Why don't all free particles lose their kinetic energy? I'm currently studying Action. I've been reading about how a particle has particular probabilities of ending at an infinite number of events.
Say I have a free particle that isn't experiencing any external forces (no potential, or friction). I give it a particular kinetic energy, whilst it is at some arbitrary event A. Why doesn't the particle stay at position A, "lose" its kinetic energy and reduce the action to a minimum? Why does it travel in a particular straight line? You can't argue from conservation laws since they are dependent on the idea of the stationary action principle. 
I'm curious to know the answer, cheers!
 A: Particles do not minimize their action. Instead, they minimize their action given certain boundary conditions. We can only apply the action principle when we know the start and end points ahead of time.
If we know that a particle will be at location $x_i$ at time $t_i$ and that it will be at location $x_f$ at time $t_f$, then the particle takes a path of least (or stationary) action between those two points. Standing still generally isn't an option because $x_i \neq x_f$ in most cases. The way the problem is set up, the particle is forced to move, simply by hypothesis, before we even being trying to minimize the action. 
If $x_i = x_f$ and you have a free particle, then indeed the action is minimized by staying still, which is exactly what the particle does. 
The free particle problem can be solved with relativity. We can transform into a frame where $x_i = x_f$, and in this frame the particle is stationary. Transforming back, the particle moves at constant velocity in the original frame.
A: Since it's a quiet Sunday morning let me dust off my brain cells and see if I can remember how to do this. We start by noting that if the initial and final positions for the particle as $\mathbf{r}(t_1)$ and $\mathbf{r}(t_2)$, then the action is:
$$ S[\mathbf{r}(t)] = \int_{t_1}^{t_2} \tfrac{1}{2}m\dot{\mathbf{r}}^2 $$
We'll make the change $\mathbf{r} \rightarrow \mathbf{r} + \delta\mathbf{r}$ in which case the action becomes:
$$ S[\mathbf{r} + \delta\mathbf{r}] = \int_{t_1}^{t_2} \tfrac{1}{2}m \left(\dot{\mathbf{r}}^2 + 2\dot{\mathbf{r}}\cdot\delta\dot{\mathbf{r}} + \delta\dot{\mathbf{r}}^2\right) $$
So the change in the action is:
$$ \delta S = \int_{t_1}^{t_2} \tfrac{1}{2}m \left(2\dot{\mathbf{r}}\cdot\delta\dot{\mathbf{r}} + \delta\dot{\mathbf{r}}^2\right) $$
and we do the usual trick with infinitesimals of ignoring squared and higher power terms to give:
$$ \delta S = m\int_{t_1}^{t_2}  \dot{\mathbf{r}}\cdot\delta\dot{\mathbf{r}} $$
The next step is a sneaky trick that only the more able physicists will guess without being told (I didn't :-). We use integration by parts:
$$ \int u\dot v\, dt= uv\bigg|_{\text{end points}} - \int \dot u v\, dt $$
with $u = \dot{\mathbf{r}}$ so $\dot u = \ddot{\mathbf{r}}$, and $\dot v = \delta\dot{\mathbf{r}}$ so $v = \delta\mathbf{r}$, and this gives us:
$$ \delta S = \left[ m\dot{\mathbf{r}}\cdot\delta\mathbf{r} \right]_{t_1}^{t_2} - m \int_{t_1}^{t_2} \ddot{\mathbf{r}}\cdot\delta\mathbf{r} $$
But the endpoints are fixed so $\delta\mathbf{r}(t_1) = \delta\mathbf{r}(t_2) = 0$ and the first term is zero giving:
$$ \delta S = -m \int_{t_1}^{t_2} \ddot{\mathbf{r}}\cdot\delta\mathbf{r} $$
At the extremum of the action $\delta S = 0$, so the integral must be zero. However $\delta\mathbf{r}$ can be anything as any variation of $\mathbf{r}$ is allowed. This means the integral can only be zero if $\ddot{\mathbf{r}} = 0$, and after all this pain we end up with Newton's first law:
$$ \ddot{\mathbf{r}} = 0 $$
So the free particle has a constant velocity and therefore constant kinetic energy.
Since someone will surely mention it (though it's somewhat beyond my personal level in this area) Noether's theorem tells us that if the action is not time dependent then the energy is conserved. So we didn't really need all this working to conclude that the kinetic energy can't change.
