I found myself puzzled with some very basic physical concepts and I hope to get enlightened with your help.

Initially my confusion arose in connection with Maxwell's equations and Lorentz transformations. It is often said that Maxwell's equations are not invariant under the Galilean transformations. However, as far as I understand it is implicitly assumed that not only coordinates transforms according to the Galilean law but also the electric and magnetic fields are transformed as vectors w.r.t. to the Galilean transformations. My question is:

1) When transforming equations from one frame to another one have to change both coordinates and fields. Aren't there field transformations that together with the Galilean transformations preserve the form of Maxwell's equations?

More generally, one can often hear that some equations are relativistic/non-relativistic based on their behavior under the corresponding coordinate transformations. After some thinking I came up with another question:

2) Can one derive coordinate transformations from the relativity principle and known form of the equations of motion?

I can not see a clear way to do it. Below I give my thoughts at the example of the harmonic oscillator in Newtonian mechanics.

In non-relativistic physics harmonic oscillator obeys the following equation of motion $$\frac{d^2x(t)}{dt^2}=-\omega^2(x(t)-x_0(t))$$

where $x_0(t)$ is the coordinate of the equilibrium position which generally can change with time. Suppose that the oscillator is moving along $x$-axis at constant speed $v$. Then we have $$x_0(t)=x_0(0)+vt$$

Now, consider another reference frame co-moving with the oscillator. Denote coordinates there by $x',t'$. They must be some functions of $x,t$ $$x'=X(x,t),\quad t'=T(x,t)$$. In this reference frame the oscillator is at rest and therefore the following equation must hold $$\frac{d^2x'(t')}{dt'^2}=-\omega'(x'(t')-x_0')$$

The form of these equations follows from a relativity principle. Moreover, one must have $\omega'=\omega$ and without loss of generality $x_0'=x_0(0)$.

Now the question is for which functions $X(x,t),T(x,t)$ equation in the 'moving' reference frame is satisfied provided that $x(t)$ solves equation in the reference frame 'at rest'? If one denotes $$\hat{X}(t)=X(x(t),t),\quad \hat{T}(t)=T(x(t),t) $$ then one must have $$\frac{\hat{X}''(t)\hat{T}'(t)-\hat{X}'(t)\hat{T}''(t)}{\hat{T}'(t)^3}=-\omega^2(\hat{X}(t)-x_0(0))$$

It's easy to see that with the Galilean choice of $X, T$ $$X(x,t)=x-vt,\quad T(x,t)=t\\\hat{X}(t)=x(t)-t,\quad \hat{T}(t)=t$$ equation above is satisfied. But I can hardly imagine that the Galilean transformations are the only solution to this equation (although after some straightforward attempts time I didn't find others, but I believe that this should be possible).

So this brings me to my third question, which is a specific version of the second. I also consider this as a toy example for the Maxwell's equations and Lorentz transformations.

3) Can one derive Galilean coordinate transformations from the relativity principle and equations of motion for the harmonic oscillator? If this is possible, then how? If not, then which ingredients should one add? Or, maybe, there is no derivation based on equations of motion available at all?

Thank you for taking an interest! Any help is highly appreciated!

  • $\begingroup$ Since there are Galileo-invariant as well as Lorentz invariant models of harmonic oscillators, it's not clear to me what you are trying to do. A harmonic oscillator model is, at best, a good clock. At worst it's not even a clock (because it may couple to a background field that may break the symmetry of the vacuum). In the best of those cases (when the oscillator behaves like an ideal clock in the postulated metric would), one can only recover what was already put into its equations of motion. The only instance that matters is nature, which simply doesn't make Galilean clocks. $\endgroup$
    – CuriousOne
    Sep 27, 2014 at 20:58
  • $\begingroup$ @CuriousOne when I referred to the harmonic oscillator I meant the Galilean one. I thought it is conventional. Whatever, the only thing that matters is its equations of motion which I explicitly stated in the post. My question is not about which transformations laws are realized in Nature, but how do we construct transformation laws from the equations of motion and the relativity principle if we really do. $\endgroup$ Sep 27, 2014 at 21:24

1 Answer 1


The basic assumptions on the space-time structure in classical mechanics are:

(1) Time intervals beetween events are absolute.

