How can one always be standing still when compared to the speed of light? I was thinking if I built a device with 7 clocks, synchronized to each other, one in the middle, one up, down, left, right, behind and in front of me, say 1 meter away, and I fired a laser from the middle and split it into six pieces would the light always arrive at all six "corners" at the same time?
If we accelerated (say to the right) up to half the speed of light and fired the laser again, wouldn't the laser have to travel farther to reach the right sensor (as it's moving at half the speed of light) and travel less distance (as it is approaching the light at half the speed of light)?  So the time the light arrives at the right sensor should be greater than the time it takes to arrive at the left sensor.
I'm assuming you will tell me that this is not true, the light will still arrive at all the sensors at the same time.  Why would this be true?
Isn't the whole apparatus moving at half the speed of light in one direction?  Why would that make no difference?  In that case, why couldn't I just accelerate to half the speed of light again, to the right?  (Then I will be going at the speed of light - half the speed of light, twice, but in each instance I was considered to be standing still.)
 A: In relativity the notion of simultaneity is relative to the obeserver. While one observer (the one "standing") see all the rays of light arrive at the same time, another observer (the one going near the speed of light) will see one ray arriving before another. The paradox here is you think simultaneity can be defined in an absolute way independent of the observer, the true is that notion is frame dependent.
EDIT:
Suppose there are three observers, observer 2 go at a half of the speed of light relative to observer 1 ($ v_{21} = \frac{c}{2}$), and observer 3 go half of speed of light relative to observer 2 ($ v_{32} = \frac{c}{2}$),the observer 3 simulate what you said about accelerate to half of the speed of light again. Then, at what speed observer 3 is moving respect observer 1? the speed of light? No, because the Galilean law for adding velocities is not true in special relativity, instead you have
$$
v_{31} = \frac{v_{21}+v_{32}}{1+\frac{v_{21}v_{32}}{c^2}} = \frac{\frac{c}{2}+\frac{c}{2}}{1+\frac{1}{4}} = \frac{4}{5}c<c
$$
So is not problem, you can still accelerate to near of speed of light again, and still be under the speed of light for all inertial observers . All inertial observers are on equal footing.
