I have some doubts about the way frames of references are introducted in Arnold's mathematical methods of classical mechanics.

It is said that, given a set $M$, then $\phi_1:M \rightarrow \mathbb{R} \times \mathbb{R}^3$ is a bijection from $M$ to the space of coordinates $\mathbb{R} \times \mathbb{R}^3$, and the map $\phi_1$ is a galileian coordinate system. $\phi_2$ is said to be moving uniformusly respect $\phi_1$ if the map $\phi_1 \circ \phi_2^{-1}$ is a trasformation of the galilean group.

I think I see the point of how the problem it is posed, if the "chart transition map" (using manifolds terminology) is a translation or a galileian boost... then the cooridnates change accordingly and I can say that the two frames are moving uniformously one respect the other. But at the same time I'd like to separate the argument of frames of reference from that of coordinates, as the last is an additional strcuture added to the frame, but I guess isn't strictly necessarly to define the relative motion of the frames. So the question is how indeed this can be done formally, how to define the relative motions of frames of refernce (and so idenitfy the class of inertial frames) independently by the coordinate system.

I think in Arnold's book such a description is given because the action of the galilean group is given explicitely only as a representation of the group on $\mathbb{R} \times \mathbb{R}^3$, so I think maybe what I am looking for is a representation of the group on the affine space itself $\mathbb{A^4}$.

there is an additional thing I can't understand which in Arnold's book follows immediately. The fact that $\phi_1$ and $\phi_2$ defines on $M$ a unique galileian structure ( the galileaian structure is the time map $t : \mathbb{A^4} \times \mathbb{A^4} \rightarrow \mathbb{R}$ and a euclidean distance $\rho : \mathbb{A^3}\times \mathbb{A^3} \rightarrow \mathbb{R}$ on the spaces of contemporary events).

  • $\begingroup$ The first chapter of Mathematical Aspects of Quantum Field Theory (the chapter on Newtonian mechanics, viewable on Google Books) gives a treatment where the Galilean structure on spacetime is defined first, and a Galilean transformation is then an affine transformation preserving the Galilean structure. You could then probably say that two frames move uniformly wrt each other if $\phi_1^{-1}\circ\phi_2$ (a map $M\to M$, in contrast to $\phi_1\circ\phi_2^{-1}$, which is $\mathbb R^4\to\mathbb R^4$) is a Galilean transformation on spacetime. $\endgroup$ – Vercassivelaunos Jan 3 at 23:43
  • $\begingroup$ @Vercassivelaunos thanks a lot for the reference $\endgroup$ – asdru Jan 4 at 11:04

You are using a too implicit formalisation where the physical content is not evident.

In classical physics the spacetime is made of the following structures

(a) a smooth four-dimensional manifold $M$, the spacetime of classical physics,

(b) a surjective non-singular smooth map $T : M \to \mathbb{R}$ called absolute time. $T$ is defined up to additive constants,

(c) a structure of $3$-dimensional Euclidean space (= affine space + positive scalar product in the space of translations) on each embedded submanifold $$\Sigma_t := \{ p \in M \:|\: T(p) =t\}$$ called absolute space at time $t$. It is finally required that the said structure on $\Sigma_t$ are compatible with the structure of manifold indiced by $M$. (In practice orthonormal Cartesian coordinates must belong to the atlas of $\Sigma_t$ induced from the atlas of $M$.)

A reference frame $R$ is a pair $(E_R, \pi_R)$, where

(1) $E_R$ is a $3$-dimensional Euclidean space called the rest space of $R$,

(2) $\pi_R : M \to E_R$ is a smooth map satisfying rhe following requirements

(2i) It is surjective (Every event has a place in the rest space of $R$.)

(2ii) $\pi_R|_{\Sigma_t} : \Sigma_t \to E_R$ is an isometry for every $t\in \mathbb{R}$. (The metrical properties of bodies are absolute, as they do not depend on the choice of $R$.)

It is easy to prove that the map

$$M \ni p \mapsto (T(p), \pi_R(p))$$ is smooth and bijective. This map identitfies $M$ with $\mathbb{R} \times E_R$ and there are infinitely many such identifications depending on the reference frame one uses.

