Wednesday, 19 October 2016

homework and exercises - Equation of motion of a free particle


We know that the equation of motion of particle can be derived from the respective action. But in the book I am reading, the author is saying:




... timelike worldline of a massive particle is parametrised by proper time, $x^a=x^a(\tau)$. The velocity (tangent) vector is $u^a \equiv \dot{x}^a(\tau)$. This is normalised as $u^au_a =-c^2$.



How can one derive e.o.m?




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