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Let's say we a free particle of mass $m$ with have the Lagrangian

\begin{equation} L_0 = \frac{m}{2} g_{\mu\nu}(x) \frac{dx^\mu}{d\lambda}\frac{dx^\nu}{d\lambda} \end{equation}

where $g_{\mu\nu}$ is the metric and $\lambda$ is an arbitrary parameter. The equations of motion are

\begin{equation} m\frac{D}{d\lambda} \frac{dx^\mu}{d\lambda}=0 \end{equation}

where $\frac{D}{d\lambda}=\frac{dx^\nu}{d\lambda}\nabla_\nu$ is the covariant parameter derivative along the worldline. This equation means that $\lambda$ is an affine parameter. so the 4-velocity satisfies the normalization condition

\begin{equation} g_{\mu\nu}(x) \frac{dx^\mu}{d\lambda}\frac{dx^\nu}{d\lambda}=c \end{equation}

where $c=-1$ would correspond to $\lambda$ being proper time. Here I'm using a signature $(-1,+1,+1,+1)$ for the metric. My question is: If we now add a perturbation to the particle, so we get a new Lagrangian

\begin{equation} L = \frac{m}{2} g_{\mu\nu}(x) \frac{dx^\mu}{d\lambda}\frac{dx^\nu}{d\lambda} - q \Phi(x) \end{equation}

so the equations of motion are

\begin{equation} m\frac{D}{d\lambda} \frac{dx^\mu}{d\lambda}=-q\partial^\mu \Phi \end{equation}

will $\lambda$ still be an affine parameter and the 4-velocity still be normalized to be

\begin{equation} g_{\mu\nu}(x) \frac{dx^\mu}{d\lambda}\frac{dx^\nu}{d\lambda}=c \end{equation}

Qmechanic
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P. C. Spaniel
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  • What wrote down is not the correct form of the Lagrangian. The correct form contains a square root: $\mathcal{L} = \frac{m}{2} \sqrt{ g_{\mu \nu} \dot{x}^{\mu} \dot{y}^{\nu} }$ – Prof. Legolasov Oct 03 '21 at 16:51
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    Thanks for your comment! I don't think that's true. The Lagrangian you mention is the reparametrization invariant one, but if you use the einbein formalism, you can obtain a whole family of Lagrangians, where each one just corresponds to fixing the gauge symmetry of the problem. In my case I work with Hamiltonians so this Lagrangian is useful because it gives a nice Hamiltonian function. The reparametrization invariant Lagrangian that you use gives $H=0$ identically. – P. C. Spaniel Oct 03 '21 at 16:54
  • Sure, you can use your Lagrangian to solve certain technical problems, but it doesn't mean that it will tell you about how to couple the particle to external fields. The possible couplings are very restricted by reparametrization invariance. If you want to explore the possible external field couplings, think about how to write down a reparametrization invariant action, and then think about how to gauge-fix it. It is always possible to gauge-fix such that $\lambda$ is an affine parameter unless the geodesic is light-like, almost by definition. – Prof. Legolasov Oct 03 '21 at 16:56
  • It is not possible, AFAIK, to consistently couple a GR particle to a scalar field. You can however couple it to a 1-form field (electromagnetism), the interaction term is just the integral of the 1-form along the worldline, which is manifestly reparametrization invariant. You can gauge that by choosing $\lambda$ to be the affine parameter that doesn't depend on the electromagnetic field at all, only on gravity. – Prof. Legolasov Oct 03 '21 at 17:01
  • Related: https://physics.stackexchange.com/q/651962/2451 – Qmechanic Oct 11 '21 at 22:20

1 Answers1

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To study $g_{\mu\nu} \frac{d x^ \mu}{d \lambda} \frac{d x^ \nu}{d \lambda} =: X^ 2$, we study $\nabla_X X^ 2$:

$$ \nabla_X X^2 = 2 X_\mu X^ \nu \nabla_\nu X^ \mu=\\ -2q X^ \mu X^ \nu \nabla_\nu \nabla_\mu \Phi $$

where we used the equations of motion in the last step. Hence, unless the second derivative of $\Phi$ is 0 (or proportional to the metric if we want null trajectories), $\lambda$ will no longer be the affine parameter. In fact, as commented in your question, it might be difficult or impossible to define proper time with that action.

Filipe Miguel
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