In this sense, it is strongly wrong saying that the spacetime is the Cartesian product $\mathbb{R} \times E^3$, as this decomposition depends on the choice of a reference frame. The existence of different kinematics (the fact that the motion is relative) is the same as this fact.

If $\gamma : I \mapsto M$ is a regular curve such that it meets every $\Sigma_t$ once, i.e., $T(\gamma(t)) = t+c_\gamma$ for some constant depending on $\gamma$, we say that $\gamma$ is a worldline. It describes the history of a particle.

The velocity of $\gamma$ with respect to $R$ at time $t$ is a notion defined in the rest space of $R$:

$${\bf v}_\gamma(t)|_R := \lim_{h\to 0} \frac{1}{h}\left(\pi_R(\gamma(t+h)) - \pi_R(\gamma(t))\right)$$

However, this vector can be exported in $\Sigma_t$ using the inverse isomorphism $(\pi_R|_{\Sigma_t})^{-1}: E_R \to \Sigma_t$. There, velocities of the same $\gamma$ computed with respect to different reference frame can be compared.

I stress that up to now I did not use coordinates. The notion of reference frame and relative kinematical notions are independent from the choice of coordinates.

Now we pass to introduce some natural coordinate systems useful in explicit computations. These coordinates are natural because they describe isometries in the rest Eucliedan spaces and along the temporal axis.

Consider an orthonormal Cartesian system of coordinates $x^1,x^2,x^3$ in the rest space $E_R$, with origin $O\in E_R$ and exes $e_1,e_2e_3$. Take a constant $c\in \mathbb{R}$ and define $t(p) := T(p)+ c$.

The composed map

$$M \ni p \mapsto (t(p),x^1(\pi_R(p)), x^2(\pi_R(p)),x^3(\pi_R(p))) \in \mathbb{R}^4$$ is a coordinate system in $M$. (It is possible to prove that it is compatible with the initial differentiable structure of $M$ in view of the properties of the introduced geometrical objects).

What is the relation between that coordinate system on $M$ and another one costructed referring to a different reference frame $R'$ $$M \ni p \mapsto (t'(p),x'^1(\pi_R(p)), x'^2(\pi_R(p)),x'^3(\pi_R(p))) \in \mathbb{R}^4 \quad ?$$

The answer is easy. First of all, for some real constant $k$ it must be $$t'(p) = t(p) + k$$ since the temporal coordinate is nothing but the absolute time up to additive constants.

Secondly, since $$(\pi_{R'}|_{\Sigma_t}) \circ (\pi_R|_{\Sigma_t})^{-1}: E_{R} \to E_{R'}$$ is an isometry of $3$-dimensional Euclidean spaces when $t$ is fixed (composition of isometries) in ortonormal coordinates it must take the standard form of isometries: $$x'^k =b^k(t)+ \sum_{j=1}^r {R(t)^k}_j x^j$$ for some differentiable maps $b^j=b^j(t)$ and $R=R(t) \in O(3)$.

In summary, the general transformation of coordinates beteen coordinates at rest with reference frames in the classical spacetime is

$$t'= t+ c\:, \quad x'^k =b^k(t)+ \sum_{j=1}^r {R(t)^k}_j x^j\:. \tag{0}$$

Now let us consider the so-called relative uniform motion of a pair of reference frames $R$ and $R'$. It means that: a point at rest in $R$ is seen at velocity in $R'$ which (a) is constant in time and (b) this velocity does not depend on the position of the point (in $R$)

I assume $k=1$ by simply changing the origin of the clocks, the constant $k$ can be restored at the end of computations (taking derivatives in $t$ or $t'$ is howvere the same thing as $t'=t+k$).

Let $P$ be a geometrical point at rest in $R$ with spatial coordinates $x^1,x^2,x^3$ there. In $R'$ its evolution is:

$$x'^k(t) = b^k(t) + \sum_{j=1}^3 R^k_j(t) x^j\:.$$

Let us place $P$ at the origin of coordinates in $E_R$. We therefore have $$x'^k(t) = b^k(t)\:.$$ Since the velocity must be constant in time, it has to be $b^k(t)= v^k t + c^k$ for some constants $v^k$ and $c^k$.

