# Translation invariance of point particles as a field theory

The case of point particles, relativistic or not, can be treated as a field theory in general, ie for the $$(1+1)$$-dimensional case this is the theory of a field theory on the vector bundle

$$\pi : \mathbb{R}^2 \to \mathbb{R}$$

with $$x(t)$$ a section of that bundle. I thought therefore to look at the set of transformations on such a field via the automorphisms $$\text{Aut}(\mathbb{R}^2)$$ that respect the structures here. As far as I can tell (and with some help from @mikemiller), the automorphisms on this bundle are

$$\Phi(t, x) = (\alpha(t), \Lambda(t) \cdot x)$$

with $$\alpha(t)$$ a diffeomorphism on the base manifold and $$\Lambda(t)$$ a time-dependent linear transformation ($$\Lambda(t) \in \text{GL}(\mathbb{R})$$) on the fiber. This preserves the manifold, bundle and vector space structure of our vector bundle.

Applying it to the usual action, ie

$$\int_{t_a}^{t_b} \dot{x}^2(t) dt$$

It's not too hard to show that the action is invariant under the action of the base space diffeomorphism (it's just reparametrization invariance), and $$\Lambda(t) = \pm 1$$ (this is just $$\text{O}(1)$$). On the other hand, translation invariance seems notably absent. Even trying to fiddle with $$\Lambda(t)$$ to turn it into a translation it's obviously not going to work because for a start $$\Lambda(t)$$ can't be $$0$$ to remain invertible. If we look at the automorphisms on just the manifold $$\mathbb{R}^2$$,

$$\Phi(t,x) = (\alpha(x,t), \beta(x,t))$$

We can see that we had to sacrifice this possibility to preserve the rescaling of the vector space, ie, $$(t, \beta(\lambda x)) = (t, \lambda \beta(x))$$.

Is it normal that we cannot get translation invariance out of such a formulation of the theory? Should it normally be so, or can we get it back by sacrificing the preservation of the vector space slightly, or would that bring itself unfortunate consequences? Are we supposed to use an affine bundle instead, perhaps.

• Your problem is the assumption that the configuration bundle is a vector bundle. In general the configuration space $Q$ is a manifold and the configuration bundle $\mathbb R\times Q\rightarrow\mathbb R$ is a general fibre bundle without structure group. – Bence Racskó Sep 3 '19 at 13:21
• @BenceRacskó Should I drop all linearity and also allow non-linear transformations? – Slereah Sep 3 '19 at 16:15
• A name for the field theory you are describing is “sigma-model” (or $\sigma$-model), this should help with literature search (see eg. here). – A.V.S. Sep 3 '19 at 16:40
• A nonlinear transformation in this case corresponds to a (in general, time-dependent) diffeomorphism of configuration space, which is allowed in Lagrangian mechanics. So basically, yes. – Bence Racskó Sep 3 '19 at 19:53

Is it not obvious that $$x(t) \mapsto x(t)+a$$, with $$t$$ independent $$a$$, leaves the action invariant? And is this not translation invariance?
• It does indeed. The point is more that it doesn't seem to be contained in $\text{Aut}(E)$, if considered as a vector bundle. From investigations into the usual EM gauge connection, ie with gauge $A \to A + d\alpha$, it does seem that considering it as an affine bundle might be the solution. – Slereah Sep 3 '19 at 12:29
• I think you are vastly over-mathematicising something simple, and bundles are just not the appropriate language. As @A.V.S says a simple sigma model of maps $x:{\mathbb R}\to {\mathbb R}^N$ describes motion of a particle in ${\mathbb R}^N$ – mike stone Sep 3 '19 at 22:14