Coordinates systems and frames of reference in classical mechanics

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 $tin mathbb{R}$. (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 $R$:

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

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 $Oin E_R$ and exes $e_1,e_2e_3$. Take a constant $cin mathbb{R}$ 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: $$v'^k = v^k + sum_jfrac{dR^k_j(t)}{dt} x^j:.$$ 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 $Rin O(3)$.

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.