Noether identities and the relativistic point particle

Question

I am trying to better understand Noether identities, i.e. relations between equations of motion in the presence of gauge symmetries for the example of the relativistic point particle.

Formally, a Noether identity arises because the variation of an action leading to an equation of motion requires identical manipulations as the variation under a gauge symmetry. Thus, if an action is invariant under a gauge symmetry, we have the relation

$$ \delta_g S[\phi] = \int d^nx\,\frac{\delta L}{\delta \phi}\, \delta_g\phi = 0,\tag{1} $$ where $\phi$ is a generic field, $\delta_g$ the variation under a gauge symmetry and $\delta L/\delta \phi$ is the equation of motion for the field $\phi$. Substituting the variation and isolating the (continuous) variation parameter then results in the constraint.

However, I cannot make sense of it applying this logic to the action of the relativistic point particle $$ S[x] = m \int d\tau \sqrt{-\dot{x}_\mu \dot{x}^\mu}.\tag{2} $$ The gauge symmetry here is parametrization invariance $\tau \to \tau'$ or infinitesimally $$ \delta \tau = \varepsilon(\tau),\tag{3} $$ for $\varepsilon$ an arbitrary (continous and monotonous) function. Calculating the variation on the coordinates gives $$ \delta x^\mu = - \dot{x}^\mu \varepsilon.\tag{4} $$ Correspondingly, the Noether identity reads $$ 0 = \delta S[x] = m \int d\tau\,\frac{d}{d\tau} \left( \frac{\dot{x}_\mu}{\sqrt{- \dot{x}^2}} \right) \delta x^\mu \\ = -m \int d\tau\,\frac{d}{d\tau} \left( \frac{\dot{x}_\mu}{\sqrt{- \dot{x}^2}} \right) \dot{x}^\mu \varepsilon,\tag{5} $$ i.e. $$ \frac{d}{d\tau} \left( \frac{\dot{x}_\mu}{\sqrt{- \dot{x}^2}} \right) \dot{x}^\mu = 0.\tag{6} $$ However, I find the result hard to interpret as the relation seems to be completely arbitrary.
Can anyone maybe expand on the interpretation of this result?

Answer 1

1. More generally, an infinitesimal gauge quasi-symmetry $$\delta q^k(t)~=~\int\! dt^{\prime}~ R^k(t,t^{\prime})\epsilon(t^{\prime})\tag{3.5a}$$ of an action functional $S[q]$, $$\begin{align} \text{possible}&\text{ boundary terms}\cr ~=~&\delta S\cr ~=~&\sum_k\int\! dt\int\! dt^{\prime}~\frac{\delta S}{\delta q^k(t)} R^k(t,t^{\prime})\epsilon(t^{\prime})\cr ~+~&\text{possible boundary terms}\end{align}\tag{3.6a}$$ leads to an off-shell Noether identity $$ \sum_k\int\! dt~\frac{\delta S}{\delta q^k(t)} R^k(t,t^{\prime})~=~0, \tag{3.6b}$$ cf. Noether's 2nd theorem and e.g. Ref. 1.

2. The Noether identity (3.6b) can be interpreted as the EOMs are perpendicular to the gauge quasisymmetry.

3. The Noether identity (3.6b) is a generalization of OP's last eq. (6) with the following change of notation $$t~\to~\tau, \qquad q^k~\to~x^{\mu}, \qquad R^k(t,t^{\prime})~\to~-\dot{x}^{\mu}\delta(\tau\!-\!\tau^{\prime}).$$

References:

1. M. Henneaux & C. Teitelboim, Quantization of Gauge Systems, 1994; eqs. (3.5a) + (3.6a) + (3.6b).

Answer 2

Looking at the Noether identity now, the interpretation is of course immediately clear.

We can multiply equation (6) of my question with $2 m^2 / \sqrt{-\dot{x}^2}$, and identify the two terms as the relativistic momenta. This then amounts to saying that $$ 2\frac{d}{d\tau} \left( p_\mu\right) p^\mu = 0 \implies \frac{d}{d\tau}\left(p^2\right) = 0. $$ This is of course simply the on-shell-condition of relativity $$ p^2 + m^2 = 0. $$