Hence $$x'^k(t) = v^k t + c^k + \sum_{j=1}^3 R^k_j(t) x^j\:.\tag{1}$$ Now we impose the other constraint: the velocity in $R'$ cannot depend on the position of the chosen point at rest in $E_R$.

From (1), the velocity of the said point in $R'$ has components: $$v'^k = v^k + \sum_j\frac{dR^k_j(t)}{dt} x^j\:.$$ Independence of $x^r$ implies $$0 = \frac{\partial v'^k}{\partial x^r} = \sum_j \frac{dR^k_j(t)}{dt}\delta^j_r,$$ that is $$\frac{dR^k_r(t)}{dt}=0 \quad r=1,2,3\:.$$ In summary, the motion of $R$ and $R'$ is uniform if and only if (the reasonig above is easily reversible) $$t'=t+c, \qquad x'^k = c^k + tv^k + \sum_{j=1}^3 {R^k}_j x^j\:,\tag{2}$$ for any choice of Cartesian orthonormal systems of coordinates at rest with the two reference frames.

With that formula, it is not difficult to prove that to be in uniform motion is an equivalence relation. In other words, the above class of trasformations form a well known group named the Galileian group when $c, c^k, v^k \in \mathbb{R}$ and $R\in O(3)$.

Let us come to the affine structure of $M$ and its dynamical nature.

When introducing inertial reference frames, one sees that the Galileian group is exactly the class of transformations beteween inertial reference frames. I stress that it is a dynamical fact, relying on the formulation of dynamics due to Newton.

Notice that the transformations (2) are affine trasformations from $\mathbb{R}^4$ to $\mathbb{R}^4$, contrarily to the transformation in (0) (due to the non-affine dependence on $t$ of some term in the right-hand side).

It is easy to see that there is a unique (up to isomorphisms) affine structure in $M$ that admits the transformations (2) in the class of affine transformations.

Hence dynamics select the affine stucture. It is interesting to notice that an affine space has a preferred affine connection, so prefereed geodesics. Indeed the worldlines which are geodesics with respect to this strucutre describe exactly the intertial motion of free bodies. To this respect classical physics and general relativity are not so far to each other. The crucial difference is that in GR the connection is also metric. In classical physics there is no such a common metric structure encompassing metrical properties in space and time.

  • $\begingroup$ Regarding $T$, wouldn't $\lambda\cdot T$ be a valid absolute time for $\lambda>0$ (I think this would correspond to the choice of different units of time)? $\endgroup$ – Filippo Jan 4 at 16:10
  • $\begingroup$ Yes, It would be, I assumed that the units were given, and also a better formalisation would be to assume that the temporal axis is an affine space equipped with a distance in order to get rid of the origin. The spacetime is a fiber bundle with basis that one dimensional affine space and whose fibers are the absolute spaces... $\endgroup$ – Valter Moretti Jan 4 at 16:19
  • $\begingroup$ I'll soon need to study fiber bundles for my bachelor thesis, hopefully I'll be able to understand that afterwards :) In the meantime I'd have a question regarding the following part: "It is finally required that the said structure on $\Sigma_t$ are compatible with the structure of manifold indiced by $M$. In practice orthonormal Cartesian coordinates must belong to the atlas of $\Sigma_t$ induced from the atlas of $M$" - Can the "compatibility" requirement be formulated in mathematically rigorous way or is that a rather intuitive concept? $\endgroup$ – Filippo Jan 4 at 16:28
  • $\begingroup$ Yes, I wrote the meaning after "in practice": that is the rigorous definition. There must exist a Cartesian orthogonal coordinate system on $\Sigma_t$ (thus constructed out of its Euclidean space structure) that is in the atlas of the embedded submanifold strucure of $\Sigma_t$ induced from the differentiable structure of $M$. (If it happens for a Cartesian system it happens for all). $\endgroup$ – Valter Moretti Jan 4 at 17:16
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    $\begingroup$ I really appreciate the effort put in the answer, this is far more than what I expected. I still have to understand it properly, but I am sure this will solve many of my doubts. Since this is exactly the type of arguments I am looking for, both rigouros and conceptual, do you have a reference that deals with such topics of classical mechanics, with this approach? (english or italian references would be both good) $\endgroup$ – asdru Jan 4 at 21:25

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