Relative speed when getting close to the speed of light I was thinking about the relative speed of an observation reference frame and an object which has been accelerated to a speed close to the speed of light. I'm by no mean an expert and the last physics class I took was more than 20 years ago so my question could be silly...
If we accelerate a particle, let's say an electron, to 99.99% of the speed of light and then we start moving in the opposite direction and reach about the 0.01% of the speed of light in the opposite direction, by the original point of observation, using a reasonable amount of energy, we should be dilating our time, so the time in the original frame of reference should pass faster then the time in our moving frame, that means that a relative velocity observed from the original frame should increase from ours. Doesn't that mean that we will observe the $e$ passing $c $?
 A: There are at least two misunderstandings in your argument.  The most fundamental one is that you tried to compare times in two different reference frames, but you can't do that in relativity.  For example, if Bob zooms off at $\frac{1}{2}c$ to the right, and Alice stays where she is, then both of them will see "time dilation" when they look at the other person.  Suppose that they are both carrying clocks.  Both of them will think that the other person's clock is running slower than their own.
This comes into your scenario because I think you are conflating time dilation observed from the particle moving at $0.01\% c$ and time dilation from the rest frame.
The second misunderstanding here is that observers will not only see time dilation, but also space dilation, so this also makes figuring out the relative velocities a little bit harder.
Doing this calculation properly leads to the velocity-addition formula.  Suppose that you have $v_1=0.9999c$ and $v_2=0.0001c$.  Then the velocities that these two particles will perceive each other to be moving at (if they could perceive things and take measurements!) would be
$$v_\text{rel} = \frac{v_1+v_2}{1+\frac{v_1 v_2}{c^2}} = \frac{c}{1+0.9999 \cdot 0.0001} \approx 0.99990002 c $$