(2) Space intervals beetween contemporary events are absolute.

We may refere to this two properties as to the "galilean space-time structure". In the first chapter of Arnold's "Mathematical Methods of Classical Mechanics" you can find a mathematical formulation of space-time structure, and one of the problems is to prove that

All the affine transformations of space time which preserve time intervals and distances beetween contemporary events are compositions of rotations, translations and uniform motions.

The principle of relativity alone allows a much wider class of transformations (for example, it allows Lorentz's transformations).

So I think that your second and third questions can be rephrased in this way:

Can the galilean structure of space-time be obtained from known equations of motions (and the principle of relativity)?

I'm not able to be cathegorical about this point, but my intuition says that the answer is no, without at least some other strong assumptions.

Here's a simple derivation of galilean transformations. Let $(t,\mathbf r)$, $(t',\mathbf r'$), denote the coordinates of an event in two frames of reference (not necessarily inertial frames). In first place, the invariance of time intervals beetween events implies that $$t'=t+t_0,$$ for some constant $t_0$.

Now, the invariance of space intervals beetween simultaneous events implies that, for a fixed $t$, $\mathbf r \mapsto \mathbf r'$ is an isometry of $\mathbb R ^3$. The most general form of such isometry is: $$\mathbf r'= \mathbf s+G\mathbf r,$$ where $GG^T=I$ and both $\mathbf s$ and $G$ may depend on time.

To establish that $\dot {\mathbf s}=0$ and $\dot G = 0$ if the two frames are inertial we notice that, by the principle of relativity, the equation: $$\ddot {\mathbf r}=0$$ for an isolated body must be covariant; a direct calculation shows that this is possible only if the above conditions are satisfied.

  • $\begingroup$ Thank you for your answer! So your point is that we actually know how should, say, electric field $E$ transform since we know that in any reference frame we have $m\ddot{x}=qE$ and we know the transformation law of $\ddot{x}$. In this way the relativity principle is satisfied. However, as far as I can see this approach does not put any restrictions on coordinate transformations. In a sense we've said "whatever the coordinate transformation is, let's transform the fields so that the equation remains invariant". But this doesn't say anything about uniqueness of such transformations. $\endgroup$ Sep 25, 2014 at 19:23
  • $\begingroup$ You are welcome! The last paragraph was in response to your sentence “it is implicitly assumed that [...] also the electric and magnetic fields are transformed as vectors w.r.t. to the Galilean transformations”. For the electric field to represent the force, it is necessary that it transforms as the coordinates. Ofcourse this doesn't actually put a restriction on the coordinate transformations themselves. $\endgroup$
    – pppqqq
    Sep 26, 2014 at 8:23
  • $\begingroup$ However, I would like to point out the $\ddot{x}$ is not necessarily a three-vector (if $t$ transforms non-trivially it is not so, Lorentz transformations are an example). Moreover, in pure Maxwell's equations we don't have a clear relation of the fields to coordinates, as in equation $qE=m\ddot{t}$. However we have the latter equation on physical grounds. It would be interesting if it is the combination of the two equations (field eq.+particle eq.) that is needed in order to derive Lorentz transformations. $\endgroup$ Sep 26, 2014 at 15:38
  • $\begingroup$ I understand your point. Also, on second thought, the force law implies how $E$ and $B$ should transform under rotations, but doesn't say how they should transform under boosts, which is the core of the problem, so the last part of my answer was quite imprecise. $\endgroup$
    – pppqqq
    Sep 26, 2014 at 16:43
  • $\begingroup$ Awarded bounty to your answer since there are no competitors and the bounty is expiring today -) $\endgroup$ Oct 2, 2014 at 10:24

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