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The action for the relativistic point particle with mass $m \geq 0$ in a curved background is given by:

\begin{equation} S[X] = \int_{\tau_0}^{\tau_1} d\tau \left[ e(\tau)^{-1} g_{\mu \nu}(X(\tau)) \dot{X}^\mu \dot{X}^\nu - e(\tau) m^2 \right].\tag{1} \end{equation}

Is there a difference in the canonical/constraint/gauge structure between the cases $m=0$ and $m>0$?

I am asking, since I have the feeling, that in the massive case the gauge choice corresponding to an affine parametrization is natural, while in the massless case there exists no natural gauge choice.

Qmechanic
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warpfel
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1 Answers1

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  1. The action (1) has a world-line (WL) reparametrization gauge symmetry in both the massive and massless case.

  2. An important difference is that in the massless case one cannot eliminate/integrate out the einbein field $e$, cf. e.g. this Phys.SE post.

    In contrast such elimination in the massive case leads to the well-known square-root action, cf. e.g. my Phys.SE answer here.

  3. If we gauge-fix $e={\rm const}>0$, then the Euler-Lagrange (EL) equation for the gauge-fixed action (1) is in both cases a geodesics equation (without an inhomogeneous term, cf. e.g. this Phys.SE post.) This means that two different parametrizations are affinely related.

    Whether the geodesic is time-like or light-like then comes down to boundary conditions (BCs).

    Affine parametrization (in the sense that spacetime arclength is affinely related to the WL parameter) holds in the gauge-fixed case for time-like geodesics, but does not make sense for light-like geodesics.

    See also e.g. this related Phys.SE post.

  4. For the corresponding Hamiltonian formulation, see e.g. this Phys.SE post.

Qmechanic
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