# Parameterisation of the equation of motion for a relativistic massive point particle

The equation of motion for a relativistic massive point particle is given by:

$$\frac{dp_{\mu}}{d \tau} = 0,$$

where $p_{\mu}$ is the four-momentum defined by $p_{\mu} = m \frac{dx_{\mu}}{ds/c}$, where $m$ is the invariant mass, $x_{\mu}$ is the space-time four-vector, and $ds/c$ is the proper time.

How can I show that the equation of motion is parameterisation invaraint, i.e., the equation of motion retains its form as $\frac{dp_{\mu}}{d \tau'} = 0$ with $\tau' = \tau'(\tau)$?

From what I know the easiest way to do this is to go back to the action from which you derive the momentum $$I = m \int \sqrt{g_{\mu\nu} dx^\mu dx^\nu} = m \int \sqrt{g_{\mu\nu} \frac{dx^\mu}{d \lambda} \frac{dx^\nu}{d \lambda}} d\lambda$$ Now taking $\lambda'= \lambda'(\lambda)$, thus $d\lambda'= \frac{d \lambda'}{d\lambda} d\lambda \ \to \ d\lambda = \left( \frac{d \lambda'}{d\lambda}\right)^{-1} d\lambda'$ and $\frac{dx^\mu}{d\lambda} = \frac{dx^\mu}{d\lambda'} \frac{d \lambda'}{d\lambda}$ such that the action becomes $$I = m \int \sqrt{g_{\mu\nu} \frac{dx^\mu}{d \lambda'} \frac{dx^\nu}{d \lambda'} \left(\frac{d \lambda'}{d\lambda}\right)^2 } \left(\frac{d \lambda'}{d\lambda}\right)^{-1}d\lambda' = m \int \sqrt{g_{\mu\nu} \frac{dx^\mu}{d \lambda'} \frac{dx^\nu}{d \lambda'}} d\lambda'$$ Now you can again set $\delta I =0$ and derive the EoM, which will obviously be exactly the same with $\lambda \to \lambda'$.
• You can go through the entire procedure for $\frac{d p}{d\tau} = \frac{d}{d\tau} \left( \frac{d x}{ds} \right)$ as well of course, but you will see that it requires some more algebra than doing it via the action. Just use the fact that $ds = \sqrt{g_{\mu\nu} dx^{\mu} dx^{\nu} }$ and it should after some algebra give you the required result. – 123hoedjevan Jun 12 '15 at 10:49
$$I \equiv \int \mathcal{L}_\lambda d\lambda, \ \ \ \text{for} \ \ \mathcal{L}_\lambda \equiv\sqrt{g_{\mu\nu} \frac{dx^\mu}{d\lambda} \frac{dx^\nu}{d\lambda} } \\ ds = \mathcal{L}_\lambda d\lambda \\ \frac{d}{d \lambda} \left(\frac{dx}{ds}\right) = \frac{d}{d \lambda} \left(\frac{1}{\mathcal{L}_\lambda} \frac{dx}{d\lambda }\right) = - \frac{1}{\mathcal{L}_\lambda^2} \frac{d \mathcal{L}_\lambda}{d\lambda} \frac{dx}{d\lambda} + \frac{1}{\mathcal{L}_\lambda} \frac{d^2x}{d\lambda^2} = 0$$ so to prove $\frac{dp}{d\lambda} =0\to \frac{dp}{d\lambda'}=0$ under $\lambda \to \lambda'(\lambda)$ we must prove that equation $A_\lambda$ behaves as $A_\lambda \to A_{\lambda'}$ with $A_\lambda$ \begin{equation} A_\lambda : \ \ \frac{d^2x}{d\lambda^2} = \frac{1}{\mathcal{L}_\lambda} \frac{d \mathcal{L}_\lambda} {d\lambda} \frac{dx}{d\lambda} \end{equation} Now under $\lambda \to \lambda'(\lambda)$ we know that $\mathcal{L}_\lambda \to \mathcal{L}_{\lambda'} \frac{d \lambda'}{d\lambda}$ so
$$\frac{d^2x}{d\lambda^2} \to \frac{d}{d\lambda'}\left( \frac{d x}{d\lambda'} \frac{d\lambda'}{d\lambda}\right) \frac{d\lambda'}{d\lambda} = \frac{d^2x}{d\lambda'^2} \left(\frac{d\lambda'}{d\lambda} \right)^2\\ \frac{1}{\mathcal{L}_\lambda} \frac{d \mathcal{L}_\lambda}{d\lambda}\frac{dx}{d\lambda} \to \frac{1}{\mathcal{L}_\lambda'} \frac{d \mathcal{L}_\lambda'}{d\lambda'} \frac{dx}{d\lambda'} \left( \frac{d \lambda'}{d\lambda} \right)^2$$ thus indeed under $\lambda \to \lambda'(\lambda)$, $A_\lambda \to A_{\lambda'} \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \square$