# Noether identities and the relativistic point particle

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?

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 $$p^2 + m^2 = 0.$$

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).