Let us define an inertial frame as a frame of reference where the laws of physics take their usual form, as opposed to non intertial frames where one has to introduce pseudo-forces.

We can further define an equivalence class which contains other inertial frames as the class of frames of reference moving of constant velocity with respect to an initial inertial frame.

Is it possible to show formally that there do not exist two inertial frames which cannot be put in the same equivalence class -- namely two inertial frames which are not moving at constant relative velocity one with respect to the other?

I am aware of the fact that in a frame accelerated with respect to an inertial one there are pseudo forces, but I am concerned in particular in showing that there is no possible configuration in which they would all cancel out.


2 Answers 2


What follows is a version of the statement you want to prove which assumes that any two frames are related by a spacetime transformation that leaves time invariant up to translation and that preserves Euclidean distances. Because of these hypotheses, the statement below is a Newtonian answer to the question. I'm confident, however, that a similar special-relativistic answer can be constructed.

The following theorem essentially says that if even just Newton's First Law is required to be preserved in the transformation between friends, then the frames must be related by a Galilean transformation, so they must be in the same equivalence class.

Theorem. Let a twice-differentiable, orientation preserving, time-dependent Euclidean isometry $T:\mathbb R\to \mathrm{ISO}(3)$ be given, and let $t_0$ be a real number. If the spacetime transformation \begin{align} G(t,\mathbf x) = (t+t_0, T(t)(\mathbf x)) \end{align} preserves Newton's First Law, then $G$ is Galilean.

Proof. $T(t)$ can be written as a time-dependent rotation plus a time-dependent translation: \begin{align} T(t)(\mathbf x) = R(t)\mathbf x + \mathbf c(t). \end{align} Therefore, under the action of $G$, a straight line $\mathbf x_0 + t\mathbf v_0$ gets mapped into the following curve: \begin{align} R(t)(\mathbf x_0 + (t+t_0)\mathbf v_0) + \mathbf c(t) \end{align} If $G$ preserves Newton's First Law, then this transformed curve must have zero acceleration no matter which $\mathbf x_0$ and $\mathbf v_0$ we choose. Thus, \begin{align} \frac{d^2}{dt^2} \big[R(t)(\mathbf x_0 + (t+t_0)\mathbf v_0) + \mathbf c(t)\big] = \mathbf 0, \end{align} for all $\mathbf x_0,\mathbf v_0\in\mathbb R^3$. Distributing the derivatives on the left hand side gives \begin{align} \ddot R(t) \mathbf x_0 + \ddot R(t) (t+t_0)\mathbf v_0 + 2\dot R(t) \mathbf v_0 + \ddot{\mathbf c}(t) = \mathbf 0 \end{align} Choosing $\mathbf x_0 = \mathbf v_0 = \mathbf 0$ gives $\ddot{\mathbf c}(t) = 0$ which implies that there exist constant vectors $\mathbf c$ and $\mathbf v$ such that $\mathbf c(t) = \mathbf c + t \mathbf v$. Using this gives the reduced constraint \begin{align} \ddot R(t) \mathbf x_0 + \ddot R(t) (t+t_0)\mathbf v_0 + 2\dot R(t) \mathbf v_0 = \mathbf 0. \end{align} Now, picking $\mathbf v_0 = \mathbf 0$ gives $\ddot R(t) \mathbf x_0 = \mathbf 0$ for all $\mathbf x_0$, and this in turn means that $\ddot R(t) = 0$. Using this further reduces to \begin{align} \dot R(t)\mathbf v_0 = \mathbf 0 \end{align} for all $\mathbf v_0$, and this implies that $\dot R(t) = 0$. This means that there is a constant rotation $R$ such that $R(t) = R$. Putting this all together, we find that our original isometry $T(t)$ takes the following form: \begin{align} T(t) = R\mathbf x + \mathbf c + t\mathbf v \end{align} and hence $G$ is Galilean, namely it consists only of a time translation, a constant spatial translation, a Galilean boost by constant velocity, and a constant rotation. $\blacksquare$.

  • $\begingroup$ Weinberg lays down some basic structure of such a proof in the relativistic context in his GR book as far as I can remember. Will update with a properly detailed reference if I can. $\endgroup$
    – user87745
    Apr 3, 2019 at 15:52

Consider time reversal or parity transformations.

Because some physics doesn't have this symmetry, there are separate in-equivalent classes of inertial frames.

In each class, the physics will appear the same, and you can rotate, translate, or boost from one inertial coordinate system to another in the same class. But the physics will appear different compared to the other class of inertial frames.


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