that I never said spacetime is equal to that trivial fiber bundle. In the definition I gave, it is only isomorphic (in the almost-Galilean sense). The fact that there must exist an isomorphism which trivializes the bundle and maps the inertial worldlines as we expect is also what one encounters in the physics presentations (see eg the L&L quote (which I’m sure you’ve already read)). $\endgroup$
    – peek-a-boo
    Jun 28 at 18:25
  • 1
    $\begingroup$ oh and regarding references for pre-GR stuff in such a description, unfortunately I don’t know any. My understanding of the spacetime structures, as presented here, comes essentially from reading that chapter of Penrose’s book (and his corresponding article Structure of Space-Time, which is page 121 of Battelle Recontres, 1967 Lectures in Mathematics and Physics). $\endgroup$
    – peek-a-boo
    Jun 28 at 18:55
  • 1
    $\begingroup$ The rest of the details I gave are essentially just math-tidbits I’ve picked up while studying math (once everything is laid out, it’s the same story: define an object, define what it means to map between such types of objects, and study isomorphisms between said objects, and to go further, see how this object interacts with other types of objects, and finally, apply this to a problem you have in mind (Physics in this case)). Just think about how the usual presentations for group theory, or vector space theory, or ring theory or field theory (in algebra, not physics) go. It’s the same story. $\endgroup$
    – peek-a-boo
    Jun 28 at 18:58
2
$\begingroup$

Some possibly useful comments.

  • "What are these 6 parameters?"
    (See my version of Fig. 17.3 below.)
    A [timelike] line can be determined by two events, A and B, where A is on the 3-d fiber over $t_1$ and B is on the fiber over $t_2$. Thus, a [timelike] line is determined by 3+3=6 parameters.
    The "set of all timelike lines" has dimensionality $\infty^6$.
    (In the (1+1)-case, since the fibers are 1-dimensional, a timelike line is determined by 1+1=2 parameters.)

  • "Affine space"
    From Ch 14.1, an affine space "is a space in which there is a consistent notion of parallelogram." From Ch 14.2 and 14.3 and 14.4, the parallelism feature of an affine space is associated with a connection whose associated curvature is zero.
    The geodesics associated with this connection are associated with the "set of all timelike lines" above.

  • "What does any of these have to do with our standard Galilean transformation between observers we all love and cherish?"
    Among the timelike lines in my version of Fig. 17.3, there is one line that corresponds to the base-space, which can be thought of as the worldline of the inertial observer who drew this Galilean spacetime diagram.
    A Galilean transformation can slide each 3-d fiber (modeling a hyperplane of simultaneity) so that another line can be associated with the base-space, while mapping each line to another line and each parallelogram to another parallelogram of equal area. Pictorially, the transformation is like beveling a deck of playing-cards to make the candidate line vertical, then sliding the beveled-deck so that this now-vertical candidate line passes through the spatial-origin of the spacetime diagram.

robphy-Penrose17.3var

$\endgroup$
1
  • 1
    $\begingroup$ minor remark: “The geodesics associated with this connection are associated with the "set of all timelike lines" above.” I would specify ‘the timelike geodesics…’ $\endgroup$
    – peek-a-boo
    Jun 29 at 2:10

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.