This sounds like a daft question, but I'm serious.

Translation and rotation are clearly different -- the symmetry between them is broken by Newton's Laws. But in the Lagrangian/Hamiltonian frameworks, they look extremely similar! The Lagrangians for free rotation and free translation are exactly the same, up to the replacement of some letters. Working entirely with the Lagrangian framework, it's unclear when and where the symmetry breaking happens.

Despite this, there are many clear asymmetries between translation and rotation:

  • There is absolute rotation, but not absolute translation. (At least, I believe this is the orthodox position.)
  • In space, starting with zero linear and angular momentum, it's possible to change your angular position but not your translational position (you can turn yourself around, but can't move your center of mass).
  • In quantum mechanics, free particles can have continuous values of linear momentum but have quantized angular momentum.

I know why the third point holds: localization causes quantization, and the set of possible angular positions is compact, while the set of possible positions is not. In fact, I feel like this is the only difference, a priori, between translation and rotation. In layman's terms, if you keep rotating, you'll get back to where you started, but if you keep translating, you won't.

Is it possible to use this reasoning to extract the first and second bullet points above? If not, what exactly is the difference between translation and rotation?


2 Answers 2


I am not sure exactly how to approach this question, but I feel a little side information could help both you or other potential answers narrow down some ideas.

Let us take the group $SE(3)$ the special Euclidean group in three dimensions. Here we will denote group multiplication as $(\tilde{\mathcal O},\tilde{r})(\mathcal O,r)=(\tilde{\mathcal O}\mathcal O,\tilde{\mathcal O}r+\tilde{r})$. Translations in the subgroup $\mathbb R^3\subset SE(3)$ act by vector addition. \begin{equation} \mathbb R^3\times\mathbb R^3\rightarrow \mathbb R^3: (I,\tilde r)(I,r)=(I,r+\tilde r) \end{equation} While rotations act by composition, \begin{equation} SO(3)\times SO(3)\rightarrow SO(3):(\tilde O,\boldsymbol 0)(O,\boldsymbol 0)=(\tilde O O,\boldsymbol 0) \end{equation} The point I am trying to make here is that at the group structure level, rotations act differently than translations.

Classical mechanics in its most general form considers the group action on an algebra of smooth commuting observables.

Im my opinion (stressing here it is opinion), since analytical dynamics generalises mechanics to a much larger degree than Newtonian physics does, then the level of abstraction causes similarities in the way that we approach translations and rotations. However at a more mathematical level their group actions are different.

I don't know if this answer helps/is obvious/ is on-topic/off-topic but it may be of some use? :)

  • $\begingroup$ That's basically it, very nice. Are you sure you have a tensor product on the left hand side rather than just a composition? $\endgroup$
    – gented
    Dec 16, 2015 at 13:21
  • $\begingroup$ @GennaroTedesco Whoops, my mistake, thank you very much :) $\endgroup$ Dec 16, 2015 at 13:36
  • $\begingroup$ Translations composed with translation are still translations - that's closure under composition; rotations composed with rotations are still rotations - again the same property. $\endgroup$ Dec 16, 2015 at 13:47
  • $\begingroup$ Translations can act on a space, as can rotations - I'm not sure I really follow the distinction between rotations and translations that is being made here. $\endgroup$ Dec 16, 2015 at 13:49
  • 1
    $\begingroup$ Well, translations close the $\mathbb{R}^3$ group, rotations close the $\textrm{SO}(3)$ group. The two groups are different things, hence translations are rotations are different things. $\endgroup$
    – gented
    Dec 16, 2015 at 15:06

Good question; I don't have an answer for you, however

The symmetry is broken by Newtons Laws

Formally, I supposed Newtons laws expressed linearly looked very similar to it expressed rotationally; ie in the first law, a particle in uniform motion remains in uniform motion - ditto for a rotating body, when it's made to spin - it keeps on spinning; and the same for the other two laws.

So in a way - so it doesn't surprise me that they look similar in the Lagrangian or Hamiltonian picture.

As you say:

There is absolute rotation but not absolute translation (at least I think this is the orthodox position)

And as I'm sure you're aware, this was demonstrated by Newtons rotating bucket.

I feel like this is the only a priori difference between them

This observation is implicit in Aristotles Physics; his general observation is:

Every moving object moves either in a circle, or in a straight line or in a motion compound of the two VIII.8.

Then more specifically, a particle moves either in place, or from place to place; and the only way it can move in place is rotation.

(I'd suggest it also holds for gauge groups, thinking of them as generalised rotations - U(1), for example - which is a circle).

In QM, free particles have quantised values of angular momentum, and continuous values of linear momentum

Ok, but does it still hold for free particles on a circle - or do you get quantised positions there? I suppose the exact parallel there would be the compact line, though.

But now I wonder what happens with the semi-infinite line; which is neither non-compact, nor compact; but does appear to have a crucial difference with the line infinite on both sides.


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