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Let's concentrate on point particle mechanics on a one dimensional manifold for simplicity. The action is $$S [q,\dot{q}]=\int dt L(q,\dot{q},t).$$ Time reparameterization would involve $t \to t'=f(t)$. So the action becomes $$\tilde{S}[q,\dot{q}]=\int dt' L(q,\tilde{q},t')$$ where $$\tilde{q}=\frac{dq}{dt'}.$$ Since we are integrating over all $t$ and this is equivalent to integrating over $t'$ (but with different limits), why isn't $\tilde{S'}=S$? In other words, if definite integrals don't change under change of variables, why does the action does in general?

I know this is not true as a fact, because not all systems are constrained (gauge) systems, and not all systems have zero naive Hamiltonian, cf. e.g. this related Phys.SE post. So where did the reasoning go wrong?

Qmechanic
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Sanjana
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2 Answers2

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  1. In the Lagrangian mechanics a time-reparametrization quasi-symmetry of a local action functional $S=\int \! L \mathrm{d}t$ usually refers to that the correponding Lagrangian 1-form $$L(q,\frac{dq}{dt^{\prime}},t^{\prime})\mathrm{d}t^{\prime}~=~L(q,\frac{dq}{dt},t)\mathrm{d}t+\mathrm{d}[\ldots]$$ is form-invariant$^1$ up to an exact 1-form term under a time-reparametrization $t^{\prime}=f(t)$.

  2. Counterexample: The Lagrangian 1-form for a free non-relativistic particle $$L\mathrm{d}t=\frac{m}{2} \left(\frac{dq}{dt}\right)^2\mathrm{d}t$$ does not have time-reparametrization quasi-symmetry.

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$^1$ More generally, a corresponding form-invariance is usually implicitly implied for other types of quasi-symmetries as well.

Qmechanic
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I) Every action can be expressed equivalently with a different reparametrization. This is done as follows:

Suppose we have a trajectory $t\mapsto q(t)$ from $t_0$ to $t_1$, then the action is $$ S[q]=\int_{t_0}^{t_1}L(q(t),\dot q(t),t)\mathrm dt. $$

Now let $p(s)$ denote a reparametrized trajectory (here $p$ is not canonical momentum and $s$ is not arc length), i.e. we have an invertible relation $t(s)$ such that $p(s)=q(t(s))$. We have $$ q(t)=p(s(t)) \\ \dot q(t)=p^\prime(s(t))\dot s(t), $$ (here and in the future the dot is $t$-derivative and the prime is $s$-derivative) so we can write $$ S[q]=\int_{t_0}^{t_1}L(p(s(t)),p^\prime(s(t))\dot s(t),t(s(t)))\mathrm dt, $$where we are still using $t$ as the independent variable.

We may now use the usual rules of integration to switch to $s$: $$ S[q]=\int_{s_0}^{s_1}L(p(s),p^\prime(s)\dot s(t(s)),t(s))t^\prime(s)\mathrm ds .$$

This integral has precisely the same value as the previous one, but it does not look like OP's formula.

II) I would argue that reparametrization invariance has no particular meaning unless one gives a transformation rule for the dynamical variables under a reparametrization. The relation $q(t)=p(s(t))$ supposes that the coordinate $q$ is a "scalar" under the coordinate transformation of the time coordinate.

Here it might actually be easier to consider field theory via differential geometry. Let $\pi:Y\rightarrow X$ be a fibered manifold whose sections are the dynamical fields. A Lagrangian of some fixed order $r$ can be thought of as a bundle morphism $L:J^r(\pi)\rightarrow\Lambda^n(X)$, where $J^r(\pi)$ is the $r$th jet prolongation of $\pi:Y\rightarrow X$.

Let now $f\in\mathrm{Aut}(\pi)$ be an automorphism of the fibered manifold with $f_0\in\mathrm{Diff}(X)$ the projection of this map. A section $\phi\in\Gamma(\pi)$ pulls back along $f$ by $$ f^\ast\phi:=f^{-1}\circ\phi\circ f_0. $$ The evaluation of the Lagrangian $L$ on the section is defined as $$ L[\phi]:=L\circ j^r\phi, $$i.e. as the composition by the $r$th prolongation of the section. Note that $L[\phi]$ is a fixed $n$-form on $X$ ($n=\dim X$).

Then the Lagrangian $L$ is said to be reparametrization invariant if for any automorphism $f$ and section $\phi$, we have $$ (f_0)^\ast L[\phi]=L[f^\ast\phi]. $$

This is essentially an equivariance condition and is not trivially satisfied. Alternatively, sometimes there is a subgroup $\mathcal G\le\mathrm{Aut}(\pi)$ that nonetheless has the property that "enough" $X$-diffeomorphisms can be obtained by projecting from $\mathcal G$, and then we also say that $L$ is reparametrization-invariant if it satisfies the above property with respect to the subgroup $\mathcal G$. The specification of $\mathcal G$ should be part of the theory.

My earlier remark regarding how reparametrization invariance is meaningful only if a transformation rule is given for the dynamical variables appears in a few places:

  • One has to specify fibered manifold $\pi:Y\rightarrow X$ by giving the allowed transition functions. This is closely related to the specification of the subgroup $\mathcal G$ mentioned before. Without this the theory isn't really well-defined.
  • When one talks about reparametrization invariance, one often considers only the group $\mathrm{Diff}(X)$ of diffeomorphisms of the base space. But the cold shower is that $\mathrm{Diff}(X)$ acts on neither $Y$, nor $J^r(\pi)$, nor $\Gamma(\pi)$. One might consider some (possibly only infinitesimal) embedding $\mathrm{Diff}(X)\rightarrow\mathrm{Aut}(\pi)$ of diffeomorphisms into fibered automorphisms. This has to be specified by hand.
  • If $\pi:Y\rightarrow X$ is a so-called "natural fibre bundle", then there is a canonical functorial lift from $\mathrm{Diff}(X)$ into $\mathrm{Aut}(\pi)$. Then the property $f^\ast L[\phi]=L[f^\ast\phi]$ is meant under reparametrization invariance, where now $f\in\mathrm{Diff}(X)$ and $f^\ast\phi$ is the natural pullback of $\phi$ under this lifting. This happens for eg. tensor field theories or when the dynamical variable is a linear connection on $X$. But then one must still - from a "classical" point of view - specify which class of geometric objects do the dynamical field $\phi$ belong.
Bence Racskó
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