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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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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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