Acceleration of free particle in Minkowski space!

Question

In Minkowski space, the equation of motion is $$a^c=\frac{d^2}{d\tau^2}x^c(\tau)=0.$$ How is this derived?

Answer 1

In general relativity the equation of motion of a freely falling object is given by the geodesic equation:

$$ {d^2 x^\mu \over d\tau^2} + \Gamma^\mu_{\alpha\beta} {dx^\alpha \over d\tau} {dx^\beta \over d\tau} = 0 $$

The symbols $\Gamma^\mu_{\alpha\beta}$ are the Christoffel symbols that are related to the spacetime curvature. In flat spacetime i.e. Minkowski spacetime (and assuming we aren't using a curved coordinate system) the Christoffel symbols are all zero so the geodesic equation becomes:

$$ {d^2 x^\mu \over d\tau^2} = 0 $$

which is the equation you quote.

If you're happy to accept the geodesic equation we need go no farther. If you now want to know how the geodesic equation is derived then it's done by varying the action. The details are on Wikipedia, but they are somewhat involved.