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Lagrangian Approach for point-particle motion

Consider a point particle with mass $m>0$ moving along a worldline $x^\mu(\lambda)$ where $\lambda$ is an arbitrary parametrization. The action principle in the Lagrangian formulation of the problem is

\begin{equation} \begin{aligned} S&=-m\int d\tau\\ &=-m\int \sqrt{-g_{\mu\nu}\frac{dx^\mu}{d\lambda}\frac{dx^\nu}{d\lambda}}d\lambda. \end{aligned}\tag{1} \end{equation} So we get a Lagrangian function $$L=-m\sqrt{-g_{\mu\nu}\frac{dx^\mu}{d\lambda}\frac{dx^\nu}{d\lambda}}.\tag{2}$$ The euler-Lagrange equations give the dynamics for the arbitrary parameter $\lambda$, which are \begin{equation} \frac{D}{d\lambda}\frac{dx^\mu}{d\lambda}-\kappa \frac{dx^\mu}{d\lambda}=0.\tag{3} \end{equation} Here, $$\frac{D}{d\lambda}=\frac{dx^\mu}{d\lambda}\nabla_\mu\tag{4}$$ is the covariant time derivative and $\kappa$ measures the failure of $\lambda$ to be an affine parameter of the worldline $x(\lambda)$

\begin{equation} \kappa\equiv \frac{1}{\sqrt{-g_{\mu\nu}\frac{dx^\mu}{d\lambda}\frac{dx^\nu}{d\lambda}}}\frac{d}{d\lambda}\sqrt{-g_{\mu\nu}\frac{dx^\mu}{d\lambda}\frac{dx^\nu}{d\lambda}}.\tag{5} \end{equation} This equation can also be rewritten in terms of the proper time parameter $\tau$, in which case we simply get \begin{equation} \frac{D}{d\tau}\frac{dx^\mu}{d\tau}=0.\tag{6} \end{equation}

The important takeaway is that the action principle is explicitly reparametrization invariant and the equations of motion are obtained in terms of an arbitrary parameter $\lambda$.

Hamiltonian Approach for point-particle motion

We can also write down a Hamilton principle

\begin{equation} S=\int p_\mu dx^\mu-\int H ds\tag{7} \end{equation} where the Hamiltonian function is \begin{equation} H=-\sqrt{-g^{\mu\nu}p_\mu p_\nu}.\tag{8} \end{equation} In this case, Hamilton equations are

\begin{aligned} \frac{dx^\mu}{ds}&=\frac{p^\mu}{\sqrt{-g^{\mu\nu}p_\mu p_\nu}},\\ \frac{Dp_\mu}{ds}&=0. \end{aligned}

From the first equation, it is clear that $s$ is proper time, since $\frac{dx^\mu}{ds}$ is automatically normalized. From this, I conclude that:

  • When I wrote Hamilton's principle, I fixed the reparametrization invariance and I chose $s$ to be proper time.

  • Hamilton equations are then derived directly in proper time.

Question

Am I understanding correctly that the Hamiltonian approach loses reparametrization invariance and expresses dynamics directly in proper time? As opposed to the Lagrangian formulation, which preserves reparametrization invariance and gives the Euler-Lagrange equations in an arbitrary time parametrization.

Extra comments

I know that if one starts from the square root Lagrangian and tries to derive the Hamiltonian formulation using a Legendre transformation, one gets $H=0$ identically. I also know that this is because the Lagrangian formulation is reparametrization invariant. I'm not sure if this relates to my question or not, it probably does!

I know there are several questions on this forum about the Lagrangian/Hamiltonian formulations for a point particle in General Relativity, but I haven't seen the question of which time parameter each formulation uses, specifically.

Qmechanic
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P. C. Spaniel
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1 Answers1

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  1. On one hand, starting from the natural Lagrangian formulation (1) & (2) of a relativistic point particle, its (singular) Legendre transformation yields a well-known constrained Hamiltonian formulation, which is world-line (WL) reparametrization covariant, cf. e.g. this, this & this Phys.SE posts.

  2. On the other hand, OP's Hamiltonian formulation (7) & (8) can be extended with an einbein field $e>0$ $$\widetilde{L}~=~p\cdot \dot{x} -\frac{p^2}{2e}+\frac{e}{2}, \qquad p^2~:=~p_{\mu}g^{\mu\nu}(x)p_{\nu}.\tag{A}$$ Eliminating/integrating out $e$ brings back OP's Hamiltonian Lagrangian (7) & (8) $$ \widetilde{L}\quad\stackrel{e}{\longrightarrow}\quad p\cdot \dot{x} +\sqrt{-p^2}. \tag{B}$$ In eq. (A) we can perform an inverse Legendre transformation by eliminating/integrating out the momenta $p_{\mu}$: $$ \widetilde{L}\quad\stackrel{p_{\mu}}{\longrightarrow}\quad \frac{e}{2}(\dot{x}^2 +1), \qquad \dot{x}^2~:=~\dot{x}^{\mu}g_{\mu\nu}(x)\dot{x}^{\nu}.\tag{C}$$ Eq. (C) is not the natural Lagrangian formulation (1) & (2) of a relativistic point particle. The einbein field $e$ is here a Lagrange multiplier that imposes proper time parametrization $(dx)^2=-(d\tau)^2$.

Conventions: The metric sign convention is here mostly plus $(-,+,\ldots,+)$ and the speed of light $c=1$ is one.

Qmechanic
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