# Acceleration of free particle in Minkowski space! [duplicate]

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

• What have you done to figure this out? From what starting point do you want to derive this? – ACuriousMind Jan 18 '16 at 18:08
• Any starting point, I have posted a question previously where the author just puts this result there. So I am supposing the author assumes this is well-known to the reader. Any starting pt that can convince me that this is it in Minkowski spacetime. I know that in Euclidean spacetime, I cann take $L=1/2(m\dot{x}^i)^2$ and use EL equations to find it. However, in Minkowski space, this is giving me a hard time @ACuriousMind – PhilosophicalPhysics Jan 18 '16 at 18:20
• Just take the standard action for the relativistic point particle, $S = k\int \mathrm{d}s = k\int\sqrt{g_{\mu\nu}x^\mu x^\nu}$, cf. this post.. – ACuriousMind Jan 18 '16 at 18:27
• In non-relativistic mechanics, how can we prove $F=ma$? well, in principle, we cannot: it is a (sensible) postulate. We can postulate, instead, a lagrangian, from which $F=ma$ follows, but there has to be a postulate sooner or later. $a^c=0$ is a nice (and sensible) postulate. If you want to derive it, you can postulate a lagrangian instead. – AccidentalFourierTransform Jan 18 '16 at 18:44
• @AccidentalFourierTransform Yes, what is the lagrangian? Do you know a link or book that looks more into this? More than it being a postulate as you names it? – PhilosophicalPhysics Jan 18 '16 at 18:59

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.

• I would prefer we start from the beginning not the end. We study special relativity and then general relativity not vice versa. I know about this shortcut, though I am asking how would we know that free massive particle would have that equation of motion starting from scratch. – PhilosophicalPhysics Jan 18 '16 at 18:33