Free Motion

A free particle in Special Theory of Relativity follows a time-like straight line.

• There is a choice of the parameter for which the coordinates are linear functions.
• Then, the tangent vector is constant. Then, we can define “acceleration” as $a^{\mu }=\frac{d{V}^{\mu }}{d\lambda }=V^{\nu }{\partial }_{\nu }{V}^{\mu }=0$
• Then, ${\eta }_{\mu \nu }{V}^{\mu }{V}^{\nu }$ is constant along the curve.
• So, the choice of the parameter for which the coordinates are linear is proper time up to a possible rescaling, $\lambda =a\tau +b$ Alternatively, a straight line is an extremum of the distance between two points.
• To extremise the proper time, we solve the Euler-Lagrange equations by taking variations of the Lagrangian, $\tau =\int \sqrt{-d{s}^2}=\int \sqrt{-{\eta }_{\mu \nu }{\mathring{x}}^{\mu }{\mathring{x}}^{\nu }}d\lambda =\int \mathcal{L}d\lambda$
• Then, $\frac{d}{d\lambda }(\frac{\partial \mathcal{L}}{\partial {\mathring{x}}^{\mu }})=\frac{d}{d\lambda }(\frac{{\eta }_{\mu \nu }{\mathring{x}}^{\nu }}{\sqrt{-{\eta }_{\mu \nu }{\mathring{x}}^{\mu }{\mathring{x}}^{\nu }}})=0$
• However, since the choice of the parameter does not matter, we can choose the parameter to be the proper time itself. Hence, $\frac{d}{d\lambda }({\eta }_{\mu \nu }{\mathring{x}}^{\nu })=0$ which is similar to the first definition.