glassjester said:
and you flew in a spaceship going the opposite direction, also at 51% the speed of light...
Would each of us be breaking the light barrier, relative to the other?
If not, why not?
You have not made it clear who measures the spaceships to be traveling at 51% of the speed of light (0.51c). Let us assume that ship A is moving at 0.51c according to observer C and ship B is moving at -0.51c also according to observer C which we can state as:
vAC = velocity of ship A according to observer C.
vBC = velocity of ship B according to observer C.
Now what you want to know is what is the speed of ship A according to ship B (vAB) or vice versa. The regular equation of relativistic velocity addition can now be written int terms of these variables as:
vAC = \frac{vAB+vAC}{1+vAB*vAC/c^2}
This equation can be rearranged to isolate the relative velocity we are interested in:
vAB = \frac{vAC-vBC}{1-vAC*vBC/c^2}
Plugging in the numbers using units of c=1 now gives:
vAB = \frac{0.51-(-0.51)}{1-0.51*(-0.51)} = 0.809c
Of course if you look at the above equations it can be seen if two particles are moving in opposite directions according to a single common observer (C), then the relative velocity of one particle according to an observer at rest with the other particle can be expressed in terms of velocity magnitudes (speed) as:
|vAB| = \frac{|vAC|+|vBC|}{1+|vAC|*|vBC|/c^2}
Ether way, if the speeds of any two particles are individually less than the speed of light according to any given observer, then the speeds of the particles relative to each other (or relative to any other inertial observer) is always less than the speed of light.
Also, just to confirm what Pengwuino said, objects can have "closing velocities" that are apparently greater than the speed of light, but this does not violate relativity because information is never transmitted from one observer to another faster than the speed of light and the velocity of one observer relative to another observer is never greater than the speed of light.